A true solution to Russell's paradox
Hi all
I won't explain the paradox. If you know the paradox then continue.
Is the set of all sets that are not members of themselves, a member of itself?
We're trying to non-paradoxically define the set of all sets that are not members of themselves.
There is the set of all sets. Call this x.
x includes all sets that are not members of themselves, as well as all sets that are members of themselves because x is the set of all sets.
There is the set of all sets that are members of themselves. Call this z. The set of all lists, is one set (a member of x). The set of all things that can be members of themselves (lists, folders, sets), is one set and this set is a member of x. x is a set. It contains all sets. Thus, only x can contain x. Therefore, clearly z = x as only x can be the set of all sets that are members of themselves.
There is the set of all sets that are not members of themselves. Call this y. If x does not exist and y is not a member of itself, then we have no set of all sets that are not members of themselves precisely because y (a set that is not a member of itself) is outside of the set of all sets that are not members of themselves. To say x does not exist is to be paradoxical. Thus, clearly, x exists. x encompasses all sets. Suppose y is the most encompassing set after x. y is not a member of itself because y is a set and is thus a member of the set of all sets (x). Thus, x is the set of all sets that are not members of themselves. Clearly, the original y in this paragraph, is x. y = x.
x, the set of all sets that are not members of themselves, is thus a member of itself.
All we need is a non-paradoxical set of all sets that are not members of themselves. We have this. We have x and x is clearly not paradoxical.
I won't explain the paradox. If you know the paradox then continue.
Is the set of all sets that are not members of themselves, a member of itself?
We're trying to non-paradoxically define the set of all sets that are not members of themselves.
There is the set of all sets. Call this x.
x includes all sets that are not members of themselves, as well as all sets that are members of themselves because x is the set of all sets.
There is the set of all sets that are members of themselves. Call this z. The set of all lists, is one set (a member of x). The set of all things that can be members of themselves (lists, folders, sets), is one set and this set is a member of x. x is a set. It contains all sets. Thus, only x can contain x. Therefore, clearly z = x as only x can be the set of all sets that are members of themselves.
There is the set of all sets that are not members of themselves. Call this y. If x does not exist and y is not a member of itself, then we have no set of all sets that are not members of themselves precisely because y (a set that is not a member of itself) is outside of the set of all sets that are not members of themselves. To say x does not exist is to be paradoxical. Thus, clearly, x exists. x encompasses all sets. Suppose y is the most encompassing set after x. y is not a member of itself because y is a set and is thus a member of the set of all sets (x). Thus, x is the set of all sets that are not members of themselves. Clearly, the original y in this paragraph, is x. y = x.
x, the set of all sets that are not members of themselves, is thus a member of itself.
All we need is a non-paradoxical set of all sets that are not members of themselves. We have this. We have x and x is clearly not paradoxical.
Comments (309)
That right there is the contradiction. x is a member of itself if and only if it's not a member of itself.
Since the assumption that there is a set of all sets leads to a contradiction, we must therefore reject that assumption, and admit that no such set exists.
It's always worth pointing out in these discussions that self-containing sets are not contradictory. We generally assume an axiom, the axiom of regularity, that outlaws self-containment as well as circular chains of containment such as [math]a \in b \in c \in d \in a[/math]
If we instead choose to allow self-containing sets, the resulting system is logically consistent. The study of such systems is called non well-founded set theory.
In modern set theory we avoid Russell's paradox by saying that we can not form a set merely out of the extension of a predicate, like [math]x \notin x[/math]. Rather, we must start with a set we already know to exist, and then apply a predicate to it. This is the axiom schema of specification. It's a schema because it stands for an infinite collection of axioms, one for each predicate.
As an example of how this works, what is the set of all natural numbers that are not members of themselves? Well, [math]0 \notin 0[/math], so [math]0[/math] is in the set. [math]1 \notin 1[/math], so [math]1[/math] is in the set. [math]2 \notin 2[/math], so [math]2[/math] is in the set. Continuing in this manner we see that the set of all natural numbers that are not members of themselves is ... drum roll ... the set of natural numbers! No paradox. Specification saves the day.
Thank you for replying and I understand where you're coming from. I will try to convey to you my understanding more specifically hoping that specification saves [s]naive[/s] true set theory. I will ask questions to see exactly where it is that we are in disagreement.
For the sake of argument, assume we have the set of all sets. Call this x. x is a member of itself because it is a set. No paradoxes so far, agreed?
Since x contains all sets, do we agree that x contains all sets that are not members of themselves?
The set of all penguins, is a set. This set is not a member of itself precisely because it is a member of the set of all sets. By this I mean It is specifically a set, not a penguin. Are we sill in agreement? The set of all animals, is one set that contains the set of all penguins. This set is also not a member of itself, precisely because it is a member of the set of all sets. Thus, by definition, any set that truly is a set (as opposed to a penguin or animal), and is not a member of itself, is not a member of itself precisely because it truly is a set and is thus a member of the set of all sets. Agreed?
If agreed, then can you see how the set of all sets that are not members of themselves, can only be (by definition) x? I will specify this some more: All sets that are not members of themselves, truly are sets. What is the set of all these sets? Can the answer be anything other than x? The set of all sets which contains all these sets, is a member of itself (because it truly is a set).
Where do we have a paradox in what I have proposed?
Quoting Philosopher19
Ok. But the core of the issue is the same as Russell noted in 1901. The "set of all sets that are not members of themselves" both is and isn't a member of itself, a contradiction. Therefore there is no such thing. This is not going to change. It would be helpful if you would carefully review the argument yourself and I urge you to do so.
Quoting Philosopher19
I'm perfectly willing to agree, as this is the assumption that will soon lead to a contradiction.
Quoting Philosopher19
Ah, perhaps I see the problem. Why do you think a set is a member of itself? Are you possibly confusing set membership with the subset relation? It's true that every set is a SUBSET of itself. But no set, in the presence of the axiom of regularity, is an element of itself. And even without regularity, a set can be a member of itself. But it doesn't have to be.
Is this the confusion? The set of natural numbers [math]\mathbb N[/math] is a SUBSET of itself; but not a MEMBER of itself. That is, [math]\mathbb N \subset \mathbb N[/math], but [math]\mathbb N \notin \mathbb N[/math]. If this resolves your question, we're done. Because saying that x is a set does NOT in any way imply that it is a member of itself. That's an error. So if this is the problem, we're done. I'll continue, but let me know if this was the issue.
Quoting Philosopher19
Yes, certainly. x contains all the sets that are members of themselves and all the sets that are not members of themselves. This will lead to the conclusion that x both is and isn't a member of itself.
Quoting Philosopher19
I have a problem. This isn't even naive set theory; it's high school set theory. Penguins are not elements of sets. In math, sets are generally "pure" sets, meaning that their only elements are other sets. There are alternate versions of set theory in which sets can contain urelements; that is, things that are elements of sets that are not themselves sets. However even in set theories containing urelements, I do not believe that penguins or any other natural objects can be urelements. I have to plead ignorance though, I don't know much about sets with urelements.
If you ask a biologist, they'll tell you that a penguin belongs to the family Spheniscidae, of the order Sphenisciformes, class Aves, phylum Chordata, kingdom Animalia. "Today I learned," as they say. Biology isn't my thing. But the classifications in biology are not mathematical sets. We could call them high school sets, but they are not even naive sets, let alone axiomatic sets. So this analogy is going in the wrong direction already. The collection -- not set, collection -- of all penguins is, as we all just learned, the family Spheniscidae. It's not a set. Whatever point you're trying to make, I'd prefer if you make it with actual mathematical sets. The set of all real numbers, or the collection of all topological spaces, which ISN'T a set (for the same reason as the set of all sets isn't a set).
Quoting Philosopher19
Now this I do not understand, even if I grant that the collection of all penguins is a set. Because the set of all sets contains all the sets that ARE members of themselves, AND all the sets that ARE NOT members of themselves. So even if I grant, for sake of argument, that the collection of all penguins is a set, that does not make it not a member of itself. Even granting your example, this statement makes no sense. Some sets are members of themselves (in non well-founded set theories) and some aren't. So just because something's a member of the set of all sets doesn't mean it's not a member of itself. How did you conclude that??
NOTE (after I've been through this a few times and think I understand your argument). If we assume no set is a member of itself, that still doesn't show that the collection of all non-self-containing sets is a set. You haven't shown that. And if you assume it is, you get Russell's paradox.
That is: You assume x is the set of all sets. You assume (as we all do all the time) that no set contains itself. Therefore x is also the set of all sets that don't contain themselves. HOWEVER! You still have your ASSUMPTION hanging around. And it falls to Russell's paradox. You assumed x is a set. What you have actually shown is that the collection of all sets is the same as the collection of all the sets that don't contain themselves. But you haven't shown that these collections are sets; and in fact they are not.
Quoting Philosopher19
I agree that the collection of all penguins is not a penguin. But I agree NOT because the collection of all penguins is a set; but rather, because the collection of all penguins doesn't happen to be a penguin. Your logic is off the rails at this point.
Quoting Philosopher19
As we've seen, a biologist would not agree. The concept of penguin is a subconcept or subcategory of the concept or category of animals. But they are not sets. The predicate "is a penguin" does imply the predicate "is an animal," I agree with that. But these are not sets. Still, for sake of discussion I'll grant your premise. I still fail to see your point and your claim that these classes do or don't contain themselves simply by virtue of being sets, is wrong.
Quoting Philosopher19
This is just wrong. The set of all sets (which we assume for sake of argument exists) contains some sets that DO contain themselves; and other sets which DON'T contain themselves. Just because some object (the collection of all penguins) happens to be a member of the set of all sets, doesn't let us conclude which is the case. It might contain itself or it might not. In this case the collection of all penguins does not happen to be a penguin; but that's a fact of nature and NOT just because it's a set.
Quoting Philosopher19
No, and I don't follow your reasoning. If the set of all sets contains all sets, then it contains all the sets that ARE members of themselves along with all the sets that AREN'T members of themselves. Given a set, we have to examine it carefully to determine whether or not it's a member of itself. Of course in standard set theory we have the axiom of regularity so no set is a member of itself. But if we drop regularity, then some sets ARE and some sets AREN'T member of themselves. We can't determine which is the case merely by knowing something is a set. Even granting all your premises I don't follow your reasoning. The set (if we call it that) of all penguins is not a penguin; but not because it's a set; but rather, because it doesn't happen to be a penguin!
NOTE: See Summary at the end. If you are assuming that by definition no set is an element of itself, that's perfectly fine and is the standard assumption in math. But that does not mean that the collection of all these non-self-containing sets is a set. You haven't shown that. What your argument shows is that the COLLECTION of all the sets that don't contain themselves, is identical to the COLLECTION of all the sets that there are. This is true. But you haven't shown that either of these collections are sets.
Quoting Philosopher19
I'm afraid not. Even granting your premises there is something terribly wrong with your reasoning. Some sets are and some sets aren't members of themselves. So even if you convince me that the collection of all penguins is a set, that doesn't tell me whether it's a penguin or not. I have to consider the specific case.
Quoting Philosopher19
No. And the problem is that I have no idea how you got to this point. x includes all the sets that aren't members of themselves, AND all the sets that ARE members of themselves. x contains all possible sets, right?
NOTE (now that I think I understand your argument): You are right that if we assume no set is a member of itself (as we normally do), then the COLLECTION of all non-self-containing sets is identical to the COLLECTION of all sets. But you haven't shown that these collections are sets. You're only assuming that; and your assumption is wrong, as shown by Russell.
Quoting Philosopher19
Well yes, because you said they're sets. All the fish that are left-handed are fish. Why? Because we said they're fish! A set that's painted green is a set, because we stipulated that it's a set. You've made a vacuously true statement. A set is a set, so of course a set that is painted green, or flies through the air, or is not a member of itself, is a set. We haven't said anything!
Quoting Philosopher19
Yes. The set of all sets that are not members of themselves is different from the set of all sets; because x contains all the sets that are not members of themselves AND all the sets that ARE members of themselves. I don't follow why you don't see this.
A set either contains itself or it doesn't. The collection of ALL sets contains the ones that do and the ones that don't. But that collection isn't necessarily a set, and can't be. And you haven't shown that it is.
NOTE (all these notes were written after I came to understand your argument, apologies for all these interspersed notes). YOU ARE CORRECT. If no set contains itself and x is the collection of all sets, then x is also the collection of all sets that don't contain themselves. BUT the claim that x is a set was an ASSUMPTION, which you haven't justified. And Russell's paradox shows your assumption is wrong.
Quoting Philosopher19
No, it's not. Russell shows it's not a set. It's a collection that's not a set.
NOTE. You have CORRECTLY shown that the collection of all sets is the same as the collection of all non-self-containing sets. But you ASSUMED that this collection is a set and Russell shows that it's not. You made an assumption but never justified it.
Quoting Philosopher19
You didn't get there yet. You have shown that the collection of all sets is the same as the collection of all sets that don't contain themselves, under the axiom of regularity. But then you assumed that x is a set, and you never justified that. And if you then apply Russell, you'll see that x can't be a set. Just a collection that isn't a set.
Summary:
* One issue as I mentioned is that you may be confusing subsets with elements. This is a fairly common point of confusion and if that's the case, let's focus in on that.
* The business with the penguins was not helpful to me, it added confusion.
* You are assuming the axiom of regularity (perfectly normal, we all do that all the time in standard set theory) and therefore saying that everything that's a set, doesn't contain itself. This, I agree with. But that doesn't mean that the collection of all the sets that don't contain themselves is a set! That's a logic error. This is the part that I didn't realize on my first couple of readings.
That is: Let us adopt the axiom of regularity, so that no set contains itself. We can therefore form the COLLECTION (not yet proven to be a set) of all the sets that don't contain themselves; and this will indeed turn out to be the COLLECTION (not yet proven to be a set) of all possible sets. I think this is the argument you're trying to make.
But you haven't shown that either of these collections is a set! And you can't, because any such attempt runs into Russell's argument.
* So the bottom line is that we assumed the set of all sets exists, and, assuming that no set contains itself, you are correct that it must be equal to the set of all sets that don't contain themselves. But then we just apply Russell's argument to show that the set of all sets that don't contain themselves both does and doesn't contain itself, a contradiction.
* So what you need to do is, AFTER you have made your argument: that x is the set of all sets and also the set of all sets that don't contain themselves; you have to apply Russell's paradox to see that x both is and isn't a member of itself. Therefore x isn't a set. It's merely a collection, the extension of a predicate. It's not a set.
To my knowledge, Russell's paradox concludes that you cannot have a set of all sets because he fails to non-paradoxically define a set of all sets that are not members of themselves.
No one is disputing that there can be no set of all sets that are not members of themselves that is itself, not a member of itself (call this absurd set y). But this paradox in no way logically amounts to saying that there is no set of all sets. x is the set of all sets. This set contains all sets including itself. No paradoxes.
For the sake of argument, let's say x and y are not the same. I have no problem in saying that y is absurd. But there is still a set that contains all sets that are not members of themselves. x contains them all does it not? If it does contain them, then it contains them. Why does x have to be not a member of itself??? Why are we trying to force a paradox where there is none? A set is a set. It doesn't matter if it's a member of itself or not. If it truly is a set, then it is clearly a member of the set of all sets.
Yes, x also contains other sets (actually it only contains one other set...which is itself). But it still contains all sets that are not members of themselves. Regarding sets, Russell misunderstood semantics and logic. I understand he was an important philosopher, but he made a mistake.
It is absurd to say that there is no set of all sets. Now are we in agreement?
Rather, it is fundamental, and perfectly intuitive.
Is there an existing thing that contains all existing things? Is it not blatantly paradoxical to deny that there is an existing thing that contains all existing things? If you accept that there exists an existing thing that contains all existing things (which I will label Existence), then you must also accept that there is a set of all sets.
I don't see how you can reject without being paradoxical. To say there isn't an existing thing that contains all existing things. logically implies things can come into or go out of Existence.
What you say is true of all non-infinite things. It is not true of that which is actually infinite. The actually infinite has no beginning and no end.
The infinite contains the infinitesimal and the infinitesimal contains the infinite. What's the alternative, that we deny that there exists and existing thing that contains all existing things? Is this better or accepting actual infinity as being infinite through and through?
Consider the following thought experiment:
On a computer, y is the folder of all folders. Your starting/beginning position is within y and you see y amongst an infinite number of other folders. If you click y, you get the same thing (an infinite number of folders with y amongst them). This continue ad infinitum. If you go up one level of folder trying to get to the root folder, again, you get the same thing (an infinite number of folders with y (the one you just came from) amongst them). This continues ad infinitum. We cannot say y fully contains itself if you cannot go up another level ad infinitum or go down another level ad infinitum. Your starting/beginning point changes nothing. Such is the nature of true infinity, it has no beginning and no end. Mathematicians confuse potential infinity with actual infinity and you get infinity paradoxes where there really are no such paradoxes. Just poor labels chosen for semantics.
The problem with the above thought experiment, is that you can see y. You cannot see infinity or infinitesimal. In other words, you cannot see Existence, you can only see non-infinite things that are in it. The infinite is in the infinitesimal and the infinitesimal is in the infinite.
If you knew more about infinity, you would not say this. Me trying again will do nothing. Try looking at my example again.
Kudos to fishfry and fdrake, resident experts in set theory. Both demonstrate great patience in unraveling the queries on that subject that crop up on TPF. :up:
You're turning the argument on its head to confuse yourself. If we can't define such a set without creating a contradiction, then there is no such set.
Quoting Philosopher19
If you call x the set of all sets, you quickly get a contradiction. You find that x both is and isn't a member of itself. Therefore there is no such set. You keep claiming there is but you have not provided proof.
Quoting Philosopher19
There is no set of all sets. You keep claiming there is, but you have not provided a proof. On the contrary, the assumption that there is a set of all sets leads to a contradiction. Therefore there is no set of all sets.
Quoting Philosopher19
The CLASS, or COLLECTION that contains all sets may indeed be formed. It just turns out to not be a set. And you haven't shown that it can be. You keep claiming it without proof.
Quoting Philosopher19
Not in this instance.
Quoting Philosopher19
You trollin' me?
The following is proof:
I find that if I say x isn't a member of itself, I am being paradoxical because x is a set. I find that if I say x is a member of itself, I am not being paradoxical because x is a set. Do you see?
Quoting fishfry
How do you get to this???? x = the set of all sets. Is x a set? Yes. Thus x is a member of itself. Is x not a member of itself? Yes it is a member of itself because it is a set! Let's try the alternative. x is not a member of itself. Why not? No reason can be given. The set of all penguins is not a member of itself. Why not? Because a penguin is not a set. See?
I find that x is a member of itself. That is all I find.
Quoting fishfry
Not at all. If I am, then I'm an idiot. I just want efficiency and truth.
Patience is a virtue.
Quoting SophistiCat
Yes I agree. But please bear in mind that all sets, are members of the set of all sets (including the set of all sets itself). So the set of all great white sharks, is a member of the set of all sharks as well as a member of the set of all sets.
Quoting SophistiCat
The set of all great white sharks is a member of at least two sets. The set of all sharks, and the set of all sets. You cannot have a set that is not a member of itself encompass/include all sets that not members of themselves precisely because (as Russell pointed out) it cannot both contain itself and not contain itself at the same time. This is clearly paradoxical. But you have the set of all sets. It includes all sets that are not members of themselves. Unlike the set of great white sharks which is a member of at least two sets, the set of all sets that are not members of themselves, is only a member of one set. That set is the set of all sets. Please look at the part in bold very carefully and recall that the set of all sets, is a member of itself, and that all sets are a member of the set of all sets (including great white sharks).
[i]The set of all sets that are not members of themselves, is a set. Thus it is a member of the set of all sets. It is in fact the set of all sets. The set of all sets is a member of itself.[/I] No paradoxes here whatsoever.
Russell claims that there is no set of all sets that are not members of themselves whatsoever. Russell is clearly wrong. This paradox is now clearly fixed. Can we move on in a unified manner? Because academics seem to think that it's unsolvable. I'm not at a uni to get my voice heard. I don't like seeing true set theory being called [s]naive[/s] set theory.
Yes. Off the first page of TPF. :roll:
Your bump pushes it to the front page.
Quoting Philosopher19
You claim x is a set but it isn't. You have no proof that x is a set.
Quoting Philosopher19
Is x the collection of all sets that are not members of themselves? If so then x is not a set. You claim it is but haven't proved it is. If x is the collection of all sets that are not members of themselves, then x also isn't a set even though you claim it is. You've used x to mean both of those things at various times. In both cases they are collections, or classes, or extensions of a predicate. But they're not sets.
Quoting Philosopher19
By Russell's paradox. Say x is the set of all sets. Then let y be the set of all sets that are not members of themselves. Is y a member of itself? If it is, then it's NOT a member of y. But if it isn't, then it IS a member of y. So y is a member of itself if and only if it isn't a member of itself. That's a contradiction.
If x is the set of all sets that are not members of themselves, just run the same argument on x.
Quoting Philosopher19
No, it can't be. I just showed that if x is the set of all sets, then we can form y and derive a contradiction. If x is the set of all sets that don't contain themselves, then x itself leads to a contradiction.
Quoting Philosopher19
You need to work through the proof of Russell's paradox carefully. You're just repeating incorrect ideas.
Quoting Philosopher19
You need to work through the argument.
Quoting Philosopher19
I have made no characterizations. It's a tricky argument. You need to go through it for yourself carefully.
Suppose x is the set of all sets that don't contain themselves. Then we ask if x contains itself? If it does, it's NOT a set that doesn't contain itself, so it DOESN'T contain itself.
See what just happened? If x contains itself then x doesn't contain itself.
On the other hand suppose x doesn't contain itself. Then it must be a member of x. So if it doesn't contain itself it does contain itself.
Having just shown that x contains itself if and only if it doesn't contain itself, we have a contradiction. Therefore there is no such set as x.
Now if you prefer to let x be the set of all sets, we let y be the set of all sets that don't contain themselves and we get a contradiction from y. So again, x can't be a set.
I'm not the one who has a problem with it. Russell's paradox is a deep argument. Frege was a smart guy (he invented the universal and existential quantifiers) and he missed it. It's worth discussion. I daresay @Philosopher19 is not the only person who's ever experienced confusion about the subject. SEP has 7448 words on it, they must think it has some importance. It's the paradigmatic example of all the self-referential arguments such as Cantor's diagonal argument (which influenced Russell to think of it), Gödel's incompleteness theorems, and Turing's Halting problem. It's worthy of discussion IMO regardless of the circumstances.
This website is not the proceedings of the Royal society. I often wish for a more high toned conversation around here, especially on mathematical topics; but we take what we can get. You should see the politics forums. I hope my participation in this thread doesn't inconvenience or distress you too much. It's the forum software that bumps active threads, I have no control over that. I couldn't help calling out the irony that someone who doesn't want to see this thread on the front page, bumped it to the front page themselves. How self-referential.
Quoting SophistiCat
I am not the one embracing a paradox. That would be unreasonable/nonsensical/paradoxical.
Quoting SophistiCat
Ok.
Ok then. A set is a set and you can have sets within sets.
Quoting fishfry
If y is not x, then y is absurd. How can a set that is not a member of itself, contain itself? It cannot. Say x encompasses all sets. Say z is the most encompassing set of sets after x. z is not a member of itself but it contains almost all sets that are not members of themselves. Because it does not contain itself, z is not the set of all sets that are not members of themselves because it does not contain itself despite being a set that is not a member of itself. z is contained in x because it is a set. y being absurd or z not containing itself takes nothing away from x containing all sets that are not members of themselves as well as itself. y is absurd because it claims to contain all sets that are not members of themselves whilst not being x. Why does x have to be treated as being absurd when y is absurd? The set of all sets that are not members of themselves is x and x is a member of itself. If you still think the last sentence is paradoxical, read on.
Quoting fishfry
Quoting fishfry
Again, this is rooted in confusing y for x. y is absurd but x contains itself and it contains all sets that are not members of themselves because all sets that are not members of themselves, are still just sets at the end of the day. x does not have to be not a member of itself in order to contain sets that are not members of themselves. The set of all sets does not have to be a penguin to contain the set of all penguins. [b]It just has to be the set of all sets.
Change "penguin" for "not a member of itself". Is a set that is not a member of itself, a set? If yes, then you know at least one set that it is a member of: The set of all sets. Are all sets that are not members of themselves, sets? If yes, then you know at least one set that they are all a member of: The set of all sets. There can be no other set that contains them all.
Quoting fishfry
Again, any set that is not a member of itself, is a member of the set of all sets. This is because it is a set. The set of all sets, is a member of itself. This is because it is itself a set. If y is absurd, that does not mean that x is also absurd. It just means y is not in x because y is not a set. Nor does it mean that x does not contain all sets that are not members of themselves when they clearly are just sets. Just for one second, consider the hypothetical possibility that all these famous philosophers after Russell, were wrong, and I'm right. I know how it sounds, yet it is still a hypothetical possibility. Just look at the proof that I am presenting without bias and without preconceived notions.
Let R be any 2-place relation. It is a theorem of logic that:
There does not exist an x such that for all y, y R-relates to x if and only if y does not R-relate to itself. Symbolically:
~ExAy(Ryx <-> ~Ryy)
In particular, where R is the membership relation, there does not exist a set x such that for all y, y is a member of x if and only if y is not a member of itself. Symbolically:
~ExAy(y in x <-> ~ y in y)
Proof is simple:
Suppose, toward a contradiction, that there is an x such that for all y, y R-relates to x if and only if y does not R-relate to itself. Then x itself R-relates to x if and only if x does not R-relate to x, which is a contradiction. Symbolically:
Suppose ExAy(Ryx <-> ~Ryy). Then Rxx <-> ~Rxx, which is a contradiction.
For set theory:
Suppose ExAy(y in x <-> ~ y in x). Then x in x <-> ~ x in x, which is a contraction.
Thus trying to deny Russell's paradox by appealing to one's personal notion of the concept of 'sets' fails, since the structure of the contradiction does not rely on any concept of set. The principle that there is no set of all sets that are members of themselves is an instance of Russell's paradox, but, as I've shown, that principle does not rely on the any particular concept of 'sets'.
This is witnessed by Russell himself where he explains the paradox by reference to an arbitrary 2-place relation such as 'shaves'. It matters not whether the relation is 'is a member of', 'shaves', 'loves', or 'billwingadoobadoodles'. Seeking to dispute Russell's paradox by recourse of arguing over the concept of 'set' misses the point and is ill-conceived.
However, it is a correct that there is no 'set of all sets' is a corollary from Russell's paradox that DOES depend on a set theoretic notion that is expressed by the Axiom Schema of Separation which says that for any any formula F, and set s, there is the set y whose members are all and only those members of s such that F holds of y. Symbolically:
AxEyAz(z in y <-> (z in x & Fz))
From that axiom we derive that there is not a set of which all sets are members. Symbolically:
~EsAy y in x
Proof is simple:
Let F be the formula '~ z in z'. Suppose, toward a contradiction, that EsAy y in x. By the Axiom Schema of Separation we have EyAz(z in y <-> (z in s & ~ z in z)). So y in y <-> (y in s & ~ y in y). But, by the supposition that Ay y in s, we derive y in y <-> ~ y in y.
So one can deny that there is no set of all sets only by denying that for any property expressible by a formula and for any set, there is the subset of the set whose members are all and only those in the set and having said property.
Of course it doesn't. Your posts are uniformly excellent.
Quoting SophistiCat
That's all I meant.
A number of posters around here would beg to differ, but thanks for the kind words.
I have been doing that. I can't add anything to what I've said other than that you should carefully examine the proof of Russell's paradox. And you should carefully examine your own argument, to see that you repeatedly claim that x is a set but you never present an argument to that effect.
Consider the actual reality (not just a hypothetical possibility) that the mathematicians thoroughly studied the subject matter down to its finest details and understand its rigorous axiomatization, including that set theoretic proofs are machine checkable, while on the other hand, it appears you have not read the first page in a textbook on mathematical logic or set theory.
(1) There exists a set such that every set is a member of it.
However, it does contradict the claim that:
(2) For any property and for any set, there is the subset of that set with the members of the subset being those with said property.
You can have (1) or you can have (2), but you can't have both. That is the basic upshot of Russell's paradox applied to sets.
Perhaps you could humor me and provide "much of an argument." Such a set would violate regularity/foundation. Are you perhaps making reference to Quine's New Foundations or some other alternate axiomitization of set theory? Reading the Wiki article in its entirety, or at least to the point where they said the category of NF sets is not Cartesian closed (*), convinced me that "much of an argument" must indeed be given to put your remark into its proper context. I could be wrong, curious what's in your mind with this post.
(*) This means that the sets of NF lack products and exponentials. There's not always a Cartesian product of two given sets; and/or there is not always the set of functions from one set to another. In computer science terms, you can't always curry functions. These are not sets as generally understood except perhaps by specialists in NF, as I understand it.
But I said ONTO ITSELF.
The formula
ExAy yex
is not a contradiction. It is consistent. Trivially, it has a model.
However, indeed it is in contradiction with certain axioms of set theory, as I mentioned in particular it contradicts the axiom schema of separation.
You surely said no such thing, in particular or in general. Here is a quote of your entire post.
Quoting GrandMinnow
I wonder if you can explain what you have in mind. You can't just say there's a set of all sets and that "One doesn't have to provide much argument" to justify it, and provide no argument, and then claim you said something you didn't say. Really wondering what your post was about. I did give the example of Quine's NF but evidently you're not talking about that. So what is the context of your remark? Not giving you a hard time for the sake of it but trying to get you to explain your cryptic remark, which is false without additional qualification. If you deny specification you haven't got a theory of sets, unless (as in NF) you stratify your sets. But why do you think you made a point about specification when you clearly didn't? Am I being unfair in challenging you here?
Separation and specification are subtly different according to Wiki. It's surely not the case that "one doesn't need to provide much argument" here. Quite the contrary IMO.
Quoting GrandMinnow
Where? Those words are clearly not in your post. Am I missing other posts of yours perhaps?
As the Wiki article on NF points out, your claim falls to Russell's paradox unless you stratify your sets, and that DOES require some explanation.
And I did not say that there is a universal set. I said that "there is a universal set" is not onto itself a contradiction.
And I didn't claim that I said something I didn't say:
You QUOTED me saying that a universal set contradicts:
(2) For any property and for any set, there is the subset of that set with the members of the subset being those with said property.
And (2) IS the axiom schema of separation.
So surely I did mention that it contradicts the axiom schema of separation.
/
If someone says to me that his concept of set demands that there is a set of all sets, but he has not made fully clear what else is in his concept of sets, then I say "Fine, it is not onto itself contradictory that there is a set of all sets, but you will incur contradiction once you add certain other principles to your concept, such as the subset principle (which is given more explicitly as separation)."
That it doesn't take much argument to see that "There exists a U such that everything is a member of U" is not BY ITSELF contradictory, we just need to show a model M.
Let M be the model for the language with the 2-place relation 'e' as follows:
The universe of M is {0}, and e is interpreted as {<0 0>}.
The reason I didn't pedantically spell out that argument is that it takes but a nanosecond of reflection to see that yes, of course, "ExAy yex" has models.
Granted, that doesn't capture an ordinary concept of sets, but I am at first allowing, for sake of argument, that one my have whatever concept of set one may wish to have. And toward that end, I am mentioning that "there is a universal set" is not ONTO ITSELF a contradiction.
The point in the context of this thread is that we find someone who claims that there is a universal set, and I wish to emphasize that explaining what is wrong with that claim may be best understood as a two step process: First, being as generous as possible to the person that he may have any concept of set he wishes to have, we recognize that, ONTO ITSELF, "there is a universal set" is not self-contradictory, then Second, mention however that "there is a universal set" does contradict other plain and well agreed upon aspects of sets, such as separation.
/
Whatever Wiki may say, 'the axiom schema of separation' and "the axiom schema of specification" are two names for the same schema, as I mentioned that schema previously. And I don't know what difference it makes for my comments, as I specifically used only "axiom schema of separation" anyway.
But without further qualification, those models are in no way sets. And you refuse to provide such context. And if you mean something like NF, that's a pretty sophisticated concept that does involve stratification in order to avoid Russell's paradox. I have given your posts quite a bit more than a nanosecond of reflection and I don't believe you are making your case in the least.
Quoting GrandMinnow
We're in agreement. You just refuted your own point. I'll leave it at that. We're not going to reach mutual understanding because you're stretching a point for its own sake, failing to provide context, and you have already refuted the point you claimed to be making. You said "There exists a set such that every set is a member of it," but in the end you agree that your claim is not about sets as commonly understood; and you have failed to provide any context in which your claim could be taken about sets.
Ironically I gave you such a context, NF, but you don't want to go down that road. There's no other road you can take.
Quoting GrandMinnow
I most definitely have. I gave the NF article a pretty good read. You should do the same to get a clue about what you think you're talking about. You've already agreed you're not talking about sets, which refutes your own point.
I didn't refute my own point. I never claimed that a demonstration of the consistency of ExAy yex would be faithful to ordinary concepts of set. Go back and read exactly what I posted: I said that it doesn't take much to show that "There exists a set such that every set is a member of it" is not self-contradictory; I did not claim that showing that fact would adhere to ordinary concepts of set.
And you say that I fail to provide context, when I just spent my time tying out for you the context per this thread.
Quoting fishfry
No I did NOT. I said the sentence is not self-contradictory. I did not say or even suggest that it is true or even compatible with ordinary concepts of set. Really, it is becoming egregious of you that you're putting words in my mouth even after I just asked you not to do that.
I'm not picking up on your remarks about NF, because I don't need recourse to NF to support my own remarks.
/
Angle brackets, indicate ordered pair, as is ordinary notation. Also, incorrectly, you put the question about the meaning of angle brackets within my own quote.
/
To recap:
(1) ExAy yex
is consistent. And it takes little argument to see that it is consistent. And I did not claim such an argument would use only our ordinary concept of set. And I did not claim ExAy yex. And indeed I pointed out that ExAy yex contradicts separation (which, goes without saying, is part of our ordinary concept of set).
(2) The point of mentioning (1) in the context of this thread was described in a previous post.
Clearly there is a problem with the manner in which you are reading. My very first sentence in this thread:
"One doesn't have to provide much argument that the following claim onto itself is not self-contradictory."
And you even quoted me saying that.
/
And I don't know why you say that Wiki claims a subtle difference between separation and specification. The Wikipedia (which I don't rely on as authoritative in mathematics anyway; for example see the wildly incorrect article on the rule of existential instantiation) article says they are the same. Nor, as I mentioned, do I know why you even mention the matter.
I am being sincere when I say I have carefully examined his argument. I also don't think I can do more by way of furthering our discussion on this matter except perhaps to highlight the following:
All existing things exist. They cannot exist in non-existence/nothingness. Call that thing in which all things exist in, Existence. Call the set of all existing things, Existence. Existence is the set of all existing things (including itself because it Itself exists).
I find this outrageously paradoxical/absurd to deny. Do you not? Russell's paradox (which again, I would say is a misunderstanding of semantics and poor usage of labels) denies this very obvious truth.
I do not think it in any way reasonable to believe in the following absurdities:
Things can exist in non-existence
Not everything exists in Existence
There is no set of all existing things
It is paradoxical/unreasonable/absurd/irrational of us to believe in the 3 aforementioned absurdities. It is paradoxical to believe or embrace or accept any absurdity. From triangles having four sides, to married-bachelors existing.
I sort of thought I should tame myself and not say that I'm right and many famous mathematicians and philosophers are wrong. If I am wrong, then I am an idiot and I apologise for wasting people's time (including yours). I would like to highlight the following:
All existing things exist. They cannot exist in nothingness/non-existence. They all exist in something. Call this thing Existence. Call the set of all existing things, Existence. Existence is the set of all existing things (including Itself because it Itself exists).
I find this outrageously paradoxical/absurd to deny. Do you not? Russell's paradox (which again, I would say is a misunderstanding of semantics and poor usage of labels) denies this very obvious truth.
I do not think it in any way reasonable to believe in the following absurdities:
Things can exist in non-existence
Not everything exists in Existence
There is no set of all existing things
It is paradoxical/unreasonable/absurd/irrational of us to believe in the 3 aforementioned absurdities. It is paradoxical to believe or embrace or accept any absurdity. From triangles having four sides, to married-bachelors existing.
My solution to the paradox, put differently;
There exists sets that are not members of themselves. All these sets, are sets. They are therefore a member of the set of all sets. The set of all sets need not be not a member of itself to contain all sets that are not members of themselves. Why should it? The set of all sets contains itself because it is a set. Semantically/logcailly there is only one set that is a member of itself, that set is the set of all sets: Existence. There is only one Existence. We are not Existence, we are members of Existence. We are not members of ourselves. We are members of Existence. We are ourselves and Existence is Itself. I will explain:
[b]We exist because Existence exists. Existence exists because Existence exists.
The set of all sets contains all sets that are not members of themselves (which includes us), as well as itself. There is no set of all sets that are members of themselves because there is only one Existence. You cannot have a set of Existences because it is absurd for non-existence to separate two Existences from each other. You have a set of existing things existing in Existence, which itself exists.[/b] If a set is not a member of itself and not a member of the set of all sets, then that set is absurd. That set is not a set. The empty set that Frege detailed, contains all absurdities such as round-squares. They are not members of Existence. They are not members of the set of all sets.
Kind regards,
Nyma
I'm a plain language person. So given the level of technical detail from fishfry et al, I'm jumping in here with more than a bit of trepidation. But fools rush in - I'll try to analyze what I think you're saying in my own clumsy way:
Quoting Philosopher19 Seems like a tautology to me, but just for completeness we need to extend the property of existence to energy fields & spacetime as well. Spacetime exists.
Quoting Philosopher19 Not quite sure what you're getting at here - it seems like you're saying "Things do not have the property of non-existence"? But this falls out of the definitions of the words. So at best you're simply re-stating your first sentence in different words. Nothing wrong with that. :smile:
Quoting Philosopher19And here is where we go astray. I'm seeing two inter-related problems. The first is calling this "thing" Existence.. Using the word Existence leads to confusion - let's use the word universe. So now we have:
They all exist in something. Call this thing the universe.
Next using the word "something" looks wrong to me. The universe is not a thing. The universe does not contain itself. So to make this work, let's rephrase these two sentences like this:
They all exist in the universe.
This works. In fact we can now combine the two revised sentences:
All existing things exist in the universe.
That works. Then finally we have this:
Quoting Philosopher19
And here is the third point where we disagree. I am far from an expert in Set Theory - but the basic concepts are clear and comprehensible to the average person. Sets are not real. Sets do not have the property of existence - they are mathematical constructs with mathematical properties according to various mathematical rules. The universe is not the set of all things that have the property of existence.
The universe IS all things.
Or put differently by this very smart philosopher guy I once read :razz:
The universe is all that is the case.
Quoting Philosopher19
I agree with the hypothesis of that sentence, I disagree with the conclusion.
I see. All triangular things are triangular because they have the property of being triangular.
All existing things exist because they have the property of existing. If x exists, then x has the property of existing. Do we agree on this?
Quoting EricH
Things that do not exist, do not have the property of existing. They are therefore not members of Existence. They are non-existent.
Quoting EricH
From what I gather, the universe had a beginning. It could not have had a beginning in non-existence/nothingness. Thus it had a beginning in some existing thing/entity/being. Do you agree with this? I would also assert that this existing thing/entity/being, is necessarily actual infinity or truly infinite. I say necessarily because if this thing/entity/being was finite, we'd run into the paradox of something coming from nothing. For example if this thing was temporally finite, then that implies that it came from nothing. If this thing was spatially finite, then that implies it is surrounded by non-existence (which implies that non-existence exists). Since non-existence does not exist, the notion of being surrounded by non-existence is absurd. Hence why this thing/being/entity must necessarily be truly infinite.
Quoting EricH
Since the universe itself is an existing thing or whatever we choose to call it, it is an existing thing or whatever we choose to call it. It is existing. It is not nothing. It is not a non-existent thing like a married bachelor. The universe does not denote the whole of Existence because it had a beginning. True infinity necessarily denotes the whole of Existence because it logically ensures there is no non-existence. By this I mean, true/actual infinity cures us of the paradox of something coming from nothing or non-existence existing. Since actual/true infinity is infinite through and through, it becomes meaningful and non-paradoxical to say that infinity contains itself. By this I mean the infinitesimal is infinite and it is contained within the infinite. This is essentially saying infinity contains infinity.
Quoting EricH
I've come here and claimed that I've truly solved Russell's paradox. If I said I think I may have solved Russell's paradox and I was wrong, then maybe I'm not an idiot. But if I said I've truly solved Russell's paradox and claim that all famous philosophers and mathematicians were misguided whilst I am not (which is what I did), and I am actually wrong, then I am arrogant and therefore an idiot. But if I am right, then I'm neither arrogant nor an idiot. I'm truthful and accurate with regards to the description of what I am, and what other philosophers and mathematicians were regarding this matter.
Quoting fishfry
I am GrandMinnow. I hadn't gotten around to answering the above.
Here is a model of "ExAy yex":
<{0} {<'e' {<0 0>}>}>
And its domain is a set, and the model itself is a set.
If one is familiar with mathematical logic, then one should recognize that is a model of "ExAy yex". It is quite trivial. But, to me, the very fact that it is so trivial is interesting.
But if one is not familiar with mathematical logic, then one would not understand the above. I am going to explain it exactly as I can and with as much detail as is feasible (even if painstakingly pedantic) in the context of posting.
(1) My original point was that "ExAy yex" is consistent but that it is not consistent with the axiom schema of separation. Note that "ExAy yex" is not consistent with this instance of the axiom schema of separation: AzEyAx(xey <-> (xez & ~xex)).
I mentioned that to point out that ruling out a universal set is not just a matter of looking alone at the notion of a universal set but rather that the notion of a universal set is not consistent with the notion of comprehension.
Then I said that "ExAy yex" has models. Note that if a sentence has a model, then the sentence is consistent.
(2)
(a) Df: a set of formulas T is consistent iff T does not prove a contradiction.
(b) Df: a formula P is consistent iff {P} is consistent.
If I recall correctly, (b) is fairly standard, but perhaps some people prefer to provide only (a).
I mention that only to ward against quibbles that originally I said:
"ExAy yex" is consistent
instead of
{"ExAy yex"} is consistent.
(3) Let 'N' stand for the set of natural numbers.
Df: a first order language L is determined by a signature
Df: L is a language for (or 'of') a set of formulas T iff L is a language that has at least all the function symbols and relation symbols that occur in T, and with the same arity they have in the formulas. Of course, this makes sense only if the formulas in the set don't have a symbol s that occurs in one formula as a function symbol of arity n but in another formula as a function symbol of arity m not equal n, or as a relation symbol; and mutatis mutandis for relation symbols. .
By "the language of set theory" we mean the language determined by the signature <0 {'e' '="} {<'e' 2> <'=' 2>}>. (Note: the appearance of '0' there does not imply that '0' is a symbol in the language, but rather that the set of function symbols is empty). This language has no function symbols and only two relation symbols: 'e' and '='. Any other function symbols or relation symbols used for doing set theory are not in the language for set theory but rather they are in a language of set theory extended by definitions. With the method of definitions, any formula that has defined symbols can be reverted mechanically to a certain formula that does not have the defined symbols.
(4)
Df: a model M for a language L with signature
So for first order logic with identity: Except for the relation symbol '=', V may map any function symbol to any function on U as long as that function is the arity for the function symbol, and V may map any relation symbol to any relation on U as long as that relation is the arity for the relation symbol. And '=' is always assigned the identity relation on U.
So a model M for the language of set theory is a tuple such that U is a nonempty set and V('e') is a 2-place relation on U.
Nota bene: A model for the language of set theory is not required to map 'e' to the membership relation on U. If a model M doesn't map 'e' to the membership relation, then M does not adhere to our intuitions of what set theory is about, but M is still a model for the language of set theory.
(5) Df: a sentence P is true in a model M for a language L iff [fill in the recursive definition here that is too detailed for this post].
Df: a set of sentences T has a model iff there is a model M for the language of T such that every member of T is true in M.
Df. a sentence P has a model iff there is a model M for a language for {P} such that P is true in M.
So "ExAy yex" has a model iff there is a model M for a language with 'e' such that "ExAy yex" is true in M.
If a sentence has a model then there is no upper bound to the number of models the sentence has, so, a fortiori, if a sentence has a model then it has at least two models. So "the sentence has a model" implies "the sentence has models". Moreover, I will trivially show two models for "ExAy yex" anyway. I mention this only to ward against a quibble that originally I said 'models' plural.
(5) If a sentence has a model, then the sentence is consistent. So to prove that a sentence is consistent, it suffices to prove that the sentence has a model.
(6) There is a model M such that "ExAy yex" is true in M.
'=' does not occur in "ExAy yex". So a model of "AxEy yex" is:
<{0} {<'e' {<0 0>}>}>
I did not fill in the definition of 'true in the model' previously in this post, because it is too detailed for this post. But here is an intuitive account regarding the above:
The universe is {0}.
The symbol 'e' maps to the relation {<0 0>}.
Nota bene: This is not an interpretation of 'e' that we have in mind for our intuitive meaning of 'e'. But a model does not have to conform to our intuitive meanings of the symbols. For the purpose of modeling the sentence, we can interpret 'e' as standing for any 2-place relation on the domain.
Uninterpreted, "ExAy yex" says that there is an x in whatever is the domain, such that every y in whatever is the domain bears whatever is the relation symbolized by 'e' to x.
With the interpretation above, the domain is {0} and the relation symbolized by 'e' is {<0 0>}.
And every y in the domain (the only y in the domain is 0) bears the relation {<0 0>} to 0. So there is an x (viz. 0) in the domain, such that every y in the domain bears the relation {<0 0>} to x.
Or, including '=' as a symbol, here is a model for the language of set theory that is a model of "AxEy yex":
<{0} {<'e' {<0 0>}> <'=' {<0 0>}>}>
Nota bene: Trivially there are theorems of set theory that are false in this model. So this model is not a model of set theory. My claim has never been that there is a model of "ExAy yex" that is a model of set theory. The model I show is a model for the language of set theory, but it is not a model of set theory. It doesn't need to be a model of set theory.
And to make it 'models' plural, trivially here's another:
<{1} {<'e' {<1 1>}> <'=' {<1 1>}>}>
And another that is not isomorphic with those:
<{0 1} {<'e' {<0 1> <1 1>}> <'=' {<0 0>}>}>
Quoting fishfry
'those models' there refers to models I claimed to exist, I but I had not specified them.
The domains of the models are sets. And the models themselves are sets:
For example, <{0} {<'e' {<0 0>}>}> is an ordered pair, and ordered pairs are sets.
What is your definition of 'all-inclusive' in this context?
The theorem is:
~ExAy yex
"It is not the case that there exists an x such that every y is a member of x."
or, in context, "There is no set of which every set is a member."
I said that the statement "There is a universal set" is not, onto itself. a contradiction.
I should have said that the statement "There is a universal set" is not, in and of itself, a contradiction.
The correct English there is not 'onto itself' but 'in and of itself'.
It seems that the other poster might have thought I meant 'onto' in the sense of a surjection, which I did not mean and would not even make sense in that context. But I don't blame the other poster on that particular point, since it was my mistake in English.
However, my substantive point, as I explained it clearly in other passages not with the word 'onto', stands:
ExAy yex
(read in context as, "There is a set of which every set is a member").
is not a contradiction, but it is inconsistent with the axiom schema of separation.
The theorem is:
~ExAy yex
"It is not the case that there exists an x such that every y is a member of x."
or, in context, "There is no set of which every set is a member."
That is not a definition. That is a statement that there does not exist a set having a certain property (the property of having every set as a member).
But that theorem is also equivalently:
Ax~Ay yex
"For every x, it is not the case that for every y, y is a member of x."
or, in context, "For every set, it is not the case that every set is a member of it."
That also is not a definition, but it is a universal quantification. For any domain of discourse for a model of the languge, the quantifier ranges over all members of that domain of discourse.
And, yes, that domain of discourse is a set. So, naturally, and fair enough, one might ask, "But then isn't that domain of discourse the set of all sets?"
The answer is 'no'. Given any domain of discourse, it is a set, but we don't stipulate that every set is a member of it. Indeed, we can't do that, since there is no set that has every set as a member.
Moreover, using a given set theory (for example, ZFC), we cannot, using only that particular set theory, specify a particular model (nor domain of discourse for the model) of the theory (per the second incompleteness theorem). But, for example, using a theory such as ZFC+"exists an inaccessible cardinal", we can specify a model (with a domain of discourse) of ZFC (but not of ZFC+"exists an inaccessible cardinal"); and that model would be a set, but still not a set that has every set as a member.
In sum: The universal quantifier ranges over some domain of discourse that is a set; but that set itself does not have every set as a member. So when informally we couch the universal quantifier as "For all sets", to be more accurate, we actually must mean, "For all sets in the domain of discourse" (and, as mentioned, that domain of discourse does not have every set as a member).
As a layman, it is interesting to hear that no domain of discourse is truly universal.
So when we say, explicitly or otherwise, that «not everything is within this domain of discourse»,
that which is inside that domain and that which is not, when taken together still do not form a universal domain of discourse but only a larger one.
Your question is insightful. You're thinking along the right lines. But there is no set of all things not in the domain of discourse.
We take only relative complements of sets:
df. T\S = {y | yeT & ~yeS}
That's fine. The set of all things in T but not in S.
But we don't have:
{y | ~yeS}
That is, we don't have the set of all things not in S.
In set theory, given any set S, there does not exist the set of all sets that are not in S. Because if there were, then the binary union of S with the set of all sets that are not in S would be the set of all sets:
thm: ~ExAy(yex <-> ~yeS)
proof: Suppose ExAy(yex <-> ~yeS). Then it would be {y | ~yeS}. Then S u {y | ~yeS) = the set of all sets.
A domain of discourse is a set. And there is no set of all sets that are not in that domain of discourse.
So we might be tempted to say there are more things outside the domain of discourse than in it. Except, since there is not a set of all things outside the domain of discourse, we can't even give the totality of things outside a domain of discourse a cardinality even to use expressions such as "more".
On the contrary, domains of discourse are often truly universal. They're just not always sets.
Let's take as an example the simplest and most important thing we can say about sets.
Two sets are equal if and only if they have exactly the same elements.
That is, the sets [math]\{1,2, 3\}[/math] and [math]\{2, 3, 1\}[/math] are exactly the same set. Sets are characterized by their membership, without regard to order.
What domain are we quantifying over when we make this statement? We are saying, "For all sets X, and for all sets Y, if X and Y have the same elements, then X = Y."
Well, we are quantifying over the collection of all sets. Twice, for that matter. And as Russell showed us, the collection of all sets is not a set. It is a perfectly well-defined collection. It just isn't a set.
So in fact we can, and commonly do, quantify over domains that are not sets.
What are these collections that are "too big" to be sets? They're called proper classes.
First, a class is any collection defined by a property, or predicate. So, "Is a set" is a property that's true or false about any given individual. The collection of all things for which the property is true, is a class.
From the Wikipedia page on Classes:
Some classes are sets. Others are "too big" to be sets, such as the class of all sets. Those are the proper sets.
Yet, we can still use a proper set as a domain of discourse. We do that every time we state a property of sets. "For every two sets, their union exists." Quantifying over the proper class of all sets. "Every set has a power set." Quantifying over a the proper class of all sets.
It's a perfectly everyday occurence in math to use a proper class as the domain of discourse. It's so commonplace that we don't even notice ourselves doing it.
In ZF (standard set theory), there are no official classes, so the usage is informal. There are set theories in which classes are formalized, but those set theories are not usually encountered except by specialists.
Of interest to this thread is the Russell class, [math]R = \{x : x \notin x \}[/math].
Russell showed that [math]R[/math] can not be a set. But it's a perfectly well defined proper class. It's the collection of all the things that satisfy the property [math]x \notin x[/math].
To sum up: Sometimes domains of discourse are sets, as when we say, "All positive integers other than 1 have a unique factorization into prime powers." Here, the domain of discourse is the positive integers, because we explicitly stated that. The statement becomes false if we change the domain to the real numbers, for example.
Other times, the domain of discourse is a proper class that is too big to be a set. For example, when we say, "Every set is completely characterized by its elements," we are quantifying over the universe of sets; which as Russell showed, is not a set. But it's still a proper class, and may be spoken of and used as a domain of discourse.
Quoting TonesInDeepFreeze
I'm afraid I can't agree. Many obvious counterexamples come readily to mind. We literally could not do math without quantifying over domains that are not sets.
* Every set is in bijective correspondence with itself, via the identity map. Quantifying over the proper class of all sets.
* The identity element of every group generates a 1-element subgroup. Quantifying over the proper class of all groups
* Every vector space has a basis. Quantifying over the proper class of all vector spaces.
* Every topological space contains at least two open sets. Quantifying over the proper class of all topological spaces.
Quantifying over proper classes is so common that we don't even notice it.
Quoting TonesInDeepFreeze
Correct. But there is a class of all such sets. That class is not a set. It's a proper class, characterized by the property of being "not in that domain of discourse" that was being discussed.
A final example involving complements of sets, relative and otherwise, that shows how you can annoy your teacher.
You are asked, "What is the complement of the set of even numbers?" You answer, "The odd numbers, of course." Being a good student, you implicitly assumed that the domain of discourse is the set of integers.
But the literally correct answer is: "Everything that is not an even number." That includes the odd numbers as well as Captain Ahab, the Andromeda galaxy, and the Mormon Tabernacle Choir. The complement of the even numbers, without any further domain restriction, is the proper class of everything in the universe, abstract entitied included, that are not even numbers.
The unrestricted complement of a set always exists. It just may not be a set. If the teacher wanted the complement of the even numbers relativized to the set of integers, they should have said so!
The official, formal definition of a 'model' is that the domain of discourse is a set.
The context of my remark was set theory. In that context, there is no operation of absolute complement but only relative complement.
In mathematical logic, a domain of discourse is a set. You may look it up anywhere.
Not formally. Formally, any model of set theory has as a set, not a proper class, as its domain of discourse. For any model, the universal quantifier ranges over the members of the domain of discourse, which is a set.
Right. But even with those theories, the domain of discourse for a model for the language of the theory is a set.
Even a class theory such as NBG has only models that have a domain of discourse that is a set.
Moreover, if we tried to allow a proper class to be a domain of discourse, we'd get a contradiction:
For example, suppose we are doing model theory in a class theory in which there are proper classes. Okay, so far. Now suppose U is a proper class and, for simplicity, we have a language with just one nonlogical symbol. And let R be the relation on U that, per the model, is assigned to the nonlogical symbol.
Then we have the structure . But then, unpacking the ordered tuple by the definition of tuples (such as Kuratowski), we get that U is a member of a class, which contradicts that U is a proper class.
If you look at textbooks in mathematical logic, model theory, and set theory, you will see that [s]without exception[/s] the definition of a model stipulates that its domain of discourse is a set.
Note: I put strikethrough there to accommodate the following post:
a model of ZFC. That entails the consistency of ZFC relative to the consistency of ZF. But I am rusty here, so I may be corrected.
Agreed. We're not talking about models here, though. We're talking about domains of discourse.
I'll admit that for me, domain of discourse is an informal phrase meaning, "the collection over which we are quantifying," rather than a formal or technical definition. So there may be subtleties I'm missing.
But model theory is not relevant to this conversation as I understand it.
My main point in posting was to address this concern of @Sunner:
Quoting Sunner
I wanted to assure @Sunner that indeed, there is a collection defined by the phrase, "the collection of all sets." The collection is just not a set.
Quoting TonesInDeepFreeze
Fair enough. But in general, there is an operation of unrestricted complement. I pointed that out and gave an example.
Even within set theory, there are unrestricted complements. The complement of the set {1,2,3}, within set theory, is the collection of all sets that are not {1,2,3}. That complement is a well-defined collection, but it's not a set. Of course the relativized complement of {1,2,3} in the powerset of the integers is a set. But the unelativized complement is NOT a set, even in the context of set theory. It's a proper class.
Quoting TonesInDeepFreeze
I semi-agree. In prepping my post I looked up Domain of Discourse on Wikipedia, and they did indeed say a domain of discourse is a set. I assumed they were mistaken, and were simply using "set" in its everyday, casual meaning, without regard for the issues of set-hood versus proper classes.
So I agree that if I looked it up, I'd find at least one source, namely Wiki, that claims a domain is a set. I just think they're wrong, and gave many examples to show why.
Quoting TonesInDeepFreeze
Oh my goodness. I do see your point, but I can't agree with it.
You are saying that when I make a statement such as, "Every set has a powerset," I am really saying:
1. I assume ZF is consistent.
2. By Gödel's completeness theorem, if ZF is consistent it has a model, which is a set.
3. The powerset axiom is implicitly quantifying over that set.
I simply can not believe that this is the implicit chain of logic behind every universal statement about sets. Indeed, the assumption of consistency is NOT part of set theory. Set theory can not prove its own consistency. The claim that every set has a powerset is true whether or not set theory has a model. All that is required is the axiom of powersets.
Indeed, "Every set has a powerset" is NOT a semantic claim; it's a syntactic one. It follows from the axiom of powersets. There are models lacking the axiom of powersets where the claim is false.
Perhaps we're arguing about syntactic versus semantic domains. "Every set has a powerset" is a purely formal statement in the language of set theory. It does not talk about models at all. And it does quantify over the universe of sets, which we know is not a set.
Likewise it can not possibly be the case that when we say, "The binary operation of every group is associative," we are implicitly quantifying over a "set" of all groups. There is no such set, and I can not believe there's a group theorist living who would agree with your point of view. Of course I have not asked them. But nobody carries around this implicit belief that universal statements about groups quantify over a mythical set of all groups, which provably does not exist. There is no set of all groups and we are not quantifying over it when we make general statements of groups.
Rather, we are quantifying over a proper class. From where I sit, you are being a bit unreasonable in your claim that there's a set of all groups that we're implicitly quantifying over. That's just not true.
Quoting TonesInDeepFreeze
Nice to know. Not relevant to our discussion here IMO.
Quoting TonesInDeepFreeze
I haven't sufficient technical knowledge, but for sake of discussion I'll concede your point that if we work in NBG or Morse-Kelley set theory, domains are sets. But that's beyond the scope of the discussion. In everyday math, domains of discourse frequently are proper classes. "Every set has a powerset" quantifies over the proper class of sets; NOT, as you seem to claim, over a set model of sets whose existence depends on assuming the consistency of ZF. That's just not right.
Quoting TonesInDeepFreeze
I was never well-oiled enough to even aspire to being rusty in these subjects. I'm sure we're far beyond any considerations relevant to @Sunner. And in that impressively buzzword-compliant paragraph, there is no mention of the domain of discourse. So again, none of this is relevant. You're perfectly correct that ZFC is consistent if ZF is, but what has that got to do with the conversation?
Quoting fishfry
The question was:
Quoting Sunner
"all inclusive in one way or another" is not definite. My choice was to give a mathematically definite framework for the the question. I surely do not presume to guarantee that I know what the poster in particular had in mind, but I responded correctly vis-a-vis the best way I know to make the question mathematically definite.
Quoting fishfry
Not in set theory (as I see you agree). And (just to be clear) in class theory, only with sets, not proper classes.
Quoting fishfry
It is a proper class (as you mention also) in class theory. In set theory, it doesn't exist. Clearly, the context of my posts was set theory.
Quoting fishfry
Wikipedia is unreliable for mathematics (and other subjects). But it happens to be correct on this matter. And you don't need to wonder whether it's just a fluke of Wikipedia. Look in any textbook in mathematical logic or model theory. Or any PDF book or class notes on the Internet. The universe (aka 'the domain of discourse') for a model is a set.
Quoting fishfry
No, I am definitely not saying that.
Phrases such as "quantifier ranges over" are not definite. To pin them down to a definite mathematical formulation, we turn to the mathematical logic. The method of models gives us definite formulations. A quantifier ranges over a universe. But what is that universe? Well, a universe is the carrier set for a model. So, per any given model, the quantifier ranges over the universe of that model.
In what I said, there is no need for an assumption that set theory has a model or is consistent. There is a crucial between (1) a model for the language of a theory and (2) a model of the theory:
(1) M is a model for a language L iff [fill in the definition here, in outline: M is a non-empty set together with a mapping of the non-logical symbols to (elements of the universe, n-ary functions on the universe, and n-rary relations on the universe)]
(2) M is a model of a theory T iff (M is a model for the language of T & every theorem of T is true in M)
Of course, if a theory is inconsistent, then there are no models of that theory. But whether or not a theory is consistent, there are models for the language of the theory.
To mention that a quantifier ranges over a universe per a model requires only (1) and no consideration whether any given model is or is not a model of the theory.
Quoting fishfry
Yes, if set theory is consistent, then set theory does not prove that set theory is consistent. Nothing I've said contradicts that.
Quoting fishfry
No, a sentence is true or not depending on what model we're looking at. Truth is defined per models. The power set axiom is true in some models but false in other models (as you agree). It is true in any model of set theory (since set theory includes the power set axiom) but it is false in other models (ones that are not models of set theory).
Quoting fishfry
A sentence is an uninterpreted syntactical object. Aside from what the word 'claim' means, a sentence is interpreted per models.
And the phrase "models lacking the axiom of powersets" doesn't make sense. Models don't have axioms. Rather, axiomatizations of theories have axioms. What you might mean is "models in which the power set axiom is false".
And "every set has a powerset" is just an English way of saying the power set axiom.
Quoting fishfry
I have never read of a "syntactic domain". I don't know what you mean by it.
On the other hand, we can always define unary predicate symbols. For example, in set theory, we can have the predicate symbol 'G' defined by
Gx <-> [fill in the requirements for x being a group]
Then we have universally quantified conditionals:
Ax(Gx -> P)
Note the quantifier ranges over the universe, but it happens that the formula it applies to is a conditional in which x being a group is the antecedent.
So that is a relativization of P to groups.(That's a simplification. Relativizations are recursive so that P and its subformulas are themselves relativized.]
For example, we define a unary predicate symbol 'L' and 'V':
Lx <-> [fill in the requirements for x being constructible]
Vx <-> x=x
So the relativizations such as:
Ax(Lx -> P)
read as "P holds for constructible sets".
and
V = L
for
Ax Lx
And per a given model, 'L' will map to a subset of the universe, and 'V' will map to a subset of the universe (and if the model is a model of "Vx <-> x=x" then V maps to the subset of the universe that is the universe itself).
Quoting fishfry
I didn't use the terminology to impress anyone with buzzwords. And they are not buzzwords. They are terminology of mathematics.
Quoting fishfry
It's about models. A model is a domain of discourse along with a function on the nonlogical symbols. So where I mentioned a model, there is a domain of discourse associated with that model. And the paragraph was not meant to address the original question, but rather to give an idea of how notions of proper classes as models are (as I hope I recall correctly) reducible to the syntactical approach of relativizations.
Quoting fishfry
It was added merely as an illustration of an important theorem that comes from relativizations.
Ax(Bx -> (x is even or x is odd))
And in the intended interpretation, 'B' maps to the set of natural numbers that is a subset of the universe for the model. But for another model, it might be a different universe, and 'B' might map to a different set from the set of natural numbers.
Or consider first order PA.
In the intended interpretation, the universe is the set of natural numbers, so the quantifier ranges over the natural numbers. But there are non-standard models (not just for the language but even of PA), so the quantifier ranges over a set very different from the set of natural numbers.
It is only by per a model that the domain of discourse is definite, and so it is only per a model that it is definite what the quantifier ranges over.
Well yes, we agree on that. But there is no set of all groups! The class of groups is a proper class. So you seem to be conceding my point.
In any event, much of the rest of your post is pretty technical and I'm not sure how it bears on the question. I did find this MathSE thread:
How do I quantify over the proper class of all the cardinal numbers? where there's a lot of learned back and forth about quantifying over proper classes.
The consensus seems to be that when we quantify over all sets or all cardinals or whatever, we are really restricting to the particular predicate that defines the object in question. Which doesn't help us, since the extension of a predicate is a class and not necessarily a set.
And then there is a comment from Asaf Karagila, a professional set theorist and prolific SE contributor. He says:
As far as I'm concerned, if Asaf says we can quantify over classes, we can quantify over classes. That is good enough for me.
As far as whether a domain of discourse must necessarily be a set, this seems like a matter of which definition we choose. When we make general statements about sets or groups or cardinals, we are quantifying over proper classes. Whether you call that a domain of discourse or not seems like a question of semantics.
Quoting TonesInDeepFreeze
This post crossed paths with mine. As far as I'm concerned if I have Asaf on my side I'm happy. And when we say "every set has a powerset," we are quantifying over the proper class of sets. And again, as far as what a domain of discourse is, that seems like a matter of semantics. If you say it has to be a set, then so be it. When we say every set has a powerset, we are quantifying over a proper class; but if you don't want to call that a domain of discourse, that's ok by me.
Why are you exclaiming that to me? Of course I agree.
Quoting fishfry
There is no class of all groups in set theory. Set theory has only classes that are sets. In class theory, there is the class of all groups.
To say I "concede" that is like saying I concede that 0+0=0.
Quoting fishfry
My post is nothing that isn't in, or discernible from, introductory mathematical logic.
And it bears on the questions in exactly the way I explained.
/
I can't comment on that quote without a link to it.
Except, as you've presented it alone, "quantifiers quantify over everything", I say:
Please rigorously, mathematically define "quantifier over" and "everything".
I mentioned a rigorous, mathematical usage. My posts are correct in that context. And that context is definitely relevant to addressing the original question in a rigorous, mathematical way.
Anyway, if we have a model of class theory (or even of set theory for that matter), then, yes, the sentences "Ax x is a class" or "Ax(x is a set or x is a proper class)" are true in the model. But, ironic though it may be, a model of class theory has a set as its domain of discourse. Class theory is a first order theory; and a model of a first order theory has a set as its universe.
Quoting fishfry
I explained that exactly in my post.
Moreover, in an earlier post, I proved that allowing a universe for a model to be a proper class implies a contradiction. You can go back to look at that.
Quoting fishfry
I guess you mean "question of semantics" in the sense of how we use words.
I don't presume to say what people may mean by 'domain of discourse'. I just gave an answer to the original question by using 'domain of discourse' in the sense of a universe for a model. The poster himself didn't use 'domain of discourse'. Rather, I first used it, and in the sense of rigorous mathematics as developed in mathematics. If someone else means something else by 'domain' of discourse' or 'quantify over' then of course I can't ensure that my use agrees, while meanwhile I would say, "Then what are your rigorous, mathematical definitions of 'domain of discourse' and 'quantifies over'?"
Quoting TonesInDeepFreeze
I hotlinked it. Here's the link.
https://math.stackexchange.com/questions/2724236/how-do-i-quantify-over-the-proper-class-of-all-the-cardinal-numbers
Quoting TonesInDeepFreeze
Yes.
Quoting TonesInDeepFreeze
As I stated up front earlier, my usage is informal.
And my rigorous, mathematical and standard use and explanations are not refuted (or whatever your disagreement is supposed to be) by your own informal usage.
The original question was informal. The original question was in invitation to explain a seeming contradiction. That merits a response that is rigorous and definitive, in order to appreciate that mathematics indeed does not tolerate such a contradiction, not just by informal hand waving, so that when we look at the matter with exactness, we do show that the seeming contradiction actually is avoided.
As I'm sure I've agreed several times. If you don't want to call quantification over a proper class a domain of discourse, I'm fine with that. We frequently quantify over proper classes, however you call it.
Quoting TonesInDeepFreeze
I thought your responses to the recent OP @Sunner were too detailed and technical to be of use at the level the question was being asked. And, frankly -- not really wanting to get back into this -- wrong. The Russell class DOES define a perfectly good collection. That was the question. @Sunner had the impression (rightly or wrongly) that you said it didn't. I pointed out that it did. Perhaps OP misinterpreted what you said. In that case you were right, and I added clarity. So everyone can be happy, yes?
Not sure what contradiction "mathematics indeed does not tolerate." The referent of this paragraph is unclear.
(1) I don't know exactly which post(s) and passage(s) he is saying "no" to.
(2) I understood everything in that thread prior to his post. And, modulo details of phrasing, I agree with the replies that, in set theory, we simply use the predicate 'is a cardinal'. That is what I said myself in my posts above.
(3) "bounded quantifiers" there is just another way of saying what I said: The antecedent of the conditional stipulates that a set has a certain definable property. And I said (though I don't know whether that poster agrees) that the quantifier itself ranges over the universe, not just those sets having the property specified in the antecedent of the conditional.
So, if I understand that poster, he's mostly in agreement with me, not you.
You had the notion that the quantifier ranges over groups. But the quantifier ranges over the universe.
As I understand him, his remarks mostly coincide with mine:
"quantifiers quantify over everything."
If 'everything' means 'everything in the domain of discourse' then I agree.
"All sets."
I don't know what 'all sets' means. But if it means 'all sets in the universe', then I agree.
"Bounded quantifiers are shorthands to make it clearer that we are only interested in a particular set."
That seems to be what I'm saying about the antecedent of the conditional.
"But you can bound them to a class just as easily."
I don't know what he means by 'bound them to a class'. But, as I keep saying, we can relativize (use a conditional) to a proper class through a defined unary predicate. Also, another poster there suggested using a defining formula, which is tantamount to what I've said.
That's your informal understanding. I can't comment with real definiteness, because your informality doesn't provide a clear, definite meaning of 'quantify over'. Meanwhile, I have given the rigorous sense in which I use 'quantify over'.
Quoting fishfry
So we've moved from your substantive criticisms, which were incorrect, to a criticism of the pedagogy.
My answer to the poster's question is somewhat (though not terribly) technical to avoid confusions from oversimplification or vague hand waving. Meanwhile, you quoted me about that, why I was as technical as I was, but replied merely to say "too technical". Argument by mere assertion is what that is. Moreover, if a poster lacks familiarity with set theory and mathematical logic, then we can bet that there is no ideal answer for him; no answer that wouldn't be either too vague to be responsible to the mathematics or too technical that he wouldn't quickly understand it. (Though, this particular poster did basically grasp the explanation.) I choose to err on the side of being correct, and then to leave it to the poster to follow up himself by learning more about the subject.
Quoting fishfry
It refers to the original question, as I said exactly that's what it refers to - in the very first sentence of the paragraph.
First you challenged me on substantive points in my posts, and you were incorrect. Most particularly, a universe for a model is a set.
Then you griped that my answer to the poster was too technical and not helpful, which contradicts what the poster himself said.
So, I really don't know what your trip is.
Yes but he understood the opposite of the correct answer!
Quoting TonesInDeepFreeze
I've been feeling the same way about you.
I'll quit while I'm behind here.
You haven't shown that my answer is incorrect. Nor has anyone said what other answer is "the" correct answer.
The problem that deserved an answer (in my words):
Set theory says, "For all sets, it is not the case that it has every set as a member" (i.e. there is no set of all sets). But that refers to all sets. So isn't what is referred to by "all sets" a set that has every set as a member, which contradicts that there is no set of all sets?
My answer is basically: The meaning of "for all" is per models, such that for any given model there is a universe that is a set, but it is not the non-existent set of all sets.
That is perfectly correct and it shows that the seeming contradiction is not actually a contradiction.
To refute my answer would require, refuting at least one of these claims:
(1) "for all" is understood per models. Correct. Just look it up in any textbook in mathematical logic.
(2) The universe for a model is a set. Correct. Just look it up in any textbook in mathematical logic.
(3) The universe for a model is not the non-existent set of all sets. Correct. There is no such set, so no set, including a universe for a model, is the non-existent set of all sets.
Quoting fishfry
I haven't, as you have, gone on and on challenging someone who is making correct mathematical statements. I have no trip, so nothing for you to wonder about.
What does that mean?
Do you mean that there is no sentence that is true in all models?
But there are sentences that are true in all models.
Quoting Sunner
No, there are sentences that are true in all models, sentences that are false in all models, and sentences that are true in some models but false in other models. And, for any given model, there is not sentence that is both true and false in that model.
Your question is incoherent.
The list of all lists (Call this L)
The list of all lists that list themselves (Call this LL)
Both the above lists list themselves (put differently, both lists are members of themselves).
Importantly, they are only members of themselves in their own respective lists. As in the list of all lists only lists itself in the list of all lists (L is only a member of itself in L). It does not list itself in the list of all lists that list themselves (L is not a member of L/itself in LL precisely because it is a member of LL).
The point I'm trying to make:
You cannot have a set of all sets that are members of themselves with all its members actually being members of themselves whilst they are members of it.
This should resolve the subset issue and we should no longer contradictorily say "the set of all sets is contradictory".
To my understanding, the subset issue was because you could have a set of all sets that are members of themselves. Since you could have this you should also have been able to have a set of all sets that are not members of themselves so that there are no inconsistencies in the subset level. I have shown that you cannot have a set of all sets that are members of themselves.
Wanting to have a set of all sets that are not members of themselves that is itself not a member of itself is a contradictory thing to want.
Only the set of all sets can contain all sets that are not members of themselves precisely because all sets are members of it and not themselves (of course it is a member of itself)
My full writing on Russell's paradox can be found here if interested:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
I don't see how that follows.
Quoting Philosopher19
That's exactly what Russell was proving.
Quoting Michael
So let's say he proved you could not have a set that contains all sets that are not members of themselves and nothing more.
Did he prove that the set of all sets is contradictory?
Did he prove that you could not have a set that contains all sets that are not members of themselves? Because by definition/logic, the set of all sets contains all sets that are not members of themselves (as well as itself).
If Russell proved anything, it's that you can't create a new set that contains absolutely all sets that are not members of themselves and nothing more.
Again, the set of all sets contains absolutely all sets that are not members of themselves as well as itself.
Your own conclusion is the answer to this question.
Quoting Philosopher19
If all sets are contained in the set of all sets (that are not members of themselves and nothing more), then still, no sets are on any other level than the other sets, all of them now being contained together inside one container (that is what contained in means)... except the set of all sets, itself also being a set. But if it is a set, it would have to be contained along with the other sets. But it's not. But it is. This is paradoxical.
I don't think there is a way completely out of the paradox. You don't undo this paradox. It manufactures itself as we speak about it from any direction.
To skip to the end explained a bit further below, the set of all sets becomes the definition of what a set is, hiding in an example of what a set is. The paradox arises from the fact that the set of all sets serves as an example of one of the sets, and the definition of any/all of the sets at the same time. The set of all sets, is itself a setting of objects into a mentally constructed container. But when we are seeking to contain all sets (mentally constructed containers) in a set, it could equally be said we are seeking to define what a set is. If we say what a set in-itself is (if we define a set), we say something of all sets. We have created a set whose members include something of all sets, or we have created a definition that applies to all sets, namely the definition of all sets.
This doesn't resolve the logical problem; it merely restates it. But I think shows how the assertion "set of all sets" points to an edge or limit where logic itself and the language used to communicate mental constructs and logic, are distinct from each other, and here, unable to overlap. Basically, we can't say what we mean here, but we somehow still know something and know what we mean.
Stepping back, a 12-year-old child who understands what a set is, can look at 5 different sets on a blackboard (numbers, letters, shapes, etc) and can easily point to all 5 of the members of the "set of all sets on the blackboard". Thus the concept of "the set of all sets" is simple, easy, useful, logical, functional, even for children. But then you ask the child "But what about the set of all sets itself? Isn't that now one of the sets on the blackboard? What happens when you point that one out too? You've just added another set to the blackboard, making your prior answer of "5" wrong. Or you failed to show how the set of all sets on the blackboard is itself one of the sets on the blackboard and given the answer '6' in the first place." Now a grown, seasoned, pioneering mathematician and logician is perplexed. Thus, the concept of the set of all sets is both really child's play, and seemingly impossible to penetrate for a wise old professor.
But I think if you look at it from other directions, (like a child perhaps), I think we start to see why even the child can make easy use of the "set" (one example being the set of all sets) despite the fact that these sets can be made to appear and disappear both within and without themselves when we say "the set of all sets."
What is a set? A set is both 1) a membership (usually of multiple members but not necessarily, but comprised of membership nonetheless), and 2) their gathering as one containing reference. Four penguins and four seals in a zoo - the set of penguins, which is a unity as one containing reference, is the multiplicity of four member individual penguins. This set is not a penguin itself, because it is a set, and this set must not be a penguin because it has to sit beyond the penguins in order to contain them all as a set. But this set evaporates if we remove all of the penguins, because it is a set of four penguins.
These are the moving parts here. Sets must be distinct from their members in order to be sets of any members. And when looking for a set of penguins, we see that the set is not only distinct from its members, the set is distinct in kind - it's not a penguin. Sets are not their own members. But sets must have members, or be comprised of membership. (Please ignore the empty set here, or pretend the empty set has one member, the object "nothingness".)
So what are we actually doing when we say "set of all sets"? Are we taking all sets, turning them into member objects like penguins, and then stepping outside these objects to make something that is different in kind to those objects, calling it a set, namely a set of all sets? Are we just misuing the word "set" somewhere when we say "set of all sets"? Or have we left the sets alone and created a new class of set so that the set of all sets is different in kind from all of the sets that are its members?
Crack it open again. What is a set? A set is a form of "all". You have four penguins and four seals in a zoo, and someone asks "how many penguins are in the zoo?" The answer can be to count the members of the set of penguins, or it can be to count all of the penguins. You don't need to clarify the "set of all penguins" to come up with same answer. You can count "the set of penguins" or "all penguins" and conduct the same operation. All of the penguins is the same thing as the set of penguins. Therefore, a "set" is a semantically distinct but nonetheless an equivalent form of "all". "All" seals means the same thing as "the set of" seals here. Part of the essence of "set" is the notion of "all" or part of the notion of "all" is to create a "set".
Now apply this to the proposition "the set of all sets." It becomes the "all of all alls". This just sounds like poetry in need of analytics to clarify. The set of all sets is the be-all end-all of alls, cried the poet!
But I think there is some analytic clarity here. Think now of encircling members as an action we will call "setting" things; instead of fixing a set as a stagnant "x", think of it as an action of "containing". Setting as an action can be made distinct from a stagnant "all" which the setting action constructs. (I could do this by all-ing a stagnant set, but did I just actually say "all-ing"? Hope I don't have to do that to this conversation! But the fact of this temptation shows how we are at an edge or limit between what is logical and clear, but what can't be communicated in language.).
Now, the definition of "setting" is "the act of identifying all members as a set." When we say "the set of" in reference to anything, we are in the act of drawing a container, we are containing members by distinguishing those members from non-members, but we are acting, we are "setting" the membership. We have to sift through the 8 animals in the zoo, identify each individual uniquely, and then by drawing the container, by setting the membership, we claim "the set of penguins has four members."
This becomes as metaphysical as it is logical/mathematical. Now we are talking about "the all" and the "the individual identity of a single member" and "sameness" of membership and "distinctness" from non-membership and the action versus the thing acted upon versus the thing thereby constructed, namely the "set" which is the same as the "all".
Step back one more time. What is a set? It's a construction. It's a mental construction. It takes even physical objects (penguins) as members, but, of them, (as in "set of"), makes a mental construct. So a set of, or the act of setting, becomes the equivalent of making an idea of, or defining a limit or container. Now, we can analogically see that the "set of penguins" is equivalent to the definition of one of those penguins, equivalent to those defining characteristics that both identify each individual penguin as they do place all penguins as member of the set of penguins.
Setting becomes defining, or a set becomes a sort of quantifiable, demonstrable way of making a definition.
Applying this to setting itself, as when setting "sets", the container for all containers, therefore, is also the definition of all containers.
So the set of all sets means the same thing, or serves the same purpose as the definition of any set. A set, is like a definition; a definition is a statement about all of example members; so the set of all sets, is a mathematical way of denoting the definition of all sets. This is why the child blows right through this. If you understand what a set is, you can easily populate the set of all sets.
This doesn't resolve the paradox. It maybe explains how we, like the 12-year-old above, already live with it. Setting is defining, so when setting all sets, in a practical sense, we are defining what all sets are. In a logical sense, we are still creating a set that can't be a member of itself, but at the same time is a member of itself.
Quoting Fire Ologist
But I am not saying this. I am saying the set of all sets semantically/logically/rationally contains all sets and it is a member of itself (because it is itself a set). So the set of all sets consists of one set that is a member of itself (itself) and many sets that are not members of themselves (all sets other than the set of all sets). Where is the paradox in this?
Consider semantics. We did not create or make up the semantic that something can be a member of itself or a member of other than itself (or that triangles are triangular or that the set of all sets encompasses all sets). We are aware that such is the nature of Existence (that triangles are triangular and that the set of all sets encompasses all sets) and have expressed awareness of it. The contradiction lies in wanting a set that contains all sets that are not members of themselves that is itself not a member of itself (which appears to be the set that Russell was talking about when he asked is the set of all sets that are not members of themselves a member of itself or not). Such a thing is by definition, contradictory.
Quoting Fire Ologist
Quoting Fire Ologist
Quoting Fire Ologist
I don't see anything wrong with the "all of all alls". You have alls of various sizes with one all encompassing absolutely all alls. By definition, this all that contains absolutely all alls has to be infinite.
Problems occur if you consider the elements of a set to not be themselves sets. Set theory only talks about sets. It does not, for example, talk about individuals.
The lists only list other lists...
Yes. But a set, by definition, cannot contain itself. The set is the act of containing. The set doesn't come to be until something else (members) are contained. Triangles and triangularity aren't equal beings. Triangularity can be predicated of something that is not a triangle. So we aren't trying to force the set of all sets to be a member of itself, just as much as we are not trying to force a sub-set as a member set to simultaneously be the set that includes itself with other member sub-sets. We just keep recognizing and restating that the set of all sets already has to not be itself a container because it is one of the members, while it has to not be one of the members, because it is a container.
I say this is because, as we keep drawing sets, and keep getting bigger, by the time we get to the most inclusive set, the set of all sets, if we want to use logic, we need to stop thinking of the set of all sets as a set. At that point, we've reached a new kind of thing, at the end of members, just like we do when we create any set (We go, penguin, penguin, penguin, and then new thing, set of penguins). When those members are sets themselves, we reach the definition of all sets, or the concept that all sets share. The set of all sets is an empty way of exemplifying the definition of "set".
Quoting Philosopher19
Agree, but wouldn't it also, in a naive sense, have to also be finite, because it is now an "encompassing" container? A container that ever-grows because its members ever-multiply is not a container at all. And we arrive where we started. Again. Or I guess I'm now saying a set of infinite, ever-increasing members, never gets to be a set.
I admit that this subject clearly needs careful focus that I've never done. Russell himself didn't resolve this - I doubt I can.
But I don't think it can be resolved because resolutions are logical, and with the assertion "set of all sets", we stand at the edge of all things logical facing, raw assertion - if we retreat, we remain logical and ignore the issue; but if we press on beyond the edge, trying to explain the shape of this edge, we find that logic alone, so reliable on the way to the edge, no longer works and the things we say are difficult to make sense of.
Quoting Fire Ologist
I think I understand where you're coming from. I agree that something cannot be the container of itself in the way that you mean "container of itself". But I see a meaningful difference between something being a member of itself, and something containing itself in the way that you mean. To highlight this difference to you, consider the following:
I make a list of all things in my room. The list is in my room so I list the list in the list. This is clearly meaningful is it not? The list meaningfully lists itself does it not? This is what I mean by contain itself (as in in the sense of being a member of itself). Not in the sense of a box that contains a box which is itself.
Quoting Fire Ologist
I don't think it follows that because x is encompassing, x is finite. If everything is in the Infinite, this does not mean that there is an end to the Infinite, but it does mean the Infinite encompasses all things.
Quoting Fire Ologist
If something (like a number sequence) goes no forever, it does not mean that it reaches infinity or that it is an infinite set/sequence. Does it reach infinity for it to be classed as an infinite set? Can we count to infinity to say it reaches infinity to be classed as an infinite set? So we cannot say 1, 2, 3 ad infinitum is an infinite set, precisely because infinity will not be reached for it to be classed as an infinite set.
There is no beyond infinity. Infinity is complete. It is precisely because Infinity is Infinite that something can go on forever or increase increasingly. But again, this does not make it Infinite/Complete. It just means it's a part of the Infinite/Complete.
Quoting Banno
That's like saying a list can't list itself. Is this in itself not enough to conclude that the Z-F set theory is inadequate/incomplete?
I'd say it's not just incomplete. It's contradictory in the sense that it logically implies "a list can't list itself". That's like saying a shape can't be a triangle (which is contradictory because a shape can be a triangle).
But you don't solve a paradox or contradiction by seeking refuge in another.
It is more damning/problematic to reject the set of all sets than to accept ZFC, but any statement, belief, or theory that has a contradiction in it, is wrong by definition.
In any case, I believe I presented a true solution to Russell's paradox. The full writing is here if you are interested:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
Russell's paradox is a mathematical proof that the axiom schema of unrestricted comprehension leads to a contradiction. As such, early naive set theories had to be abandoned.
Zermelo–Fraenkel set theory is the most used replacement, and doesn't allow for a universal set.
New Foundations is an alternative replacement that does allow for a universal set.
This isn't really anything to do with philosophy. It's just about the internal consistency of some set of mathematical axioms. Some lead to a contradiction, as Russell's paradox shows, and so their axioms must change.
If you're trying to argue that a "correct" set theory must allow for a universal set then I don't think you really understand mathematics.
Yes, the axiom of regularity and the axiom of pairing entail that no set is an element of itself.
The question of what "a set" is, the definition of "set", becomes an issue when you consider the possibility of an empty set. If a "set" is taken to be the type, category, or definition which indicates the criteria of membership, then an empty set is possible. If "set" is supposed to refer to the group or collection of elements itself, then an empty set is impossible because that would be a non-existent collection of elements. The "non-existent collection" could not be understood by that definition, because that would mean "set" is understood by the category, not the elements in membership, So it is at this point, when we consider the possibility of an empty set, that we need to make a judgement about the relationship between individuals and sets.
Quoting Michael
I believe I understand Russell's paradox very well, but I am not a mathematician.
So you understand the below?
Axiom of extensionality:
[math]{\displaystyle \forall x\,\forall y\,(\forall z\,(z\in x\iff z\in y)\implies x=y)}[/math]
Axiom schema of unrestricted comprehension:
[math]{\displaystyle \exists y\forall x(x\in y\iff \varphi (x))}[/math]
Substitute [math]{\displaystyle x\notin x}[/math] (the Russell set) for [math]{\displaystyle \varphi (x)}[/math]:
[math]{\displaystyle \exists y\forall x(x\in y\iff x\notin x)}[/math]
Therefore:
[math]{\displaystyle y\in y\iff y\notin y}[/math]
The conclusion is a contradiction. Therefore the premises are inconsistent. In this case, the problematic premise is the second premise, which is the axiom schema of unrestricted comprehension.
I don't understand the notation you have used. If you were to put it into words, I could reply in kind.
The axiom of extensionality
Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B:
[math]{\displaystyle \forall x\,\forall y\,(\forall z\,(z\in x\iff z\in y)\implies x=y)}[/math]
The axiom schema of unrestricted comprehension
There exists a set B whose members are precisely those objects that satisfy the predicate [math]?[/math]:
[math]{\displaystyle \exists y\forall x(x\in y\iff \varphi (x))}[/math]
Russell's paradox
Let [math]?(x)[/math] be [math]{\displaystyle x\notin x}[/math] ([math]x[/math] is not a member of [math]x[/math]):
[math]{\displaystyle \exists y\forall x(x\in y\iff x\notin x)}[/math]
Therefore:
[math]{\displaystyle y\in y\iff y\notin y}[/math]
This is a contradiction. Therefore the axiom of extensionality and the axiom schema of unrestricted comprehension are inconsistent.
ZFC replaces the axiom schema of unrestricted comprehension with the axiom of regularity and the axiom of pairing and as such is consistent. This doesn't allow for a universal set.
New Foundations restricts the axiom schema of comprehension by allowing only stratifiable formula for [math]?(x)[/math]. This allows for a universal set.
Are you saying that A and B only contain X as members of themselves and they contain nothing other than X as members of themselves? If you are, I don't see it included in the above. If you are not, then either A or B can contain members other than X such that A is not equal to B despite both A and B have all Xs as members of themselves
It says that A and B are equal if every member of A is a member of B and every member of B is a member of A.
Quoting Michael
Ok
Quoting Michael
is predicate ? "A and B are equal if every member of A is a member of B and every member of B is a member of A"? If not, what is it?
In Russell's paradox, [math]?[/math] is "sets that are not members of themselves".
Thanks.
So when you say: Quoting Michael are you essentially saying "there is a set that contains all sets that are not members of themselves"? If not, can you clarify?
Yes. Given the axiom schema of unrestricted comprehension there exists a set B whose members are sets that are not members of themselves.
This leads to a contradiction. If B is not a member of itself then it is a member of itself.
Therefore, we must reject the axiom schema of unrestricted comprehension (or the axiom of extensionality, but that would be far too problematic).
Quoting Michael
There is a difference between:
1) There exists a set whose members are sets that are not members of themselves
2) There exists a set that contains all sets that are not members of themselves
2 is not contradictory at all whereas 1 could be contradictory.
2 is not contradictory because by definition, the set of all sets contains all sets that are not members of themselves and it is a member of itself. Where is the contradiction here?
1 is contradictory if you say set B only contains all sets that are not members of themselves. Again, the only set that by definition can contain all sets that are not members of themselves, is the set of all sets and it does not just contain all sets that are not members of themselves, it contains itself too.
All sets that are not members of themselves have to be a member of something don't they? It's like saying all existing non-self-contingent things have to be contingent on something don't they?
Yes, that's the premise. Let R be the set of all sets that are not members of themselves. This is the Russell set.
If R is not a member of itself then it is a member of itself. This is a contradiction. Therefore we must abandon the axiom schema of unrestricted comprehension.
ZFC replaces the axiom schema of unrestricted comprehension with the axiom of regularity and the axiom of pairing. These entail that no set is an element of itself, and so doesn't allow for a universal set.
Whereas NF restricts the axiom schema of comprehension by allowing only stratifiable formulas. This doesn't allow for a Russell set but does allow for a universal set.
There's really nothing to argue here. Russell's paradox is an undeniable proof that naive set theory is inconsistent. The "resolution" is to replace the problematic axiom with others, which is what has been done. ZFC does it one way, NF another way, and others in other ways.
( A is a subset of A ) = ( {1,2,3} = {1,2,3} )
(A is a member of A ) = ( {1,2, A} = {1,2, {1,2, {A} }, } = {1,2,{1,2,{1,2,{1,2{...}}}}}
From this example, you can see that defining a set as a member of itself immediately leads to infinite recursion.
Let A be the set of all integers.
Let B be the set of all positive integers.
Every member of B is also a member of A, but some members of A are not members of B (i.e. the negative integers).
B is a subset of A.
A is a superset of B.
But you haven't addressed my point.
We are in agreement that p) you cannot have a set that only contains all sets that are not members of themselves. I am saying this has nothing to do with the fact that you can have q) a set that contains all sets that are not members of themselves. The set of all sets contains all sets that are not members of themselves (plus itself). Again, where is the contradiction in this?
It is clear that p is contradictory. You have not shown how this logically obliges us to view the set of all sets as contradictory (you have only shown how p is contradictory. See your post). And you have also not shown any contradiction in q.
This is where you show that you don't understand the problem.
Naive set theory accepts both the axiom of extensionality and the axiom schema of unrestricted comprehension.
The axiom schema of unrestricted comprehension entails that there is a set that only contains all sets that are not members of themselves (the Russell set). In conjunction with the axiom of extensionality this is a contradiction, and so naive set theory is shown to be inconsistent.
This is all Russell's paradox does.
In response to this, naive set theory had to be changed. Specifically, it had to reject the axiom schema of unrestricted comprehension.
ZFC replaces this axiom with two others; the axiom of regularity and the axiom of pairing. These axioms entail that no set is a member of itself, and so that there is no universal set.
NF restricts the axiom schema of comprehension to permit only stratifiable formulas. This does allow for a universal set.
It's not entirely clear what you're trying to argue. Is it that Russell's paradox doesn't prove naive set theory is inconsistent? Is it that ZFC allows for a universal set? You'd be wrong on both counts. Is it that NF is "better" than ZFC? I'm not entirely sure such a claim would make sense.
Or are you trying to argue that, independently of any set theory, a set that contains all sets that are not members of themselves is possible, therefore only set theories that allow for this are "correct"? That's putting the cart before the horse.
Quoting Michael
I followed your original notation and tried to get clarity on it. We came to the following:
Quoting Philosopher19
To which you answered "yes". To which I highlighted to you a clear difference:
Quoting Philosopher19
Again, 1 is contradictory. Put it in clear language as to why the contradictoriness of 1 obliges us to reject 2 or to view the set of all sets as contradictory.
The axioms of naive set theory entail (1). Therefore, the axioms of naive set theory are inconsistent.
This is all Russell's paradox shows. Again, you're showing that you don't even understand the problem.
When you say the axioms of naive set theory, are you referring to those notations that I asked you to put in clear language. If so, it seems to me you left half way through trying to clarity on it.
Yes.
Quoting Philosopher19
I'm not interested in teaching you mathematics. I am simply explaining to you that Russell proved that the axioms of naive set theory are inconsistent. That's it. It's not a debatable issue.
Peace
Quoting Philosopher19
Then have a look here:
Quoting Open Logic: Complete build
Banno's Law says that it is easier to critique something if you begin by misunderstanding it. That is what your OP does.
You, too,
, you may find the Open Logic text useful.
Again, I have said this multiple times. I recognise and acknowledge the following:
Quoting Open Logic: Complete build
What has not been shown to me is how this logically obliges us to view the set of all sets as contradictory.
Again, it is clearly contradictory for any set to contain all sets that are not members of themselves and no other set. But this does not mean that it is contradictory for the set of all sets to contain all sets that are not members of themselves. The set of all sets contains absolutely all sets that are not members of themselves and it is a member of itself.
Just because one set is contradictory (In the above case R), doesn't mean another set is also contradictory when they are not the same sets.
The set of all sets and R are not the same sets.
And this shows that you have not understood R = {x : x ? x}:
:wink:
To me it looks like no meaningful answer has been given to me and that what I have posted has not been paid attention to.
Because the choice is between the whole of the remainder of that project being founded on an error unnoticed by more than a century of study by logicians world wide; and your being mistaken.
Which is most likely?
Below are three theorems, using only intuitionistic (thus classical too) rules of inference, applied to certain axioms:
One may wish to conceive mathematics in some other way, but if the discussion pertains to the rules of inference and axioms of ordinary mathematics, then below are English renditions of formalizations that have formalized proofs that are are machine-checkable from the stated axioms and rules of inference; that cannot be rationally disputed.
(1) There is no y such that for all x, x is a member of y if and only if x is not a member of itself.
Proof is by pure logic alone, no set theory need be invoked, as it is merely a case of:
There is no relation M such that there is a y such that for all x, x bears M to y if and only if x does not bear M to x.
(2) There exists a unique y such that for all x, x is member of y if and only if x is a member of x. That y is 0 (the empty set).
Proof uses the axiom of regularity, the schema of separation for existence, and the axiom of extensionality for uniqueness.
(3) There is no z such that for all x, x is a member of z.
Proof uses (1) and the axiom schema of separation.
EXPLANATIONS:
'The set of all sets that are not members of themselves' refers to a set, call it 'R', such that for any set x, x is a member of R if and only if x is not a member of x.
Russell's paradox is this contradiction:
The set of all sets that are not members of themselves is a member of itself, and the set of all sets that are not members of themselves is not a member of itself.
A proof* is obvious:
Suppose there is a set R such that for all sets x, x is a member of R if and only if x is not a member of x.
So substituting R for x, we have:
R is a member of R if and only if R is not a member of R.
Suppose R is a member of R. So R is not a member of R. So both R is a member of R and R is not a member of R.
Suppose R is not a member of R. So R is a member of R. So both R is member or R and R is not a member of R.
But either R is member of R or R is not a member of R. And, in either case, as above, R is a member or R and R is not a member of R. So R is a member of R and R is not a member of R. QED.
/
So, the supposition 'There is a set R such that for all sets x, x is a member of R if and only if x is not a member of x' yields a contradiction. So we infer the negation of that supposition, viz. 'There is no set R such that for any set x, x is a member of R is and only if x is not a member of x'.
But the earliest version of set theory had an axiom schema we all 'unrestricted comprehension', which is (read 'P(x)' as 'x has property P'):
For any property P that we can formulate, we have the axiom:
There is a set y such that for all sets x, x is a member of y if and only if x has property P.
Then let P(x) be 'x is not a member of x'.
Then we have:
There is a set y such that for all sets x, x is a member of y if and only if x is not a member of x.
But above (using 'R' instead of 'y') we proved that there is no such set y.
So the axiom schema of unrestricted comprehension yields a contradiction.
/
So set theory was reformulated to not have the unrestricted axiom axiom schema of comprehension but instead a restricted axiom schema (called the 'separation schema' or the 'specification schema' or the 'subset schema'):
For any property P that we can forumulate, we have the axiom:
For all sets z, there is a set y such that for all sets x, x is a member of y if and only if (x is a member of z and P(x)).
And to that schema we add the axioms: Extensionality, Pairing, Union, Power Set, Infinity and Regularity. (This is known as 'Z set theory').
In Z set theory (and in the extended theories, ZF, ZFC, etc.), there is no known proof of 'There is a set R such that for all sets x, x is a member of R if and only if x is not a member of x', and no known proof of any contradiction.
Moreover, in Z set theory there is a proof of 'There is not a set R such that for all sets x, x is a member of Y if and only if x is not a member of x'. It's just Russell's argument again:
Suppose there is a set R such that for all sets x, x is a member of R if and only if x is not a member of x.
So substituting R for x, we have:
R is a member of R if and only if R is not a member of R.
Suppose R is a member of R, so R is not a member of R. So both R is a member of R and R is not a member of R.
Suppose R is not a member of R, so R is a member of R. So both R is member or R and R is not a member of R.
But either R is member of R or R is not a member of R. And, in either case, as above, R is a member or R and R is not a member of R. So R is a member of R and R is not a member of R.
NOTICE that that argument didn't even need any axioms of set theory anyway. Moreover, we don't even have to mention the notion of 'set' nor 'member of'. We have (read 'M(x y) as 'x bears property M to y'; with one instance being 'x bears the property of being a member of y'):
There is no 2-place relation T such that there is a y such that for all x, T(x y) if and only if it is not the case that T(x x). I.e., 'There is no 2-place relation T such that there is a y such that for all x, x bears T to y if and only if x does not bear T to x.'
It's purely logical, not needing to mention sets nor membership:
Suppose there is a T such that there is a y such that for all x, T(x y) if and only if it is not the case that T(x x).
So, substituting y for x, we have T(y y) if and only if it is not the case that T(y y). And, as seen above, we then derive a contradiction.
Indeed, this may be seen as the lesson of the Barber Paradox in its purest form (don't even have to mention 'village' or 'barber'). There, instead of the 2-place property 'is a member of' we use the property 'shaves':
There is no y such that for all x, y shaves x if and only if x does not shave x.
/
Moreover, even in set theory, we don't have to mention the term 'set'. Rather, the only primitive is 'member of'. Indeed, though most people don't bother to do state these trivial steps, we can actually define 'set' ('iff' stands for 'if and only if'):
First with the axiom of extensionality, we define a unique y such that for all x, x is not a member of y. We dub that unique y as '0':
Axiom of extensionality:
For all x and y, if for all z, z is a member of x iff z is a member of y, then x equals y.
Also, from the logic of identity, we already have:
For all x and y, if x equals y, then for all z, z is a member of x iff z is a member of y.
So putting that together with the axiom of extensionality we have:
For all x and y, x equals y iff for all z, z is a member of x iff z is a member of y.
Then:
From the schema of separation we have:
For all z, there is a y, such that for all x, x is a member of y iff (x is a member of z and (x is a member of x and x is not a member of x)).
Let z be any object. (There is at least one object, as provided by the logic itself).
So there is a y, such that for all x, x is a member of y iff (x is a member of z and (x is a member of x and x is not a member of x)).
By logic, we can reduce that to:
There is a y, such that for all x, x is a member of y iff (x is a member of x and x is not a member of x)).
By the axiom of extensionality, such a y is unique, since there is no x such that x is a member of x and x is not a member of x.
Then:
df. 0 is the unique y such that for all x, x is a member of y iff (x is a member of x and x is not a member of x).
df. x is a class iff there is a y such that y is a member of x or x is 0.
df. x is a set iff x is a class and there is a z such that x is a member of z.
So, since the context here is set theory, we can drop mentioning 'set' throughout. Though, we use the nickname for 0 as "the empty set".
/
Next, using only the schema of separation we prove that there is no z such that for all x, x is a member of z:
Suppose there is a z such that for all x, x is a member of z:
So, by the schema of separation:
There is a y such that for all sets x, x is a member of y iff (x is a member of z and x is not a member of x).
With some routine logic steps, we get:
There is a y such that for x, x is a member of y iff x is not a member of x.
Thus, there is no z such that for all x, x is a member of z.
NOTICE, we need no invoke extensionality nor regularity, only an instance of the axiom schema of separation.
/
* That proof uses excluded middle, but we can prove it intutionistically too:
Instead of proving that both R is a member of R and R is not a member of R, we prove a different contradiction, viz. that both R is not a member of R and it is not the case that R is not a member of R:
Suppose there is a set R such that for any set x, x is a member of R if and only if x is not a member of x.
So substituting R for x, we have:
R is a member of R if and only if R is not a member of R.
If R is a member of R then both R is a member of R and R is not a member of R. But it is not the case that both R is a member of R and R is not a member of R. So, by modus tollens, R is not a member of R.
If R is not a member of R then both R is a member of R and R is not a member of R. But it is not the case that both R is a member of R and R is not a member of R. So, by modus tollens, it is not the case that R is not a member of R.
But both 'If R is a member of R then R is not a member of R' and 'If R is not a member of R then R is a member of R'. So both R is not a member of R and it is not the case that R is not a member of R. QED.
Quoting Banno
That sounds like bias and dogma to me as opposed to actual discussing of the argument with attention to detail.
If ZF is consistent then ZF is incomplete, in the sense that there are sentences in the language of ZF such that neither they nor their negations are theorems of ZF. But any consistent, recursively axiomatized theory that expresses a certain amount of arithmetic is incomplete.
'inadequate' is not defined here.
/
Regarding the claim that set theory is contradictory.
A contradiction is a sentence P along with not-P.
There is no known proof of a contradiction in set theory.
One may find that the theorems of set theory go against one's own intuitions or conceptions about mathematics, but that does not establish a claim that the theory is contradictory, since 'contradictory' doesn't mean "Goes against the way I conceive things" but rather means "Proves both a sentence P and not-P".)
It doesn't. This is your misunderstanding. Russell's paradox only shows that the axiom schema of unrestricted comprehension leads to a contradiction, and so that naive set theory is inconsistent.
There is no y such that for all x, x is a member of y iff x is not a member of x
nor
There is no z such that for all x, x is a member of z.
At least personally I would take 'list' to mean a function whose domain is either a natural number or the set of natural numbers.
A finite list has a natural number as its domain (recall that every natural number is the set of its predecessor natural numbers).
A denumerable list has the set of natural number as its domain.
I guess one might say that any ordinal could be the domain of a list, including countable ordinals other than the set of natural numbers or even uncountable ordinals.
/
Then, what would be involved here is not lists being members of lists, but rather lists being in the RANGE of lists. So it is not "a list being a member of itself" but rather a list being in the RANGE of itself. That is, to say "the list lists itself" doesn't mean "the list is as a member of itself" but rather "the list is in the RANGE of itself".
For example, If I have a list L of three lists, then it looks like this
L = {<1 List-a> <2 List-b> <3 List-c>}
List-a, List-b, and List-C are not in L. Rather they are in the RANGE of L.
/
In any case, there is no known contradiction in set theory arising from questions of lists
In set theory, from any set, we can take subsets defined by any property.
So, let B be a set that has as members all the sets that are not members of themselves plus some other members. Now take the subset of B that has only the sets that are not members of themselves, and we are back to the contradiction.
So it is a theorem of pure logic:
There is no y such that for all x, x is a member of y iff x is not a member of x.
And it is a theorem from the subset axiom schema:
There is no y such that for all x, if x is not a member of x, then x is a member of y.
/
Moreover, given the subset axiom schema, it is a theorem:
There is no z such that for all x, x is a member of z.
/
Again, one may wish to conceive of mathematics and sets in some other way than as usual in mathematics, but then we may ask, "What are your primitives, your rules of inference and your axioms to express your alternative conception and to prove things about it?"
There is no set y whose members are all and only the sets that are not members of themselves.
Then, from Russell's paradox, along with the subset schema we show:
There is no set z that has every set as a member.
and
There is no set such z such that for every x, if x is not a member of itself then x is a member of z.
R = {x | x not-e x}
has not been explicated.
'{ x | }' is the abstraction operator. It is variable binding notation that takes any formula P and names a set {x | P(x)} that is the set of all and only the x such that P(x).
However, proper use of the abstraction operator requires first proving that indeed there is such a set for a given property P.
And we prove that for the property P, where P(x) iff x is not a member of x, there is no such set of all and only the x such that x is not a member of x.
There are different ways of handling such situations (Russellian vs Fregan, e.g.).
In ordinary discourse, mathematicians would say that
R = {x | x not-e x} is an "illegal" definition, as there is no such object described by the right side of the equation.
With the Russell treatment, names with abstraction operators are regarding not standalone but rather contextually, so that we evaluate the truth value or satisfiability of formulas without regarding the abstraction operator expression as naming but rather as a nested component indicating an assertion about existence of an object having certain properties.
With the Fregean treatment, if we have at least one constant symbol, then we may set all non-referring abstraction operator expressions to that constant.
I suspect it is in vein, and that the issue here is not logical so much as pedagogic.
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
Peace
Where I said things like 'EyAx xey' is not inconsistent onto itself', I meant 'EyAx xey' is not inconsistent in and of itself' (though it is inconsistent with the axiom schema of separation).
And a post of mine was deleted in another thread. So I listed the reasons why it and the posts in this thread should not be deleted. Then that post was deleted and an admin made a post saying where discussions of deletions should be, but when I came back to read the admin's post and take note of where the admin had in mind for discussion of deletions, that post by the admin was also deleted. Thus, I didn't get to save my own post about the deletions to instead put it in the location suggested, and I didn't even get to read the admin's post about the suggested location. That doesn't even include all the posts by me and others in this thread that were wiped out, without notification or stated reason.
Good question. This thread has been a particularly vexing one and has to a large extent been a conversation, sometimes a bit heated, between two of TPF's top people in math (set theory in particular). It seems to have resulted in one of these, @fishfry, leaving the forum. I question why such a technical thread on set theory didn't go to the Lounge. I suppose the obvious fact that the initiator was not conversant with the subject might have argued for keeping it on page one, seeing what interesting philosophical ideas might emerge.
The reason technical content should not be shunted elsewhere is that if the philosophical discussion is ABOUT the mathematics, especially a critique of the mathematics, then the mathematics should not be misrepresented, otherwise (1) Misinformation about mathematics is permitted to thrive and (3) The philosophical claims themselves are errant for not even being correctly about what they are supposed to be about.
Anyway, the recent deletions were not about the earlier exchanges with fishfry.
You are correct. Coincidence is not proof.
Quoting TonesInDeepFreeze
So true. The OPs lay out belief systems in one form or another, and sometimes they don't budge. Which I find acceptable in @Metaphysician Undercover's pronouncements, for he dwells with the ancients as they ponder space, time, and points and curves - although he balks at 1+4=5 and has little patience with Weierstrass and his limit ideas: admittedly useful, but fundamentally flawed. But I see where he is coming from there. Others, like this thread, are more or less unmovable in their opinions, which clash with standard mathematics. How you deal with the frustration of offering knowledge to those unwilling to accept it is admirable.
Now, now, let's have fair representation. I balk at the claim that "1+4=5" implies that "1+4" refers to the exact same thing as "5" does. And, I have no patience for people like fishfry who simply assert over and over again, that the axiom of extensionality proves that "1+4" and "5" must refer to the exact same thing in common applications. Further, although I am very interested in the fundamental incompatibility between the proposal of non-dimensional points, and the proposal of a continuous one-dimensional line, I believe no one until today has brought Weierstrass to my attention.
(I am one of his 35K math descendants) I think you have indicated that the limit concept is useful but doesn't tell the whole story. Now's the time to elaborate. :chin:
- We need a meaningful distinction between "member of self" and "not member of self"
- We need a set of all sets (math/logic would be incomplete without it, if not contradictory)
So:
The set of all sets encompasses all sets that are not members of themselves (precisely because they are members of it and not themselves) as well as itself (precisely because it is a set).
A subset of "all sets that are members of themselves" and "all sets that are not members of themselves" is contradictory because whether something is a member of itself or not, is purely dependent on the single set/context/reference that it is in. Here is the proof for this:
Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different standards/contrexts as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?)
There aren't actually any sets within the set of all sets. There are only members.
For example, if we take the set of all numerical values (0 through 9 - ten members) and the set of all alphabetical values (a to z - twenty-six members), we could make the set of these two sets and call it the Set of all Characters (for sake of argument). So does this set of all characters (including 1, 2, 3, etc., and a, b, c, etc.) now contain a total of 36 members? I say yes. Or does it contain 2 more members, total of 38, being the prior sets called "numerical" and "alphabetical" plus their members? If we call it the set of two sets, do we need to add in the set designated "numerical" and the set designated "alphabetical" and count the members separately from the sub-sets, when we join the sub-sets into a new super-set?
I say no. We are smart animals, so we can look at the single set of all Characters which has 36 members and simultaneously see that there are sub-sets of 10 numbers and 26 letters, but we are not counting those two distinctions as 2 additional members of the set of all Characters. The set of numbers has 10 members, and the set of letters has 26 members, and any set containing only these two sets would have 36 members.
When you bring sets together, under a new super-set, the sub-set distinctions that were named for example "numerical set" and "alphabetical set" no longer exist - these distinctions are irrelevant or non-existent to the set of all characters.
Apply this when joining all sets into the "set of all sets". We take the set of numbers, the set of letters, the set of atomic particles, the set of forces, etc., etc., until we take up all sets, list them on the blackboard, and fashion the idea of "the set of all sets". Aren't we just overcounting this new "set of all sets" if we count the sets within it, and not just the members of those sets (ignoring the prior set distinctions themselves)?
If so, then we have no need to call it "the set of all sets" - the set of all sets (incoherent paradoxical term) becomes the set of all members, or just a misapplication of the term "all".
This doesn't solve the paradox. It just shows that the paradox is where logical process, which occurs between multiple things (as in between sets and members), borders on the identity of single things (like what is a member, or what is a set). Logic lives between things, in their relations. Once we say the set of all sets, we have a logical problem if we lose site of the members of all those sub-sets and just look at those sub-sets themselves as if those sub-sets could be members without their own sub-members in the first place. As we gather up sets into bigger sets, the distinctions between this set and that set are no longer relevant and do not count in the membership, and only the new set exists.
The set of all sets is really the set of all members, which is also how we use the word "all" in the first place.
That is confused. A set of all sets has as members all the sets. Let that sink in. ALL sets are members of the set of all sets. So whether a set x is a member of itself or not a member of itself, if there is a set of all sets then x is a member of that set of all sets.
You claim that a set cannot be a member of itself and also be a member of another set.
I've refuted that claim. You skip the refutation.
Quoting Philosopher19
Maybe you mean that no set x can be both a subset of a set of all sets that are members of themselves and a subset of a set of all sets that are not members themselves.
That is true of all sets except the empty set.
Quoting Philosopher19
"The z of all zs" has no apparent meaning to me. But if you mean a set z such that every set is a member of z, then that is said as "the z such that for all x, x is a member of z". Let me know that you've fixed that, then maybe I'll go to your next sentence.
Quoting Philosopher19
We have it.
Quoting Philosopher19
Again, you skip my whole explanation to you about incompleteness.
Again, you skip my refutation of the claim that without a set of all sets we have a contradiction.
Again, discussion goes in a circle with you as you skip the counterpoints given you. And, again, you've been given a refutation that it is your interlocuters who are circular, so skipping that refutation and instead yet again claiming the fault is with your interlocuters would be yet more circularity from you.
Members of what?
Of course, every set is the set of all and only its members.
And in ordinary set theory, every set is a member of another set.
Quoting Fire Ologist
If by "within" you mean 'member of', then the above nonsense. If x is the set of all sets, then, by that description alone, every member of x is a set and every set is a member of x
And in ordinary set theory (without urelements and without proper classes) every object is a set.
.Quoting Fire Ologist
The set {0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z} has 36 members.
Quoting Fire Ologist
We call it 'the union' of two sets.
{0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z}
is the union of
{0 1 2 3 4 5 6 7 8 9} and {a b c d e f g h i j k l m n o p q r s t u v w x y z}
But let N = {0 1 2 3 4 5 6 7 8 9} and L = {a b c d e f g h i j k l m n o p q r s t u v w x y z}.
Then we also have the set:
{0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z N L}
which has 38 members.
Quoting Fire Ologist
No.
Quoting TonesInDeepFreeze
Yes, this is true by definition.
Quoting TonesInDeepFreeze
I am not skipping. You say:
Quoting TonesInDeepFreeze
This point implies that a set v can have more than one set that is a member of itself, as a member of itself/v.
I refuted this point with my z example of which you perhaps fairly said "the z of all zs has no apparent meaning to me". Here is another form of the refutation:
L = the list of all lists
LL = the list of all lists that list themselves
Is L a member of itself in L? Yes. Is L a member of itself in LL? No, because in LL, it is a member of LL as opposed to a member of itself/L. Does this not prove that L is not a member of itself in LL?. Just because a set is a member of itself in its own respective set, doesn't mean it is a member of itself in another set.
My original refutation with the z example was, I believe, more complete. So:
z = any set that is not the set of all sets
v = any set
The v of all vs = the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets
Quoting Philosopher19
Quoting Fire Ologist
And by definition all those members are sets.
Quoting Fire Ologist
I get where you're coming from. I believe the issue lies in correctly determining what it is for something to be a member of itself and what it is for something to be not a member of itself. This is something I address in my last reply to TonesInDeepFreeze (which is literally the post that is above/before this post) in case you're interested.
You're confused. What you skipped is my refutation (posted twice) of your claim that a set can't be both a member of itself and a member of another set.
Quoting Philosopher19
"is a member of itself, as a member of itself" has no apparent meaning to me.
Quoting Philosopher19
Yes, so fix it.
Quoting Philosopher19
I gave you guidance before in how the notion of 'list' is couched in set theory. But you skip it. To avoid incoherence with a clash between 'lists itself' and 'member of itself', you could go back to what I wrote.
Quoting Philosopher19
You said that it is perhaps fair to say that such locutions have no apparent meaning, but then you proceed to post them again, as, for the second time, you've ignored my suggestion of the exact way you could reformulate so that you make sense.
We don't say what 'is a member of' means. Rather, 'member of' is the primitive relation of set theory. What happens then with that primitive is determined by the axioms. But the meaning of the atomic formula "x is a member of x" in and of itself can't be explicated more than to say that x has bears the membership relation with x. Then, "x is not a member of x" is merely the negation of "x is a member of itself".
You seem to have an ardent interest in the subject. So why not read a book on the subject to find out about it?
You agreed with me that a set cannot be both a member of itself and not a member of itself. You said it was an important point in your refutation to me. The above logically implies you are rejecting it. First look at the part I underlined in the quote above. Then look at how there is no difference between:
1) A member of other than itself (which is the same as saying a member of another set that is not itself)
2) Not a member of itself
If it's not a member of itself, it's a member of other than itself. If it's a member of itself, it's not a member of other than itself. You need to show a meaningful difference between 1 and 2 since you're the one claiming that on the one hand a set can't be both a member of itself and not a member of itself, and on the other hand, a set can be both a member of itself and a member of another set (to be a member of another set is to be a member of other than itself)
Quoting TonesInDeepFreeze
This is perhaps why you didn't get my point/refutation. Perhaps if you try to reply to my above point, you will start to get my point.
Quoting TonesInDeepFreeze
No, z and v were clearly defined. What I did not say (which is why I said "perhaps fair to say") was that the v of all vs = the set of all sets and that the z of all zs = the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. I thought they could be inferred from v and z, but still, in my next post to you I clearly defined them, yet you did not address that main refutation of mine.
Wrong. You cannot produce a valid demonstration that
There is an x and y such that x is a member of x and x is a member of y
implies
There is an x such that both x is a member of x and x is not a member of x.
Quoting Philosopher19
You insist on that without basis. It's an idea you have stuck in your head, but it doesn't follow from any commonly held ideas about sets, let alone from actual axioms. I mentioned that previously, but you SKIPPED it.
Quoting Philosopher19
Previously I made the argument that x is a member of {x} and, without the axiom of regularity, it is not precluded that there is an x such that x is a member of x. That's correct, but now I see it's not the argument I should make, since in that case x = {x}.
So revised: Let x not equal b. Let b not be a member of x. Let x be a member of x. But x is a member of {x b}. So x is a member of x and x is a member of a set different from x, viz {x b}.
That refutes your claim.
Quoting Philosopher19
You are not replying on point. I did not say I object to z and v. I said there is no apparent meaning to me in the locutions "the z of all zs" and "the v of all vs". I don't know what you think you mean by that.
I offered you an actually intelligible phrase that possibly does capture what you mean. But you SKIPPED that.
Learn some logic then some set theory. You're not making sense without at least a bit of understanding of them. The information I'm giving you is wasted on you, as you lack the needed basic logic skills or even familiarity with common notions about sets such as simple pairing.
I'm not sure I get what I'm saying.
Help me out. Besides the set of all sets, what is an example of a set that is a member of itself?
Quoting Philosopher19
No, we do not.
Suppose, toward a contradiction, that there is an x such that for all y, we have that y is a member of x if and only if y is not a member of y.
If x is a member of x, then x is not a member x.
If x is not a member of x, then x is a member of x.
So both x is a member of x and x is not a member of x. Which is a contradiction. So it is not the case that there is an x such that for all y, we have that y is a member of x if and only if y is not a member of y.
Notice that that is pure logic. It doesn't even need any reference to the notion of 'set' (indeed, it doesn't even mention 'set') or 'member of'. We could replace 'member of' by any 2-place relation R and still get the result:
It is not the case that there is an x such that for all y, we have y bears R to x if and only if y does not bear R to y.
Quoting TonesInDeepFreeze
How does this show there is a meaningful/semantical difference between 1 and 2? Therefore, how has this refuted my claim? It is clear that "not itself" and "other than itself" mean the same thing.
Quoting TonesInDeepFreeze
I did. I think it was very clear. Here it is again in a numbered format. Tell me which number doesn't follow from (or is irrelevant to) which number or tell me which number is wrong if you are sincere in this discussion.
1) There is no meaningful/logical difference between "not a member of itself" and "a member of other than itself".
2) A set cannot be a member of itself and not a member of itself
3) If a set is a member of itself, it is not a member of other than itself (precisely because it is a member of itself)
4) If a set is a member of itself, it is not a member of another set (precisely because it is a member of itself. If it is a member of another set, it is not a member of itself precisely because it is a member of another set).
The 1-4 point I'm making is clear. If I get a reasonable/meaningful response to it, I believe I will respond to that response.
I think your instincts/intuition is in the right place (or at least trying to get to the right place)
Quoting Fire Ologist
I recommend the following: http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
I read the article. I certainly get how this discussion needs to be analytical to be precise, but I am not fluent enough in the symbolic language to keep up.
Is there any way you can provide an example of a set that IS a member of itself, other than the set of all sets, in plain language? I can't think of one.
Quoting Fire Ologist
That's because speaking in absolute terms, only the set of all sets is a member of itself.
In short, if we were to focus on absolutely all sets, then only the set of all sets is a member of itself. However, if we were to only focus on all sets other than the set of all sets, then another set is a member of itself (but then we are not speaking in absolute/complete terms). I will try and show this to you:
Call any set a v.
Call the set of all sets the v of all vs.
If we were to focus on all sets, we would be focused on all vs.
Now call any set that is not the set of all sets a z.
If we were to focus only on all sets other than the set of all sets, we would be focused on all zs as opposed to all vs.
Since z = any set that is not-the-set-of-all-sets, the z of all zs means "the not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets. The z of all zs, is a member of itself as a z, but it is not a member of itself as a v. As a v, it is a member of the v of all vs.
So if we were to talk about all vs, then only the set of all sets is a member of itself.
If we were to only talk about all zs, then only the z of all zs is a member of itself.
Let S be any set other than the set of all sets. Let T be the set of all sets that are not S. T is not the set of all sets, and the set of all sets is a member of T.
Another:
Let M be the set of all sets with at least two members. M is not the set of all sets, and the set of all sets is a member of M.
There are many more.
But don't forget:
By logic alone, we prove that there is no set of all sets that are not members of themselves.
With some set theory axioms, including the axiom of regularity, we prove that the set of all sets that are members of themselves is the empty set.
With an instance of the subset axioms, we prove that there is no set of all sets.
I think you can understand this if, for a few moments, you clear your mind of the voice in it that keeps saying "I am right. I know I am right. I must be right. All the logicians and mathematicians are wrong and I am right", then very carefully, very slowly, consider:
P: There is a y such that x is not y and x is a member of y
Q: x is not a member of x
You say that P and Q are equivalent.
But P and Q are equivalent if and only if P implies Q, and Q implies P.
So if P does not imply Q then P and Q are not equivalent.
To show that P does not imply Q, it suffices to show that P and the negation of Q are consistent together.
Here is a situation in which both P and the negation of Q hold:
Let x not equal b. Let b not be a member of x. Let x be a member of x. Let y = {x b}. x is not y, since b is a member of y but b is not a member of x; but x is a member of y, So there is a y such that x is not y and x is a member y. And x is a member of x. But "x is a member of x" is the negation of Q.
So P and the negation of Q are consistent together.
So P does not imply Q.
So P and Q are not equivalent.
https://thephilosophyforum.com/discussion/comment/879811
You have been confused about the same thing you were confused about three years ago.
No, don't call it that. It's, at best, confusing notation.
The set of all sets is:
{x | x is a set}
If you want, call it z:
z = {x | x is a set}
But with an instance of the subset axiom, we prove that there is no such set.
Quoting TonesInDeepFreeze
I tried to look at your reply, but it is unclear to me as to what it's doing. I believe it deliberately strays from what is clear simple language to try and force something that cannot be forced (perhaps due to dogma).
A) When a set is not a member of itself, it is a member of a set other than itself.
B) When a set is a member of itself, it is not a member of another set.
Rejection of either A or B is blatantly contradictory, yet, you seem to be arguing that rejection of B is not contradictory. Until you acknowledge that the rejection of B is contradictory, I don't see how we can progress.
I stated exactly what it clearly does:
It shows that P and Q are not equivalent.
Quoting Philosopher19
The language is as simple as it can be while being exact, rigorous and not skipping the details.
And previously I gave you a more simply worded version, not belaboring every detail. So, since you didn't understand it, this time I gave you every exact detail.
Moreover, that you don't understand such a straightforward, rudimentary proof is not the fault of the proof, but rather your fault for your utter unfamiliarity with basic logic and basic, common notions about sets.
And there is no dogma. I used only basic logic and utterly common notions about sets.
Quoting Philosopher19
THAT is dogma. You have no proof of it, and I gave an exact disproof of it.
You are hopeless.Three years in which you have made not a mote of progress in understanding anything in this subject.
False dichotomy. A dichotomy insisted upon only by your idiosyncratic dogma, enabled by your inability to turn off in your head for even one moment your own voice telling you that you are right, that you must be right, so that you don't step back for even a second to question yourself, to consider that the dichotomy that you so obdurately cling to might actually not be true when you think about it just a bit more.
You are merely restating your dogma that being a member of itself is mutually exclusive with also being a member of another set.
I proved that that dichotomy is false.
These are two different claims:
1. A is not a member of itself
2. A is a member of some other set
Given this:
[math]A=\{A\}\\B=\{A,0\}[/math]
(1) is false and (2) is true.
Regarding Russell's paradox, it is simply this:
1. [math]x[/math] is a member of [math]R[/math] if and only if [math]x[/math] is not a member of [math]x[/math].
Is [math]R[/math] a member of [math]R[/math]?
Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent.
Quoting TonesInDeepFreeze
My proof was here:
Quoting Philosopher19
Quoting Philosopher19
As if "the rejection of the set of all sets is by definition contradictory" is not proof enough.
Or as if "a set is not a member of itself when it is a member of another set" is not proof enough.
Quoting Michael
Evidently, in A, A is a member of itself.
Evidently, in B, A is not a member of itself because A is a member of B.
So you have not shown that 1 and 2 are two different claims.
I repeat:
A) When a set is not a member of itself, it is a member of another set
B) When a set is a member of itself, it is not a member of another set
I understand Russell's paradox. Here is what I say in response:
Quoting Philosopher19
Quoting Philosopher19
A is a member of both A and B.
I'll explain it to you in non-math terms:
I am a member of the football team and a member of the tennis team.
These are two different claims:
1. I am not a member of the football team
2. I am a member of a non-football team
(1) is false and (2) is true.
1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R
(3) is a very obvious contradiction. You don't even need to know maths to see that.
So at least you're engaging me in clear meaningful language. I will respond in kind.
You are not a set. You can never be a member of yourself. But a set can either be a member of itself or a member of other than itself. And it is logically the case that when a set is a member of itself, it is a member of itself. And when it is a member of other than itself (as was the case with A in B), it is not a member of itself.
So your non-math example doesn't apply because you are talking about something that by definition can't be a member of itself, whereas in your A and B example, you were talking about something that by definition was a member of itself in its own set whilst a member of other than itself in another set.
Quoting Philosopher19
A set can be a member of more than one set. You just don't understand the basics of set theory.
You should really take a few math lessons before you start telling mathematicians that they're wrong about maths.
I believe you're not paying attention to what I'm saying. I'm not saying a set can't be a member of more than one set. I am saying:
Quoting Philosopher19
See again my replies to you:
Quoting Philosopher19
Quoting Philosopher19
And this is a fundamental misunderstanding of set theory.
If A = {A} and if B = {A, 0} then A is a member of A and a member of B.
Yes. But what you're not responding to is the following:
In the case of A = {A}. A is a member of itself.
In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself?
Also, don't forget that your non-math example doesn't apply because you are an x that by definition cannot be a member of itself, whereas a set is an x that can be a member of itself as well as other than itself.
Both
Quoting Michael
But in B, A is a member of B. In B, A is not a member of both A and B. So once again, in B, is A a member of itself or not a member of itself?
Both a member of itself and a member of B.
Quoting Philosopher19
Yes it is.
Look at what you're saying:
In B, A is both a member of A and B.
"In B" does not equal to "in both A and B".
Do you see your contradiction? You have treated "in B" as the same as "in both A and B".
It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A".
[math]A=\{A\}\\B=\{A,0\}=\{\{A\},0\}[/math]
Quoting Michael
And now you have strayed from clear language. What I have given you is clear should you choose to pay attention to it:
I asked you:
Quoting Philosopher19
to which you said:
Quoting Michael
to which I said:
Quoting Philosopher19
Quoting Michael
And in B (as opposed to in both A and B), A is not a member of itself because it is a member of B (and not A). This is basic.
N is the set of natural numbers.
R is the set of real numbers.
Every natural number is a member of both N and R (every natural number is both a natural and a real number).
We don't say "in R, 1 is not a member of N".
Quoting Michael
I addressed a similar point in my reply to your non-math example. Every natural number can be both a member of N and R. I am not denying this. But what a natural number cannot be (and what N and R cannot be) is members of themselves. When you talked about A, you talked about something that was by definition a member of itself precisely because A = {A}. But in the case of B = {A, 0}, A is not a member of itself because now, by definition, A is a member of B.
Quoting Michael
I never said A can't be both a member of A and B. I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself.
Scenario 1
B = {0, A}, where A = {1}
Scenario 2
B = {0, A}, where A = {A}
I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself.
I believe I have already said more than enough. Check my replies to you because I believe I am now repeating myself where I shouldn't have to.
And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else.
It’s the case that in scenario 2, A is a member of A and B.
Quoting Philosopher19
Quoting Philosopher19
Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself?
These are blatantly obvious things. As far as I can see, only dogma and bias and insincerity to Truth would cause one to fail to see/recognise them.
Brain; (mathematical objects)
These come in different forms with specific properties.
Like fixed,
Brain; (fixed mathematical objects)
Things like pi, i, e, trig functions, ?2, ?3.
Or defined,
Brain; (defined mathematical objects)
Like defining sets, setting parameters, variables in functions.
Not sure if this will help with Russell's paradox but in general, applying universal form and understanding mathematical objects will work in understanding paradoxes or contradictions
I might give it a try later.
I cover the basics of mathematical objects in my post on universal form. There is an example of the contradictions in time perseption given.
Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it.
1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R
You haven’t proved that (3) is not a contradiction. And you can’t because it is.
The solution to the paradox is known: construct a set theory with axioms that do not entail (1).
ZFC does this by not allowing a set to be a member of itself. New Foundations does this by restricting which sorts of sets can be members of themselves.
Quoting Michael
I believe I addressed this point both in my reply to your non-math example and in your N and R example and in my other replies. No point doing in doing it again. Plus there is my z and v example.
1. x is a member of A if and only if x is a member of x
2. Let x = B
3. B is a member of A if and only if B is a member of B
But according to your reasoning, (3) is a contradiction. Therefore (1) is a contradiction.
Your axioms lead to an inverse Russell paradox.
We can resolve this either by allowing that B is a member of both A and B, or by not allowing a set to be a member of itself.
Quoting Michael
I have been saying to you that x is not a member of A if x is a member of x. I have not been saying 1. 1 is blatantly contradictory. So it seems to me that you have misrepresented/misunderstood what I have been saying to you.
You've argued that there is a set of all sets, U.
If A is the set {A} then A is a member of both A and U.
That's a common misconception.
Yes, ZFC has the axiom of regularity that implies that no set is a member of itself. But that doesn't avoid Russell's paradox.
Rather, Russell's paradox is avoided by not having unrestricted comprehension.
Even with the axiom of regularity, and even if there are no sets that are members of themselves, with unrestricted comprehension we would have the set of all sets that are not members of themselves, thus a contradiction.
Explicity:
1. ExAy(yex <-> ~yey) ... instance of unrestricted comprehension
2. Ex(xex <-> ~xex) ... from 1
Having the axiom of regularity, or any other axiom, does not block getting 2 from 1.
The way to not have 2 is not to have 1, irrespective of the axiom of regularity.
I would classify the Russell set as a defined mathematical object. That means it is subject to a determination of if it exists or does not exist. The fact that paradoxes develop means it is a defined mathematical object that does not exist.
This is basic theory of predication. Fixed mathematical objects are determined by predication. Defined mathematical objects are arbitrary and defined by rules and ultimately may not exist.
In short....the Russell set never exists as the defining process procedes and once the problems are discovered the conclusion should be the Russell set is non-existent.
In a theory with unrestricted comprehension we define the set whose members are all and only those sets that are not members of themselves. And having that set implies a contradiction.
In modern set theory, we do not have unrestricted comprehension and there is no way to define a set whose members are all and only those sets that are not members of themselves.
Never had the set in the first place.
Who or what never had the set? And what is "the first place"? You, personally? People who don't work with set theory?
The math procedes in a way that assumes an ultimate set will exist.
I'm saying ultimately the Russell set does not ultimately exist.
Edit: rather never exists and ultimately does not exist...
"The math" refers to what mathematics?
After Russell discovered the paradox, mathematicians replaced the systems that allowed the paradox with systems that don't allow the paradox. With set theory since the early 20th century, we have systems that don't have the paradox, as indeed the systems prove that there does not exist a set whose members are all and only the sets that are not members of themselves.
Quoting Mark Nyquist
Quoting Mark Nyquist
Then you agree with set theory.
You misunderstand the paradox.
Naive set theory allows the Russell set. The Russell set is a contradiction. Therefore, naive set theory is inconsistent.
That's all it is.
We don't need to bring up mathematical realism.
You seem to be arguing for the paradox after the paradox has been dismissed.
I'm arguing that Philosopher19 doesn't understand set theory, and that his attempted "solution" to the Russell paradox makes no sense.
The Russell paradox has already been "solved": see ZFC, for example.
Okay.
I don't know what you mean by "dismissed" but, mathematics came up with a system in which the paradox does not occur.
Great. That's progress.
That is incoherent and has no apparent meaning.
"A is a not a member of of both A and B" does not take a qualifier "In B".
In general: Let P be any formula of set theory and B be any set. We don't say "In B, P". Putting "In B" before "P" has no meaning and indeed is not even grammatical in this context.
Philosopher19 has some word salad dogma tossing around in his head. His first step should be to at least learn how to coherently express whatever it is he's trying to say.
Proposed defined mathematical object
Confirmed defined mathematical object...exists
Rejected defined mathematical object ... non-existent
So the Russell set as a rejected defined mathematical object does not exist and cannot be a paradox.
A paradox occurs when there is a contradiction or highly counter-intuitive statement that
follows from premises or principles that we regard as themselves true, reasonable or intuitive.
I suggest that the supposed Russell set is not itself a paradox.
But rather, the paradox is that from the premise "for any property of sets, there exists the set whose members are all and only those sets that have that property" we get "There exists the set whose members are all and only those sets that have the property of not being members of themselves (i.e., the supposed Russell set)", which implies the contradiction "The Russell set is a member of itself and the Russell set is not a member of itself".
There, the premise that up to 1901 we thought was true, reasonable and intuitive is "for any property of sets, there exists the set whose members are all and only those sets that have that property" but the contradiction that follows from it is "The Russell set is a member of itself and the Russell set is not a member of itself".
So we see that what we thought back around 1901 to be true, reasonable and intuitive, and that we moreover had as an axiom around that time, is actually worse than false as it implies an outright contradiction. Therefore, since Russell's paradox was announced, mathematicians have eschewed using the aforementioned premise, have eschewed using it as an axiom.
Note that this can be seen to be a matter even more basic than set theory. We can see formulate it even without mentioning 'set' or 'member':
For any 2-place relation R, there is no x such that for all y, y bears R to x if and only if y does not bear R to y.
For example, famously:
There is no Mr. X such that for everyone, they bear the relation with Mr. X of being shaved by Mr. X if and only if they bear the relation with themselves of not being shaved by themself.
More simply: There is no one who shaves all and only those who do not shave themself.
Or:
There is no one who loves all and only those who do not love themself.
Doesn't matter what the binary (2-place) relation is: 'member of' or 'shaves' or 'loves', etc.
You suggest the Russell set exists only based on the process of defining it.
I'm saying it does not exist.
Which is it.
In my development of the issues these are mathematical objects that don't exist outside of brain state.
One method of establishing existence is predication such as for fixed mathematical objects.
Defined mathematical objects are not based on predication....because we set the definition process. I'm saying for an object to be a legitimate mathematical objects it needs to pass an existence test. Creating paradoxes is a failure of an existence test.
You accept that than the Russel set exists and is legitimate. I don't think it has a sound basis. It's based on definition and that's not proof of existence. You have a burden of proof.
The term "naive set theory" is used in various ways. In one usage, naive set theory is a formal theory, that is formulated in a first-order language with a binary non-logical predicate [math]\in[/math], and that includes the axiom of extensionality:
[math]\forall x \, \forall y \, ( \forall z \, (z \in x \iff z \in y) \implies x = y)[/math]
and the axiom schema of unrestricted comprehension:
[math] \exists y \forall x (x \in y \iff \varphi(x))[/math]
for any formula [math]\varphi[/math] with the variable [math]x[/math] as a free variable inside [math]\varphi[/math]. Substitute [math]x \notin x[/math] for [math]\varphi(x)[/math] to get:
[math] \exists y \forall x (x \in y \iff x \notin x)[/math]
Then by existential instantiation (reusing the symbol [math]y[/math]) and universal instantiation we have:
[math]y \in y \iff y \notin y ,[/math]
a contradiction. Therefore, this naive set theory is inconsistent.
But we are trying to dispell the contradiction, not prove it.
If the Russell set doesn't exist there is no contradiction.
This is what you're failing to understand.
According to naive set theory, the Russell set "exists".
The Russell set doesn't "exist", because the Russell set is a contradiction.
Therefore, naive set theory is inconsistent.
The solution is to fix the inconsistencies with naive set theory, which was done in ZFC (and others).
Finally, yes, that is what I was getting at.
It's my first look at anything in set theory so I don't have background.
Then you really shouldn't comment, because by your own admission you don't understand the problem.
Quoting Mark Nyquist
This was explained in the very first comment of this discussion by fishfry.
Okay. I'll go over and sit in the corner and read some set theory textbooks.
It seems evident to me that you've not been paying attention to what I've been saying. I've already given more than one relevant reply to this. I'm not going to repeat myself.
Quoting Michael
p) In A, A is a member of itself/A.
q) In U, A is not a member of itself/A.
p and q are true by definition.
If definitional truths are not enough (and they really should be), here is a demonstration of how a set can only be viewed as a member of itself in its own respective set:
v = any set
The v of all vs = the set of all sets
z = any set that is not the set of all sets
The z of all zs = The not-the-set-of-all-sets-set of all not-the-set-of-all-sets-sets
In the z of all zs, the z of all zs is a member of itself. But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the v of all vs. We can't treat two different things/references/contexts/standards as one (as in are we focused on the context/items of vs or the context/items of zs to determine what is a member of itself or not a member of itself?)
No, it's nonsense. That's not how set theory works.
1 is a member of N and R.
A is a member of A and B.
That's it.
Just take a few actual lessons in set theory.
That is tautological. As pointed out before there is no such as "in A". Yes, within the set A, A is a member of A. That is a tautalogy, cut it out to "Yes, A is a member of A" and you get the same meaning. So it is wrong to say the quote above because it is the same as saying that A can't be both a member of A and B.
Your argument seems to be that A is not a member of B in A because B is not defined in A. That doesn't make sense because sets do not include definitions of other sets. Even if they did somehow, by your argument: even in B, A would not be a member of B because B is not defined within B either.
I'm saying whether something is a member of itself or not is determined by whether it is in its own set or not. This matter of pure reason/definition seems to have been disregarded and faulty notation/language has then been used to falsely conclude there is no universal set. It seems to me foundationally corrupt theories have been built on a misunderstanding and so much has been invested in them that there is either unwillingness to let them go, or there is so much dogma that the obviously true is not paid attention to.
Quoting Deleted user
My argument is that A is not a member of itself in B because A is a member of B in B. To preserve this truth and take into account what set theorists seem to be focusing on, I wrote the following:
Quoting Philosopher19
For the whole of what I'm saying, read:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
Already responded to it at least twice. Won't repeat myself.
So it's settled in your mind that you won't read even a single book or article about this subject, but that others should read your website?
Oh, yes you will. I guarantee it.
Quoting TonesInDeepFreeze
I believe I understand the matter well enough. I don't feel it sincere to Truth/Goodness to read/research any more on it than I already have. The link to what I've written on Russell's paradox and Infinity is there should anyone choose to read it. Whether they should read it or not, is not for me to say. It may be sincere to Truth/Goodness for them to read it, it may be sincere to Truth/Goodness for them not to read it. It may even be that Truth/Goodness is not their priority. I don't know, I don't have their self-awareness.
Nobody disagrees on that.
Quoting Philosopher19
Again, your argument is nonsensical. It does not mean anything in mathematics.
In B, A is not a member of anything, A simply exists. Because it exists in B, it is a member of B. But that has no bearing on Russell's paradox. It is a semantic point.
Quoting Philosopher19
This is just repeating the same semantic nonsense.
What book or article in the subject have you read/researched?
You are so very mixed up that you are getting this all completely reversed.
No, I do not at all suggest that such a set exists. Rather, we are giving you proofs that such a set does NOT exist.
Same as above. You're mixed up and have it reversed.
We very clearly do NOT think there is such a set. Rather, we PROVE that there is no such set.
Again, you're mixed up, and likely unfamiliar with proof by contradiction.
We prove that the assumption "there is a set whose members are all and only those sets that are not members of themselves" implies a contradiction thus that that assumption is false, which is to say that there does NOT exist such a set.
If you continue to insist that we're saying that such a set does exist, after three posters have already explained, in detail and in different ways, your misunderstanding, then I'm guessing you're trolling.
EDIT: I see now that after the quoted post, you replied further that you recognize that you don't understand. So, fair enough.
I watched a YouTube video amongst other things. Didn't read a book on it. Had a look at wiki and Stanford but maybe or maybe not with massive amounts of focus. Just enough to understand the problem. Also had an exchange with a math lecturer who specialises in set theory (but we disagreed with each other fundamentally, but perhaps the exchange helped me better understand the other's position)
In any case, given the responses I have seen, I see no point in continuing this discussion.
Peace TonesInDeepFreeze
Just enough to think you understand it. But more than enough for you to completely mangle it.
Quoting Philosopher19
You've said that about twenty times already, yet you continue. But you are right. If you are unwilling to carefully read toward understanding, not just skim for an opportunity to misrepresent, an article such as the one at the Stanford Encyclopedia then there is no point for you to be posting about it.
It might be jolly for you to say what you think is the very first incorrect sentence in the Stanford article.
The article says that intuitionistically:
~ReR -> ReR implies ReR.
But I don't see how that is so.
Yes, ~ReR -> ReR implies ~~ReR. But we can't infer ReR from ~~ReR.
So, the only way I could think of doing an intutionistic proof is not by way of:
ReR & ~ReR
but rather by way of:
~ReR & ~~ReR
So sorry you got mixed up about my view of the existence of the Russell set.
We both are saying the Russell set ultimately does not exist. If you missed it I was developing an alternative method using the concept of mathematical objects as proposed, existent or non-existent.
Would it be fair to say your view develops the Russell set as a proposed mathematical object and concludes that it is ultimately a non-existent mathematical object? If so we should have nothing to disagree on.
Also, as relates to the philosophy of mathematics, my view reaches back fully to physical brain state. So for me it was an exercise in exploring that and as a simple result I got the same answer you did.
I am relying on your intermediate conclusion that the Russell set does not exist to go straight to the final conclusion that the Russell set does not exist....a paradox does not exist
I'm not at all saying the Russell set shouldn't be explored. How else would you know?.
Summary,
The Russell set does not exist.
Based on the proposed defined mathematical object failing by contradiction.
Additionally,
In defining the Russell set, two or more (known to exist) defined mathematical objects are used to define the Russell set and produce a failed defined mathematical object (non-existent).....Which I think is an interesting result
Reference,
I covered mathematical objects as brain state 4 days ago in my post on Universal Form.
I didn't get mixed up. You were mixed up.
Quoting Mark Nyquist
I had no comment on that. Rather, your replies regarding the mechanics of the ordinary proof were confused.
Quoting Mark Nyquist
No. The way you say it is kind of along the lines of what I say, but I don't bring along notions of "proposed object" and "ultimately non-existent".
Rather, as I explained, it is a proof by contradiction, keeping it precise and simple, without the extraneous "proposed object" and "ultimately non-existent":
Toward a contradiction, suppose there is an x such that for all y, y is a member of x if and only if y is not a member of y.
Then x is a member of x, and x is not a member of x.
Therefore there is no x such that for all y, y is a member of x if and only if y is not a member of y.
Quoting Mark Nyquist
What? What you call my "intermediate conclusion" is not intermediate, and it is the same as what you call the "final conclusion".
The theorem ("the final conclusion") proved is:
There is no x such that for all y, y is a member of x if and only if y is not a member of y.
Put more informally, "There is no set whose members are all and only those sets that are not members of themselves".
Put another way, the name "The Russell set", which is supposed to name a set whose members are all and only those sets that are not members of themselves, does not properly refer to anything.
Quoting Mark Nyquist
The paradox was that, from the ordinary view (once given as an axiom) that for every property there is the set of those things having that property, we derive a contradiction.
When we eschew that view (and the axiom that captures it) that for every property there is the set of things having that property, we are not burdened with the contradiction that that view (and axiom) entails.
Quoting Mark Nyquist
That's not how I would say it, but it's close enough. I won't split hairs to quibble with it.
Quoting Mark Nyquist
I don't know what you're referring to.
The definition is:
R = {y | y is not a member of y}
i.e.
R is the set of all and only those y such that y is not a member of y.
But it's an improper definition, because there is no set whose members are all and only those y such that y is not a member of y.
I used intermediate because there are two questions to this problem based on two phases.
Phase one, discovery,
Does the Russell set exist?
Requires exploration.
No.
Phase two, end point,
Is the Russell set a paradox?
Given: the Russell set does not exist.
Since the Russell set does not exist we now know it cannot be a paradox.
You can say my method is extraneous because you resolve it using your own method.
But you are using the ideas subconsciously.
And you default to 'not a paradox' when you reach your intermediate conclusion (is a contradiction therefore non-existent).
Also, there is word confusion in contradiction and paradox so be careful of that.
A little more.....You have,
A discovery phase where the question is does a proposed mathematical object exist or not exist.
And
An end point phase where you are given the state.... a(n) existent mathematical object or non-existent mathematical object. Is an existent mathematical object a paradox? No. Is a non-existent mathematical object a paradox? No.
So I don't see how a paradox can exist. The contradiction in your intermediate result is only a basis for determining non-existence.
Also the problem is misnamed because no true paradox ever exists. Because the discovery phase is hypothetical.
I thought you meant an intermediate step in the proof. I have no comment on your characterization of phases.
The problem is not misnamed. I explained why it is a paradox from certain assumptions or axioms. Also, to be clear, 'paradox' is not a technical mathematical term. It's a general term for the situations I mentioned: From supposedly acceptable assumptions, principles or axioms, we derive a contradiction or even just a highly counter-intuitive implication.
I have this in my notes:
Russell discovered the paradox while studying Cantor's argument that there is no function from a set onto its power set. Zermelo had discovered the paradox earlier in 1900 or 1901 (cf. "Zermelo's discovery of the "Russell Paradox"" by Rang & Thomas in Hist Math 8 pp. 15-22).
Frege proposed the fix by using the following axiom schema (here couched set theoretically) instead of unrestricted comprehension:
For any formula, P in which x does not occur free:
ExAy(yex <-> (y not=x & P))
But the above axiom schema, with identity theory, is inconsistent with Exy x not=y, as became known to Russell and to Frege, and Lesniewski provided a proof in 1938 (cf. Fixing Frege by Burgess - pg 32-34; "On Frege's Way Out" by Quine (in Mind 1955 and in Selected Logic Papers - pg 151))
The proof is pretty cool, and it's interesting that it's much much more complicated than the easy Russell proof.
Of course it was Zermelo who provided a robust fix with:
AzExAy(yex <-> (y in z & P))
accompanied with the rest of the existence axioms.
Hi Michael
In case you are interested in discussing this further:
L = The list of all lists
LL = The list of all lists that list themselves
1) In which list does L list itself?
2) In which list is L a member of itself?
Can you answer both questions consistently and non-contradictorily?
In ordinary mathematics: A list is a sequence. A sequence is a function whose domain is an ordinal. A function is a certain kind of set of ordered pairs. So, usually we're not looking to see whether a list is a member of a list or not. Rather, usually, we would look to see whether a list is a member of the range of a list.
So, suppose there is a set S whose members are all and only the lists. (No such set exists in ordinary mathematics that does not have unrestricted comprehension, but suppose we have unrestricted comprehension.)
Let L be a sequence whose range is S. (L is a "list of all lists")
Let K be a sequence whose range is {x | x in S & x in range(x)}. (K is a "list of all the lists that list themselves")
Is L in L? No.
Is L in range(L)? Yes.
Is L in K? No.
Is L in range(K)? Yes.
Is K in L? No.
Is K in range(L)? Yes.
Is K in K? No.
Is K in range(K)? Yes.
Next question:
So what?
/
In mathematics, it makes no sense to ask, "In which set does x have the property P?" Rather, we ask, "Does x have property P?" If x has property P, then that is not qualified by "x has property P in some sets but not others". So one can't give sensical answers to nonsensical questions such as 1) and 2) in the previous post.
That is another way of saying what Michael said in the post to which the previous post is in response.
Quoting TonesInDeepFreeze
The property P is instantiated based on what set the item x is in. 1 and 2 were solid meaningful questions that highlight precisely this point. L is listed in both L and LL. In L it has the property of being a member of itself/L, in LL it has the property of being a member of other than itself/L. L is not LL. L is L.
I don't know what you mean by that. I don't know what you mean by a property being instantiated in this context. I referred to any property. As I said: If x has property P, then that is not qualified by "x has property P in some sets but not others".
In a theory that does not disallow self-membership, a set may be a member of itself and it is a member of other sets too. An example has been shown many times already:
Let x in x. Let y not in x. So x not equal {x y}. x in {x y}.
So x in x, and x in a set other than x.
This has been gone over already, but now we come back around full circle.
This is the same tautological nonsense as before:
Quoting Philosopher19
Same thing:
Quoting Deleted user
L is listed in both L and LL. In L, it has the property P of being a member of itself/L. As in the fact that L is in L is what instantiates the property P (P = self-membership) in L's case.
In terms of function and logic, there is no difference between "lists itself" and "is a member of itself". Listing oneself and being a member of oneself are both matters of self-reference. So:
Does L list itself in L?
Does L list itself in LL?
So on the one hand you say:
Quoting Deleted user
On the other hand you say:
Quoting Deleted user
Do you see how you have contradicted yourself?
And then you say
Quoting Deleted user
Yes, proper attention has not been paid to the semantics of "member of self" and "not member of self". Whether a set is a member of itself or not, is determined by what item it is and in what set it is in.
L lists itself in L. L does not list itself in LL. L is a member of itself in L/itself. It is not a member of itself in LL/not-itself.
No, I explained the difference.
I'll say it again, a list is a sequence. A sequence is a function whose domain is an ordinal. So the members of a list are ordered pairs. The members of the range of a list are the items listed by the list.
Quoting TonesInDeepFreeze
I answered those questions exactly.
Quoting TonesInDeepFreeze
I believe this has nothing to do with what I said.
Quoting TonesInDeepFreeze
I believe the exact answer to "Does L list itself in L?" is yes.
I believe the conversation won't progress beyond this point.
Peace
And I said that the exact answer to "Does L list itself?" is yes.
Progress will begin upon you paying attention to the replies you've received.
I have not because those two are different sentences.
I believe you used your notation to avoid/miss the semantical point I was trying to make. Here are the answers to the questions I asked:
Does L list itself in L? Yes
Does L list itself in LL? No
Additionally
Is L a member of itself in L? Yes
Is L a member of itself in LL? No
Quoting TonesInDeepFreeze
I don't believe it does. I am talking about the semantical implications of being a member of self and not being a member of self, I believe you are talking about something else. I believe you are in essence removing/ignoring the semantical implications of "member of self" with the way you engage.
You literally said two things that contradict each other:
In B, A is not a member of anything.
In B, A is a member of B.
If in B A is not a member of anything, then how is it a member of B in B?
That is not what I said, which is why you typed that out instead of quoting me.
Quoting Philosopher19
It is not. That means nothing in mathematics.
Here is what you said:
Quoting Deleted user
See the parts I underlined? Here they are for more clarity:
In B, A is not a member of anything
in B, it is a member of B
That is cute that you highlighted half of a sentence to make it seem it says something that it does not. Let me fix it:
Quoting Deleted user
You are having trouble figuring out the difference between a locative sentence and an explanatory one. I will leave you to it.
Ok
"L is a member of itself in L" has no apparent meaning if it does not simply mean "L is a member of L". Nothing is added by saying "in L". If L is a member of L then L is in L. Nothing is qualified by saying "in L" again.
There are two matters:
Whether L is a member of L. It is not, since the members of L are ordered pairs and L is not an ordered pair.
Whether L is a member of the range of L. It is.
I.e. whether L is one of the items listed by L. It is.
Saying 'semantical' adds nothing substantive in this context.
The notion of "member of self" is not more than "x is a member of itself if and only if x is a member of x".
And I addressed exactly whether L is a member of L. It is not. Rather, L is a member of the range of L.
Let's consider these four lists:
Months
Planets
Lists
Lists that list themselves
Q1. How many members does Months have?
A1. 12
Q2. How many members does Months have when it is a member of Lists?
A2. 12
Q3. How many members does Planets have?
A3. 8
Q4. How many members does Planets have when it is a member of Lists?
A4. 8
Q5. How many members does Lists have?
A5. 4
Q6. How many members does Lists have when it is a member of Lists?
A6. 4
Q7. How many members does Lists have when it is a member of Lists that list themselves?
A7. 4
Q2, Q4, Q6, and Q7 are redundant/confused questions. We only have to consider Q1, Q3, and Q5.
So returning to your questions, they should simply be:
1. Does L list itself?
2. Is L a member of itself?
Assuming that these mean the same thing, the answer to both is yes. L lists itself/is a member of itself.
And so it is also listed by/a member of LL.
Quoting Michael
I'm not asking how many members does x or y have. So I don't see how your example is relevant to what I asked.
Quoting Michael
My questions were reasonable/relevant/meaningful questions. I believe they were what they should have been. They highlight that a list only qualifies for the property of 'listing itself' when it is listed in its own list. The property of "being a member of self" is instantiated depending on what the item is and what set it is in.
So the questions to be asked are:
Does L list itself in LL?
Is L a member of itself in LL/not-L?
What is the difference between asking if L lists itself and asking if L is a member of itself?
Quoting Philosopher19
If L is a member of itself "in L" but not a member of itself "in LL" then L has [math]n[/math] members "in L" and [math]n - 1[/math] members "in LL".
But this makes no sense. A set is defined by its members.
If L has [math]n[/math] members then it has [math]n[/math] members "in L" and it has (the same) [math]n[/math] members "in LL".
So if L is a member of itself then it is a member of itself "in L" and it is a member of itself "in LL".
A list is a sequence. S lists x if and only if x is in the range of S. In that case, x is not a member of S, but rather it is a member of the range of S.
For example:
B = {John Paul George Ringo} is just a set.
T = {<1 John> <2 Paul> <3 George> <4 Ringo>} is a list of the members of B. Sometimes represented as the equivalent ordered 4-tuple:
or
1 John
2 Paul
3 George
4 Ringo
or
John
Paul
George
Ringo
B = range(T)
So John is not a member of T, but John is a member of the range of T = B.
Can test here. Click "Run" in top left. Bottom right will show "true".
The terminology in this context needs to be exact.
(I use 'K' instead of 'LL' because it is not good notation to use a letter 'L' as a standalone constant and also concatenated with itself to form another constant.)
L = the list of all lists
K = the list of all lists that list themselves
I hope it is recognized that these are equivalent:
x is in S
x in S
x is an element of S
x is a member of S
xeS {read 'e' as the letter epsilon)
So, these are equivalent:
L is in L
L in L
L is an element of L
L is a member of L
LeL
These are equivalent:
T lists y
T is a list & y in range(T)
These are equivalent:
T lists T
T is a list & T in range(T)
T lists itself
So, these are equivalent:
L lists L
L is a list & L in range(L)
L lists itself
And these are equivalent:
K lists K
K is a list & K in range(K)
K lists itself
/
Quoting Philosopher19
"L lists itself" is sensical.
"L lists itself in L". What does that mean other than "L lists itself"?
In other words, what exactly would need to be the case for "L lists itself" to be true while "L lists itself in L" is false? And what exactly would need to be the case for "L lists itself" to be false while "L lists itself in L" is true?
More generally, what exactly would need to be the case for "S lists y" to be true while "S lists y in x" is false? And what exactly would need to be the case for "S lists y" to be false while "S lists y in x" is true?
Even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?
Quoting Philosopher19
"L is a member of itself" is sensical.
"L is a member of itself in L". What does that mean other than "L is a member of itself"?
In other words, what exactly would need to be the case for "L is a member of itself" to be true while "L is a member of itself in L" is false? And what exactly would need to be the case for "L is a member of itself" to be false while "L is a member of itself in L" is true?
More generally, what exactly would need to be the case for "S is a member of y" to be true while "S is a member of y in x" is false? And what exactly would need to be the case for "S is a member of y" to be false while "S is a member of y in x" is true?
Again, even more generally, what exactly would need to be the case for "S has property P" to be true while "S has property P in x" is false? And what exactly would need to be the case for "S has property P" to be false while "S has property P in x" is true?
Quoting Philosopher19
What does that mean?
What does 'not-L' stand for? Does it stand for the set of all things not in L? That is, {x | x not in L}. Or does it mean the set of all things that are not L? That is, {x | x not= L}.
What does '/' stand for?
There is no difference. I asked the question to highlight a point about the property of "being a member of oneself".
Quoting Michael
Every member of LL (including L) is a member of LL because there is a list wherein which it lists itself.
So how does it follow that L has n-1 members in LL? Now note how of all the members of LL there is only one item that is actually LL. This is what instantiate LL lists itself in LL. Again, L is not a member of itself in LL (even though it is in LL because it is a member of itself in L). L is only a member of itself in L.
Quoting TonesInDeepFreeze
Quoting TonesInDeepFreeze
If I ask the question where does L list itself, the answer is in L. If I ask where else does L list itself, the answer is nowhere else. This shows that L is only a member of itself in L.
Quoting TonesInDeepFreeze
Quoting TonesInDeepFreeze
It stood for LL. I meant to highlight that LL is not-L to try and draw attention to L is not a member of L/itself in LL/not-L
Set theory does not have a "where".
My question remains:
Quoting TonesInDeepFreeze
You are using your own personal terminology for an unclear notion that no one other than you can make sense of.
You need to define your terminology in already understood mathematical phrasing. Otherwise, whatever you have in mind won't be understood by others.
You said that L is a member of itself "in L" but not a member of itself "in LL". So you're saying that L "in L" has one more member (itself) than L "in LL".
Which is nonsense.
If L has [math]n[/math] members then L has [math]n[/math] members "in L" and L has [math]n[/math] members "in LL".
Quoting Philosopher19
This is wrong.
1. L is a member of L.
2. L is a member of LL.
3. L is a member of L "in LL"
4. January is a member of Months
5. Months is a member of L
6. January is a member of Months "in L"
Let's make this simple:
A = {A}
B = {A, 1}
C = {{}, 1}
When A is a member of B, is A the empty set?
The answer is no. "In B" the set A contains itself as the only member.
Your position entails that B = C, which is false.
When in fact B = {A, 1} = {{A}, 1}
Quoting TonesInDeepFreeze
L is a member of itself in L is true.
L is a member of itself is true (but it is logically implied that L is a member of itself in L).
Quoting TonesInDeepFreeze
It makes sense to say x is a member of itself because x is in x. It makes sense to say x is a member of a set other than itself in a set other than itself. If you reject this because "set theory does not have a "where"", then our conversation cannot progress.
You didn't answer the question. I won't bother to post it yet again.
If you can find even one mathematician who can say what you mean by "L is a member of itself in L" then I'd welcome hearing about it.
And if you can define 'where', in whatever sense you mean by it, from the terminology of set theory, then I'd welcome hearing about it.
That is not what I'm saying. Nor do I see what I'm saying as amounting to that.
Quoting Michael
L does not have n members in LL. L is one item in LL. And of all the items in LL (of which L is one), only one item is LL.
Whether an item in a set is a member of itself or not is dependent on what it is and in what set it is in.
LL is an item.
It is in both L and LL.
In L it is not a member of L/itself.
In LL it is a member of LL/itself.
Quoting Michael
I do not see how it does. Again, L does not have n members in LL. L has n members in L, and LL has n members in LL.
If x in x, then x in x no matter what other sets x is or is not in.
If you don't agree or don't understand, then either you need to state your alternative system and terminology or you need to learn basic set theory. I recommend starting with the latter.
As I said, you have a highly idiosyncratic notion of sets and a highly idiosyncratic terminology to go with that. If you wish to be understood by other people, then you need to either state your system of notions and terminology, or show how they can be defined from notions and terminology that other people do already understand.
/
You can start at the very beginning:
'is an element of'
'is a member of'
'is in'
'in'
'e'
are just variants of the same primitive relation of set theory.
The symbol 'e' itself is a primitive relation symbol. It is not defined. Rather, there are axioms that determine what theorems are derived with 'e' in them. However, from 'e' we define a number of other symbols that intuitively stand for various concepts, including 'list'. With those definitions we formulate yet more theorems (though they all are in principle reducible to just the primitive 'e').
Your notions though, as so far stated, don't resolve to that process. Therefore, if you want to be understood, then you need to either show how your notions do resolve to those of set theory, or you need to state your own primitives, axioms and definitions that do explicate your own notions, or at least give some coherent outline about that.
/
Meanwhile, at least you should pay attention to the fact that there is a difference between 'set' and 'list' as a list is a certain kind of set. For example:
{John Paul George Ringo} = {George Paul Ringo John}
but
{<1 John> <2 Paul> <3 George> <4 Ringo>} not= {<1 George> <2 Paul> <3 Ringo> <4 John>}
and
{John Paul George Ringo} is the set whose members are John, Paul, George and Ringo, in whatever order you want to mention them.
{<1 John> <2 Paul> <3 George> <4 Ringo>} is a set theoretic list, which is itself a set whose members are <1 John>, <2 Paul>, <3 George> and <4 Ringo>, in whatever order you want to mention them.
The range of {<1 John> <2 Paul> <3 George> <4 Ringo>} is {John Paul George Ringo}. Or, we can say that the entries of the list are John, Paul, George and Ringo.
Crucial takeway: A set is not in any particular order, except a list is a special kind of set that does convey a particular order. And the members of a list are ordered pairs, not the members of the RANGE of the set.
So, I've given you information that is at least a start for you to use common mathematical terminology, so that you may be understood by people other than yourself.
Quoting TonesInDeepFreeze
Let me see if I've understood what you mean by RANGE.
v = any set
z = any set other than the set of all sets
V = the set of all v
Z = the set of all z
Am I right in saying that the RANGE of V is greater than Z because there are more v than z? Is this what you mean by RANGE?
That doesn't mean anything.
We don't say "the set of all z". We say "the set of all z such that [fill in some property here]"
Examples:
the set of all z such that z is an even number
which is
{z | z is an even number}
the set of all z such that z is a subset of the set of natural numbers
which is
{z | z is a subset of the set of natural numbers}
/
'domain' and 'range' are defined:
domain(S) = {x | Ey
range(S) = {y | Ex
That is most applicable when S is a relation (a relation is a set of ordered pairs).
And every function is a relation.
The notion of 'list' is captured by a certain kind of function.
{<1 John> <2 Paul> <3 George> <4 Ringo>} is a list. It maps
1 to John
2 to Paul
3 to George
4 to Ringo
It tells you not just the members of the set {John Paul George Ringo} but also a particular ordering of that set.
The set of objects that the function maps to is the range of the function. In this case
the domain is {1 2 3 4} and the range is {John Paul George Ringo}.
Think of it this way:
Suppose I tell you that B is the set whose members are the Beatles. I haven't told you an order of the musicians in the Beatles, just that B is the set of them no matter in what order.
There are 24 ways to list the Beatles.
Set theory conveys the notion of a list with functions:
{<1 John> <2 Paul> <3 George> <4 Ringo>} is one list.
{<1 George> <2 John> <3 Paul> <4 Ringo>} is another list.
and there are 22 more.
In other words, we exhibit the order of the list by putting the number of the item in the list before the item.
Again:
the members of {John Paul George Ringo} are John, Paul, George and Ringo, in no particular order.
the members of one LIST of the Beatles are <1 John>, <2 Paul>, <3 George> and <4 Ringo>.
the members of another LIST of the Beatles are <1 George>, <2 John>, <3 Paul> and <4 Ringo>.
but all those lists have the same RANGE, which is {John Paul George Ringo}, whose members are John, Paul, George and Ringo.
/
So, if L is the list of all lists, then L is a function. And the items listed by L are the range of L. In this case, the range of L is the set of all lists. So, since L is itself a list, L is in the range of L. But L is not in L, since L is not an ordered pair.
Quoting Philosopher19
Quoting TonesInDeepFreeze
So I think I understand what you mean by RANGE, but I feel any further conversation will not be fruitful. I feel like I've already said all that I should have said. And anything that could have added to what I should have said I think would be in the post to which I provided the link to before. In the event that I've not done justice to what you've been saying to me, I apologise.
Forgot to say thank you for the detailed reply. Thanks.
When a set is a member of another set it is still a set with members of its own.
Given this:
A = {A}
B = {A, 1}
One of these must be true:
1. In B, A isn't a set
2. In B, A is a set with 0 members
3. In B, A is a set with 1 member, and that member is itself
4. In B, A is a set with 1 member, and that member isn't itself
5. In B, A is a set with more than 1 member, one of which is itself
6. In B, A is a set with more than 1 member, none of which is itself
So which of these claims are you making? It must be one of them.
The correct answer is (3):
A = {A}
B = {A, 1} = {{A}, 1}
Quoting Michael
I understand/agree.
Quoting Michael
Let's focus on 3. There is a difference between:
3) In B, A is a set with 1 member, and that member is A/itself. (Correct)
3a) A is a set in B that is a member of itself in A/itself (Correct)
3b) A is a set that is a member of itself in B
Here is why 3b is incorrect. Compare:
p) In B, A is a set that is a member of itself in A
q) In B, A is a set that is a member of itself in B
p and q are not the same and they both highlight a meaningful difference. p is correct (because in B, A is a member of A/itself in A). q is false (because in B, A is not a member of A/itself in B)
To my understanding, you either wrongly view p and q as the same thing, or, you wrongly view q as correct. But how can you view q as correct given that p is truth and p and q are opposing things?
So my question to you is:
Which is correct: p or q or both or neither?
Your p and q make no sense in set theory. Only 3a is meaningful in set theory (although as already mentioned, the “in B” part of the sentence is vacuous). And using that meaning, the set of all sets that don’t contain themselves is a contradiction. Russell proved this.
Whatever you’re trying to argue has nothing to do with Russell’s paradox and nothing to do with set theory.
Assumption: S is the set of all sets that are not members of themselves.
Option 1:
S = {}
S is not a member of itself. But, as per the assumption above, it ought be a member of itself.
Option 2:
S = {S}
S is a member of itself. But, as per the assumption above, it ought not be a member itself.
Neither option 1 nor option 2 work. Therefore, the assumption is a contradiction.
Maybe I'm missing something, but that seems to rely on the premise that either S = 0 or S = {S}, which is a premise we are not allowed.
I was being lazy because this discussion has gone on long enough. Should have been:
S = {S, …}
and
S = {x1, x2, …} where no xn = S
But, yes, we do have that either S in S or S not in S.
Either S in S or S not in S.
Suppose S in S. Then S not in S. So S not in S.
Suppose S not in S. Then S in S. So S in S.
So both S in S and S not in S.
Quoting Michael
I believe it makes perfect sense to say set x is only a member of itself in its own set, and I tired to prove it to you, and we even narrowed it down to p and q, but you have ceased to engage on the grounds that "p and q make no sense in set theory". This truth (the truth of a set only being a member of itself in its own set) has logical implications that would have shown (contrary to those who reject the universal set because they believe a set can be a member of itself outside of its own set) that the universal set is not contradictory in any way.
Quoting Michael
I believe if I respond to this, we'll just go around in circles, and then perhaps we'll come back to the p and q point again. But instead of doing that, I will just say that my question remains unanswered:
Quoting Philosopher19
I answered it. Neither p nor q make sense. @TonesInDeepFreeze has explained to you in depth that the sentence "A is a member of B in C" is meaningless in set theory.
Quoting Philosopher19
This isn't about the universal set. This is about the Russell set. The Russell set is contradictory. It can neither include nor exclude itself without defying its own definition.
There are a number of set theories with a universal set, such as New Foundations and positive set theory.
No one knows what you mean by such locutions as "x is a member of itself only in its own set". You have not defined what it might mean.
Someone might as well say "x is not a member of itself not only outside itself" or similar nonsense and then require you to understand it though it is impossible to understand.
./
The assertion that there exists a set of which every set is a member allows Russell's paradox.
Yes, going back to your p and q would be going back full circle yet again.
To break the circle requires that you give serious consideration to the fact that no one understands your phraseology but you. That would lead to you learning more about the subject so that you could communicate your ideas about it with other people.
No problem. If it makes sense to you, that's OK. Mathematicians don't have to agree with you. It's not like the fate of the world hinges upon this. It's OK to feel good about your own creation. Why argue with others?
Quoting jgill
It could be that arguing with someone is what it takes for them to be sincere to a truly perfect existence/being (or God/Goodness/Truth). It could be that that is what it takes for them to favour good/truth over evil/falsehood in a particular instance/moment. It could also be the opposite. I'd like to think I got into this discussion as a part of my efforts to favour/prioritise Goodness/Truth (or to strive in the cause of a truly perfect existence/being).