The overlooked part of Russell's paradox
If x, y, and z are sets that are not members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}.
If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}.
You cannot have a set of ALL sets that are not members of themselves because it will result in at least one set not being included in the set. In other words, x will have to be in x, but it can't.
You cannot have a set of ALL sets that are members of themselves because it will result in at least one set being a member of itself twice. In other words, x will have to not be in x, but it can't.
For the solution to Russell's paradox:
http://philosophyneedsgods.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z}.
You cannot have a set of ALL sets that are not members of themselves because it will result in at least one set not being included in the set. In other words, x will have to be in x, but it can't.
You cannot have a set of ALL sets that are members of themselves because it will result in at least one set being a member of itself twice. In other words, x will have to not be in x, but it can't.
For the solution to Russell's paradox:
http://philosophyneedsgods.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
Comments (155)
If y and z are members of x, then you actually can write it (if a set can be a member of itself). (I'm refering to the part where you say x is a member of itself).
Quoting Philosopher19
The idea that such a set can't be constructed to solve the paradox does seem correct, as Russell's own solution (the theory of types) maintains.
But your reason for holding that there can't be such a set seems inadequate, all you are doing is to re-state the paradox: If x can't be in x then, by definition, it would have to be in x, since it would be a set that doesn't contain itself.
It's easier to understand with Grelling's paradox, which is similar to Russell's:
Suppose we define an adjective that describes itself as an autological adjective. For example: “polysyllabic” is a polysyllabic adjective, therefore it is autological.
And suppose we define an adjective that does not describe itself as a heterological adjective. For example: “monosyllabic” is not monosyllabic, therefore it is a heterological adjective.
But then the question arises: Is “heterological” a heterological adjective?
Let's say we maintain that it is, that means “heterological” is an adjective that does not describe itself. But if that's the case, then the adjective in question must not be named “heterological”, for that would necessarily imply that it describes itself. Yet we know that it is named “heterological”, therefore it can't be heterological.
Let's now maintain that it isn't heterological (that it's autological), meaning it does describe itself: That implies that it is heterological, since otherwise it would not describe itself. But obviously it is a contradiction to maintain that “heterological” both is and is not a heterological adjective. Therefore it can't be the case that it is autological either.
That is incorrect.
Of course, to countenance sets being member of themselves, we have to delete the axiom of regularity. With that done:
Suppose xex, yey, and zez
Suppose x = {x y z}
Those suppositions together are consistent. For example by letting:
y = {y} and z = {z}
Quoting Philosopher19
With the axiom of regularity, that is correct, since such a set would have all sets as members, from which we derive a contradiction.
The Russell principle though is:
There is no set of all and only the sets that are not members of themselves.
And that is derived from first order logic alone:
For any relation R, ~ExAy(Ryx <-> ~Ryy)
Quoting Philosopher19
That is incorrect. With the axiom of regularity, that set is the empty set. And without the axiom of regularity, it would still be consistent for there to be a non-empty set of all sets that are members of themselves. For example, allow that there is just one set S that is a member of itself. Then the set of all sets that are members of themselves is {S}.
Second this (almost). Here are the relevant links. Turns out that self-containing sets are an object of interest. However I do not believe there could be a set of all sets that contain themselves, even in the absence of regularity. Such a set would be subject to Russell's paradox.
https://en.wikipedia.org/wiki/Non-well-founded_set_theory
https://plato.stanford.edu/entries/nonwellfounded-set-theory/
With regularity, It's the empty set.
And we can't derive a contradiction by dropping an axiom, so such a set is consistent also without regularity. But it would be inconsistent with set theory without regularity if every set were a member of itself, but "every set is a member of itself" is already inconsistent with set theory (even without regularity).
Quoting fishfry
Nope.
Hmmmm. Don't think this can work. If there's a set of all sets that contain themselves, then its complement is the set of all sets that don't contain themselves; which both does and does not contain itself. So this can't work.
Why not? Well to form a set via the axiom schema of specification, you need an existing set: [math]\{x \in X : P(x)\}[/math] where [math]X[/math] is a set and [math]P[/math] is a unary predicate.
You would like to form the set [math]R = \{x : x \notin x \}[/math] but you haven't got an existing set to start with, so this is not a legal set specification.
Of course you COULD form the set [math]R = \{x \in \mathbb Z : x \notin x \}[/math] or [math]R = \{x \in \mathbb R : x \notin x \}[/math], for example. Given any particular set, you can form the set of all its elements that are not elements of themselves, and the result (under regularity) will be the empty set. But you cannot form the set of ALL such sets, because you have no universal set with which to invoke specification.
What say you? I admit this is a tricky one but I've got myself 99.9% convinced of my own reasoning.
I don't think you mean {x | ~xex}. We're talking about {x | xex}.
And yes, without regularity, we can't prove there is a nonempty set {x | xex}.
But that's not the question. The question is whether we can prove that it is not the case that there is a set {x | xex}. And we can't. It is consistent with set theory that there is a set {x | xex}. And with regularity, it is provable that there is a set [x | xex}.
I'll retract that objection and reiterate my point that you cannot form the set [math]\{x : x \notin x\}[/math] absent some existing set. Which in your case needs to be the universal set, which does not exist. At best you can form the set [math]\{x \in X: x \notin x\}[/math] for every set [math]X[/math] you can name. That's the best you can do.
Your math notation in your previous post does not format form me.
Anyway, {x | ~xex} is not at question. There is no such set.
It formats fine for me, but it took a couple of edits before it did. Is it still bad? Can't do anything about that, it looks right at my end.
Quoting TonesInDeepFreeze
Then we're in agreement and you have conceded my point, since that is exactly the set you claim exists. Write out your claim formally and you'll get exactly what you just wrote.
Quoting TonesInDeepFreeze
Yes ok, you're right, but that also is not a legal set specification.
"given a set A and a predicate P, we can find a subset B of A whose members are precisely the members of A that satisfy P."
https://en.wikipedia.org/wiki/Axiom_schema_of_specification#Statement
In order to form the set of all sets that are members of themselves, you have to start with some existing set and then apply specification to the predicate "x element of x".
I thought Russell's paradox was meant to undermine set theory. As far as I can tell, it begins as C = the set of all sets that doesn't contain itself. Either C is itself in C or not. If C contains itself, C can't contain itself. If C doesn't contain itself, C contains itself. In short, C contains itself AND C doesn't contain itself. Contradiction.
Let's look at the issue as the conditionals they're said to be
1. If C contains itself then C doesn't contain itself
2. C doesn't contain itself or C doesn't contain itself (1 Imp)
3. C doesn't contain itself (2 Taut)
4. If C doesn't contain itself then C contains itself
5. C contains itself or C contains itself (4 Imp)
6. C contains itself (5 Taut)
Ergo,
7. C contains itself and C doesn't contain itself (3, 6 Conj) [Contradiction]
But then, Russell's argument boils down to,
3. C doesn't contain itself
6. C contain itself
8. C contains itself and C doesn't contain itself (3, 6 conj)
Ergo,
9. C contains itself (deny 3)
Or
10. C doesn't contain itself (deny 6)
Both 3 and 6 can't be denied...leads to a contradiction.
It's a tautology, C contains itself or C doesn't contain itsellf. There's no contradiction!
Looks okay now.
Quoting fishfry
No, I don't. Indeed, I said the opposite in my original post.
Quoting fishfry
Howzabout you quote me where you think I claimed that there exists a set whose members are all and only the sets that are not members of themselves.
Indeed, I pointed out that there is not such a set, and even on logical principles alone.
Quoting fishfry
It doesn't depend on the abstraction operator. I could just as well write the whole conversation without the abstraction operator.
Quoting fishfry
To prove the existence of sets having a certain property, we can only use the axioms. But the axioms don't say that other sets don't exist, except as we can prove from the axioms that there do not exist sets of a certain property.
Again, the axioms don't prove that there does not a exist a set whose members are all and only those sets that are members of themselves. Indeed, with regularity, the axioms prove that here does exist such a set.
It undermined Frege's idea that a set is defined by a predicate, as in "the set of things satisfying such and so." That leads to a contradiction. The fix is to require that the predicate must be applied to an already existing set, as in "the set of things in some set X that satisfy such and so." This is the meaning of the axiom schema of specification.
"given a set A and a predicate P, we can find a subset B of A whose members are precisely the members of A that satisfy P."
https://en.wikipedia.org/wiki/Axiom_schema_of_specification
In symbols, if we know we have some set [math]X[/math], we can form the set [math]\{x \in X : P(x)\}[/math] where P is some predicate.
But we can NOT form the set [math]\{x : P(x)\}[/math], that gets us into Russell trouble.
Good, thanks.
Quoting TonesInDeepFreeze
My head hurts but I think I'm right. Let's see what comes next ...
Quoting TonesInDeepFreeze
Then do so. Let me see it. Writing it in words doesn't help. "The set of sets that are members of themselves." It's true that there aren't any such sets (under regularity). But it does NOT mean there is a set of them. So if you claim there is a SET of them, let me see a legal set specification.
Again: Yes I agree that under regularity, there are no sets that are members of themselves. But there is NOT a SET of all sets that are members of themselves, not even the empty set. Though I agree that it's a bit of a puzzler. To solve it you have to get very formal. Show me a valid set specification that supports your claim and I'll believe you.
Quoting TonesInDeepFreeze
You can't form the set you claim exists. So you are saying that some set exists that's not given by the axioms? That's a stretch. Explain this a bit more.
Quoting TonesInDeepFreeze
So you agree you can't prove that your set exists, but you claim it exists anyway? Is that your argument? You might almost have an argument, I want to make sure I'm understanding you.
Let me restate your argument. The empty set is the set of all flying purple elephants. And the empty set is the set of all sets that contain themselves under regularity.
Perhaps you've convinced me. Is that what you mean? You could be right.
In order to object to this argument, I have to say that the set of all flying purple elephants is also not a legal set. The set of all flying purple elephants that are animals is the empty set. And if I get pedantic that way about the axiom schema of specification, then your argument fails. But I admit it's not an entirely bad argument.
Russell's paradox shows the contradiction in set theory with unrestricted comprehension. After Russell's note, we moved to a set theory that does not have unrestricted comprehension.
Quoting TheMadFool
That there does not exist such a set.
If you can formalize that for mathematics, fine. But meanwhile, ordinary mathematical logic and set theory do not "ban" such a formulation.
(1) In set theory, we don't ban writing "ExAy(yex <-> ~yey)". Rather, we prove ~ExAy(yex <-> ~yey).
(2) In mathematical logic, we don't dispute that some theories may formalize a liar-like sentence. Rather, we prove that any theory that can do so is inconsistent.
Yes correct! Exactly. But you are also claiming that "the set of all sets that are members of themselves" exists and is the empty set. That's not consistent with what you just said! You can't invoke unrestricted comprehension. All you've done is remind me that when I tell people that "the empty set is the set of all purple, flying elephants," I'm violating the axiom schema of specification. But at least I'm doing it for a good cause :-)
Are you serious? Come on, you know how to do it yourself:
ZF |- ~ExAy(yex <-> ~yey)
ZF |/- ~ExAy(yex <-> yey).
ZF-R |/- ~ExAy(yex <-> yey)
ZF |- ExAy(yex <-> yey)
ZF-R |/- ExAy(yex <-> yey)
/
Quoting fishfry
With regularity, yes the empty set 0.
E!yAx(xey <-> yey_
In English: There exists a unique set whose members are all and only those sets that are members of themselves. Proof:
xe0 <-> xex
So Ax(xe0 <-> xex)
So E!y(xey <-> xex)
So 0 = the-y Ax(xey <-> xex) = {x | xex}
Quoting fishfry
No, I am not. Read what I said. I said that we can only prove existence from the axioms. But we can't prove non-existence from the axioms except when the axioms actually prove the non-existence. And in the case of ZF-R, the axioms don't prove the non-existence of a set whose members are all and only those sets that are members of themselves. And with ZF, the axioms do prove there is such a set; it's the empty set.
I didn't.
Quoting fishfry
No, I'm not violating separation. Separation and extensionality were used to prove the existence of 0 (in an axiomatization where the existence of an empty set is not taken as an axiom).
To prove the existence of a set, we don't always have to do it directly from separation. We have union, power set, etc. But in this case I did prove it from separation, extensionality, and whaever other axioms prove ~Ex xex from regularity, as separation was used in the previous result that ExAy ~yex.
I'll try to go through your proof later. I'd prefer it if you'd mark it up but I'll slog through the ASCII later maybe. I think you are making an interesting point. You might be right after all, I'm not sure.
I don't like using markup. The text is plenty clear enough.
No, you know it.
(1) is set theory proving there is no set whose members are all and only those sets that are not members of themselves.
(2) is Tarski's theorem.
But my point is that x, y, and z are not members of themselves, whereas x = {x, y, z} means that x is a member of itself. Hence why I have to write something like p = {x, y, z} to represent x, y and z not being members of themselves.
But the key part of my post is that you cannot have a set of all sets that are members of themselves because it will result in at least one set being a member of itself twice. This is the overlooked part of Russell's paradox.
I'm going back to the root of the matter. The problem occurred because there wasn't enough clarity with regards to what it is for a set to be a member of itself, and what it is for a set to not be a member of itself.
Consider the following:
Call any set that is not a member of itself a -V. Call any set that is not the set of all sets a V'. Call any set that's simply a set, a V (the V of all Vs = the set of all sets).
Is the V of all -Vs a member of itself? It is impossible to have a V of all -Vs that contains all -Vs and no other sets. You cannot have a set of all sets that are not members of themselves that is itself not a member of itself. You cannot have a -V as a V of all -Vs.
No V, or V’, or -V, can contain all -Vs and nothing more, but one V can contain all -Vs and something more.
Two Vs contain all V's:
One (which is a V') contains all V's and nothing more. The other (which is not a V') contains all V's and something more. The latter is the set of all sets (the V of all Vs), the former is the V' of all V's (the not-the-set-of-all-sets set of all not-the-set-of-all-sets sets). Thus, only one V can
contain all Vs and nothing more (the V of all Vs). Only one V' can contain all V's and nothing more (the V' of all V’s).
The above shows that whilst there can be no -V that contains all -Vs, there is a V that contains all -Vs. Whilst there can be two Vs that contain all -V’s, there can only be one V’ that contains all -V’s. -V and -V’ are semantically not the same. -V’ = any V’ set that is not a member of itself.
Rejecting the set of all sets is the last thing we should be doing, and it is a member of itself. See the link in the OP for a more detailed proof of this.
What if you view it this way:
The set of all sets encompasses all sets. It encompasses all sets that are not members of themselves, and it is a member of itself (because it encompasses itself). No contradictions here. See?
If interested, see the link in the OP for a more thorough discussion and solution of Russell's paradox.
How about the following argument. Please go easy on me.
Suppose C = the set of all sets that don't contain itself
1. A set can contain itself like so A = {1, A} but there seems to be problem when you list down the elements. Suppose B is a set that contains itself:.{1, 2, {1, 2, {1, 2,..the task can't be completed. In other words, no set can contain itself. Ergo ALL sets are sets that don't contain themselves. Can we construct a set of all sets that don't contain themselves? Why not? Of course we can because ALL sets can't contain themselves. However, this set can't can't contain itself.
1. All sets don't contain themselves.
2. The set of all sets that don't contain themselves = The set of all sets.
3. The set of all sets is impossible because it can't be member of itself and so it can't be the set of all sets. (from 1)
4. The set of all sets that don't contain itself is also impossible (from 2 and 3).
5. For Russell's paradox, the set of all sets that don't contain itself must be a set.
6. The set of all sets that don't contain itself is impossible i.e. it isn't a set.
Ergo,
7. Russell's paradox is not a paradox.
Quoting Philosopher19
Here you say they are members of themselves. If they are members of themselves, then x can be contained in x, right? So if set y and set z are members of x, what’s the problem with writing that the set x contains sets x, y and z?
If y and z are not members of x, then obviously they cannot be contained in x, but x can still be contained in itself.
Quoting Philosopher19
Suppose x is the set of all ideas: since the set of all ideas is itself an idea, it contains itself. What’s wrong with that? I guess you are saying that if the set of all ideas contained itself, then the set of all ideas contained in the set of all ideas would also have to contain the set of all ideas again, and so on in infinitum. Well yes, but what’s wrong with that?
Using a mathematical analogy: the golden ratio’s infinite radical and its continued fraction contain themselves infinitely many times, and yet that does not give rise to any contradictions:
... and so on forever.
With regularity, there is a set whose members are all and only those sets that are members of themselves. That set is the empty set. And if we drop regularity, then the statement that there is a set whose members are all and only those sets that are members of themselves is independent of the axioms.
Quoting Philosopher19
'member of itself twice' has no apparent mathematical meaning.
That's the part I believe everyone has overlooked. I will try and show this clearly:
x, y and z, are sets that are not members of themselves. I am trying to form a set of these three sets that are not members of themselves.
A) It cannot be the case that x = {x, y, z} because that implies x is a member of itself, precisely because x is in x. But it can be the case that p = {x, y, z} precisely because x is not in x.
Now consider this: x, y, and z, are sets that are members of themselves. Consistency with A would have us say it cannot be the case that x = {x, y, z} precisely because x is in x. This either means that y and z are not members of themselves, or it means that y and z are members of themselves once, whilst x is a member of itself twice. Do you see my point with consistency? If I wrote p = {x, y, z} then you could say x, y, and z, are members of themselves precisely because x is not in x.
A set of all sets that are not members of themselves is impossible precisely because one set will have to be a member of itself. In other words, x will have to be in x, but it can't.
A set of all sets that are members of themselves is impossible precisely because one set will have to be a member of itself twice. In other words, x will have to be in x, but it can't because that would either amount to x being a member of itself twice (with all other sets being members of themselves once), or it would amount to x being a member of itself once, with all other sets not being members of themselves.
I am trying to bring consistency where I have seen inconsistency in mainstream set theory. Do you see the inconsistency in saying "you cannot have a set of all sets that are not members of themselves, but you can have a set of all sets that are members of themselves".
Once the above inconsistency is dealt with, we can see that the set of all sets is a member of itself, and it encompasses all sets that are not members of themselves (as well as itself). Note that this statement is not contradictory. Note that saying there can be no set that encompasses all sets is blatantly contradictory (and set theorists know this, they just haven't realised what they've overlooked). Again, the link in the OP covers this in more detail.
Yes I know. It's a contradiction for something to be a member of itself twice. I am saying that this is the logical consequence of a set of all sets that are members of themselves as proven here:
x, y and z, are sets that are not members of themselves. I am trying to form a set of these three sets that are not members of themselves.
A) It cannot be the case that x = {x, y, z} because that implies x is a member of itself, precisely because x is in x. But it can be the case that p = {x, y, z} precisely because x is not in x.
Now consider this: x, y, and z, are sets that are members of themselves. Consistency with A would be such that it cannot be the case that x = {x, y, z} precisely because x is in x. If we are to be consistent with A, then this either means that y and z are not members of themselves, or it means that y and z are members of themselves once, whilst x is a member of itself twice. Do you see my point?
A set of all sets that are not members of themselves is impossible precisely because one set will have to be a member of itself. In other words, x will have to be in x, but it can't.
A set of all sets that are members of themselves is impossible precisely because one set will have to be a member of itself twice. In other words, x will have to be in x, but it can't because that would either amount to x being a member of itself twice (with all other sets being members of themselves once), or it would amount to x being a member of itself once, with all other sets not being members of themselves.
In your OP you mention 2 possibilities:
1.x, y and z not being members of themselves.
2. x, y and z being members of themselves.
Then you say: if 1 is true then U follows, If 2 is true then P follows.
I'm refering only to the second possibility, in which we assume that x,y and z are members of themselves, and then we see what that would logically entail (P).
You say: Quoting Philosopher19
So you say if 2 is true, then x would be a member of itself, then we could write x={x,y,z} ,which we can't do because...well, you've given no reason to accept that yet, except that x would “contain itself twice”, which, if you mean what I said earlier (otherwise I don't know what you mean by that), is no reason for supposing that you can't have a set that contains itself as a member, in the same way as there is no contradiction about the golden ratio's continued fraction containing itself not just twice, but infinitely many times.
As for “the set of all sets that contain themselves”, regardless of whether or not such a set can be constructed, you have not yet given a good reason for maintaining that it can't be.
Usually, we have an intuitive notion that sets are not members of themselves. However, since 'is a member' is primitive, we will not have a formal explication of the notion.
This is handled typically in three ways (not necessarily in order of preference):
Z set theory includes the axiom of regularity. And the axiom of regularity proves that no set is a member of itself.
ZFC-R set theory drops the axiom of regularity. The existence of a set that is a member of itself is independent of those axioms.
Z-R+D set theory [where D is an appropriate anti-foundation axiom] set theory drops the axiom of regularity and adds an axiom so that the theory proves that there are sets that are members of themselves.
/
Quoting Philosopher19
Your notation is mixed up (especially as you conflate 'V' as naming an object with 'V' standing for a predicate). I'll use ordinary notation:
Quoting Philosopher19
First we must ask whether ExAy(yex <-> ~yey). The answer is no. It is a theorem of first order logic:
~ExAy(yex <-> ~yey)
However, toward a contradiction, we do start with the assumption:
ExAy(yex <-> ~yey)
Then we show xex & ~xex.
Thus ~ExAy(yex <-> ~yey).
Quoting Philosopher19
Correct. ~ExAy(yex <-> ~yey).
Quoting Philosopher19
Correct, because there is no set of all sets that are not members of themselves anyway.
Quoting Philosopher19
Incorrect. ~ExAy(~yey -> yex). The "and something more" doesn't get around that fact.
That's enough for now; I'm not at this time going on with the rest of your argument, especially to untangle your poorly chosen notation.
You didn't go far enough in the argument:
If there is a set of all sets, then it has the subset that is the set of all sets that are not members of themselves. But the claim that there is the set of all sets that are not members of themselves implies a contradiction. Therefore, the claim that there is a set of all sets implies a contradiction.
Set theory doesn't prove things in this kind of context by saying "the task cannot be completed".
However, the axiom of regularity disallows infinite descending membership chains and
circular memberships (so also no sets being members of themselves).
Quoting TheMadFool
Because the axioms don't provide for such a construction while also the axioms prove that there is no such set nor such a construction.
Quoting TheMadFool
It is correct that with regularity, no set is a member of itself. But that doesn't imply that there is a set whose members are all and only the sets that are not members of themselves.
Quoting TheMadFool
1. Correct. With regularity, ~Ex xex
2. You cannot refer to "the set of all sets" without first proving there is such a set. And we prove that there is not such a set.
3. Correct. If there were a set of all sets then that would violate regularity. But, we don't need regularity (used for your item 1.) to prove there is no set of all sets.
4. Correct. But again, we don't need regularity (used for your item 1.) to prove there is no set of all sets.
5. What do you mean "for Russell's paradox"? The role of Russell's paradox is this:
Suppose ExAy(yex <-> ~yey).
Then xex <-> ~xex.
Therefore, ~ExAy(yex <-> ~yey)
6. Correct.
7. What do you mean by "it's not a paradox". The paradox (the contradiction) comes from the claim that there is a set of all sets that are not members of themselves. That contradiction implies that there is not a set of all sets that are not members of themselves.
Wrong. It is possible that all these are the case: x = {x y z} (so x is a member of itself), and y is a member of y, and y is a member of x, and z a member of z, and z is a member x.
That's enough for now. At this time, I'm not going further with your argument.
Quoting Philosopher19
It's not inconsistent. It does not imply a contradiction. A contradiction is a statement and its negation. If you claim to point out an inconsistency, then you need to show a statement and its negation that are both implied.
Quoting Philosopher19
Nope. You have not shown that it implies a statement and its negation.
"member of itself twice" has no apparent mathematical meaning.
Why don't you look up a text in set theory so you would know how set theory axiomatically, clearly and unambiguously proves theorems and defines terms?
FWIW I think I've convinced myself that you're right. Under regularity, the collection of sets that are members of themselves is indeed the empty set. In fact the empty set is often defined as [math]\{x : x \neq x \}[/math]. It's always bothered me that there is no containing set as required by specification. But this is a standard definition. So maybe it's ok. I don't know why it's ok. I should look into this.
I didn't go through your proof yet but perhaps I'll make a run at it.
Ok. There are a couple of different ways sets can be written down as:
1. Set notation: S = {x | x is a prime number less than 10}
2. In words: S = {all prime numbers less than 10}
3. As a list of members of the set: S = {2, 3, 5, 7}
1. Set notation, C = {x | x is a set that contains itself}. Nothing seems amiss
2. In words, C = The set that contains itself. Doesn't seem to be problematic
3. As a list of members of the set: C = ??? [Problem. No such set can be]
There is no rule of set theory that we must name a set with set abstraction notation.
Moreover, for a finite set with members that are members of themselves, we can do it by extensional notation. For example:
Suppose
y = {y z}
z = {z j}
x = {x y z}
So, extensionally, we named the set whose members are x y z only.
The fact that we can continue to "expand" the notation in ways like this:
x = {x {y z} {z j}}
and
x = {{x y z} {{y z} {z j}} {{z j} j}}
ad infinitum, ad nauseum
doesn't prove without regularity that ~Ex x = {x y z} and doesn't prove ~Ex xex.
To prove something in set theory, you have to do if from the axioms. And set theory without the axiom of regularity does not prove ~Ex x = (x y z} and does not prove ~Ex xex.
Quoting TheMadFool
It is problematic, because we cannot use set abstraction properly without first proving that there is a set that satisfies the defining property.
Here's a set that doesn't contain itself: A = {7, &, troll}
Here's a set that contains itself: Your turn
I could only do that by using anti-foundation (or non-wellfounded) set theory.
These are different:
(1) The theory proves that there is not a set that is a member of itself.
(2) The theory does not prove that there is not a set that is a member of itself.
Look at those closely to see that they are different.
These hold:
Z proves that there is not a set that is a member of itself.
ZFC-R does not prove that there is not a set that is a member of itself.
/
So to make it plainly clear:
I have never claimed that ZFC-R proves that there is a set that is a member of itself.
What I have said is that ZFC-R does not prove that there is not a set that is a member of itself.
Put another way:
Ex xex is consistent with ZFC-R.
On the other hand, if you claim that ZFC-R does prove that there is not a set that is a member of itself, then the burden is on you to show such a proof. Hint: Don't bother. There is no such proof.
A set that doesn't contain itself: {1, y, $}
A set that contains itself: ???
x = {x}
Read my previous post regarding proof and lack of proof of existence.
I don't think there are any common or obvious examples, but they are studied. Perhaps there are some clues here, I didn't read through this.
https://plato.stanford.edu/entries/nonwellfounded-set-theory/
No examples here either.
https://en.wikipedia.org/wiki/Non-well-founded_set_theory
It's interesting that there are articles about non well founded sets, but no specific examples. The SEP article does have some clues but nothing particularly simple.
So, set X = {X} = X1
That means, I can substitute X with {X}.
Here goes, X = {{X}} = X2
X1 = X2 (should be) because both are X but {X} =/= {{X}}.
Without the axiom of regularity, you can't prove
~Ex x = {x} = {{x}} = {{{x}}} ... for as finitely many iterations you want to make.
As I said, if you think there is such a proof, then try to show it.
I'm not sure, but I think a place to look might be proofs showing the relative consistency of anti-wellfoundeness. Maybe there is a construction in such proofs.
Then, there can't be a set that contains itself.
1. All sets are sets that don't contain themselves
2. No sets that contain themselves are sets [from 1]
In other words, sets that contain themselves aren't sets which simply means the sentence, "sets that contain themselves" is meaningless.
Let's visit Russell's paradox now.
Suppose C = set of all sets that doesn't contain itself
1. If set C doesn't contain itself then, set C contains itself.
On the face of it, statement 1 looks reasonable but, as shown above, the consequent of the conditional 1 (above) is meaningless ["set" and "contains itself" in the same sentence is a contradiction]. Ergo, Russell's conditional (1 above) is gibberish.
2. If set C contains itself then, set C doesn't contain itself.
Again for the same reason as above, the antecedent of this conditional is nonsense. Ergo, Russell's conditional 2 is balderdash.
Russell's paradox can't be a paradox because the two key conditionals have no meaning at all.
You're not reading my posts.
Without the axiom of regularity, we cannot prove ~Ex xex.
And the rest of your post is more of your misunderstanding of how set theory works.
You started to learn the sentential calculus. That's good. Finish doing that. Then learn the predicate calculus. Then learn basic set theory. Then you'll be in a position to discuss these matters coherently.
You mentioned that without the axiom of regularity, we can't prove ~Ex xex. As far as I can tell that means,
Can prove ~Ex xex -> Axiom of regularity.
No axiom of regularity -> Can't prove ~Ex xex
Since I've proved a set can't contain itself, it follows that I've assumed the axiom of regularity then.
Challenge for you: Can you prove that a set contain itself? Feel free to use any axiom of your choice.
No, it is a non-sequitur to infer
ZFC-R |- ~Ex xex -> regularity
from
it is not the case that ZFC-R |- ~Ex xex
Now, clearly
ZFC-R |- regularity -> ~Ex xex
which is the same as saying
ZFC |- ~Ex xex
However, I'm not absolutely sure, but I'm really pretty sure that it is not the case that
ZFC-R |- Ex xex -> regularity.
That is, Ex xex is a clear consequence of regularity. But regularity is a more comprehensive statement than Ex xex.
Quoting TheMadFool
The problem is that your argument is not a valid proof. You didn't use the axiom of regularity; instead you just used hand-waving about [paraphrasing here:] "can't show it using braces and then infinite nesting in braces".
I will say this though: It is quite reasonable to think that our common notion of a set disallows sets from being members of themselves. And it seems you have been relying on that common notion. So it is fine to say that self-membership is not in our ordinary conception of sets and that therefore we should not countenance self-membership. Very fine. But that's not proof from axioms. To make it proof from axioms you need an axiom that ensures that that common notion abides axiomatically. And that's where the axiom of regularity comes in.
Quoting TheMadFool
(1) Again, note that I did not claim to be able to prove Ex xex. All I claimed was that ZFC-R does not prove ~Ex xex.
(2) Trivially, we can prove Ex xex, and consistently with ZFC-R by adding an axiom: Ex xex.
(3) For a non-trivial proof, we would need a non-trivial anti-foundation axiom. To see what that looks like:
https://en.wikipedia.org/wiki/Non-well-founded_set_theory
Note: I am referring to the Wikipedia article instead of the Stanford Encyclopedia of Philosophy article only because this particular Stanford article seems to start out at a pretty difficult level even from the start. I have not gone over the Wikipedia article closely, so I don't guarantee its accuracy.
Meanwhile, I hope I can take it that you now grant that you don't know how you would prove ~Ex xex in ZFC-R.
It's technical to prove that non well-founded sets are consistent. See
https://math.stackexchange.com/questions/1148634/show-that-there-are-non-well-founded-models-of-zermelo-fraenkel-set-theory
I don't claim to understand the responses, but this is how people prove consistency of non well founded sets. In this case they're not showing a set that contains itself, but rather a model of set theory with an infinite membership chain [math]c_1 \in c_2 \in c_3 \in \dots[/math]
The discussion thread consists of highly technical responses from professional mathematicians. There are no specific examples of sets that contain themselves; but rather, consistency proofs for ZF with the negation of the axiom of foundation/regularity.
ps here is a better answer.
https://math.stackexchange.com/questions/253818/example-of-set-which-contains-itself
The answer is that it's consistent with ZF (minus regularity) to have a set [math]x = \{x\}[/math]. There's no point asking, "What is it?" It's what it is as notated. The point is simply that it's logically consistent to have such a thing.
That thread also points to this article:
https://en.wikipedia.org/wiki/Aczel%27s_anti-foundation_axiom
It says:
I believe that last bit gives us the best visualization we're going to get for a set that contains itself; namely, a graph consisting of one node with a path that points from the node to itself.
The point being is that any set may be viewed as a graph, where the edges are the element-of relation. If you let the graph loop, you have a set that violates Regularity/Foundation.
"Let B be the set of all sets explicitly referred to in this book. Clearly, since B is referred to in this book (I am just now referring to it) we have
BeB."
Keep it simple for me, ok.
1. Assume whatever axiom you want to/have to assume to prove the proposition, C = sets can contain themselves.
2. I'll assume, given your knowledge, C is proven or C
3. If C then there has to be a set N ="{...} such that it contains itself
4. There has to be a set N = {...} such that it contains itself [2, 3 modus ponens]
My request to you is express N as I express the set P (prime numbers less than 4), P = {2, 3}. In other words, I want to know if it's possible to have a set N = {x, N} where x = no, one, or more members of that set.
This seems impossible for the following reason.
1. There's a set N that contains itself [assume for reductio ad absurdum]
2. Suppose, N = {x, N}
3. N is a proper subset of {x, N} [proper subset of a set]
4. If N is a proper subset of {x, N} then, n(N) < n({x, N})
5. n(N) < n({x, N}) [3, 4 modus ponens]
6. n(N) = n(N) [true for all sets]
7. n(N) < n(N) [2, 5 substitution]
8. n(N) = n(N) and n(N) < n(N) [contradiction]
Ergo,
9. There's no set N that contains itself [1 - 8 reductio ad absurdum]
Just to be clear, ZR-R+~R is relatively consistent with ZF-R.
And to underline your point: The conjunction of "ZF-R+~R is relatively consistent with ZF-R" with "ZF is relatively consistent with ZF-R" is equivalent to "ZF-R is consistent -> R is independent from ZF-R".
N = {N}
is such a set.
Quoting TheMadFool
I don't know what you mean by that.
/
Quoting TheMadFool
Wrong. N = {x N}, so N is not a proper subset of {x N}.
I'll stop there.
I don't think it's so over your head. Give it a bit of thought and you'll see it pretty clearly.
This is not what I suppose. I definitely believe in a set being a member of itself. I think it is a logical necessity.
Quoting Amalac
I understand why it looks as though I haven't. Again, I'm suggesting that this has been overlooked for more than a hundred years by the likes of Russell and Frege (amongst others). It's not immediately recognisable. But I've seen what's been overlooked/misunderstood clearly. if you can see it too, I believe you will see that that there is no alternative.
A set is only a member of itself if it contains itself as a member. So when I write x (x, y, z), I mean to say set x contains items x, y, and z. Set x is a member of itself purely because it contains x (itself). WIth this in mind, consider the following:
1) x (x, y, z). Here, when we say x is a set that is not a member of itself, we get a contradiction. (because x is in x).
2) x (x, y, z). Here, when we say x is a set that is a member of itself, we get no contradiction (because x is in x).
3) x (x, y, z). Here, when we say x is a set of three sets that are members of themselves, we get a contradiction (it amounts to saying one set can be a member of itself twice, compare again with 1 and 2. It amounts to x being a member of itself twice because x is in x and the context is sets that are members of themselves. Either y and z are not members of themselves (making x the only member of itself), or x is a member of itself twice whilst y and z are members of themselves once. It cannot be the case that x is a set of three sets that are members of themselves).
Suppose I wrote: x (p, q, y). Now, x can be a set of three sets that are members of themselves (precisely because it does not include itself for it to amount to being a member of itself twice). But try having a set of all sets that are members of themselves. You cannot avoid ending up with a set that is a member of itself twice (which is absurd).
Do you see how this is like the inversion of Russell's paradox?
Everyone recognises the impossibility of a set of all sets that are not members of themselves, but nobody seemed to have recognised the impossibility of a set of all sets that are members of themselves.
It does not have such a subset as evidenced by such a subset being clearly contradictory. However, it is the set that contains all sets (rejecting this is also clearly contradictory). So it contains all sets that are not members of themselves (because they are all members of it, precisely because they are all sets), and it is a member of itself (precisely because it is a set and is therefore a member of itself). You cannot show me a contradiction in this. But a contradiction is clear in the following two statements:
1) There is a set that contains all sets that are not members of themselves that is itself not a member of itself.
2) No set can logically encompass all sets.
How do you possibly allow yourself to reject the set of all sets? It's the last thing one should be doing.
In any case, I don't think you read my post with enough attention to detail, otherwise I reckon you would have seen where I was coming from. But as it stands, your belief system is contradictory precisely because it sees 2 as being rational/consistent, whereas 2 is clearly contradictory.
In conclusion:
There's no such thing as a set of all sets that are not members of themselves [i]that is itself not a member of itself[/I]. But there is a set of all sets that are not members of themselves [i]that is itself a member of itself[/I]. That set is the set of all sets. It encompasses all sets that are members of themselves, as well as itself (hence why it is a member of itself).
There is no subset of all sets that are not members of themselves, or all sets that are members of themselves.
How is it logically necessary? (With an ordinary understanding of 'logically necessary'.)
Quoting Philosopher19
I have said:
(1) "member of itself twice" has no apparent set theoretic meaning. Of if you think it does, then you should define it set theoretically.
(2) A contradiction is a statement and its negation. You have not shown that x = {x y z} implies a contradiction.
Quoting Philosopher19
I have said, it's not impossible. It is correct that there is the set of all sets that are members of themselves. It is the empty set.
You just keep repeating yourself without coming to grips with the key points that refute you.
I believe I've understood your point, but I don't think you have understood mine. So from my point of view, this is what you are doing.
Quoting TonesInDeepFreeze
I have shown that x = {x y z} results in a contradiction when we say:
1) x is not a member of itself (because x is in x along with y and z. Thus, x is a member of itself whilst y and z are not members of themselves).
2) x is a member of itself as well as y and z (because x is in x along with y and z. Thus x = {x y z} has to be interpreted as either meaning x is a member of itself twice with y and z being members of themselves once, or x is a member of itself, with y and z not being members of themselves).
Do you see that if I wrote p = {x y z}, then it is not contradictory for me to say x is a member of itself as well as y and z?
Look at the word "IF" [emphasis added] I wrote.
IF there is a set of all sets, then it has a subset that is the set of all sets that are not members of themselves.
So, indeed, since there is no set of all sets that are not members of themselves, there is no set of all sets.
Quoting Philosopher19
By deriving a contradiction from the assumption that there does exist a set of all sets. I made that clear.
Quoting Philosopher19
Yes {1) 2)} is an inconsistent set of statements, since {1)} itself is an inconsistent set. It is inconsistent since "There is a set that has as members all sets that are not members of themselves" implies a contradiction, so, a fortiori, "There is a set that has as members all sets that are not members of themselves and that set is not a member of itself" implies a contradiction.
Quoting Philosopher19
I read it in such detail that I pointed out exactly its fatal errors.
Quoting Philosopher19
We prove from the axioms that there is no set that has all sets as members. If you want to have a set that has all sets as members, then state your axioms and prove from those axioms that there is a set that has all sets as members..
No, you skipped my points.
Quoting Philosopher19
You see, right there, you skipped my point, posted at least three times now, that "member of itself twice" has no apparent set theoretic meaning. Or if you think it does have meaning, then define it in set theoretic terms.
Moreover, even if you did define it, you would still need to show that "member of itself twice" implies a contradiction.
Quoting Philosopher19
So? There's no contradiction with x = {x y z} and ~yey and ~zez.
[Of course, we always bear in mind, per context, whether we are working with or without the axiom of regularity.]
You're talking about Russell's paradox in context of sets, and about set membership, and using extensional braces.
But your posting shows that you have no understanding of set theory. So, contrary to your claim, of course you don't understand my points.
Do you the know primitive of the language? Do you know the axioms? Do you know the proof system? Do you know the method of definitions? Do you know any of the basics? Do you know anything at all about it?
1. Sets can contain themselves. [assume for reductio ad absurdum]
2. Suppose N is a set that contains itself
3. Let N = {x, N} where x is either no elements or x is some elements
4. N = {x, {x, N}} [substituting N with {x, N}]
4. N - N = {x, N} - {x, N} = { } [N = {x, N} from 3]
5. N - N = {x, {x, N}} - {x, N} = {{x, N}} = {N} [N = {x, {x, N}} and N = {x, N} from 4 abd 3]
6. N - N = { } and N - N = {N} [from 4 and 5] [contradiction]
Ergo,
7. Sets can't contain themselves [1 - 6 reductio ad absurdum]
What do you think?
That's not phrasing I've ever seen in set theory. I already told you I don't know what that means. I don't know why you present it again without saying what it means.
Quoting TheMadFool
Wrong.
{x, {x, N}} - {x, N} = {y | y e {x {x N}} & ~ y e {x N}}.
But {x N} = N, so {x N} e {x N}, so ~ {x, {x, N}} - {x, N} = {{x, N}}.
It fallacious to argue that, because N is written as '{x N}' in '{x {x N}}' but as 'N' in '{x N}', we have that {x N} is not in {x N}. It is fallacious to argue from the mere happenstance of two different means of notating the set in the way you have.
Ok. I've run out of options. Let's get straight to the brass tacks.
A set that doesn't contain itself = {1, &, :sad: }
Now, consider the set {...} = N1 where "..." stand for element(s). Suppose {...} = {N}
Let's try and make N1 contain itself, {...{...}} = {N, {N}}
But N2 =/= N1 as {...} =/= {...{...}} or {N} =/= {N, {N}}
In other words, N1 couldn't be made to contain itself.
Let's try to make N2 contain itself, {...{...{...}}} = N3
But N2 =/= N3 as {...{...}} =/= {...{...{...}}} or {N, {N}} =/= {N, {N, {N}}}
So and so forth.
In other words, no set can be made to contain itself as trying to do that results in the new set being different from the old set i.e. the (say) set A that contains the set A is different from the set A.
The only option then is to try out the empty set N1 = { }.
Let's try and put N1 inside N1, {{ }} = N2
Unfortunately, { } =/= {{ }}
So and so forth...
I've proved here that N = {N} isn't possible
First, do you understand my explanation that you just quoted?
Sorry, if it seemed as though I hadn't paid attention to it but my argument in the previous post seems to make your well-meaning explanation moot.
1. Take N = {N}, the set that contains itself.
2. N contains itself = {N} contains itself
3. {N} doesn't contain itself.
QED
Please stop using '=' to stand for the biconditional.
(1) N = {N} premise
(2) N e N <-> {N} e {N} from (1)
(3) ~ {N} e {N} non sequitur
Why do you waste our time?
Everyone recognises such a set is contradictory. What you and mainstream set theorists seem to think is that this logically entails that the set of all sets is contradictory, whereas it does not (and I have provided proof of this).
Quoting TonesInDeepFreeze
It seems that you are not reading what I'm writing with enough attention to detail. You seem to have not understood why I have said "member of itself twice" (which is my proof of why a set of all sets that are members of themselves is as contradictory as a set of all sets that are not members of themselves).
Quoting TonesInDeepFreeze
You've rightly recognised that a set of all sets that are not members of themselves that is itself not a member of itself is contradictory. What you've wrongly concluded is that this means the set of all sets is contradictory.
Only one set contains all sets that are not members of themselves (the set of all sets). Again, there is nothing contradictory about the set that encompasses all sets that are members of it (because they are sets), and it is a member of itself (because it is a set). There is nothing contradictory about the set of all sets. It's rejection is blatantly contradictory and all set theorists know this (including you). ZF is either inconsistent, or not comprehensive enough (which ultimately means it is inconsistent).
This fierce dogma needs to die.
I don't just think it. I prove it from the axioms of Z set theory. You are welcome to present your own system and axioms in which there is a set of all sets.
By the way, there are alternative systems in which there is a set of all sets, but their formation syntaxes, inference rules, and axioms are very different from Z set theory. If you were sincere in sustaining having a set of all sets, then you would look up such alternative systems instead of flaunting your utter ignorance of what proof is.
Quoting Philosopher19
You provided muddled argument that depends on an undefined notion of "member twice". Proof, on the other hand, is from axioms with rules of inference.
Quoting Philosopher19
You ignored my request - more than once - that you define it.
Quoting Philosopher19
It's contradictory, a fortiori, since a set of all set that are not members of themselves is contradictory.
Quoting Philosopher19
it contradicts the axioms of Z set theory. You don't have to accept Z set theory, of course. But then to prove that there is a set of all sets in some other system, then you would need to state your axioms and inference rules for that system.
Quoting Philosopher19
ZF has not been shown to be inconsistent. And lack of comprehensiveness does not imply inconsistency. Indeed, an inconsistent theory is ultimately comprehensive since it proves every statement.
It is typical of cranks to use the word 'inconsistent' in their own personal way. Usually for the crank it just means the crank disagrees with, or dislikes, something.
Quoting Philosopher19
I have not asserted any dogma. Rather, I have stated the fact that the axioms of set theory prove that there is no set of all sets. That is a finitistic fact that can be objectively, mechanically checked. Meanwhile, I am quite happy to allow that one may present alternative theories that do prove that there is a set of all sets. That is the opposite of dogmatism. And it is not dogmatism to point out that your own arguments are handwaving and rely on undefined terminology. Indeed, it is your own conviction that is dogmatic.
In this case, we have the slight twist that it's not so much the paradox itself that has the crank's objection but rather that the paradox entails that there is no set of all sets.
There is. It can be any set whatsoever.
Here's a proof:
AyExAz(zex <-> (zey & ~z=z)) instance of axiom schema of separation
(zex <-> (zey & ~z=z)) ?UI, EI, UI
z=z identity theory
~zex sentential logic
ExAz ~zex UG, EG
(Az ~zex & Az ~zey) -> x=y extensionality
E!xAz ~zex from definition of '!'
x=0 <-> Az ~zex definition
Thanks for the clarification and I'm sorry if I've wasted my time but I suppose for people likey yourself who have to deal with those less knowledgeable than themselves, it's part of the territory.
Work in progress.
ZF implies incompleteness in proof, theory or system. Perhaps some are happy with such standards, I am not.
In any case, your last reply to me suggests that it's a waste of time to continue this discussion with you. Also, the post that followed it suggests that you are upset, angry, or frustrated, which is not a good state to be in when discussing matters of logic or pure reason. I presented what I say is clear proof, you say that I have not. This has happened twice now, there's no point in there being a third time. I think we should just agree to disagree.
Suppose a set P
1. P = P [reflexivity. Nothing was done to set P]
2. {P}. P was made an element of the set {P}. Something was done to P]
3. P = {P} [P is the set that contains itself] [assume]
4. Nothing was done to P = Something was done to P [contradiction]
5. P =/= {P} [3 - 4 reductio ad absurdum]
Russell's paradox boils down to whether the set of all sets that doesn't includes themselves in includes itself? or, for easier comprehension of my attempt at a resolution of the paradox, does the set of all sets that doesn't contain themselves contain itself?
As you can see, one of the central issues is whether a set can contain itself or not because Russell's paradox can be rephrased as,
1. If C doesn't contain itself then C contains itself
2. If C contains itself then C doesn't contain itself
where C = the set of all sets that don't contain themselves.
If sets can't contain themselves, the consequent in 1 above and the antecedent in 2 above become meaningless for sets can't contain themselves. At least that's what I think.
If sets can contain themselves, there should be a set N ={N}
1. N = {N}
2. N contains itself, {N} contains itself
3. If {N} contains itself, {N} = {{N}}
Did you see what happened there?
The outermost curly braces "{...}", in a sense, collapsed or behaves as if it didn't exist at all: {N} = { {N} }.
Likewise,
4. N = {N} = {{N}} = {{{N}}} = {{{{N}}}} =...
Same things's happening or rather not happenning as any attempt to make the set N an element of another set like {N} returns, to use a computer terminogy, the set N itself.
In short, the set N, though defined as {N} can't be contained in another set for the reasons provided above.
The entire series N = {N} = {{N}} =... is an illusion so to speak.
Thus, a set N such that N = {N} can't exist. In other words, no set can contain itself and so Russell's paradox is a none issue.
You're welcome.
Quoting TheMadFool
I think I get what you're saying, and I recognise that what you say holds true when N is finite. But when N is not finite, what you say does not hold true:
Imagine having four folders on your computer. These are folders '1', '2', '3', and 'all folders on this computer'. The last folder must contain itself in order to meaningfully qualify as a folder of all folders on this computer.
If it is the case that when you open 'all folders on this computer', you get the following folders: '1', '2', '3', and 'all folders on this computer', and then you click the last folder from these four folders and you again get the following folders: '1', '2', '3', and 'all folders on this computer', and you do this again and again forever and this holds true, then arguably the folder 'all folders on this computer' contains itself.
But you're looking at this from a metaphysical perspective (in which case the computer at hand would have to be non-finite because a finite computer does not have the capacity to really contain a folder that contains itself, because it does not have endless energy or potential, and endlessness is needed to sustain a self-containing folder as discussed in the previous paragraph).
I'm approaching this from a purely logical angel:
A) Assume that the letters a-z were representative of all sets that are not members of themselves. You cannot have a set of all sets that are not members of themselves: How are you going to logically write this? a = {a b c...} Here, a is a member of itself. Whatever letter you choose from a-z, it will be a member of itself, so it can't be a set that is not a member of itself.
B) Now assume that the letters a-z were representative of all sets that are members of themselves. You cannot have a set of all sets that are members of themselves: How are you going to logically write this? a = {a b c...} Here, a is a member of itself twice (and whatever letter you choose from a-z, this problem will occur), and such a thing is as contradictory as a set that is not a member of itself, that is in fact a member of itself (as highlighted in A).
Do you see how B is the overlooked part of Russell's paradox?
2. {N} does something to N
3. N = {N} [a set can contain itself][assume for reductio ad absurdum]
4. If N = {N} then {N} does nothing to N
5. {N} does nothing to N [3, 4 modus ponens]
6. {N} does something to N and {N} does nothing to N [2, 5 conj]
7. False that N = {N} or no set can contain itself [3 - 6 reductio ad absurdum]
I see your argument. What would you say to the following:
The list of all lists, lists all lists (including itself). So this list contains itself as an element. So this list is a member of itself. Do you agree?
Call the list of all lists L. L = L. Given my interpretation of your argument, it's not the case that L = {L}.
So I don't think it's a case of {N} doing something to N. I think it's a case of N being N and N being such that it contains itself as an element (like the list of all lists).
I strongly recommend you have a read of this:
http://philosophyneedsgods.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/
(1) Set theory is incomplete, therefore set theory is consistent.
(2) Any consistent, recursively axiomatized, arithmetically adequate theory is incomplete.
(3) There are alternative theories to set theory.
(4) You have not given even a whiff of an indication of your own alternate theory.
Quoting Philosopher19
There is plenty in this world to be angry and frustrated about. Your ignorant, arrogant, stubborn dogmatism is hardly one of them.
Quoting Philosopher19
You have not shown any point on which my posts have not been clear and correct. Meanwhile, your ignorance, confusion, and stubbornness don't help you in discussing logic.
Quoting Philosopher19
I don't know what you think the operative meaning of that is. In any case, when you post nonsense and misinformation, I will decide for myself whether to rebut it.
"something was done to it" is not a set theoretic predicate.
/
S = SuS
With S, "nothing was done to" S. With SuS, "something was done to S".
/
2 = 2*1
With 2, "nothing was done to 2". With 2*1, "something was done to 2".
So, am I to think that putting, say, a list of items e.g. 1, w, # inside curly braces like so, {1, w, #} amounts to doing nothing?
What you think reflects a profound ignorance of logic. Namely, that the logic is monotonic. Adding the theorem that no set is a member of itself does not eliminate a contradiction otherwise. Once again: The logic is monotonic.
The import of Russell's paradox for set theory is this:
1 ExAy(yex <-> ~yey) [premise]
2 xex <-> ~xex [EI, UI]
3 ~ExAy(yex <-> ~yey) [2]
Then, we derive ~ExAy yex as follows:
1 ExAy yex [premise]
2 Ay yex [EI]
3 AxEzAt(tez <-> (tex & ~tet)) [instance of axiom schema of separation]
4 zez <-> (zex & ~zez) [UI, EI, UI]
5 zex [2 UI]
6 zez <-> ~zez [4 5]
7 ~ExAy yex [6]
Meanwhile, with regularity, we have ~Ex xex. But that doesn't eliminate the proof above. The logic is monotonic.
With regularity, we may prove:
E!yAx(xey <-> xex)
thus justifying abstraction notation:
{x | xex}
and the theorem:
{x | xex} = 0 ['0' stands for the empty set]
If I say I did x, and you say I did not do x, and neither of us changes his position, this means that we agree to disagree on this. I hope that's clear to you now.
Quoting TonesInDeepFreeze
If that's what you want to believe, then believe. Rejection of the set of all sets is blatantly contradictory. I say I have provided proof, you say I have not, except you accuse me of arrogance but seem to not apply it to yourself. This is despite the fact that you are defending a contradiction.
Quoting TonesInDeepFreeze
You are like child in your reasoning and manner of discussion. I shouldn't have to spoon-feed you, but they say feeding the needy is good, and per the dictates of pure reason, Karma is real (I've provided proof of this in another thread).
The list of all lists lists itself. In this list, one item is a member of itself whilst all other items are not members of themselves (precisely because they are members of it, and the reference is [b]it[/b]).
By definition, the set of all sets encompasses all sets. This includes itself. Thus in the context of sets, the set of all sets encompasses all sets that are not members of themselves, as well as itself. Because it encompasses itself, it is a member of itself. Because all other sets are encompassed by it, they are members of it, and not themselves. This is when the reference is sets (as opposed to lists).
If you directly show a contradiction in the above, I might reply to you. If not, I'm done trying to spoon-feed you. You are in need because your belief system is contradictory, yet you act like you are not, and you complain about the world like some spoilt child (you set these standards for discussion. I am reciprocating). You are contradictory/unreasonable/inconsistent. I suggest you reconcile.
1. Take a set {P}. If it's impossible to make this set a member of another set, then {P} always. Any attempt {{P}} will result in {P}
2. P = {P} where P is the set that contains itself. Let's try to make {P} a member of another set like this, {{P}}. But {{P}} = {P}. In other words, {P} always
So,
A set that cannot be made a member of a another set (see 1 P always) is the same as a set that can be made a member of itself (see 2 P always).
Contradiction.
Hence, Impossible/false that P = {P}.
You are to think that "doing something" is not a set theoretic predicate.
And you skipped my counterexamples to your incorrect reasoning.
Circular. I asked what is the operative import of "agree to disagree". Your answer is that we "agree to disagree".
Quoting Philosopher19
It's not just what I believe, it's easily proven, basically from the definitions of 'incomplete' and 'consistent'.
Quoting Philosopher19
You keep repeating that, but without showing an actual contradiction. That is dogmatism.
Quoting Philosopher19
You've shown nothing childish. Especially nothing childish in informing you of the exact mathematical formulations that rebut your ignorant dogmatism.
Quoting Philosopher19
Instead, you shove nonsense down the throat.
And I shouldn't have to spoon feed you clear and correct information about the subject on which you ignorantly espouse, but I do.
By the axioms, there is no set x such that every set y is a member of x.
That's not childishness; it's axiomatic mathematics.
{P} is a member of other sets.
Meanwhile, as I already mentioned, without regularity, it is not inconsistent that
P = {P} = {{P}} = ... for finitely many iterations as you like.
A statement of set theory is inconsistent it it implies both a statement S in the language of set theory and the statement ~S.
x = {x} is inconsistent in set theory.
It is not the case that x = {x} is inconsistent in set theory without the axiom of regularity.
"something done" is not in the language of set theory, so it's not part of any statement S in the language of set theory. "always" is not in the language of set theory, so it's not part of any statement S in the language of set theory.
/
One thing ridiculous about your arguments is that they simply overlook that you get everything you want anyway just by recognizing that set theory has the axiom of regularity. That is, set theory agrees with you that no set is a member of itself. But it agrees with you via an axiom. But, as I have been making this point that you refuse to understand, if we drop that axiom, then it is not inconsistent for a set to be a member of itself. Of course, a set being a member of itself is intuitively incorrect to most people, which is fine. But inconsistency is not determined by what is intuitively incorrect, but rather by the definition: a set of formulas is inconsistent if and only if the set proves a contradiction, where a contradiction is the conjunction of some statement (in the language) and its negation.
I'm all for self-scrutiny, but not so overzealous that I chase demons that there is no reason to think exist. I have plenty of faults, but being a crank is not one of them.
~ExAy(yex <-> ~yey) is proven from first order logic alone; we don't need any set theory axiom for that.
Quoting tim wood
Everything east is in the class, not set, of all sets. However, in Z set theories, that class itself is not an object; rather we refer to that class as an informal locution from "outside" the theory. But in class theories (such as NBG) the class of all sets is an object, but then there is no class of all classes that is an object in the theory.
Right.
Assume that it's impossible to make P a member of another set i.e. {{P}} is not possible or, more relevantly, always {P}.
Now if sets csn contain themselves, {P} = {{P}} = {{{P}}} =...
In other words, always P.
That means a set {P} that can't be contained in another set (always {P} never {{P}}) is the same as the set {P} that contains itself (even if {{P}} always {P}).
You see the point don't you? A set {P} that contains itself is the set that can't be a member of another set!
That is so blazingly incorrect that it scorches the core of this planet.
:ok: Thank you for your time.
So you figured why this is not the case?:
without regularity
Ax({x}e{x} -> Ay(~y={x} -> ~{x}ey))
No.
If there is a set of all sets, then that set has a subset that is the set of all sets that are not members of themselves, which implies a contradiction.
You thought about it for at least half a minute?
One day, you're at your desk and a prison guard comes up to you and announces the arrival of a new prisoner, his name is K. K is a very unique prisoner. What's unique about fae? Well, fae has the following relationship with prisons, K ={K}. In other words, K inside the prison {K} is equal to (is the same as) K outside the prison.
You don't think that's too much of a problem and imprison K like so, {K} and feel quite content with the arrangement. In what seemed almost instantaneous, a guard informs you that K is no longer in prison because K = {K}.
You're angry and surprised but you give the matter some thought and realize K = {K}. You decide that one "solution" is to build another prison outside your prison and if so {K} will become {{K}} and K will become {k}. K is now in prison or so you think.
{{K}} = {K} but {K} = K. The prison guards are mortified as K is again outside the prison (it couldn't be contained in a set).
You then decide to build a third prison like so {{{K}}} and feel confident that you've finally managed to solve the problem...once and for all.
Unfortunately for you, {{{K}}} = {{K}} = {K} = K. Again, K is no longer in prison.
You hold a meeting with your colleagues and after many, many hours of brainstorming you soon realize the gravity of the situation K = {...{...{...{K}...}...}...} i.e. even if you build an infinite number of prisons, K can't be imprisoned.
Well, sorry. I have issues. I hope you'll cut me some slack and let my impudence slide. G'day. You've been very helpful.
Then the demon is not allowing the subset operation. So the collection would not be one recognizable as serving an ordinary set theoretic role. But you do continue to say we'll disallow certain subsets:
Quoting tim wood
Nope, if have taking of subsets, but then stipulate that we are not allowing in particular a set of all sets that are not members of themselves, then we could still derive a contradiction from "the set of all sets except the set of all sets that are not members of themselves".
Moreover, this demon would not be making his decisiond by algorithmic determination; his sniffing is purely magical, unlike set theory.
Set theory does not have a predicate "inside braces".
I can't imagine you'd have issues so severe that you couldn't see that self-membership does not imply impossibility of being in another set.
I had a typo of omission there. I fixed it now.
And I said that even if the demon throws out the set of all sets that are not members of themselves, then there would still be a contradiction, if we allow subsets. I should add that what happens is that we get an infinite process of forming subsets and then the demon throwing them out ad infinitum. But then that is not analogous to set theory, even if we grant a premise of infinite process.
Which logicians and what formulations do you have in mind?
Quoting tim wood
I later corrected my typo of omission there.
It should be:
If we have taking of subsets, but then stipulate that we are not allowing in particular a set of all sets that are not members of themselves, then we could still derive a contradiction.
Do you want the proof? [I'd probably need to use some symbols, since otherwise it may be unwieldy, but I'd try to keep the symbolization brief.]
Quoting tim wood
Yes, you are. The axiom schema of separation is not inconsistent itself. It is inconsistent with the claim that there is a set of all sets.
A sentence (such as set existence assertion) is found to be contradictory with other sentences by being found to be a self-contradiction (logically false) or by being found to contradict previously proved sentences. So what the demons keeps or throws out, will be based on what he's already kept.
But set theory can't do that and be recursively axiomatized, because there is no algorithm to determine whether a given sentence contradicts a previous set of statements. That's why I say the demon is magical, because he's able to make immediate determinations that are not calculable even in principle.
It is true that there are maximally consistent sets of sentences. But membership in such sets is not algorithmically decidable.
If you're referring to Russell's paradox, it is a matter of logic and is not peculiar to set theory, but rather applies to any 2-place relation R:
For any 2-place relation R, there is no x such that, for all y, y bears relation R to x if and only if y does not bear relation R to itself. (With set theory, R happens to be 'is a member of'.)
You have not proven this. You have just stated it. Here's my response:
Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.
Again, you are defending a contradiction. That contradiction being "there is no set of all sets". Consider that it is you who is being dogmatic and not me. Who is possibly dogmatic here? The one that is defending a contradiction, or the one that is against it?
Ok.
Quoting tim wood
Ok, so that means that D necessarily contains A, B, and C.
Quoting tim wood
Ok.
Quoting tim wood
D not containing itself or not being a member of itself is a contradiction because you said:
[i]Consider three sets A, B, C that do not contain themselves. There are a lot of such sets, but let's suppose these three are all of them.[/I]
If A, B, and C are ALL of them, then by definition, D is a member of itself because D is the set that contains ALL sets that do not contain themselves. D can only be such a set if it contains itself (and it does).
You cannot have a set of all sets that are not members of themselves that is itself not a member of itself. This does not mean that the set of all sets is contradictory. Rejecting the set of all sets is blatantly contradictory. It is the last thing western philosophers should have done with regards to being sincere to the semantic of "set". I reckon because this has gone on for a 100 years, it has become a fierce dogma.
That is false.
https://thephilosophyforum.com/discussion/comment/546798
That is not a proof. There are a few problems with it, but most glaring:
There is no set D such that for all y, y is a member of D if and only if y is not a member of itself. So, it makes no sense to say that S "chokes" on D (whatever a clear definition of 'chokes' might be) since there is no such D for S to choke on.
AGAIN, you have not shown that "there does not exist a set of which all sets are a member" is contradictory. Au contraire, I have shown a proof that "there does exist a set of which all sets are a member" is inconsistent with taking subsets.
Quoting Philosopher19
I considered it:
(1) I refer to axioms and inference rules such that it is objectively, publicly, mechanically verifiable whether a given sequence is or is not a proof from the axioms with the rules. I show certain proofs with those axioms and inference rules.
(2) I use standard definitions in mathematical logic. I point out when criticisms of mathematical logic are based in misunderstandings of the definitions. But I allow that other people may have offer different definitions, though they should be clear, non-circular, and properly formulated, and as long as the context is made clear and there is not confusion by mixing contexts with conflicting definitions.
(3) I happily embrace that other people may propose a wide range of different axioms and rules and show proofs relative to those axioms and rules. These include, among others, constructivist and intuitionist, finitist, strict finitist, predicativist, multi-valued, relevance logic, free logic, and even para-consistency. Even ersatz proposals and even non-axiomatic sketches, though I may criticize if they are not coherent or are presented with ignorant, confused, and incorrect criticism of abiding mathematics.
(4) I don't opine whether mathematics is or is not to be regarded merely as application of axioms and inference rules. And without commitment to a philosophical stance, I do countenance considerations in a wide range of alternatives to mere extremist formalism (which itself I don't opine to be necessarily incorrect), including truth regarded in different ways such as reasonable formalism, intuitiveness, correspondence, coherence, realist, structuralist, fictionalist, consequentialist, contextualist, operationalist, pragmatist. constructivist and intuitionist, common sense everyday notions, and I even allow the legitimacy of interest in para-consistency, and even (brainstorming) contrarianism. So not only am I not dogmatic, but to the contrary, I am liable to criticism for being too agnostic and. lacking philosophical commitment, too philosophically timid.
(5) I happily admit that set theory and mathematical logic themselves have certain difficulties (call them even 'mysteries') that I can't completely explain.
(6) I happily admit that my knowledge of set theory, mathematical logic, mathematics, philosophy of mathematics, and philosophy only extends to some basics and that I am not an expert. And I welcome being corrected on anything I've posted that is indeed incorrect, and am happy to post recognition of the correction and to retract as needed. And, when I have myself noticed that I made mistaken remarks or claims that I realize are not on solid footing, then I post a correction.
You:
(1) Keep demanding that your position is correct and that mathematics is incorrect, by repeating your argument while ignoring the specific points demonstrating the errors in your argument.
(2) Tendentiously make incorrect denunciations of mathematics and mathematicians that you don't know anything about.
(3) Accuse non-dogmatists of dogmatism while not facing your own dogmatism.
It is in the definition of the semantic of "set" that you can have a set of ALL things of which you can have more than one of. Examples of such things include: Numbers, people, shapes, trees, sets. If you can have more than one set, you can have a set of all sets. It is clearly contradictory to say: [i]You can have more than one X, but there's no such thing as a set of all Xs.[/I]
Quoting TonesInDeepFreeze
That which you described to me as an axiom I showed as being false. You have not addressed this. Again:
You said: [i]By the axioms, there is no set x such that every set y is a member of x.[/I]
To which I replied: Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.
To which you replied: [i]AGAIN, you have not shown that "there does not exist a set of which all sets are a member" is contradictory[/I].
To which I will repeat the last part of the beginning of this post again: It is contradictory to say you can have more than one X, but there's no such thing as a set of all Xs.
You have yet to address that which I have underlined for you. I have addressed the "axiom" which you present as an objection to the set of all sets.
Bear in mind that it was me who suggested that we agree to disagree, to which you decided to hurl insults at me, but still decided to provide your "axiom" as a refutation of the set of all sets.
You're posting a reply to another person. Do you not see that you have hurled insults at me accusing me of not giving you proof and suggesting that you gave me proof and then provide a link to something that you said to someone else? Do you not see the problem with this?
In any case, I checked the link (in an attempt to be charitable). I am neither rejecting a set that is a member of itself, nor a set that is not a member of itself. I am rejecting the rejection of the set of all sets. I suggest you read the OP carefully.
It is not unreasonable to desire a set theory that upholds our everyday notion of 'set'. However, certain difficulties arise. For example, our everyday notions might include that for every property there is the set of all and only those sets having that property. But that doesn't work, as Russell's paradox reveals.
And people also like to have a theory grounded axiomatically. Trying to have both - our everyday notions and axiomatization - presents more difficulties.
So, for those inclined to axiomatics, we have to admit that not all everyday notions will be preserved. But that is not contradiction; rather it is modesty.
Quoting Philosopher19
Which axiom do you claim is false?
Quoting Philosopher19
As long as we have taking of subsets, the inconsistency comes with the assumption that there is a set of all sets. I proved that.
Quoting Philosopher19
In context of the mathematics you are denouncing, a contradiction is a sentence and its negation. If you cannot show that set theory implies both a sentence and its negation , then you have not shown that set theory is inconsistent.
You may have your own notion of 'contradiction'. Your own notion of 'contradiction' may be that something is to you incompatible or counterintuitive or not in accord with everyday notions. And if that is the definition of 'contradiction' that we use, then, of course, I cannot deny that set theory has results that you find incompatible or counterintuitive or not in accord with everyday notions, so under that notion of 'contradiction', set theory is contradictory. But, again, that is not what mathematicians mean by 'contradiction'. And as 'contradiction' is regarded in mathematics as a sentence and its negation, you have not shown set theory to be contradictory.
"you can have more than one X"
Exy ~x=y
"there's no such thing as a set of all Xs"
~EyAx xey
Observe that Exy ~x=y and ~EyAx xey is not a statement and its negation, nor have you shown how they imply both a statement and its negation.
Quoting Philosopher19
The axioms I used is an instance of the axiom schema of separation. You have not addressed that.
Also, you putting 'axiom' in scare quotes is silly and jejune.
(1) Pointing out that you have not proved something is not an insult.
(2) You began the volley regarding 'dogmatism' as you claimed that mathematicians are dogmatic (arguably a general insult).
(3) Whatever insults I might post do not make me dogmatic.
Quoting Philosopher19
I didn't say that I addressed a proof to you personally. You said that I had not proved my claim. So I correctly said that is false and I gave you (and whomever else is reading) a link to a post in which I did prove the claim.
Quoting Philosopher19
It's no charity to me. It's your own improvement that would be gained by understanding the proof.
Quoting Philosopher19
And I proved that there is no set of all sets, using only an instance of the axiom schema of separation.
Essentially, I was looking for a reply to:
Call the set of all sets X. Call any set that is not X, a Y. X contains all Ys plus itself. Every set Y is a member of X. Show me how this is contradictory.
Quoting TonesInDeepFreeze
The above which I have underlined. The best that I can see from your last reply as addressing the underlined is:
Quoting TonesInDeepFreeze
Then there is an issue in the manner in which you take subsets. Call any set that is not the set of all sets a V', call any set that is not a member of itself a -V, call any set that is simply a set a V.
Can you have a -V as the V of all -Vs? No. Can you have any V as the V of all -Vs? If by a V of all -Vs you mean a V that encompasses ALL -Vs and no other Vs, then you are asking for a -V as the V of all -Vs (in which case what you are asking for is contradictory). But where you are not asking for a -V as the V of all -Vs, then the V of all Vs is such that it encompasses all -Vs, and it is not a -V (because it is a member of itself). -V is only meaningful in the context of the V of all Vs. -V' (any non-set-of-all-sets set that is not a member of itself) is only meaningful in the context of the V' of all V's.
Either we say:
A) There is the V of all Vs. It encompasses all Vs. All other Vs are -Vs in this context.
Or
B) There is the V of all -Vs. It encompasses all -Vs. It is not a -V because it is a V.
If you think a V of all Vs is contradictory, then fine, but then you cannot say a V of all -Vs is contradictory. Do you see? You cannot reject both A and B and be consistent at the same time. It's either A or B. But to my understanding, you reject both A and B.
I've put effort into understanding you and trying to accommodate you in this discussion. I've also put effort into giving you explanations that are easily accessible. If I do not feel you reciprocate this with regards to the last reply I sent you, then I will stop trying.
I showed that the very first sentence is inconsistent with subsets.
Quoting Philosopher19
What? The underlined passage is not an axiom of set theory. And it's your claim, not mine, so there's no reason for me to defend it. You are extremely confused.
Quoting Philosopher19
Finally we're making progress. Yes, if you require that the theory uphold "There is a set of all sets" then you must reject the axiom schema of separation. And you are welcome to propose an alternative (otherwise, good luck doing any set theory without subsets).
And I've read your 'V' stuff before, and I commented on it with exactness. But if you wish to engage me with the additions you've made now, then, to start, you need to clean up the incoherent notation. In a previous post, I suggested how you could do that, but you ignored.
I suggest you parlay the progress you made when you identified that it is the subset axiom (axiom schema of separation) that is the root of your disagreement with set theory. But you should also check out the rest of the axioms to see which you might also reject.
I have no interest in trying to accommodate you any further. What you have said to me and what I have replied to you is clear. I think you have failed to prove your position, whilst I have proved my position left, right, and centre. Evidently you disagree with this.
It's like I said before, we'll have to agree to disagree.
I don't seek accommodation from you. Cleaning up your notation would be a favor to yourself.
Quoting Philosopher19
My main point is that ExAy yex is inconsistent in set theory. I proved it.
You made your only progress for yourself thus far when you replied that, given such a proof, you think the axiom used is false.
Quoting Philosopher19
As I asked before, what is the operative meaning of that? Your reply was the circlarity that 'agree to disagree' means to agree to disagree.
Quoting TonesInDeepFreeze
Somewhere you asked about having a set of all sets except the Russell set. I didn't send a proof that that doesn't work, but I want to now, because it's cute (basically reapplying the Russell argument.
1, EUAx(xeU <-> ~Az(zex <-> ~zez)) [premise]
i.e, assume, toward a contradiction, that there is a set of all sets except the Russell set.
2. Ax(xeU <-> ~Az(zex <-> ~zez)) [EG]
3. ERAy(yeR <-> (yeU & ~yey)) [separation]
i.e. take the subset of U that is the set of all sets in U that are not members of themselves.
4. Ay(yeR <-> (yeU & ~yey))
5. ReR <-> (ReU & ~ReR) [UI]
6. ReR [premise]
7. ~ReR [5, 6]
i.e. we have ~ReR since the premise ReR implies ~ReR.
8. ~ReU or ReR [5, 7]
9. ~ReU [7, 8]
10. ~ReU <-> Az(zeR <-> ~zez) [2 UI]
11. Az(zeR <-> ~zez) [9, 10]
12 ReR <-> ~ReR [UI]
13. ~EUAx(xeU <-> ~Az(xex <-> ~zez)) [12]
Read 'em and weep.
Suppose K can't be contained in another set like so {K}, any attempt to do so will result in K.
The set that allegedly contains itself is K = {K}. Take K and make it an element of a set thus, {K} and what happens? It's, according to how it's defined, K again. That didn't quite go as planned, did it? Let's work with {K} then; making {K} a member of another set like this {{K}} and what happens? {{K}} = {K} = K which in plain English means K can't be made a member of another set.
The whole exercise involving K = {K} is akin to claiming that b × 1 =/= b. Multiplying by 1 doesn't change b into some other number.
Suppose we have a set U whose members are all only those that are not the Russell set.
Let R be the subset of U such that R has every member of U except sets that are not members of themselves.
Some reasoning shows that R is not a member of itself, and that leads to some more reasoning that R is not a member of U. And some more reasoning leads to R is a member of itself if and only if R is not a member of itself.
So the initial supposition that there is a set whose members are all and only those that are not the Russel set is contradictory.
/
Basically, you wondered about avoiding Russell by a cutdown that deletes the Russell set from a universal set. So, instead of applying the Russell argument itself, I applied it to the cutdown, and still got the contradiction.
I already gave you multiple pointers to articles about non well-founded sets, and I showed you how a graph with a single node and an edge from that node to itself models a set that contains itself.
Here is that thread. You might be interested in rereading it and looking at the references I gave.
https://thephilosophyforum.com/discussion/comment/546388
The essential point is that there is no actual definition of a set. A set is defined by its behavior under a given collection of axioms. If you include the axiom of regularity, no set can contain itself, and also there are no circular membership chains like [math]a \in b \in c \in a[/math].
On the other hand, it turns out that the negation of Regularity is consistent with the other axioms. That's the key point. There's no a priori reason a set can't contain itself. After all the collection of all ideas is an idea, and the collection of all collections is a collection. Self-containing entities are natural.
The ONLY reason there isn't a set of all sets is that we typically adopt an entirely arbitrary axiom saying so. Drop the axiom, and sets can contain themselves.
https://en.wikipedia.org/wiki/Axiom_of_regularity
With Z set theory, there is no set that is a member of itself.
With ZFC-R, it is not inconsistent that there is a set that is a member of itself.
I gave you copious explanation about that. But you just go on your merry way ignoring the information you've been given.
No, it was as "planned", and consistent (without reguarity).
Quoting TheMadFool
Wrong. That kind of nesting is consistent (without regularity). And it does not mean that K can't be a member of another set. I informed you about that multiple times. Proof:
Suppose K = {K}. Let ~x=K and ~xeK. Then ~ {x K} = K but K e {x K}.
What is wrong with you that you can't see that a set being a member of itself doesn't prohibit the set from being a member of other sets?
Quoting TheMadFool
That is argument by analogy, which is not valid for deduction such as mathematics. And the analogy even works against your claim.
Multiplication by 1 is idempotent.
With Z, the singleton operation is not idempotent. (So your analogy is backwards.)
With ZFC-R, it is consistent that there are sets for which the singleton operation is idempotent. (So your analogy DOES work in that context, and it works AGAINST your claim.)
So often, you are plainly illogical.
Sorry, I beg to differ. K =/= {K}.
I gave you a very good reason. Please address the proof. I'll restate it here for your viewing pleasure.
1. There's a set N that can't be a member of another set. Either the attempt to make N an element of another set won't make sense or it will.
If it's the former, case closed. Like trying to multiply 2 with $. It's nonsense.
If it's the latter, all attempts to make N a member of another set will result in N, the set itself. Like multiplying 2 with 1: 2 × 1 = 2; 2 × 1× 1 = 2. Multiplying by 1 does nothing to 2.
2. You claim there's a set K = {K}. Let's try and make K a member of another set like so {K}. However, {K} = K. Let's try something different like so {{K}}. However {{K}} = {K} = K. Basically, this is analogous to multiplying 2 with 1 (see vide infra): just as "× 1" does nothing to the number 2, if K = {K}, making K a member of another set literally does nothing too.
...{...{...{...{K}...}...}...}... = K. The curly braces, an infinity of them return the same value K. Just like 2 × 1 × 1 × 1 x... = 2 in which case, multiplying by 1 changes nothing about the number 2.
In essence, there's no difference between set N, a set that can't be a member of any set (thus can't contain itself) and set K [K = {K}] defined as a set that contains itself. It's a paradox! The set K = {K} cannot exist. Period!
Quoting TonesInDeepFreeze
The analogy was meant for you and is definitely not something I would submit for publication although I just might if given the opportunity - it makes so much sense.
Since it's a restatement, I don't need to address it again, since I've replied to your "proofs" already, in quite detail. And you have not gotten back to me on my replies. Typically, all you do in reply is to restate your incorrect argument that had just been refuted.
Moreover, I just gave you a proof in my last post, and you have not addressed it.
Quoting TonesInDeepFreeze
Quoting TheMadFool
I gave you specific detail why the analogy doesn't work for you. Instead of responding to that, you merely reiterate your claim that you are right.
You haven't addressed it and that's why I'm restating it.
Quoting TonesInDeepFreeze
The analogy is perfect. There's a precise 1-to-1 correspondence between 2 and K, creating the set that contains itself and multiplying by 1, and last but not the least, making K a member of itself matches perfectly with 2 × 1 = 2.
If you link to where you first posted it, then I'll link to where I answered it.
Quoting TheMadFool
You still have not addressed my rebuttal.
Incorrect.
2 = {0 1} and has cardinality 2.
K = {K} and has cardinality 1.
You're barking up the wrong tree. The choice of number 2 is irrelevant to my argument.
Quoting TonesInDeepFreeze
The analogy is perfect for describing the silliness/inanity/vacuousness of the notion of a set that contains itself.
I pretty much figured that you didn't know what you were writing when you said that there is a 1-1 correspondence.
Quoting TheMadFool
I pointed out that you have failed to address my rebuttal. And your reply to that is to again fail to address my rebuttal.
And now I know you don't know what an analogy is.
Quoting TonesInDeepFreeze
You're assuming things. Whatever you said doesn't qualify as a rebuttal and thus I had to restate my argument.
If you think there is anything wrong in my rebuttal, then you should be able to point to it exactly.
For the nth time, your post isn't a rebuttal.