Russel's Paradox
The problem is stated as "The set of all sets that do not contain themselves as subsets."
Already we have a problem with the way the problem is stated because how do we know that "All sets that do not contain themselves as subsets" is a set? We cannot assume from the start that it is a set. And this assumption is the very cause of the problem that arises.
What is "All sets that do not contain themselves as subsets" if it is not a set?
In the following the symbol \x means 'without x' or 'excluding x'
Let Set X = "All sets that do not contain themselves as subsets"\X
I don't see anything wrong with this definition because it avoids the paradox by excluding X regardless of whether X can contain itself.
Let Set X[sub]2[/sub] = (X U X)\X[sub]2[/sub]
Now X and X are united, but in X[sub]2[/sub]\X[sub]2[/sub]
Let Set X[sub]3[/sub] = (X[sub]2[/sub] U X[sub]2[/sub])\X[sub]3[/sub]
Now X[sub]2[/sub] is included and X[sub]3[/sub] is included if we continue-
Let Set X[sub]4[/sub] = (X[sub]3[/sub] U X[sub]3[/sub])\X[sub]4[/sub]
And we can continue in this way for an infinity of sets, thereby including every set that does not include itself as a subset.
The conclusion is that "All sets that do not contain themselves as subsets" is a limit rather than a set. The limit of an infinity of sets.
Already we have a problem with the way the problem is stated because how do we know that "All sets that do not contain themselves as subsets" is a set? We cannot assume from the start that it is a set. And this assumption is the very cause of the problem that arises.
What is "All sets that do not contain themselves as subsets" if it is not a set?
In the following the symbol \x means 'without x' or 'excluding x'
Let Set X = "All sets that do not contain themselves as subsets"\X
I don't see anything wrong with this definition because it avoids the paradox by excluding X regardless of whether X can contain itself.
Let Set X[sub]2[/sub] = (X U X)\X[sub]2[/sub]
Now X and X are united, but in X[sub]2[/sub]\X[sub]2[/sub]
Let Set X[sub]3[/sub] = (X[sub]2[/sub] U X[sub]2[/sub])\X[sub]3[/sub]
Now X[sub]2[/sub] is included and X[sub]3[/sub] is included if we continue-
Let Set X[sub]4[/sub] = (X[sub]3[/sub] U X[sub]3[/sub])\X[sub]4[/sub]
And we can continue in this way for an infinity of sets, thereby including every set that does not include itself as a subset.
The conclusion is that "All sets that do not contain themselves as subsets" is a limit rather than a set. The limit of an infinity of sets.
Comments (67)
Well, a set is an unordered collection of individuals. The unordered collections of individuals that do not contain themselves is an unordered collection of individuals; therefor it is a set.
Unless of course you re-define set.
It follows from the assumptions of a certain kind of naive set theory. The point is that it's not a set, but from the naive set theory it follows that it should be. Hence the problem.
x={x}
Nonsense, however.
There is no need to redefine the set. All that is needed is to see that there are collections that are not single sets - as the paradox implies. I think Russel's Paradox is superficial and I never believed it "undermines mathematics" which strikes me as an unjustifiably dramatic statement.
In fact it is a trick question because of the way it is stated: "The set of all sets that do not contain themselves as subsets." Why are they calling it a set? I don't think something can be called a set unless it can be demonstrated to be such. And the paradox shows that it cannot be a set. The entity should be defined as "All sets that do not contain themselves as subsets" There is no ambiguity in stating it this way and intuitively I feel that such an entity is possible. But what is it if it is not a set?
If the logic of my first post is correct it is an infinite collection of sets, each nested within another.
What seems to be the case is that there can be infinite sets that are not simply a set. I say this, not because of my reasoning but also because Russel's solution involves something similar: an infinity of "types", each nested one within the other. But it may be possible to resolve the issue with sets alone (Russel's solution seems very artificial and contrived)
But the real question I am asking in my first post is: Is the logic I am using coherent? I don't see anything wrong with it, unless you can.
Set A = {a, w}
Set B = {a, x}
Set C = {a, y}
Set X = the set of sets that have {a} as an subset.
Set X = {A, B, C,...}
{a} is in X (because {a} is in A, B, C,...)
therefore X contains X
None of them do. They have "a" as an element. D={{a},z} does.
This stuff is deleterious to mental health. :scream:
No, {a} is not "in" A,B,C,...
IMO this stuff is not worth the effort.
{a} is a subset of A and A is a subset of X therefore {a} is a subset of X
Also, when it comes to set of sets, {a} can be an element.
{A} is a subset of X, not A. Ask fdrake or fishfry to explain this stuff to you. I'm done.
Who ever said that Russel's Paradox "undermines mathematics"? It undermines what is now known as "naive set theory" (an early attempt at an axiomatic set theory).
Quoting EnPassant
Because of the axiom of unrestricted comprehension, which is what had to be ditched in the wake of Russel's paradox:
There exists a set B whose members are precisely those objects that satisfy the predicate ?.
You need to understand the difference between being a member of and being a subset of.
Set X = {A, B, C} = { {a, w}, {a, x}, {a, y} }
a is a member of A, B and C, but not a member of X. {a} is a subset of A, B and C, but not a subset of X.
Subset is transitive: If A is a subset of B and B is a subset of C, A is a subset of C.
{a} is a subset of {A} and {A} is a subset of {X} ---> {a} is a subset of {X}
Quoting EnPassant
Exactly. Membership isn't.
But that's what you did.
I think this is the correct answer from Snakes Alive:
"It follows from the assumptions of a certain kind of naive set theory. The point is that it's not a set, but from the naive set theory it follows that it should be. Hence the problem"
That was one of several (prior to @tim wood's) where expertise of the reader had caused unconscious correction (and hence ignoring) of the mis-statement of the problem.
Ironically, one (Lesniewski's) response to the paradox was to try to base arithmetic on a transitive part-whole relation. One that wasn't, like subset, derived from a non-transitive one. Part of the motivation, if I understood and now recall correctly, was to separate out a distributive sense of "set" (by which to attribute properties to each member) from a collective one (for attributing to the whole collection as a thing). A bit like the possible difference between talking (distributively) about,
Quoting EnPassant
and talking (collectively) about
Quoting EnPassant
so defined.
Misstatement?
Quoting EnPassant
Hence @tim wood's and @SophistiCat's clarifications.
In 1923, Ludwig Wittgenstein proposed to "dispose" of Russell's paradox as follows: The reason why a function cannot be its own argument is that the sign for a function already contains the prototype of its argument, and it cannot contain itself.
In practice this is nonsense as well, IMO, although in some abstract sense it may have weight. In iteration the first application is f(f(z)). So the outer f acts upon a function value and not on the function itself. Fractals arise from these processes. Perhaps f(f) makes less sense. If F is a functional, then neither F(F(z(t))) nor F(F) is normally well-defined.
No, I am saying there are infinite collections of things that are not a set.
See this link https://math.stackexchange.com/questions/24507/why-did-mathematicians-take-russells-paradox-seriously
Quoting tim wood
The paradox asks the question "Is X a member of itself?"
Let's say Set X = {{a}, {b}, {c},....}
If {X} is a member of X then
Set X = {{a}, {b}, {c},....{X}}
If {X} is a not member of X then
Set X = {{a}, {b}, {c},....}
But I am excluding {X} regardless of whether it can be a member of itself.
So Set X = {{a}, {b}, {c},....}
{X} can only be a member of itself according to the definition of X. I am explicitly excluding {X} from being a member of itself by definition. X\{X} excludes {X} as a member, not as the entire set.
Then I unite Set X with {X} in
X[sub]2[/sub] = (X U {X})\{X[sub]2[/sub]} (See my next post)
Since the paradox shows that X is a kind of 'pathological' set we don't know where to put it.
I am creating X[sub]2[/sub] and putting it in there. Then the process is repeated infinitely so that all relevant sets can be contained. The result is an infinite progression of sets that contain "All sets that are not members of themselves" And this entity turns out to be an infinity of sets, each nested within the other.
(It may also be that every X[sub]i[/sub] contains every X[sub]j[/sub] but not {X[sub]i[/sub]} but I have not got this far with it yet.)
Yes, you are correct. Since we are talking about sets of sets, a better notation would be-
Set X = X\{X}
Set X[sub]2[/sub] = (X U {X})\{X[sub]2[/sub]}
Set X[sub]3[/sub] = (X[sub]2[/sub] U {X[sub]2[/sub]})\{X[sub]3[/sub]} and so on.
Apologies for the sloppy notation.
Not given the axiom of regularity and the axiom of pairing.
A set that has only the moon as a member is a distinct entity from the moon itself. A set is an abstract object.
A set is criteria, kind of like a club.
So the set of all non-penguins, NP, intuitively has itself as a member because NP is not a penguin.
It is not 'without X' it is 'without {X}' as a set. {X} is not the same as X, my bad notation in the beginning notwithstanding. X\{X} is every set in X but not the set {X} itself.
X\{X} = {{a}, {b}, {c},...} but not {X}, regardless of whether {X} can be a member of X.
Excluding {X} is not the same as excluding X.
The paradox asks if {X} is a member of X but I am disposing of the paradox by defining X as X\{X} so there is no contradiction.
That's an interesting discussion there. Most of us here are non-mathematicians, and among mathematicians only a small fraction are working in or at least interested in foundations.
Quoting EnPassant
Your notation is confusing. If you want to say that a is a member of X (a ? X), you would write that as
X = {a, ...}
which is not the same as
X = {{a}, ...}
{a} is a singleton set with a as the sole member.
X={X}
:roll:
"...and among mathematicians only a small fraction are working in or at least interested in foundations." How true!
Yes, but X is a set of sets so X = {{a}, {b}, {c},...} but {a, b, c, ...} might be correct too as long as the logic of what I'm saying holds up. Link: https://truebeautyofmath.com/lesson-4-sets-of-sets/
a = {x}
b = {y}
c = {z}
Set X is the set of sets a, b, c so
Set X = {{x}, {y}, {z}}
If X is included
X = {{x}, {y}, {z}, {{x}, {y}, {z}}}
If X is not included
X = {{x}, {y}, {z}}
So X\X is {{x}, {y}, {z}} which is what I originally meant by X\X or X\{X}
I don't see what logic could imply that {{a}, {b}, {c},...} is the same as {a, b, c, ...}
You keep making the same mistake over and over again:
Quoting EnPassant
No!
The paradox asks if X is a member of X.
X ? {X}
{X} is a set with one member: X
Quoting EnPassant
No, that's not how it works.
X = {{x}, {y}, {z}}
X' = {{x}, {y}, {z}, {{x}, {y}, {z}}}
X ? X'
X ? X'
X' ? X'
Let 'All sets that do not contain themselves as members' be
a = {x}
b = {y}
c = {z}
d = ... and these sets go on for as long as is necessary, e, f, g, h,...
Set X = {{x}, {y}, {z},...}
Suppose for some set h, h = {X}
I am saying X = {{x}, {y}, {z},...}\h
That is, X = {{x}, {y}, {z},...}\{X}
There may be h such that h = {X} or there may not.
I am saying X\h regardless and this is the definition of X.
In simple language X = "All sets that do not include themselves as members, except {X}"
You seem to be assuming that {X} is included in X but by definition it is not.
Or suppose Set V = {{x}, {y}, {z},...{X}}
Set X = V\{X} and there you have it.
No, I am saying IF X is included in X then
X = {{x}, {y}, {z},... {{x}, {y}, {z}}}
But IF X is not included
X = {{x}, {y}, {z},...}
I am only saying this to clarify things. But by definition X is NOT included so
X = {{x}, {y}, {z},...}
Precisely X = {{x}, {y}, {z},...}\{{x}, {y}, {z}}
{x}, {y}, {z} and {{x}, {y}, {z}} are different sets so excluding the set {{x}, {y}, {z}} does not exclude {x}, {y}, or {z}
Create an algorithm to list all the penguins. Then take the complement in the Universal Set.
Why are they all singletons?
a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...
X is going to be {a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...}
But for some i,
a[sub]i[/sub] = {a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...} = X
So the sets in question are-
a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...a[sub]i[/sub]...
= a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...{a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...}...
So X is to be defined as {a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...}\a[sub]i[/sub]
= X\{a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...} = X\X
That is, X is defined as not being a member of itself.
Don't worry about the notation. X is defined as not being a member of itself, that is all.
By the way, is how do I type these set symbols? Latex? Is there a guide?
Edit: Found it. [math]\in[/math] Logic and Philosophy of Mathematics sub forum
Thanks. The first thread in the Logic and Philosophy of Mathematics forum (this forum) explains how to use the math tag and how to create symbols.
Would your definition of [math] X [/math] imply
[math] X \cup X \neq X [/math]
Indeed, it seems to imply
[math] X \subset X \cup X [/math]
where containment is strict. You should consider checking out NBG class-set theory which is an alternative formulation of set theory. The theory uses the notion of classes to avoid Russell's paradox. I'm not sure I can say I'd prefer it over the standard ZFC but it's interesting to having an alternative point-of-view. There is a good dover book by Smullyan and Fitting on it as well. It can be used for self-study.
EDIT: fixed the second equation.
My idea is that it can be framed in terms of set theory alone without the invention of classes. My definition is that if X as a subset of itself is excluded this leaves X\X' where X' is the subset.
Hmm, I'm not sure I follow. ZFC does not deploy the use of classes, it is set-based. One can avoid the difficulties of Russel's Paradox without the invention of the class object. Indeed, I believe the use of classes was for alternative purposes, namely, to simplify some aspects of ZFC. From what I understand, NBG is generally regarded as more elegant than ZFC--again for more info see the book I linked in my last post. If you don't like classes, then just avoid the use of a class-set theoretic system.
Quoting EnPassant
The problem with this definition is that the set of all sets that do not contain themselves as subsets is shown, by Russell's Paradox, to be logically contradictory. Your definition requests that we posit an object which is logically contradictory, and then remove [math]X[/math] from it. This is akin to requesting the reader to take the smallest prime number with exactly three divisors, subtract it from itself, and then insist that the answer is 0.
The barber shaves those and only those who do not shave themselves. Actually this is obvious if he is a barber. So the barber's hair grows long, or another barber does shaves him. Where is the paradox with regard to the barber?
My argument is to define X without X as a subset of itself regardless of whether it can or can't be such. In this way the paradox is avoided by defining a set that contains 'All sets...' but not X. X is then included in X[sub]2[/sub] and the paradox is avoided. The same argument is then applied to X[sub]2[/sub], X[sub]3[/sub]... The result is that the 'Set of all...' is really an infinity of sets each containing the other.
What's the difference to Russell's Type theory?
I don't know enough to say but as far as I can see this question can be resolved with simple set theory alone - if my ideas are coherent that is. The problem is with the way the paradox is stated: "The set of all sets that are not members of themselves." Why assume it is a set? Because it turns out that 'The set' is not a set at all. But we can intuitively grasp the concept of 'All sets...etc'. It must be something so what is it if it is not a set? Seemingly it is an infinite collection of sets.
This is similar to Russell's types because there is also a hierarchy of types, each containing the ones below it. So why not just frame the whole thing in terms of sets alone?
I've had great difficulty in this thread with notation but the concept I'm trying to define is simple, namely X\X.
Suppose set A = {a, b} the subsets of this set are:
{{0}, {a}, {b}, {a, b}} Now redefine set A as:
A\A. That is, A\{a, b}
Now the subsets of A are:
{{0}, {a}, {b}}
This is what I mean by X\X.
I see what you mean. I'm still sticking to my previous statement on why this set is ill-defined. In any case, I don't think one can avoid Russell's Paradox proceeding in the way you have--regardless of the logical consistency of the argument.
We have to remember that Russell's Paradox established the inconsistency of Frege's set theory, in particular, the abstraction principle which loosely stated that, given any property P, there exists a (unique) set A consisting of those and only those things that have property P.
Russell suggested we consider the collection "the set of all sets which are not subsets of themselves". Note that, by Frege's abstraction principle, this is necessarily a set. Asking that we, in effect, look the other way and consider instead another set, as you've proposed, doesn't prevent us from considering Russell's set.
The paradox establishes that the abstraction principle is unsound. Historically, this lead Zermelo to suggest the limited abstraction principle to prevent the issue--one would need to clarify which set from which the elements satisfying P are being taken. Thus we cannot speak of the set of all x having property P, but we can speak of the set of all x in A that have property P. For an explanation of how this avoids the paradox, I recommend looking at pg. 12 of Smullyan's book, linked in my first post.
Ah, so this is not the correct notation. You have asked us to consider the set [math]2^X \setminus X[/math], where [math]2^X[/math] is the power set of [math] X [/math]. Note that we cannot redefine X and that this set is quite distinct from [math]X[/math] itself.
Well, Russell himself use Type Theory and basically Zermelo-Fraenkel Set Theory (ZF) was made basically to avoid Russell's paradox. If you find your ideas resembling theirs, you can be proud of yourself.
Yes, this is the difficulty with notation but it is really just a notational wrinkle. You can rename the set X' and you have it. All that is required is a set without X as a subset but with all the other sets. The same logic can then be applied to X'[sub]2[/sub] and so on. It is really a question of getting the notation right but the concept seems to be coherent.
Quoting ssu
There are immense complexities with infinite sets so I'm very unsure if my logic holds...
If I understand you correctly, you are saying "the set of all sets which are not subsets of themselves" is necessarily a set. But it turns out that it is a pathological set or not really a set at all. My argument is that it is an entity of some kind and I'm attempting to define what this entity is. It seems that there can be things that are not simple sets but are infinities of sets. Maybe all sets are really subsets of infinite sets...
If we accept Frege's abstraction principle, then yes, we are committed to the position that this is a set. As you have pointed out, Russell's Paradox illustrates that such a notion is inconsistent, which is why no one accepts Frege's abstraction principle.
You seem quite motivated to learn more about this topic (which is a very good thing! :up: ). I really recommend you check out the book I mentioned in my first post. It is actually quite good and the first few chapters are quite approachable. It's Dover so its also extremely affordable. A worthwhile investment.
If the mathematical formalism is difficult at first, then you might want to try Halmos' Naive Set Theory. While I own a copy, I can't honestly say I've sat down and spent any real time with it, but I understand it is a very good place to start.
Best of luck! Hope to see more of you in the forums.
I'll definitely check that one out. Thanks.