I decided to improve my question since there were mistakes pointed out (THANK YOU!) and I realized that I did not mention if I put any kind of axiomatic conditions on the sets which I consider. I have a good math background (numerical analysis/mathematical statistics/linear algebra), but I am a biologist, so mathematicians please consider that, try defining what you are talking about.
First of all, I operate under the assumption that my sets (any definable collections) do not obey any kind of axioms, except that they (collections) should be logically defined. For example set X=[{x∈R:0>x>1} and X ≠∅] is not defined logically for me, because it leads to a paradox: 0>1. Any set whose definition leads to a paradox is not logically defined, because we are doing mathematics, there should be no paradoxes. Thus, for me, "the set of all sets which do not contain themselves" is not a set, because its definition leads to a paradox. This set does not exist. Note: I do not operate under naive set theory. Naive set theory assumes as an axiom, to be more precise a principle, called comprehension principle: "Given any property, there is a set which consists of all objects having that property." This principle leads to paradoxes, thus it is not logical for me. For me a set is logically defined when it is clear what it is from definition and definition does not lead to paradoxes or any kind of illogical things. Lets call it "logical set theory".
Second, here is my question. Many books prove that the set of all sets does not exist using Cantor's theorem: for any set X its Cardinality is strictly less than the Cardinality of a set consisting of all subsets of X. I am here to say that I think that one can not use this theorem as a proof of this point: the set of all sets does not exist. Actually one member of the forum already asked this question: can we use this theorem to prove this fact? "Is this prove correct?" Note: nobody answered yes or no. My answer is solid NO. But I am not 100% sure. Thus, I am providing my explanation for my "NO" and I am asking members of this forum whether they think that I am wrong. And if I am wrong, I would like to know why.
If one assumes that all valid sets have to obey ZFC axioms, then one need not prove the point using Cantor's theorem, because "the set of all sets" is not a valid set in ZFC theory, because it does not obey the Axiom of Regularity, for example.
If one assumes that all valid sets have to obey "naive" theory comprehension principle: then we are not talking about something logical, we are are not doing math. So forget about it.
If one assumes that all valid sets have to obey New Foundation axioms, then, as member of this forum pointed out – the set of all sets does exist, it is ok-set, thus we do not need to prove that it does not exist.
If one assumes that all valid sets have to obey Cantor's rules, then actually as a person on this forum pointed out, Cantor's conception of sets rules out the set of all sets. Thus, we do not need to prove the point, because the set of all sets does not exist under Cantor's conception.
If one assumes that all valid sets have to obey "logical theory of sets", then it is easy to find one to one correspondence between the elements of set of all sets and elements of its power set: s=f(x)=x. So these sets are equinumerous, and one can not prove the opposite.
Finally, it is clear to see that the point (the sets of all sets does not exist) is easily proven from Cantor's theorem. Thus, if we proved that the point is not true (under the "logical set theory"), and I do think that it is not true (under the "logical set theory"), when we have to find a mistake in the prove given by Cantor, using logical set theory.
Here I am providing a proof by Cantor and point out the mistake later on:
The mistake leads to the conclusion that this theorem is true for any set X except for "a set of all sets" in the universe. Thus one can not derive that the "set of all sets does not exist" based on Cantor's theorem.
Here is the proof of Cantor's theorem:
Definitions:
a)sets are called equinumerous if there exists one-to-one correspondence between its elements.
b)cardinality of set X is considered less than cardinality of set Y if a subset of Y is equinumerous to X, but no subset of X is equinumerous to Y.
Lemma: let S be a set of subsets of X such that S is equinumerous to X. Then there exists a subset of X (call it A), such that it is not contained in S (A ∉ S).
(PS, it is easy to see that the theorem automatically follows from the Lemma: lets assume that the theorem is not true, than there exists a set X such that it is equinumerous to the set consisting of all its subsets (call it S). But such a set S satisfies the conditions of the Lemma. Thus there exists a subset of X that is not contained in S. But this is absurd, thus we have a contradiction and S can not be equinumerous to X.)
Proof of Lemma:
All we have to do is to define A and prove that is satisfies the conditions of lemma:
it should be a subset of X, such that it is not contained in S (A ∉ S).
Lets take any S, that satisfies the conditions of lemma (a set of subsets of X such that S is equinumerous to X). Let f(x)=s is the one-to-one function that images the elements of X (x) into elements of S(s).
Then, A is defined as: A={x ∈ X: x ∉ f(x)}. That is A is a subset of X, such that for every element x of A: x is not contained in s=f(x).
Then we assume that A∈S and find a contradiction (explained below). Thus A∉S and the Lemma is proven. The contradiction is the following:
If A∈S, then exists element of X (call it e), such that A=f(e). If e∉A than e∉f(e)=A – contradiction (because from the definition of A follows that if e ∉ f(e): e should be contained in A). If e∈A than e∈f(e)=A, that is not possible by the definition of A, thus there is also a contradiction.
Lemma (and the theorem) is proven. I saw other proofs on the internet and in books, but all of them define a set A={x ∈ X: x ∉ f(x)}.
I highlighted the mistake in bold.
Indeed, lets ask, is A logically defined for a set X=set of all sets?
Let X="set of all sets in the universe". Let S be a set that contains all subsets of X. Let s=f(x)=x (that is possible only for a "set of all sets in the universe").
Now look at the definition of A: the set A is not logically defined: there is no such a set with elements of X, such that x ∉ f(x), because x ∉ f(x) for all x. In fact A can not be even an empty set: because even for empty set: f(empty set)=empty set, thus
empty set ≠ A. Thus A is still undefined.
I saw another proof of the fact that the "set of all sets" does not exist: they use Russell's paradox. But I already told that I am not operating under the comprehension principle, because it is nonsense.
Any mathematicians here? please let me know if there is a flaw in my arguments. Does the set of all sets exist? Is it still debated? Thank you!
Best Answer
Let $X$ be a set, and let $f$ be a function $X \to \mathcal{P}(X)$. To prove Cantor's theorem, we want to show that $f$ is not surjective. (There is no need to assume that $f$ is injective, by the way.)
Your objection is that it is not "logical" to define the set $A=\{x \in X: x \notin f(x)\}$. To the extent that this is a philosophical criticism, it may or may not be correct, but in any case wouldn't amount to showing a "mistake" in the proof. So let's stick to the mathematics.
The formal justification for the existence of $A$ is an instance of the Separation axiom schema: for every set $X$, formula $\varphi$, and parameter $p$ there is a set denoted $\{x \in X: \varphi(p,x)\}$ whose elements are the elements $x$ of $X$ such that $\varphi(p,x)$ holds.
Your objection seems to be that this definition of $A$ leads to a contradiction if $X$ is the "set of all sets" and $f$ is the identity function. This is a good and important observation. If the conjunction of two things leads to a contradiction, then (at least) one of them must be wrong. Either Cantor's argument is wrong, or there is no "set of all sets."
After having made this observation, to ensure that one has a consistent theory of sets one must either (1) disallow some step in Cantor's proof (e.g. the use of the Separation axiom) or (2) reject the notion of "set of all sets" as unjustified. Mainstream mathematics has done (2), for the reason that you (and Russell) pointed out. It is also possible to do (1) instead, as in New Foundations set theory.
EDIT: I missed some things on my first reading. Your third-to-last paragraph, if all its claims were true, would show that set theory as used by Cantor was inconsistent. Note that this would be quite different (mathematically, if not philosophically) from showing that there was a mistake in his proof! A proof of inconsistency would be a perfectly valid proof, not to mention an important one. In any case, some of your claims are not true, as I explain below.
Letting $f$ be the identity function, you argue that $A$ must be empty:
This step is not right. There are plenty of possibilities for $x$ such that $x \notin x$. Perhaps you are confusing $x \notin x$ with $x \ne x$.
Then you argue that $A$ cannot be empty.
This step is not right either. First, as Greg Martin points out, $\emptyset \notin \emptyset$, so if $\emptyset \in X$ then $\emptyset \in A$. Second, even if $\emptyset \notin A$ it doesn't follow that $A \ne \emptyset$. Perhaps you are confusing "$\in$" with "$=$" again.