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I know it has been asked several times, but there is always one step that I don't see. My argumentation goes as it follows:

  1. Given the set $\mathbb{V}$ set of all sets, then $V \in V$.
  2. Because of the Axiom of Schema of Specfication we can define a certain property $P(x) = x \not\in x$, and prove that $\exists A \forall x, A = \{x \in V : x \not\in x\}$
  3. We can prove that this is a contradiction by checking if $A \in A$ (since $A \in V$ but $A \not\in A$, but then if $A \not\in A$ then $A \in A$, which is a contradiction).

This proves that such set $A$ does not exist, but I cannot see why can we conclude that this proves that $\nexists V$, since we could have started the analysis with any other set than $V$ (this is, the proof is independent of the initial set) that contains $A$.

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    $\begingroup$ This is a proof by contradiction: start with an assumption which we don't think it is true and then derive a contradiction, thus disprove the assumption in the first place. Within this proof, where does the assumption shows up? $\endgroup$ Commented Mar 14, 2024 at 9:31
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    $\begingroup$ Note that $A$ is defined in terms of $V$: it doesn't make sense to "start with any other set that contains $A$." $\endgroup$ Commented Mar 14, 2024 at 12:57
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    $\begingroup$ Put another way (and this is what Chad K's answer is saying), what this proof really does is construct a "class function" mapping any set $X$ to a set $A_X$ with the property that $A_X\not\in X$. As a consequence of this, no universal set can exist (consider "its $A$-set"). $\endgroup$ Commented Mar 14, 2024 at 19:02
  • $\begingroup$ No, your proof is not independent of the initial Class V. If you started with some set B. Then A not in A, doesnt mean that A is in A, because A isnt the set of all sets which dont contain themseves, A is the set of all sets in B whivh dont contain themselves. $\endgroup$ Commented Mar 15, 2024 at 2:28

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There is no other set that contains the same $A$.

Start with a set $Y$ and form $$ S=\{p\in Y\colon p\not\in p\} $$ What can we say about $S$?

  1. Is $S\not\in Y$? Possibly.
  2. Is $S\in Y$ and $S\not\in S$? No because then by definition of $S$, $S\in S$.
  3. Is $S\in Y$ and $S\in S$? No because then by definition of $S$, $S\not\in S$.

Conslusion: The first possibility must be true and $S\not\in Y$.

Corollary: For any set $Y$, $\mathcal P(Y)\not\subseteq Y$ where $\mathcal P$ denotes powerset.

Now if $V$ is a set of all sets, it must satisfy $\mathcal P(V)\subseteq V$, because all sets are in $V$.

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  • $\begingroup$ I don't see exactly where the Power Set is coming from, although I understand why it is also a way to prove that the set of all sets does not exists (the power set of V should be contained in V, but by axiom of the power set, V is contained in the power set, and there I am not sure if we can say that then they are equal through the axiom of extensionality although I know it is wrong). $\endgroup$ Commented Mar 14, 2024 at 20:09
  • $\begingroup$ @pdaranda661: $\mathcal P(Y)$ is the class of all subsets of $Y$. $S_Y=S$ is a set such $S\subseteq Y$ and $S\not\in Y$, so $S\in\mathcal P(Y)$ and $S\not\in Y$. If you start with $Y=V$ then you conclude that $S_V\not\in V$, but $V$ is all sets, so $S_V\in V$. That's the contradiction. For just some arbitrary set $Y$ there is no contradiction in saying $Y$ has a subset that is not a member of $Y$. $\endgroup$ Commented Mar 14, 2024 at 20:59
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Suppose there is a set containing all sets, say $A$. Then by the axiom of comprehension, there is a set $B$ such that $\forall x(x \in B \iff (x \in A \wedge x \notin x))$. Since this holds for all $x$, it holds for $B$ so $(B \in B \iff (B \in A \wedge B \notin B))$. But $B \in A$ as $A$ contains all sets, so $(B \in B \iff B \notin B)$, a contradiction. Hence there is no set containing all sets,

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By axiom schema of separation:

$B$ is a set iff $(\forall x)(x\in B\implies x \in A \land \phi(x))$

Now suppose there is a universal set $V$

Then we can replace $A$ in the axiom with $V$ and the axiom trivially becomes abstraction only:

$B$ is a set iff $(\forall x)(x\in B\implies x \in V \land \phi(x))\implies (\forall x)(x\in B\implies \phi(x) )$

Now plugin Russel's paradox.

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