One more time, ZF based proof that set of all sets does not exist

Question

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 $A$.

Answer 1

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 $V$.

Answer 2

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,

Answer 3

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.