ZF Proof – Set of All Sets Does Not Exist

axiomselementary-set-theoryset-theory

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 $A$.

Best Answer

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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