How can we show that there is no such a set? We want to prove that the set of all sets do not exist.
We suppose that the set of all sets, let $V$, exists.
So, for each set $x, x \in V$.
We define the type $\phi: \text{ a set does not belong to itself , so } x \notin x$.
From the axiom schema of specification, we conclude that there is the set $B=\{ x \in V: x \notin x \}$
$$\forall y(y \in B \leftrightarrow y \notin y)$$
How can we continue?
 A: The set $B$ is commonly used in Russell's Paradox. Now that $B$ exists, it is well-defined to ask if $B\in B$ or not.  Derive a contradiction from this.
A: The argument that a set cannot be an element of itself is quite convincing, but it's not the real reason that the collection of all sets isn't a set.
Assume we do allow sets to be elements of themselves (the set of all sets would necessarily be such a set). Then look at the subset $R$ of the universe defined by
$$
x\in R \leftrightarrow x\notin x
$$
i.e. it contains all sets that do not contain themselves.
Here's a question: Do we have $R\in R$? This is known as Russel's paradox, and it was a decisive factor against naive set theory.
Edit A few more details: If the universe $V$ of all sets is a set, then in particular we have $V\in V$, so we have to allow sets to be elements of themselves.
Also, for any two sets $x,y$ we either have $\color{red}{x\in y}$ or we have $\color{blue}{x\notin y}$. Now, if we let $x=y=R$, which one is it?
It cannot be the first case, $\color{red}{R\in R}$ since if that's the case then by definition of $R$ we have $R\notin R$ and we have a contradiction.
Then it must be the case that $\color{blue}{R\notin R}$. But then, by construction of $R$, we must have $R\in R$. Which means that $R$ neither is nor isn't an element of itself. This is the paradoxical result that carries Bertrand Russell's name.
A: $B \in B$ iff $B \notin B$ (this is how $B$ is defined: a set is in it, if it is not an element of itself. So $B \in B$ implies $B \notin B$, contradiction. So $B \notin B$. But then again: $ B \in B$ (as $B$ is a set, so in $V$ and it's not an element of itself..). Contradiction again. So both $B \in B$ and $B \notin B$ lead to contradictions....
So $V$ cannot have been a set (this is where the trouble started..)
