Can we conclude this from ZFC?

Question

Assume that sets exist for a given property. Then can we derive from ZFC that there exists a set containing all those (existing) sets?

Answer 1

No. In particular, in ZFC, there is no set of all sets (which would be equivalent to the set of all sets $x$ which satisfy the property $x = x$).

The proof is the famous Russell's paradox. We will prove, by contradiction, that there is no set which contains all sets. Start by assuming that $V$ is the set of all sets. Now, using the axiom of specification, define the set

$$P = \{x : x \in V, x \notin x\}.$$

Is $P$ an element of $P$? By definition, $P$ is in $P$ if and only if $P$ is in $V$ but $P$ is not in $P$. But since all sets are in $V$, we can simply say: $P$ is in $P$ if and only if $P$ is not in $P$. But that's impossible.

So, we have proven, by contradiction, that there is no set which contains itself.