Why do we assume the set of all sets that do not contain themselves exists?

Question

It seems to me that the set of all sets that do not contain themselves is very similar to the set \begin{equation} X = \{x | \ x \not\in X\} \end{equation} which is again very similar to the number $x$ such that \begin{equation} x\neq x \end{equation} For this number $x$ we simply state, \begin{equation} \mathrel{\nexists} x \in \mathbb{R} : x\neq x \end{equation} Why isn't the same done in the case of sets? \begin{equation} \mathrel{\nexists} X : X = \{x | \ x \not\in X\} \end{equation} and \begin{equation} \mathrel{\nexists} X : X \text{ contains all the sets that do not contain themselves} \end{equation}

Answer 1

We do! Assuming the existence of such a set, we can reach a contradiction. This proves that no such set exists. So, it is a theorem of set theory that there does not exist a set of all sets that do not contain themselves.

The problem (in set theory with unrestricted comprehension) is, it is also a theorem that such a set does exist. Namely, the unrestricted comprehension axiom says that for any property $P$, there exists a set whose elements are exactly the sets that satisfy $P$. In particular, if $P(x)$ is $x\not\in x$, this means that there is a set $\{x:x\not\in x\}$.

So, it is both a theorem that $\{x:x\not\in x\}$ does not exist and a theorem that it does exist. This is a contradiction which makes set theory with unrestricted comprehension inconsistent.

In modern set theory, though, the axiom of unrestricted comprehension is not accepted (since it leads to a contradiction!). So, Russell's paradox becomes simply a theorem that $\{x:x\not\in x\}$ does not exist, not an actual paradox.