ZFC: Why is the set $\{ x \mid x = x\} $ not defined? Why is the set $\{ x \mid x = x\} $ not defined?
Since, $x=x$ is always true, the set is actually "the set of everything".
But why is it illegal to be defined as a set?
 A: ZFC has no method of set construction that allows us to construct the set of all sets with a given property. If it did, we could define $\{ x | x \notin x\}$, the set of all sets that don't contain themselves, and we would have Russell's paradox; that set would have to both contain and not contain itself.
Instead, ZFC has the axiom schema of restricted comprehension. For any set $S$ and predicate $p$, we can define
$$\{ x \in S \mid p(x)\}$$
the set of all elements of $S$ such that $p$ holds for those elements. We have to restrict the scope to elements of another set; we can't perform a set comprehension over all sets.
A: While it is quite possibly to prove that this collection is not a set via Russell's paradox, here's an easier conclusion from the axiom of regularity (and pairing).
Suppose $A=\{x\mid x=x\}$ is a set, then in particular $A=A$ so $A\in A$. However this is a contradiction to the axiom of regularity, since now $\{A\}$ is a set and $A\cap\{A\}\neq\varnothing$ contrary to the requirement that $\in$ is well-founded.
A: Using the axioms of ZFC, you can't prove the existence of it because is a proper class (using the NBG terminology).
A: Because there is no pre-existing set that it is a subset of.  To construct a set from a predicate we require the axiom of comprehension/specfication.
