Vtyjkyuk

I came across the following statement:

If the subset relation $subseteq $ is defined as a set in ZFC, then it doesn't exist because $mathrmdom(Rmathrm)$ is guaranteed to exist for any relation $R$, and $mathrmdom(subseteq)$ is $V$, which doesn't exist.

I understand everything except "$mathrmdom(subseteq)$ is $V$". The domain of $subseteq$ is the set of all $A$'s such that $Asubseteq B$, but since we are defining $subseteq$ set-theoretically, $A$ is not any set, it is any set in ZFC, so I don't understand how this domain equals $V$.

Can anyone explain this?

$V$ is the common name for the collection of all sets (which doesn't exists as a set, as implied by the Russell's paradox, therefore we often refer to it as the class of all sets). As you probably know,

$$operatornamedom R = x : (exists y) , (x, y) in R . $$

Since for each set $A$ there is a set $B$ (for instance, $B = A$) such that $A subseteq B$, we have that if $subseteq$ were a set, then $operatornamedom subseteq$ would be the set of all sets $V$, which is a contradiction.

The symbol $V$ is being used, presumably, to stand for the "universe" of all sets, which is a class but is not a set. To say that "dom($subseteq$) is $V$" simply means that in the sentence $A subseteq B$, you may replace $A$ by any set, i.e. any element of $V$, and you may replace $B$ by any set, and having made those replacements the sentence will then have a truth value.

Well, $xsubseteq x$ holds for every $x$. So, if you were write down the definition of $operatornamedom(subseteq)$ you'd get $x:(exists y)(xsubseteq y)$, which is a subclass of $V$, but by the first sentence $V$ is part of $operatornamedom(subseteq)$.