If $\subseteq$ is defined set-theoretically, it doesn't exist

Question

I came across the following statement:

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

I understand everything except "$\mathrm{dom(\subseteq)}$ is $V$". The domain of $\subseteq$ is the set of all $A$'s such that $A\subseteq B$, 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?

Answer 1

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

Answer 2

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.

Answer 3

$V$ is the common name for the collection of all sets (which doesn't exist 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,

$$\operatorname{dom} R = \{ x : (\exists y) \, (x, y) \in R \}. $$

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 $\operatorname{dom} \subseteq$ would be the set of all sets $V$, which is a contradiction.