In ZFC given any set $X$ if $x\in X$ then is $\emptyset=x\cap X$?

Question

Given any set $X$ and any element $x\in X$ then is $\emptyset=x\cap X$?

Feels like it should be, I mean most of the time I don't even think of elements of sets as sets unless I'm deliberate doing so. Though I know if like I'm working in ZFC or something, then like pretty much everything I write is a set I think. Anyway an answer to the first question would be helpful.

Also if yes, what axiom(s) would it follow from? All of them? Only a couple?

Lastly if in modern axiomatic set theory essentially everything is a set wouldn't things like this:

https://en.wikipedia.org/wiki/Set_function

Be redundant?

Answer 1

Let me answer your last question

Lastly if in modern axiomatic set theory essentially everything is a set wouldn't things like this:

https://en.wikipedia.org/wiki/Set_function

Be redundant?

Yes, it's redundant. In set theory, every function is a set function.

Answer 2

For any non-empty set $X$, some element of $X$ is disjoint from it; this is the axiom of regularity. But not all $x \in X$ are required to have this property. Consider for instance $X=\{ \emptyset,\{ \emptyset \} \}$: $\{ \emptyset \}$ is an element of $X$ which is not disjoint from $X$.

The fact that not all $x$ in $X$ are required to have this property is actually important in how we might use ZF to do other math. For instance, $\{ \emptyset,\{ \emptyset \} \}$ is the "bare metal" representation of the ordinal number "2", and in general the ordinal number $n+1$ is given by $n \cup \{ n \}$.