# How to prove the existence of the empty set from other axioms in ZFC?

In another post, the top answer says:

If there are sets at all, the axiom of subsets tells us that there is an empty set: If $$x$$ is a set, then $$\{ y∈x ∣ y≠y \}$$ is a set, and is empty, since there are no elements $$y$$ of $$x$$ for which $$y≠y$$. The axiom of extensionality then tells us that there is only one such empty set.”

I am having trouble understanding the move from 'there exists a set $$x$$' to '$$\{ y∈x ∣ y≠y \}$$ is a set'. Why is it the case that there is a subset of $$x$$, $$y$$, such that $$y$$ is distinct from itself?

I may be missing something basic here.

• The Axiom Schema of Separation says (roughly) that if $\phi$ is a formula for sets, and $x$ is a set, the there is a set whose elements are precisely the $y\in x$ for which $\phi(y)$ is true. So if $\phi$ is the formula $y\neq y$ then $\{y\in x\mid \phi(y)\}$ is a set. Aug 20, 2023 at 19:53
• @ArturoMagidin That can be an answer. Aug 20, 2023 at 19:57
• Nobody said "there is a subset of $x$, $y$, such that $y$ is distinct from itself." Read the passge you quoted; that's not what it says.
– bof
Aug 20, 2023 at 20:14
• I'm just guessing, but it seems what you're missing is an understanding of what the notation $\{ y∈x ∣ y≠y \}$ means.
– bof
Aug 20, 2023 at 20:16
• @YV1999 The problem with your interpretation "... there is a subset of $x$, $y$, such that $y$ is distinct from itself?" of $\ \{y\in x\,|\,y\ne y\}\$ is in misidentifying a putative element $\ y\$ of that subset of $\ x\$ with the subset itself. A correct statement would be "...there is a subset of $\ x\$ comprising those elements $\ y\$ of $\ x\$ which are distinct from themselves. Aug 20, 2023 at 20:58