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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.

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    $\begingroup$ 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. $\endgroup$ Aug 20, 2023 at 19:53
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    $\begingroup$ @ArturoMagidin That can be an answer. $\endgroup$
    – aschepler
    Aug 20, 2023 at 19:57
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    $\begingroup$ 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. $\endgroup$
    – bof
    Aug 20, 2023 at 20:14
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    $\begingroup$ I'm just guessing, but it seems what you're missing is an understanding of what the notation $\{ y∈x ∣ y≠y \}$ means. $\endgroup$
    – bof
    Aug 20, 2023 at 20:16
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    $\begingroup$ @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. $\endgroup$ Aug 20, 2023 at 20:58

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