This is a strengthening of my previous question, here: How to prove that if two sets belong to the same sets, then they are identical?. My current question is this. Let $x$ and $y$ be sets, and suppose that $x$ belongs to every set that $y$ belongs to, in symbols: $(\forall z)(y \in z \rightarrow x \in z)$. Does it follow that $x$ and $y$ are the same set?
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Yes. By the Axiom of Pairs and the Axiom of Separation, given sets $x$ and $y$, there exists a set $z$ such that $z=\{x,y\}$.
By the Axiom of Separation, the set $w=\{a\in z\mid a\neq x\}$ is a set.
Note that $x\notin w$; hence $y\notin w$. Since $y\in z$, it follows that for all $a$, $a\neq x\implies a\neq y$. Then using universal instantiation with $a=y$ we obtain $y\neq x\implies y\neq y$; by contrapositive, $y=y\implies y=x$, hence $y=x$.
Arturo Magidin
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10A little shorter: $y\in{y}$, so $x\in{y}$, and so $x=y$. (The usual formulation of the pairing axiom doesn't require the two members of the pair to be distinct, so it produces ${y}$ directly.) – Andreas Blass Apr 12 '24 at 18:16