# Identity of indiscernibles

It follows from the axioms of identity alone that $x = y \Rightarrow \big((\forall z) x \in z \equiv y \in z\big)$ and $x = y \Rightarrow \big((\forall z) z \in x \equiv z \in y\big)$.

One of the most important axioms of presumably every set theory is the axiom of extensionality: $\big((\forall z) z \in x \equiv z \in y\big) \Rightarrow x = y$.

But what about its reverse: $\big((\forall z) x \in z \equiv y \in z\big) \Rightarrow x = y$? Does this statement have a name among mathematical logicians and/or set theorists, maybe Identity of indiscernibles? In which set theories can it be proved? And in which set theories (or models) does it not hold?

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 For the sake of formal correctness, and for readability, parentheses should be used. – André Nicolas Apr 14 '11 at 17:14 @user6312: done – Hans Stricker Apr 14 '11 at 17:22

We can prove it in any set theory with the axiom of extensionality and the axiom of pairing.

Assume $(\forall z) x\in z$ iff $y\in z$.

Given $x$, form the set $\{x,x\} = \{x\}$ by the pairing axiom. Now, set $z = \{x\}$.

We see that $x\in \{x\}$ so we must have $y\in \{x\}$ so that $y= x$.

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 I was too impressed by Leibniz' and the principle's name: it's a triviality (in the context of set theory), isn't it. – Hans Stricker Apr 14 '11 at 17:29

The statement is trivially true in any reasonable set theory. Given any $x$, let $z$ be the set whose only element is $x$. Then for any $y$, $y\in z$ iff $y=x$.

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