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To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's The Foundations of Mathematics that presents ZF set theory. I haven't gotten really far since I am stuck at the axiom of Extensionality, stated as follows:

$$\forall x,y \; [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y]$$

As far as I understood it, the purpose of this axiom is to state that every two sets that have exactly the same members are the same set. In terms of the above formula, when antecedent is true, the consequent needs to be true. What confuses me here is the case when the antecedent is false. Then, the consequent can be false or true, i.e., we can't say much about the consequent. But, don't we really want to say that in that case those two sets are not equal? More precisely, shouldn't we use equivalence instead of implication? If not, what is the reason?

Asaf Karagila
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zpavlinovic
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2 Answers2

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The reason the converse is not part of the axiom is that it already follows from the axioms of first-order logic: $x = y \implies (\varphi(x) \iff \varphi(y))$ for any formula $\varphi$. This is known as the substitution property of equality.

Trevor Wilson
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  • Isn't quantification over functions an aspect of second order, and not first order logic? – user400188 Sep 23 '20 at 01:56
  • This is an axiom schema containing a separate axiom for each formula $\varphi$, not a single axiom involving quantification over formulas. (A similar remark holds for the separation and replacement axioms of ZF.) – Trevor Wilson Sep 23 '20 at 04:11
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    Thank you for the clarification. This is actually the first time I've understood why we use axiom schemas. – user400188 Sep 23 '20 at 05:12
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Equality is often a part of the logical symbols, and if $x=y$ then for every relation $R$, and every $z$ we have $R(x,z)\leftrightarrow R(y,z)$ and $R(z,y)\leftrightarrow R(z,x)$.

In particular $\in$ is such relation.

In older texts, you can sometimes find that people don't assume that $=$ is part of the language and then we simply define $x=y$ as the formula $\forall w(w\in x\leftrightarrow w\in y)$, from which extensionality follows as a theorem. Then the language containing only the binary relation symbol $\in$ (without equality) is called the Language of Set Theory (often abbreviated as LST).

Asaf Karagila
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    Ha, I beat you for once (but only by a few seconds) – Trevor Wilson May 19 '13 at 23:09
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    @Trevor: My girlfriend is somewhat ill, and sleeping in the other room. I was trying to type slowly as to not make a ruckus with my loud buckling spring mechanical keyboard. ;-) – Asaf Karagila May 19 '13 at 23:10
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    Ah, I knew there must be a reason. Hope she is better soon. – Trevor Wilson May 19 '13 at 23:11
  • She'll be fine by morning. It doesn't work, by the way. The keyboard is still as loud as the apocalypse. I love it! ;-) – Asaf Karagila May 19 '13 at 23:11
  • I like them too. I have a collection of four model M's. – Trevor Wilson May 19 '13 at 23:14
  • I recently bought a Unicomp (with blank keys, it's great!) – Asaf Karagila May 19 '13 at 23:15
  • If you don't have "=" in the language, how can you state any version of extensionality? Perhaps you mean that you can define equality using "$\in$" in such a way that the extensionality axiom is a theorem in your definitional expansion? – Trevor Wilson May 19 '13 at 23:18
  • @Trevor: Yes, that's what I meant. Any ambiguity or mistake is only the fault of an extremely long and tedious day with too little sleep last morning. ;-) I'll clarify some more in a bit. – Asaf Karagila May 19 '13 at 23:20
  • @AsafKaragila interesting that you say one can define $x=y$ using the condition $(\forall t)(t\in x \iff t\in y)$. What then would be the axiom that follows? Currently I'm looking into a separate but related fact, that $x=y$ if and only if $(\forall t)(x\in t \iff y\in t)$, which has a kind of beautiful symmetry to the original definition. Would this fact be equivalent to the axiom of extensionality, or would it have another name? – chharvey Dec 15 '13 at 01:52
  • @TestSubject528491: I'm not quite sure what you mean. – Asaf Karagila Dec 15 '13 at 12:14
  • @AsafKaragila http://math.stackexchange.com/questions/607312/axioms-of-equality – chharvey Dec 15 '13 at 14:41
  • @TestSubject528491: Yes, I saw that. – Asaf Karagila Dec 15 '13 at 15:15