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As stated on Wikipedia, the formal definition of the Axiom of Extensionality is

$$\forall A \forall B(\forall X(X \in A \iff X\in B) \implies A=B)$$

Which left me underwhelmed, as it seemed more of a longhand for equality rather than an axiom. I then continued reading and saw that I fell into the category of people that

[...] treat the above statement not as an axiom but as a definition of equality

Hence the axiom of extensionality is rewritten to

$$\forall A \forall B(\forall X(X \in A \iff X\in B) \implies \forall Y(A \in Y \iff B\in Y))$$

I.e; if $A$ and $B$ conatin the same elements, then they belong to the same sets.

My question is, without that axiom, is it even possible to construct two sets that are identical in what they contain, but aren't in all the same sets?

Here's my thought process: Suppose $A$ and $B$ contain the same elements.

By the Axiom Schema of Specification, for some subset $S\subseteq T$, such that $\forall x[x \in T \iff (x \in S \wedge \phi(x))]$, such that $\phi(x)$ is entirely determined by what is contained in $x$, then $(A\in S)\iff (B\in S)$.

However if $\phi(x)$ is not entirely determined by what is contained in $x$, i.e; it's dependent on if $x$ is contained within another set, then it may not immediately be the case that $(A\in S)\iff (B\in S)$. But if those sets (that $\phi(x)$ depends on $x$ being contained within) were constructed by some $\phi_2(y)$ which is entirely determined by what is contained in $y$, then $(A\in S)\iff (B\in S)$ holds.

I hope you see where I'm going with this. To explain it a different way, all the sets that I can imagine that are capable of being constructed within ZF are (perhaps not directly) but inevitably dependent on only what is contained within a set, hence it seems that the axiom of extensionality is implied. Is there a flaw in my logic?

Graviton
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    I think the problem in your first argument is that you do not distinguish the equality as a logical symbol and the defined extensional equality. Some non-extensional set theories (e.g., appearing in Friedman's article about double-negation translation on set theories) differentiate these two notions. – Hanul Jeon Jul 25 '20 at 06:57
  • @HanulJeon Perhaps I'm misinterpreting your comment, but I'm not exactly concerned with equality as a logic symbol, more so my question about it being possible in ZF (without the variant of extensionality without "=") to have sets that both have identical elements yet not be within the same sets. – Graviton Jul 25 '20 at 07:04
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    There is a discussion of this topic in this answer to a related question. Performing a search on "extensionality" on this site reveals other similar questions and answers as well. – Greg Martin Jul 25 '20 at 07:23
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    Let $A$ and $B$ be empty sets, but $A$ is blue and $B$ is red. Then $A\ne B$ because they have different colours, even though $x\in A$ and $x\in B$ are both false for all $x$ – Hagen von Eitzen Jul 25 '20 at 08:15
  • @HagenvonEitzen I appreciate this analogy and understand what you're getting at, but I don't think it can be translated well into the language of first order logic. ZF (disappointingly) doesn't care about what attributes we give our variables, it only cares about the attributes of what the variables represent. – Graviton Jul 25 '20 at 08:20
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    @Graviton The statement that sets have no colours or any other properties that can be used to distinguish them besides which elements they contain is precisely what the Axiom of Extensionality formalizes – Hagen von Eitzen Jul 25 '20 at 08:23
  • @HagenvonEitzen That was a magical moment. I think it all just clicked. Thank you. – Graviton Jul 25 '20 at 08:25
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    https://math.stackexchange.com/questions/426959/axiom-of-extensionality-in-zf-pointless/ – Asaf Karagila Jul 25 '20 at 09:41
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    You might want to take a look at Krivine's work on classical realizability. He often laments how extensionality is the hardest axiom to satisfy. (Or at least he did in the few times we had a discussion on these models, as well as other models of ZF.) – Asaf Karagila Jul 25 '20 at 09:46
  • Put in another way: By adding $\in$ to our language, we merely are able to express that the things we talk about (so by intention: sets) are things that may or may not contain other things. EXT expresses that "containing some things and others not" uniquely determines a set, so that it is justified to say that a set is a collection/container of other things/sets, which is the naive origin of the terminus "set" – Hagen von Eitzen Jul 25 '20 at 10:46

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