# Question Regarding the Axiom of Extensionality

Jech's text on Set Theory states the following:

If X and Y have the same elements, then X = Y :

∀u(u ∈ X ↔ u ∈ Y ) → X = Y.

The converse, namely, if X = Y then u ∈ X ↔ u ∈ Y, is an axiom of predicate calculus. Thus we have

X = Y if and only if ∀u(u ∈ X ↔ u ∈ Y).

The axiom expresses the basic idea of a set: A set is determined by its elements.

To check my understanding, are the following true?

(1) The axiom of extension states only that if two sets have the same members, then those two sets equal.

(2) It is, on the other hand, an axiom of the language of set theory (i.e., predicate calculus) that if two sets equal, then they have the same members.

A third question:

(3) It seems that the axioms of ZFC are very much independent of the language being worked with. That is, the axioms of ZFC don't state we must be working in the language predicate calculus. Then I take it there exist languages we could be working with that don't have the property that if two sets are equal, they have the same members (so that we could only conclude if two sets have the same members, they are the same and not the reverse). Is this true? Do mathematicians ever work in languages besides predicate calculus?

(1) and (2) are correct.

As for (3), recall Leibniz's Law -- if $a$ and $b$ are one and the very same thing, then whatever property $a$ has $b$ must have too. Applied to sets, if $a$ and $b$ are one and the very same set, then whatever property $a$ has $b$ must have too. So in particular, if $a$ and $b$ are one and the very same set, if $a$ has the property of having $x$ as a member, $b$ has the property of having $x$ as a member. In symbols, if that helps, if $a = b$ then $x \in a \to x \in b$. And of course, we likewise have if $a = b$ then $x \in b \to x \in a$. So:

if $a$ and $b$ are sets, then if $a = b$ then $x \in a \equiv x \in b$

But $x$ was arbitrary, so there's implicit generalization here which we can make explicit, again borrowing logical notation

if $a$ and $b$ are sets, then if $a = b$ then $\forall x(x \in a \equiv x \in b)$

Now, in getting this far we are just using the informal Leibniz's Law, notation for set-membership, and some handy logical notation. We aren't appealing to a formal system; rather we are appealing to informal mathematical reasoning (the sort of reasoning that logic books aim to formalize using the classical predicate calculus). Because we do want the formal predicate calculus to replicate this informal mathematical reasoning, we will get -- inside the calculus, applied to a language with $\in$ available

$a = b \to \forall x(x \in a \equiv x \in b)$

But the ultimate grounds for this half of the formal version of the extensionality principle are the background informal reasoning using Leibniz's Law which we aim to regiment formally. What's going on here is not an arbitrary appeal to one formal calculus rather than another.

• Peter -- this was an excellent answer (I loved the philosophical motivation) and after looking at your biography I have to say I'm stunned that I had someone like you personally answer one of my questions. Quite a testament to math.stackexchange. Cheers. – user1770201 May 23 '13 at 9:08
• @user1770201 Thanks for the kind words: you're welcome! – Peter Smith May 23 '13 at 11:40

The first two questions are answered yes. It should be remarked that in older texts equality was omitted completely from the language, and was defined using $\in$ as to make the axiom of extensionality trivial.

To the third question, we work with whatever fits our context best. Set theory is usually thought as a theory in first-order logic, but it can be thought as a theory of second-order logic (including a nice upgrade to the axiom schemata), and it can be thought of in the context of intuitionistic logic as well.

There are several levels to formality, the logic (axioms, inference rules), the language, the axioms, the semantics. We can modify whatever we want for whatever we want to modify it. We just need to wonder whether or not it is going to be useful, possibly consistent, or interesting.

• "It should be remarked that in older texts equality was omitted completely from the language, and was defined using ∈ as to make the axiom of extensionality trivial." Does this contradict math.stackexchange.com/a/437873/76284, or am I misreading this? – user76284 Jan 17 at 2:40
• Yes, you are misreading it. My point is that you "move" extensionality to an axiom of logic, rather than an axiom of set theory. – Asaf Karagila Jan 17 at 9:38
• What do you mean by "moving it to an axiom of logic"? – user76284 Jan 17 at 10:13

Yes to (1) and (2).

As for (3): The language of set theory is $\{\in\}$. Therefore it seems to me that in any language in which you can express set membership you will necessarily be able to conclude "if two sets are equal, they have the same members". Of course, this corresponds to our intuition: two sets that are equal should contain the same elements.

Note that here the membership ranges over elements in the model. If you add elements to the sets that are not elements of the model then you can well have two sets that are equal in the model but not equal (do not contain the same elements) if seen from outside the model.

As for your last question: yes there are logical deduction systems other than first order logic (= predicate calculus). The language of sets $\{\in\}$ will still be $\{\in\}$ so that what I wrote above should still hold.