First ZF axiom.

Question

The first ZF axiom states: $(A = B) \iff ((x\in A)\iff(x\in B))$. The question is: can I use conjunction instead of double implication there between statements about X's? So, does this whole expression above is the same as the following: $(A=B)\iff((x\in A) \wedge (x\in B))$? It's difficult to me to grasp the first one that is used in textbooks. If these two expression are not the same, then please explain why. And also explain why is the first one is exactly what we are looking for when defining the equality between sets. Thank you very much.

Answer 1

These are quite different.

First of all, what you've written isn't really what you want: you're forgetting the universal quantifiers. The extensionality axiom is $$\forall A,B(A=B\iff \forall x(x\in A\iff x\in B)).$$ At this point it should be clear that we can't replace the second "$\iff$" with a "$\wedge$," since if we do we get $$\forall A,B(A=B\iff \forall x(x\in A\wedge x\in B)).$$

(intuitively: "Two sets are equal iff each contains everything").

And this is clearly silly: for example, we get $\{5\}\not=\{5\}$, since taking $x=2$ we have that $x\in \{5\}\wedge x\in \{5\}$ fails.

There is a way to phrase extensionality in terms of conjunction, though: it's $$\forall A,B(A=B\iff \forall x((x\in A\wedge x\in B)\vee (x\not\in A\wedge x\not\in B))).$$ But this is just a translation of $\iff$, since $$p\iff q$$ is the same as $$(p\wedge q)\vee (\neg p\wedge\neg q).$$