I've decided to start reading into topology, which is apparently based on Set theory. My set theory is pretty useless, as I've never gone very far with it, but it seems topology is based on it, and can have some useful applications. So, I've started reading about set theory again, and I'm having trouble with this axiom: (...Thefirstaxiom :P)

"Using the notion of equivalence of sentences we may rewrite the first axiom as:

Axiom 1b:

Consider two sets A and B. Then A = B if and only if x ∈A

⇔x ∈B."

Am I first right in thinking that for, x ∈A⇔x ∈B to be true, then both x ∈A AND x ∈B must BOTH be true, or BOTH be false? so if one is true and the other is false, then the statement is false?...In which case the operator '⇔' is much like an "XOR" (exclusive OR) operator?

If thats right, then what about the case where x DOES NOT belong to A, and x DOES NOT belong to B?...in which case x ∈A is false, and x ∈B is false, meaning the statement is true that A=B. but how can this be so? A, and B could be anything....in fact, because of that, what if we had:

A={x, 1, 3, 5}

B={x, 2, 4, 6}

even if x ∈A AND x ∈B, it doesn't mean A=B. I think I'm missing something important here. :?