Thm: [a]=[b] <--> a~b
I have proved:
[a]=[b] --> a~b
but cannot figure out how to prove:
[a]=[b] <-- a~b
I don't know if I can just write the steps backwards since I don't see how I would know: (4) --> (5)
This fact was stated without proof in a youtube playlist on equivalence relations: Equivalence relations 3 - YouTube
Yes. But isn't that a different proof? I mean if we want to make the proof in the OP correct, isn't there a need of a quantifier in order for a (an element, ultimately, a set) to be in relation with another element (which x is not, because it's a free variable)? I'm unsure.
yes, something along the lines of: let x be in [a] = [b]. we know such an x exists because both [a] and [b] are non-empty (because ~ is an equivalence, so ~ is reflexive. so at the very least, a is in [a], and b is in [b] (a and b may actually be the same, but perhaps not)).
but it's not necessary to introduce x at all. as Plato points out, we have:
a in [a] = [b], so a in [b], so b~a, and by symmetry, a~b. doing so eliminates the need to use transitivity to show [a] = [b] → a~b (we we *do* need transitivity to show the reverse implication, however).