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- Aug 16th 2012, 02:20 PM #1

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## [a]=[b] <--> a~b

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

- Aug 16th 2012, 02:48 PM #2

- Aug 16th 2012, 02:51 PM #3

- Aug 16th 2012, 03:00 PM #4

- Aug 16th 2012, 03:53 PM #5

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- Aug 16th 2012, 06:01 PM #6
## Re: [a]=[b] <--> a~b

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.

- Aug 16th 2012, 07:11 PM #7

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## Re: [a]=[b] <--> a~b

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).