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

1. ## [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:

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

Originally Posted by lamp23
Thm: [a]=[b] <--> a~b
I have proved:
[a]=[b] --> a~b
but cannot figure out how to prove:
[a]=[b] <-- a~b
Suppose that $a \sim b$ then if $x\in[a]$ we know that $x \sim a$.
Because $a \sim b~\&~x \sim a$ we know that $x \sim b$ or $x \in [b]$

Can you finish?

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

I think (3) and (4) need an existencial quantifier for the variable x. Otherwise, seems fine.

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

Originally Posted by ModusPonens
I think (3) and (4) need an existencial quantifier for the variable x. Otherwise, seems fine.
Actually no that is not necessary.
If we start with $[a]=[b]$ then because $a\in [a]$ we know that $a\in [b]$ or $a\sim b$.

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

Originally Posted by Plato
Suppose that $a \sim b$ then if $x\in[a]$ we know that $x \sim a$.
Because $a \sim b~\&~x \sim a$ we know that $x \sim b$ or $x \in [b]$

Can you finish?
Yes, thanks a lot. I was missing using the given of "Suppose $x\in[a]$"

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.

7. ## 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).