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

  1. #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:
    Last edited by lamp23; August 16th 2012 at 02:22 PM.
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    Re: [a]=[b] <--> a~b

    Quote Originally Posted by lamp23 View Post
    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?
    Thanks from lamp23
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    Re: [a]=[b] <--> a~b

    I think (3) and (4) need an existencial quantifier for the variable x. Otherwise, seems fine.
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    Re: [a]=[b] <--> a~b

    Quote Originally Posted by ModusPonens View Post
    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.
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    Re: [a]=[b] <--> a~b

    Quote Originally Posted by Plato View Post
    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]"
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    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.
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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).
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