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Equivalence Principle: A hint on how to start

  1. Sep 11, 2012 #1

    dpa

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    Equivalence Principle: A hint on how to start!!

    Hi, I have no idea where to start.

    1. Statement Problem
    Let X be a non empty set with a equivalence relation ~ on it. Prove that for all x,y[itex]\in[/itex]X,
    [x]=[y] if and only if x~y.


    2. Relevant equations
    For the Equivalence Relation to exist, it must be transitive, reflexive and symmetric.


    3. The attempt at a solution
    I have no idea where to start. May be,
    ~ exists means that, x=y. But is self evident.
    How do I prove the "only If" part as well?

    Thank You.
     
  2. jcsd
  3. Sep 11, 2012 #2
    Re: Equivalence Principle: A hint on how to start!!

    What do the square brackets mean in [x] = [y]?
     
  4. Sep 11, 2012 #3

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    Hi voko,

    I did not understand that either. That is the exact statement in the Homework question. I assumed it simply meant x=y.
    Can you help me if it is x=y?

    Thank You.
     
  5. Sep 11, 2012 #4
    Re: Equivalence Principle: A hint on how to start!!

    This cannot be proven for simple equality, because I can easily give you a counter-example.

    Can the square brackets mean the "equivalence class" of x?
     
  6. Sep 11, 2012 #5

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    Yes, that is it. Still, I have no idea how to prove equivalent classes as equal.
     
  7. Sep 11, 2012 #6

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    I will try it now.
    Thank You.
    dpa
     
  8. Sep 11, 2012 #7
    Re: Equivalence Principle: A hint on how to start!!

    Start by proving the "if" part. Use the definition of the equivalence class and the properties of equivalence.
     
  9. Sep 11, 2012 #8

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    Here is my solution,

    From the definition of equivalence class, we suppose any z such that,
    [x]={z[itex]\in[/itex]Xlx~z} holds. - - - - - - - - - - - - - - - - -(i)

    for every such z,
    since,
    x~y and x~z=z~x [property of symmetry]
    we write from law of transitivity,
    z~y exists.
    Thus,
    we can now define,
    [y]={z[itex]\in[/itex]Xly~z}, which is true.
    can also be written as
    [y]={z[itex]\in[/itex]Xlx~z} - - - - - - - - - - - - - - - - - - - - - - - - -(ii)
    Thus from i and ii, we can write that
    [x]=[y]

    Now we show that it holds true only when x~y by method of contradiction (how???).
    Suppose there exists w such that x~w, but x~y is false,
    but what next?

    Do I say since x~y does not exist,
    we cannot write,
    [y]={z[itex]\in[/itex]Xly~z} exists for every
    [x]={z[itex]\in[/itex]Xlx~z}
    ??
    Any more hints?
     
  10. Sep 11, 2012 #9

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    I got it. I have to prove if a=>b and then if b=>a.
    Is first part correct then?
     
  11. Sep 11, 2012 #10
    Re: Equivalence Principle: A hint on how to start!!

    The first part seems OK. The second part, prove by contradiction. Suppose there are x and y such that [x] = [y], but x not ~y.
     
  12. Sep 12, 2012 #11

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    is it better now?

    To Prove [x]=[y] iff x~y
    Here, First we take x~y and prove [x]=[y]. Then we take x is not ~y but [x]=[y] and arrive at x~y.

    Firstly,
    For any element zEX and zE[x] we know there exists a relation z~x
    Sincem z~x and x~y, we can write from law of transitivity,
    z~y which implies zE[y].

    Thus we can write from zE[x] and zE[y] that [x]=[y].

    Next, we assume x is not ~y but [x]=[y] exists.
    Here, [x]=[y] implies there exists an element z where zE[x] and z~x.
    There also exists zE[y] for which z~y exists.
    Thus from law of transitivity, and from z~y and z~x, we can write,
    x~y.

    Thus Prooved.
     
  13. Sep 12, 2012 #12
    Re: Equivalence Principle: A hint on how to start!!

    In the second part of the proof, instead of saying "there also exists zE[y]" you should say "because [x] = [y], zE[y]".

    Otherwise the proof is good.
     
  14. Sep 12, 2012 #13

    dpa

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    Re: Equivalence Principle: A hint on how to start!!

    Someone said zE[x] and zE[y] cannot imply [x]=[y]. It is indeed true if equivalent relation does not exist.
    What happens when the equivalence relation x~y does exist as in above example. So, is the first part of proof correct in the light of this comment?
    I am totally new and just had one/two class on this topic.
    Thank You.
     
  15. Sep 12, 2012 #14
    Re: Equivalence Principle: A hint on how to start!!

    The first part of the proof is correct. You take ANY element from [x] and show that it also exists in [y]. You could equally take ANY element for [y] and likewise show it is in [x]. So each contains all the elements of the other.
     
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