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Homework Help: Metric space and topology help

  1. Apr 1, 2010 #1
    Let (X,d) be a metric space. Show that if there exists a metric d' on X/~ such that
    d(x,y) = d'([x],[y]) for all x,y in X
    then ~ is the identity equivalence relation, with x~y if and only if x=y.

    i have:

    assume x=y
    then d(x,y)=0 and [x]=[y] which implies d'([x],[y])=0 also.

    now assume d(x,y) = d'([x],[y]) for all x,y in X
    i now need to show that this implies x=y in order to copmlete the proof but i don't know where to go. any ideas?

    thanks.
     
  2. jcsd
  3. Apr 1, 2010 #2
    Re: Topology

    If x~y, what is d'([x],[y])? How about d(x,y)?
     
  4. Apr 1, 2010 #3
    Re: Topology

    if x~y then d'([x],[y])=d(x,y)=0, right?

    was what i had above on the right lines? is it done ok so far?
     
  5. Apr 1, 2010 #4
    Re: Topology

    Yes. In your OP you proved that x=y imply x~y (i.e. ~ is reflexive). In this reply you proved that x~y imply d(x,y)=0. To complete the proof you will have to prove that x~y imply x=y, but since d is a metric you know that d(x,y)=0 imply x=y. Thus you have proven that x=y iff x~y.
     
  6. Apr 1, 2010 #5
    Re: Topology

    so in full it is:

    assume x=y (want to show x~y)

    x=y implies d(x,y)=0 which implies d'([x],[y])=0 which implies [x]=[y] which implies x~y done.

    now assume x~y (want to show x=y)

    x~y implies [x]=[y] which implies d'([x],[y])=0 which implies d(x,y)=0 which implies x=y done.

    therefore x=y iff x~y.
     
  7. Apr 1, 2010 #6
    Re: Topology

    Looks fine to me.
     
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