Metric space and topology help

[latentcorpse]

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.

 

[rasmhop]

If x~y, what is d'([x],[y])? How about d(x,y)?

 

[latentcorpse]

If x~y, what is d'([x],[y])? How about d(x,y)?

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?

 

[rasmhop]

was what i had above on the right lines? is it done ok so far

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.

 

[latentcorpse]

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.

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.

 

[VeeEight]

Looks fine to me.