Supoose that f:D -> X where D is any set and x is any metric space with metric d. We can define a function d*:DxD->R given by:

d*(x,y) = d(f(x),f(y)) for any x,y in D. If f is injective prove that d* is a metric on D!

Any help would be great!

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- Dec 9th 2009, 11:24 AMreally.smartymetrics
Supoose that f:D -> X where D is any set and x is any metric space with metric d. We can define a function d*:DxD->R given by:

d*(x,y) = d(f(x),f(y)) for any x,y in D. If f is injective prove that d* is a metric on D!

Any help would be great! - Dec 9th 2009, 11:24 AMreally.smarty
sorry its f: D -> X

and later on d*: DxD -> R.

the weird smiley faces come sometimes.. - Dec 9th 2009, 12:52 PMPlato
- Dec 9th 2009, 08:06 PMreally.smarty
if f(x) = f(y) that means the first three parts are satisfied for metrics.

1. not negative.

2. d(x,y) = 0 iff x=y

3. d(x,y) = d(y,x)

i am more concerned with how to show the triangle inequality for this question? - Dec 10th 2009, 02:59 AMPlato
- Dec 10th 2009, 09:29 AMreally.smarty
and i would go about proving that statement how?

than you again.