# metrics

• December 9th 2009, 12:24 PM
really.smarty
metrics
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!
• December 9th 2009, 12:24 PM
really.smarty
sorry its f: D -> X
and later on d*: DxD -> R.
the weird smiley faces come sometimes..
• December 9th 2009, 01:52 PM
Plato
Quote:

Originally Posted by really.smarty
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!

There is really one of the three properties that is not trival.
If $d^*(x,y)=0$ that mean $d(f(x),f(y))=0$.
Does that mean f(x)=f(y)? And what does that mean?
• December 9th 2009, 09:06 PM
really.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?
• December 10th 2009, 03:59 AM
Plato
Quote:

Originally Posted by really.smarty
i am more concerned with how to show the triangle inequality for this question?

$d^* (x,y) = d(f(x),f(y)) \leqslant d(f(x),f(z)) + d(f(z),f(y))$
• December 10th 2009, 10:29 AM
really.smarty
and i would go about proving that statement how?

than you again.