# metrics

• Dec 9th 2009, 11:24 AM
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!
• Dec 9th 2009, 11:24 AM
really.smarty
sorry its f: D -> X
and later on d*: DxD -> R.
the weird smiley faces come sometimes..
• Dec 9th 2009, 12:52 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 \$\displaystyle d^*(x,y)=0\$ that mean \$\displaystyle d(f(x),f(y))=0\$.
Does that mean f(x)=f(y)? And what does that mean?
• Dec 9th 2009, 08:06 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?
• Dec 10th 2009, 02:59 AM
Plato
Quote:

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

\$\displaystyle d^* (x,y) = d(f(x),f(y)) \leqslant d(f(x),f(z)) + d(f(z),f(y))\$
• Dec 10th 2009, 09:29 AM
really.smarty
and i would go about proving that statement how?

than you again.