# space of continuous functions on a compact space with uniform metric

• August 1st 2010, 06:50 AM
Iondor
space of continuous functions on a compact space with uniform metric
Hello,

i have read the following assertion in a book, but i can't prove it.

Let $X$ be a compact metrizable space and $Y$ a metrizable space.
We denote by $C(X,Y)$ the space of continous functions from $X$ into $Y$ with the topology induced by the sup or uniform metric
$d_{u}(f,g)=\sup_{x\in X} d_{Y}(f(x),g(x))$ , where $d_{Y}$ is a compatible metric for $Y$.

Now the assertion that i want to prove, but don't know how to:
A simple compactness argument shows that this topology is independent of the choice of $d_{Y}$

Any ideas how to prove this?
• August 2nd 2010, 03:59 PM
Jose27
Seems like a weird assertion, since the space itself depends on the metric of $Y$. For example take $X=[0,1]$ with the usual metric, and $Y=\mathbb{R}$, now if we give $Y$ the usual topology then the identity function $id:X\rightarrow Y$ is clearly continous, but it is not continous if we give $Y$ the discrete metric. So in general, if we change $d_Y$ then $C(X,Y)$ might not be the same set.
• July 8th 2011, 02:16 PM
alanmath
Re: space of continuous functions on a compact space with uniform metric
Yes, I have proved it and posted it on my blog. Here is a link to the post:

Uniform metrics define same topology on C(X,Y) | alanmath

As to Jose27's response, he hasn't understood that $d_Y$ is any compatible metric for $Y$. So that is any metric that gives the topology on $Y$. So, of course, if the metric gives the same topology, then the set of continuous functions will not change.