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

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

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 $\displaystyle d_{Y}$

Any ideas how to prove this?

2. Seems like a weird assertion, since the space itself depends on the metric of $\displaystyle Y$. For example take $\displaystyle X=[0,1]$ with the usual metric, and $\displaystyle Y=\mathbb{R}$, now if we give $\displaystyle Y$ the usual topology then the identity function $\displaystyle id:X\rightarrow Y$ is clearly continous, but it is not continous if we give $\displaystyle Y$ the discrete metric. So in general, if we change $\displaystyle d_Y$ then $\displaystyle C(X,Y)$ might not be the same set.

3. ## 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 $\displaystyle d_Y$ is any compatible metric for $\displaystyle Y$. So that is any metric that gives the topology on $\displaystyle Y$. So, of course, if the metric gives the same topology, then the set of continuous functions will not change.