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Math Help - space of continuous functions on a compact space with uniform metric

  1. #1
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    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?
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  2. #2
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    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.
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  3. #3
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    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.
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