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Let $I=[0,1]$ and $K$ is a compact space. Then could the function space $I^K$ be submetrizable, even metrizable? In other words, in general, if $I^A$ can be submetrizable (metrizable) for some space $A$, what's condition that $A$ should satisfying?

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What topology do you have on $I^A$? If $A$ is uncountable, the product topology is not metrizable. –  Michael Greinecker Dec 30 '11 at 8:29
    
Really, I want to know the results on the topology of uniform convergence, the topology of pointwise convergence and the compact-open topology. –  Paul Dec 31 '11 at 2:56

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up vote 4 down vote accepted

If $A$ is compact, $I^A$ is metrizable with the metric being the uniform norm. That is, $d(f,g):=\sup_{a\in A} d(f(a),g(a))$.

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Do you mean that if $A$ only needs to be compact then $I^A$ has this metric $d$? –  Paul Dec 30 '11 at 2:44
    
@John Correct. In general, $X^Y$ is metrizable if $Y$ is compact and $X$ is a metric space. Hatcher's Appendix has a proof of this (Proposition A.13). –  SL2 Dec 30 '11 at 3:03
    
It's nice! Do you know other conditions on $A$ such that will make the space $I^A$ metrizable? For example $A$ is countable and so on. –  Paul Dec 30 '11 at 3:19
    
$A$ sigma-compact and locally compact Hausdorff will also guarantee that all functions from $A$ to $\mathbb{R}$ in the compact-open topology will be metrizable, and this is necessary as well, for functions with codomain the reals. –  Henno Brandsma Dec 30 '11 at 12:42
    
@Henno Brandsma, could you give me a proof for it? –  Paul Dec 31 '11 at 1:37

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