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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.2011-12-30
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    Really, I want to know the results on the topology of uniform convergence, the topology of pointwise convergence and the compact-open topology.2011-12-31

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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 $d$?2011-12-30
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    @John Correct. In general, $X^Y$ is metrizable if $Y$ is compact and $X$ is a metric space. [Hatcher's Appendix](http://www.math.cornell.edu/~hatcher/AT/ATapp.pdf) has a proof of this (Proposition A.13).2011-12-30
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    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.2011-12-30
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    $A$ 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.2011-12-30
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    @Henno Brandsma, could you give me a proof for it?2011-12-31