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Let $X$ be a separable metric space and $Y$ be a second-countable $\sigma$-compact Hausdorff space. Then the compact-convergence (compact-open) topology on $C(Y,X)$ is metrizable with metric $$ d(f,g):= \sum_{n =0}^{\infty} \frac1{2^n} \sup_{y \in K_n} \min\left\{d_X(f(y),g(y)),1\right\}, $$ where $\{K_n\}_{n \in \mathbb{N}}$ is a countable compact cover of $Y$ and $d_X$ is the metric on $X$. Moreover, if $X$ is Banach then $C(Y,X)$ is a Fréchet space.

This is easy to show, but I'm looking for a reference to this result. Thanks in advance.

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    $\begingroup$ Corona prevents me from checking the bookshelf in my office. Just note that the formula is not completely correct, if the metric $d_X$ is not bounded the series may diverge. You should replace $d_X$ by $\min\{d_X,1\}$. $\endgroup$
    – Jochen Wengenroth
    Commented Mar 18, 2020 at 18:28
  • $\begingroup$ Haha, ya same here (not the biggest fan from working from home). Thanks for the tip, I made the modification :) $\endgroup$
    – ABIM
    Commented Mar 18, 2020 at 18:55
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    $\begingroup$ There are two points here, one very elementary, one rather subtle. The first one involves the metrisability. This is more transparent in the following version. If a uniformity is defined by a sequence $(d_n)$ of pseudometrics, then it can be be specifies by a single one. The standard ploy is to use $\sum \frac 1{2^n}\frac {d_n}{1+d_ n}$. Separability plays no role. $\endgroup$
    – user131781
    Commented Mar 19, 2020 at 6:23
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    $\begingroup$ This gives the metrisability condition in the second one. Here it is the completeness which is tricky. For this you need some version of the Kelley condition, i.e., that a function is continuous whenever its restriction to compacta is. I am not a point set topologist but flicking through my home library suggests that your conditions might not suffice. $\endgroup$
    – user131781
    Commented Mar 19, 2020 at 6:28
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    $\begingroup$ How does that work for the space of rationals where you take each $K_n$ a singleton (using some enumeration of $\mathbb{Q}$)? Don't you get the topolog of pointwise convergence that way? $\endgroup$
    – KP Hart
    Commented Mar 19, 2020 at 10:59

1 Answer 1

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According to Engelking (exercise 3.4E, which is based on a paper by Arens):

If $C(X,\Bbb R)$ (with the compact-open topology and $X$ Tychonoff) is first countable, then $X$ is hemicompact.

A Hausdorff space $X$ is hemicompact if there is a countable family $K_n$ of compact subsets of $X$ such that every compact $K \subseteq X$ is a subset of some $K_n$ (i.e. all compacta of $X$ ordered under inclusion has countable cofinality). For $X$ second-countable, hemicompactness is equivalent to local compactness.

So a space like $\Bbb Q$, which is not locally compact but is $\sigma$-compact has $C(X,\Bbb R)$ not even first countable, let alone metrisable.

But Arens showed in that same paper (ex. 4.2H in Engelking) that for hemicompact $X$ and metrisable $Y$ , $C(X,Y)$ in the compact-open topology is metrisable, using a metric like yours.

So the moral is: you need to add "locally compact" to your $Y$ (and the space then becomes hemicompact and all is well).

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  • $\begingroup$ Wow, thanks for the great answer Henno. I really appreciate all the references! $\endgroup$
    – ABIM
    Commented Mar 23, 2020 at 12:44

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