# Metrizability of topology of compact convergence

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.

• 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 $\min\{d_X,1\}$. Mar 18, 2020 at 18:28
• Haha, ya same here (not the biggest fan from working from home). Thanks for the tip, I made the modification :)
– AIM
Mar 18, 2020 at 18:55
• 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. Mar 19, 2020 at 6:23
• 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. Mar 19, 2020 at 6:28
• 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? Mar 19, 2020 at 10:59

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).

• Wow, thanks for the great answer Henno. I really appreciate all the references!
– AIM
Mar 23, 2020 at 12:44