2
$\begingroup$

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.

$\endgroup$
8
  • 2
    $\begingroup$ 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 $\min\{d_X,1\}$. $\endgroup$ 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 :) $\endgroup$
    – AIM
    Mar 18, 2020 at 18:55
  • 1
    $\begingroup$ 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
    Mar 19, 2020 at 6:23
  • 1
    $\begingroup$ 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
    Mar 19, 2020 at 6:28
  • 1
    $\begingroup$ 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
    Mar 19, 2020 at 10:59

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Wow, thanks for the great answer Henno. I really appreciate all the references! $\endgroup$
    – AIM
    Mar 23, 2020 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.