I would like an easy way to show that if $X$ is a compact metric space, then the group $Homeo(X)$ of homeomorphisms of $X$ with the compact-open topology (or the topology of uniform convergence, they are the same in this case) is second countable.
I have some ideas but I don't like them really much.
My first idea was to construct the countable base. If $m$ and $n$ are fixed in $\mathbf{N}$, there is a finite number $M$ (resp $N$) of balls of radius $\frac{1}{m}$ (resp. $\frac{1}{n}$) which cover $X$. Then if for each of these balls of radius $\frac{1}{m}$, I choose one of radius $\frac{1}{n}$ (there are $N^M < +\infty$ possibilities), I can choose (if it exists) some $f \in Homeo(X)$ such that the center of each ball is sent in the chosen corresponding ball... Then if I take all the balls of $Homeo(X)$ centered in these $f$, and with a radius $\frac{1}{p}$ for $p \in \mathbf{N}$, this is countable and we could show that this forms a countable base of topology... But I actually also have to take an $f$ with some uniform convergence, so this idea is not a good idea :D
Another way to do this is to cite some well-known results. It's easy to see that a compact metric space is second countable, and I have found some book that say "It is well known that the space of continuous function from $X$ to $Y$ with the compact-open topology has a countable base when $X$ and $Y$ are locally compact and have countable base. I could use that result but I would like some references... (and this is perhaps a strong result for my easy case).
I would like the reader of my text to be convinced that this result is true. If you know one, could you tell me the easiest way you know to do it?
Thank you in advance,
Nicolas
