Let $X$ is a compact metric space and $Y$ is a seperable metric space.Let $\mathcal{C}(X,Y)$ denote the space of continuous function from $X$ to $Y$,with the compact-open topology.(i.e. the space's topology has as a subbasis all the sets $$ \mathcal{S}(C,U)=\{ f \in \mathcal{C}(X,Y)|f(C)\subset U \}$$ with $C$ compact in $X$ and $U$ open in $Y$.)Show that $\mathcal{C}(X,Y)$ is,a seperable metric space.

How to find a proper dense countable subset in $\mathcal{C}(X,Y)$ from the dense countable subset in $Y$?


Here is my approach: Note that $X$ is compact implies that there is a countable basis $\mathcal{U}_X$ of relative compact open sets, i.e. $U\in \mathcal{U}_X$ has the property that $\overline{U}$ is compact. By separatedness there is a countable basis $\mathcal{U}_Y$ of $Y$ (just take the balls of rational diameter around your countable subset). The claim is that the sets

$$ \mathcal{S}(\overline{U},V) $$

form a subbasis on $\mathcal{C}(X,Y)$ where $U\in \mathcal{U}_X$ and $V\in \mathcal{U}_Y$. Since there are countable many of them this would imply that $C(X,Y)$ is separable. Just take an element out of every finite intersection of $ \mathcal{S}(\overline{U},V) $. Now every open subset of $\mathcal{C}(X,Y)$ contains such an element and you are done.

Proof of claim: It suffices to show that for an arbitrary $S(C,O)$ we find finitely many $U_i\in \mathcal{U}_X$ and a $V \in \mathcal{U}_Y$ s.t.

$$ \bigcap_{i=1}^n \mathcal{S}(\overline{U_i},V)\subset \mathcal{S}(C,O) $$

Note that we find a $V \in \mathcal{U}_Y$ s.t. $V\subset O$ and therefore

$$ \mathcal{S}(C,V)\subset \mathcal{S}(C,O). $$

Let $f \in \mathcal{S}(C,V)$ then $C \subset f^{-1}(V)$. Now choose for every $c \in C$ an element $U_c \in \mathcal{U}_X$ s.t. $U_c \subset \overline{U_c} \subset f^{-1}(V)$. Now the $U_c$ cover $C$ and by compactness choose a finite subcover $U_{c_i}$. Now you have:

$$f \in \bigcap_{i=1}^n S(\overline{U_{c_i}},V)=S\left(\bigcup_{i=1}^n\overline{U_{c_i}},V\right)\subset S(C,V)\subset S(C,O). $$

This shows the claim.

I know that this seems a little overkill for your simple question but this is the best I could come up with. I hope it helps.

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  • $\begingroup$ Sorry I'm still confused why just take one point out of each $\mathcal{S}(\bar{U},V)$ then every open set contains such an element.It seems not so clear yet. $\endgroup$ – T.Jassen Apr 13 '16 at 9:52
  • $\begingroup$ @T.Jassen Since the $\mathcal{S}(\overline{U},V)$ is a subbasis take element out of every finite intersection, say $f$. Now for $W\subset \mathcal{C}(X,Y)$ open you have $$f\in \cap\mathcal{S}(\overline{U},V) \subset W $$. $\endgroup$ – Maik Pickl Apr 13 '16 at 10:14
  • $\begingroup$ @T.Jassen I editet to add the "finite intersection" part. $\endgroup$ – Maik Pickl Apr 13 '16 at 10:14
  • $\begingroup$ @T.Jassen One could also put it this way: We just proofed that $\mathcal{C}(X,Y)$ is second countable by giving a countable basis. But every second countable topological space is separable. $\endgroup$ – Maik Pickl Apr 13 '16 at 10:55
  • $\begingroup$ I know your mean.But I'm just confused why you take points out of the intersections but not in them?Maybe I misunderstand it. $\endgroup$ – T.Jassen Apr 14 '16 at 1:01

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