# Is the space of continuous functions from a compact metric space into a Polish space Polish?

Let $K$ be a compact metric space, and $(E,d_E)$ a complete separable metric space. Define $C:=C(K,E)$ to be the continuous functions from $K$ to $E$ equipped with the metric $d(f,g)=\sup_{k\in K}\ d_E (f(k),g(k))$. Is the space $C$ separable?

The result is true when $E$ is the real line; this is Corollary 11.2.5 in Dudley's book Real Analysis and Probability.

The result is also true when $K=[0,1]$ (if I'm not being too careless) by considering $C$ as a subspace of the Skorohod space $D_E[0,1]$, which is complete and separable by Theorem 5.6 in Ethier and Kurtz's book Markov Processes: Characterization and Convergence.

For general $K$, it is not so obvious how to find an explicit countable dense set in $C$, but I suspect one could modify Ethier and Kurtz's approach and get a proof.

But surely this result is known, and stated in some book? I've searched through my library without success.

Update: This result is also Theorem 2.4.3 of S. M. Srivastava's book A Course on Borel Sets. His proof is the same as Kechris's. I have also found an alternative, but false, published proof using the "fact" that $C(K,E)$ is $\sigma$-compact. Beware!

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We have the following. Fix $X, (Y,d)$ polish spaces where $d$ is some bounded metric. Topologise $C^{0}(X,Y)$ by the metric $d(f,g)=sup_{x\in X}d(f(x),g(x))$. Then one can tweak Kechris' proof to show, that the subspace $S$ of uniformly continuous maps with bounded images, is Polish.
Is it possible to show that $C^{0}(X,Y)$ can be generated by $S$, using point-wise limits of $\omega$-sequences of functions? This would be a useful result.
This is false: take $X=\mathbb N$ with the discrete distance, and $Y=\{0,1\}$ with its usual distance. Clearly both satisfy your assumptions, yet the set you denote by $C^0(X,Y)$ is not separable: pick two different subsets $A,B$ of $\N$, and consider the distance between their characteristic functions... – Julien Melleray Dec 6 '10 at 8:10