Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|cite|improve this question

2 Answers 2

up vote 9 down vote accepted

Yes, it appears e.g. as Theorem 4.19 in Chapter I of Kechris' Classical Descriptive Set Theory. (The relevant page is visible in Google Books if it's not in your library.)

share|cite|improve this answer
Thanks! This is exactly what I was looking for. –  Byron Schmuland Nov 14 '10 at 4:22

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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