Let $(X,d)$ be a separable complete metric space. Let $K$ be a compact metric space.

I denote by $C(K,X)$ the set of continuous functions $f:K\rightarrow X$ with the metric $\rho:C(K,X)\times C(K,X)\rightarrow\mathbb{R}$

$$\rho(f,g) = \sup_{k\in K} d(f(k),g(k))$$

*This is the counterpart of the infinity norm $\|f-g\|_\infty$.

It is not hard to show that $\rho$ is indeed a metric and that the sup always exists.

I ask whether it's true that $(C(K,X),\rho)$ a separable metric space? (My guess is yes, note that when $X=\mathbb{R}$ it's not hard to prove)


It is true. It is a well-known fact that if $K$ is a compact space and $(X,d)$ a metric space, then the metric topology on $C(K,X)$ (supremum-metric!) coincides with the compact-open topology on $C(K,X)$. See any textbook on topology treating function spaces. Note that the compact-open topology has as a subbase all sets $$ M(C, W) = \{ f \in C(K,X) \mid C \subset K \text{ compact }, W \subset X \text{ open } \}. $$ Now let $K$ be compact metrizable and $(X,d)$ separable. Then $K$ and $X$ have countable bases $\mathcal{K}$ and $\mathcal{X}$, respectively. We may assume that $\mathcal{K}$ and $\mathcal{X}$ contain all finite unions of their members. Then the set $\mathcal{B} = \{ M(\overline{V}, W) \mid V \in \mathcal{K}, W \in \mathcal{K} \}$ is countable. We shall show that it is a subbase for the compact-open topology. So let $C \subset K$ compact, $U \subset X$ open and $f \in M(C,U)$. Since $C$ is compact, there exist finitely many $V_i \in \mathcal{K}$ such that $\overline{V_i} \subset f^{-1}(U)$ and $C \subset V = \bigcup V_i$. We have $V \in \mathcal{K}$ and $f(\overline{V}) \subset U$. Since $\overline{V}$ is compact, we can find finitely many $W_i \in \mathcal{X}$ such that $f(\overline{V}) \subset W = \bigcup W_i \subset U$. We have $W \in \mathcal{X}$, hence $M(\overline{V},W) \in \mathcal{B}$ and $f \in M(\overline{V},W) \subset M(C,U)$.


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