Space of Complex-Valued Continuous Functions on Compact Hausdorff Space is Separable iff Space is Metrizable

Theorem

Let $K$ be a compact Hausdorff space.

Let $\map \CC K$ be the space of complex-valued continuous functions of compact Hausdorff space.

Then $\map \CC K$ is separable if and only if the topology on $K$ is metrizable.

Proof

Necessary Condition

Suppose that $\map \CC K$ is separable.

Then by Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{\map \CC K^\ast}, w^\ast}$ is metrizable.

For each $k \in K$, define $\delta_k : \map \CC K \to \C$ by:

$\map {\delta_k} f = \map f k$

for each $f \in \map \CC K$.

In Point Evaluations are Continuous Linear Functionals on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space, it is established that $\delta_k$ is a continuous linear functional with $\norm {\delta_k}_{\map \CC K^\ast} = 1$.

So we can define $\delta : K \to \struct {B^-_{\map \CC K^\ast}, w^\ast}$ by:

$\map \delta k = \delta_k$

for each $k \in K$.

We show that $\delta$ is injective and continuous.

If $K$ is a singleton, there is nothing to show, so suppose that $K$ is not a singleton.

Let $k, \ell \in K$ have $k \ne \ell$.

By Urysohn's Lemma, there exists a continuous mapping $f : K \to \closedint 0 1$ with $\map f k = 0$ and $\map f \ell = 0$.

So $\map {\delta_k} f \ne \map {\delta_\ell} f$ for this $f \in \map \CC K$.

So $\delta_k \ne \delta_\ell$ if $k \ne \ell$.

We now want to show that $\delta$ is continuous.

By Continuity in Initial Topology, it is enough to show that for each $f \in \map \CC K^\ast$, $f^\wedge \circ \delta \in \map \CC K$ is continuous.

For $k \in K$, we have:

\(\ds \map {\paren {f^\wedge \circ \delta} } k\)

\(=\)

\(\ds \map {\delta_k} f\)

\(\ds \)

\(=\)

\(\ds \map f k\)

hence $f^\wedge \circ \delta = f \in \map \CC K$.

So $\delta$ is continuous mapping.

So $\delta : K \to \struct {\map \delta K, w^\ast}$ is a continuous bijection.

From Weak-* Topology is Hausdorff, we have that $\struct {\map \delta K, w^\ast}$ is Hausdorff.

So $\delta$ is a continuous bijection between a compact space $K$ and a Hausdorff space $\struct {\map \delta K, w^\ast}$.

From Continuous Bijection from Compact to Hausdorff is Homeomorphism, this map is a homeomorphism.

So $K$ is homeomorphic to $\struct {\map \delta K, w^\ast}$.

Since $\struct {B^-_{\map \CC K^\ast}, w^\ast}$ is metrizable, so is $\struct {\map \delta K, w^\ast}$ from Subspace of Metric Space is Metric Space.

So $K$ has a metrizable topology.

$\Box$

Sufficient Condition

Suppose that $K$ is metrizable.

From Compact Metric Space is Separable, $K$ is separable.

Let $\sequence {k_n}_{n \in \N}$ be a countable dense subset of $K$.

For $q \in \Q$ and $n \in \N$ define:

$\map {g_{n, q} } k = \max \set {q - \map d {k, k_n}, 0}$

for each $k \in K$.

From Distance Function of Metric Space is Continuous, the map:

${k, \ell} \mapsto \map d {k, \ell}$

is continuous.

From Horizontal Section of Continuous Function is Continuous, we have that:

$k \mapsto \map d {k, k_n}$

is continuous for each $n \in \N$.

From Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous, we have that:

$k \mapsto q - \map d {k, k_n}$

is continuous for each $n \in \N$ and $q \in \Q$.

Hence from Maximum Rule for Continuous Functions, we have that $g_{n, q}$ is continuous for each $n \in \N$ and $q \in \Q$.

Now let:

$\FF = \set {g_{n, q} : n \in \N, \, q \in \Q} \cup \set {1_K}$

where $1_K$ is the constant $1$ function on $K$.

Define:

$\ds A = \span \set {\prod_{j \mathop = 1}^n f_j : \set {f_1, \ldots, f_n} \in \FF, \, n \in \N}$

We will use Stone-Weierstrass Theorem, to show that $\map \cl A = \map \CC K$.

Clearly $A$ is a subalgebra, is closed under conjugation and contains $1_K$.

We show that $A$ separates points.

Let $k, \ell \in K$ with $k \ne \ell$.

We argue there exists $n \in \N$ such that $\map d {k, k_n} \ne \map d {\ell, k_n}$.

Indeed if we had $\map d {k, k_n} = \map d {\ell, k_n}$, then we would have $\map d {\ell, k} = 0$ from density, and hence $\ell = k$.

Without loss of generality suppose that $\map d {k, k_n} < \map d {\ell, k_n}$.

Then there exits $q \in \Q$ such that:

$\map d {k, k_n} < q < \map d {\ell, k_n}$

Then we have $\map {g_{n, q} } k \ne 0$ and $\map {g_{n, q} } \ell = 0$.

So $g_{n, q}$ separates points.

So from the Stone-Weierstrass Theorem, we have that $\map \cl A = \map \CC K$.

From Closed Linear Span of Countable Set in Topological Vector Space is Separable, we have that $\map \cl A$ is separable.

So $\map \CC K$ is separable.

$\blacksquare$