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 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:
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.
=== 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 $\ell = k$. 
 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 for compact spaces, 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.
Category:Space of Continuous Functions on Compact Hausdorff Space
Category:Metrizable Topologies