Stone-Weierstrass Theorem

Theorem
Let $X$ be a compact topological space.

Let $\struct {\map C {X, \R}, \times, \norm \cdot_\infty}$ be the Banach algebra of real-valued continuous functions on $X$.

Let $\AA$ be a subalgebra of $\map C {X, \R}$.

Let $\AA$ be such that it separates points of $X$, that is:
 * for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.

Suppose that $1 \in \AA$.

Then the closure $\overline \AA$ of $\AA$ is equal to $\map C {X, \R}$.

Proof
Let $\struct {C', \norm {\,\cdot\,}_{C'} }$ be the dual space of $\struct {\map C {X, \R}, \norm {\,\cdot\,}_\infty }$.

In view of the spanning criterion, it suffices to show that:
 * $\forall \ell \in C' : \ell \restriction_\AA = 0 \implies \ell = 0$

Let $w^\ast$ be the weak-$\star$ topology on $C'$.

Let $B' \subseteq C'$ be the closed unit ball, i.e.:
 * $B' := \set { \ell \in C' : \norm \ell_{C'} \le 1}$

By Banach-Alaoglu Theorem, $B'$ is $w^\ast$-compact.

Let:
 * $U := \set { \ell \in B' : \ell \restriction_\AA = 0}$

Then $U$ is also a $w^\ast$-compact convex subset of $C'$.

Let $\map E U$ be the set of extreme points of $U$.

Lemma
Each $\ell \in \map E U$ is an evaluation, i.e. there exists an $x \in X$ such that:
 * $\forall f \in \map C {X, \R} : \map \ell f = \map f x$

Proof of Lemma
there exists an $\ell \in C' \setminus \set 0$ such that:
 * $\forall a \in \AA : \map \ell a = 0$

In particular, $U \ne \O$, since $\frac \ell {\norm \ell_{C'} } \in U$.

Thus, by Krein-Milman Theorem, there exists an $L \in \map E U$.

By the above lemma, $L$ is an evaluation.

This is a contradiction, since for $1 \in \AA$:

Thus the theorem follows.

Also see

 * Weierstrass Approximation Theorem, of which the Stone-Weierstrass Theorem is a generalization.