Theorem 7.4.5 (Stone-Weierstrass).label Let $X$ be a compact Hausdorff space and $A \subset C(X; \real)$ be a closed subalgebra. If
- (S)
For each $x, y \in X$, there exists $f \in A$ such that $f(x) \ne f(y)$.
then exactly one of the following holds:
- (1)
There exists $x_{0} \in X$ such that $A = \bracs{f \in C(X; \real)| f(x_0) = 0}$.
- (2)
$A = C(X; \real)$.
Proof, [Theorem 4.45, Fol99]. For each $x, y \in X$ with $x \ne y$, the mapping
is an algebra homomorphism, and $\pi_{xy}(A)$ is a subalgebra of $\real^{2}$. By (S) and Lemma 7.4.4, there exists no $x, y \in X$ with $x \ne y$ such that $\pi_{xy}(A) = \bracs{0}$, so exactly one of the following holds:
- (1)
There exists $x_{0} \in X$ such that $f(x_{0}) = 0$ for all $f \in A$.
- (2)
For all $x, y \in X$ with $x \ne y$, $\pi_{xy}(A) = \real^{2}$.
If (1) holds, let $B = \bracs{f \in C(X; \real)| f(x_0) = 0}$. If (2) holds, let $B = C(X; \real)$. By Proposition 7.4.2, $A$ is a lattice. In both cases, $A$ satisfies condition (C) of Proposition 7.4.3 for all $f \in B$, therefore $A = B$.$\square$
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