Theorem 7.4.6 (Stone-Weierstrass (Complex)).label Let $X$ be a compact Hausdorff space and $A \subset C(X; \complex)$ be a closed subalgebra. If
- (S)
For each $x, y \in X$, there exists $f \in A$ such that $f(x) \ne f(y)$.
- (*)
For each $f \in A$, the complex conjugate $\ol f$ is also in $A$.
then exactly one of the following holds:
- (1)
There exists $x_{0} \in X$ such that $A = \bracs{f \in C(X; \complex)| f(x_0) = 0}$.
- (2)
$A = C(X; \complex)$.
Proof, [Theorem 4.51, Fol99]. By (*), for each $f \in A$, $\text{Re}(f), \text{Im}(f) \in A$ as well. Let $A_{sa}= A \cap C(X; \real)$, then $A_{sa}\subset C(X; \real)$ satisfies (S) of the Stone-Weierstrass Theorem, so 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)
$A = C(X; \real)$.
If (1) holds, let $B = \bracs{f \in C(X; \real)| f(x_0) = 0}$. If (2) holds, let $B = C(X; \real)$. For any $f \in B$, $f = \text{Re}(f) + \text{Im}(f)$, where $\text{Re}(f), \text{Im}(f) \in A_{sa}\subset A$. Therefore $f \in A$, and $A = B$.$\square$
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