Proposition 7.4.2.label Let $X$ be a compact Hausdorff space and $A \subset C(X; \real)$ be a closed subalgebra, then $A$ is a lattice.
Proof. For each $f, g \in C(X; \real)$,
\[f \vee g = \frac{f + g + |f - g|}{2}\quad f \wedge g = \frac{f + g - |f - g|}{2}\]
so it is sufficient to show that for each $f \in A$, $|f| \in A$.
Let $f \in A$ and assume without loss of generality that $\norm{f}_{u} \le 1$. By Lemma 7.4.1, there exists $\seq{p_n}\subset \real[x]$ such that $p_{n}(x) \to \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$. In which case, $p_{n}(f^{2}) \in A$ and $p_{n}(f^{2}) \to \sqrt{f^{2}}= |f|$ uniformly as $n \to \infty$. Since $A$ is closed, $|f| \in A$.$\square$
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