Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 21H / Solution

The real Stone-Weierstrass theorem states that if $X$ is compact Hausdorff and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$. Let $B$ be its uniform closure, still an algebra over a field. If $h\in B$ and $M\ge \Vert h{\Vert }_{\infty }$, scale the permitted polynomial approximation of absolute value on $[-1,1]$ to $[-M,M]$. Applying these polynomials to $h$ proves $|h|\in B$. Thereforebelong to $B$: it is a lattice.

Fix $f\in C(X,\mathbb{R})$ and $\epsilon >0$. For every $x,y\in X$, point separation and an affine adjustment give $g_{x,y}\in A$ with $g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$; if $x=y$, a constant suffices. For fixed $x$, the open sets where $g_{x,y}>f-\epsilon$ cover $X$. Select finitely many and take their maximum $h_x\in B$. Then $h_x>f-\epsilon$ everywhere and $h_x(x)=f(x)$. By continuity, $h_x<f+\epsilon$ on some neighborhood $V_x$ of $x$. Choose finitely many $V_x$ covering $X$ and take the minimum of the corresponding $h_x$. The result $h\in B$ satisfies $f-\epsilon <h<f+\epsilon$ everywhere. Since $B$ is closed and $\epsilon$ arbitrary, $f\in B$. Thus $\overline{A}=C(X,\mathbb{R})$, proving the theorem.