$C(X) = C(X,{{\mathbb C}})$ is the algebra of continuous complex-valued functions on $X$, with the usual supremum norm. ${{\mathcal A}}{\sqsubseteq}C(X)$ means that ${{\mathcal A}}$ is a subalgebra of $C(X)$ which separates points and contains the constant functions. ${{\mathcal A}}{\, \mbox{\rlap{$\sqsubseteq$}     \raise 2.3pt \hbox{\scriptsize $\mathrm{c}$}}\; }C(X)$ means that ${{\mathcal A}}{\sqsubseteq}C(X)$ and ${{\mathcal A}}$ is closed in $C(X)$. $X$ has the *Complex Stone-Weierstrass Property (CSWP)* iff every ${{\mathcal A}}{\sqsubseteq}C(X)$ is dense in $C(X)$; equivalently, iff every ${{\mathcal A}}{\, \mbox{\rlap{$\sqsubseteq$}     \raise 2.3pt \hbox{\scriptsize $\mathrm{c}$}}\; }C(X)$ equals $C(X)$.

The CSWP is easily seen to be true for finite spaces. The complex analysis developed in the 1800s shows that the CSWP is false for many compact subspaces of the plane; for example, it is false for the unit circle ${{\mathbb T}}$; the classic counter-example being the algebra of complex polynomials ${{\mathcal P}}{\sqsubseteq}C({{\mathbb T}})$. These remarks are subsumed by results of W. Rudin [@RUD1; @RUD2] from the 1950s:

\[thm-rudin\] Let $X$ be any compact space.

If $X$ contains a copy of the Cantor set, then $X$ fails the CSWP.

If $X$ is scattered, then $X$ satisfies the CSWP.

If a compact space is metrizable (equivalently, second countable), then it contains a Cantor subset iff it is 