Fall 2018 - Problem 2

Solution.

1. In light of Stone-Weierstrass, it suffices to show that $A$ vanishes nowhere and separates points.

   Let $x \in X$. Then by assumption, there exists $y \in X$ different from $x$, so let $\delta = \rho\p{x, y} > 0$. By density of $D$, there exists $z \in D$ such that $\rho\p{y, z} < \frac{\delta}{2}$. Then by the triangle inequality,

   $$\rho\p{x, z} \geq \rho\p{x, y} - \rho\p{y, z} \geq \frac{\delta}{2} > 0,$$

   so in particular, $f_z\p{x} \neq 0$. To show that $A$ separates points, let $x, y \in X$ be distinct. As before, let $\delta = \rho\p{x, y} > 0$. By density, there exists $z \in D$ such that $\rho\p{y, z} < \frac{\delta}{2}$, so $\rho\p{x, z} \geq \frac{\delta}{2}$ by the same calculation as before. Thus,

   $$f_z\p{y} < \frac{\delta}{2} \leq f_z\p{x},$$

   so $f_z$ separates $x$ and $y$, so $A$ is dense in $C\p{X}$.

2. Since $X$ is a compact metric space, it is separable. Thus, let $D$ be a countable dense subset of $X$ so that $A$ (as in the previous part) is dense in $C\p{X}$. Notice that elements in $A$ have the form

   $$\sum_{k=1}^n c_k \prod_{j=1}^{m_k} f_{z_{jk}}$$

   where $z_{jk} \in D$, $n \in \N$, and $c_k \in \R$. Hence, it is easy to see that if we replaced $c_k$ with elements in $\Q$, then the resulting set of functions $B$ is dense in $A$, hence dense in $C\p{X}$. Since $\Q$ is countable and $D$ is countable, it follows that $B$ is also countable, so $C\p{X}$ is separable.