TITLE: Choosing a subalgebra that separates points
QUESTION [1 upvotes]: Recall the (complex) Stone Weierstrass theorem:
Let $X$ be a compact Hausdorff space and let $C(X)$ denote the $\ast$-algebra of continuous maps $X \to \mathbb C$. Then any $\ast$-subalgebra of $C(X)$ is dense in $C(X)$ if and only of $A$ separates points.
Is it possible to choose some number (hopefully a countable number) of $f \in C(X)$ such that the algebra generated by these $f$ is dense in $C(X)$?
Let $1$ denote the constant $1$ function. Of course $1$ must be in this set. If for every $x\in X$ we pick an $f_x\in C(X)$ such that $f_x(x) = 2$ then it looks like the algebra generated by $1$ and $\{f_x\}_{x \in X}$ is dense (or am I missing something?). So it looks like it's always possible to choose some functions that generate $C(X)$ but can it be improved to be either countable or even finite?

REPLY [0 votes]: The subalgebra generated by a family of functions $f_i, i \in I$ is dense iff the map $X \to \mathbb{R}^I$ induced by the functions $f_i$ is injective. For any particular choice of $I$ you can find an $X$ that does not have this property because it has cardinality strictly larger than that of $\mathbb{R}^I$. 
In particular, if $X$ has cardinality strictly larger than $\mathbb{R}$ then it cannot inject into $\mathbb{R}^n$ for any positive integer $n$, so no finite family of functions can be dense in $C(X)$. But there are counterexamples even with a bound on the cardinality: any injection of $X$ into $\mathbb{R}^n$ is necessarily an embedding, and in particular if one exists then $X$ is second-countable (equivalently, metrizable). An example of a compact Hausdorff space with cardinality $\mathbb{R}$ which is not second-countable (equivalently, not metrizable) is the one-point compactification of $\mathbb{R}$ with the discrete topology.
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