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In my notes we prove Stone-Weierstrass which tells us that if we have a subalgebra $A$ of $C(X)$ such that it separates points and contains the constants then its closure (w.r.t. $\|\cdot\|_\infty$) is $C(X)$.

A few chapters later there is a lemma that if $X$ is a compact metric space then $C(X)$ is separable. The proof constructs a subalgebra that separates points by taking a dense countable subset of $X$, $\{x_n\}$, and defining $f_n (x) = d(x,x_n)$.

Question: could we treat this as a corollary of Stone-Weierstrass and say that polynomials with rational coefficients are a subalgebra containing $1$ and separating points? Thank you.

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    @Thomas It was a joint venture : )2012-08-04

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In the hope I did understand the question correctly by now: It is a corollary of Stone-Weierstrass. Add the constant functions to the algebra you constructed and you are done.

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    @MattN. For a compact metric space you can take Lipshcitz functions in place of polynomials. That is your appropiate algebra would be $ C^{0,1}(X) $ which would be dense by Stone Weierstrass theorem. All you have to show is that $ C^{0,1}(X) $ is seperable.2013-03-15