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Supose $X$ is a compact Hausdorff topological space and $Y \subseteq X$ a dense subset. If $S$ is a set of functions in $C(X)$ such that separates points in $Y$ then the Stone-Weierstrass theorem still hold?

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  • $\begingroup$ If memory serves, Stone-Weierstrass for $C(X\rightarrow\mathbb{C})$ requires self-adjointness as well, which suggests that your conjecture is false. But I'm having trouble constructing a counterexample. $\endgroup$ – Jacob Manaker Aug 11 '17 at 0:38
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No, this never works unless $Y$ is all of $X$. Indeed, suppose $x\in X\setminus Y$ and $y\in Y$. Consider the subalgebra $A$ of $C(X)$ consisting of all functions $f$ such that $f(x)=f(y)$. Then $A$ is a closed *-subalgebra of $C(X)$ and it separates points of $Y$, but it is not all of $C(X)$.

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