Any idea how to do this problem:
If $X$ is a compact Hausdorff space and $A$ a subalgebra of $C(X)$ , where $C(X)$ is the algebra of all continuous functions, such that $A$ contains the constant functions. Show that $A$ is dense in $C(X)$ iff $A$ separate points in X.
Best Answer
As you said in the comments, one direction is Stone-Weierstrass. For the other direction, suppose $A$ is a dense subalgebra of $C(X)$, and let $x,y\in X$. Take any function $f\in C(X)$ such that $f(x)\neq f(y)$ (e.g., using Urysohn's Lemma).
By density of $A$, there exists $g\in A$ such that $\Vert g-f\Vert_{\infty}<|f(x)-f(y)|/2$, where $\Vert\cdot\Vert_\infty$ is the uniform norm. Then $$|g(x)-f(x)|\leq\Vert g-f\Vert_\infty<|f(x)-f(y)|/2$$ so $$|g(y)-f(x)|\geq |f(x)-f(y)|-|f(y)-g(y)|>|f(x)-f(y)|/2,$$ thus $g(x)\neq g(y)$, so $A$ separates points of $X$.