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If we assume the Stone-Weierstrass theorem, how to prove the following statement:

If $X$ is compact Hausdorff, $ C(X \to \mathbb R)$ is the set of continuous real valued functions. If $ A \subset C(X \to \mathbb R)$ is a closed algebra without the unit and separates points, then there is a (unique) point $ x_0 \in X$ such that $ A = \{ f \in C( X \to \mathbb R ): f(x_0 ) = 0$ }.

I don't know how to get started. Any help is appreciated.

user112564
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  • If you add a non-zero constant function to $A$, the resulting algebra is dense by the Stone-Weierstrass Theorem. So $A$ is a maximal ideal. – Disintegrating By Parts Apr 16 '14 at 03:53
  • @T.A.E. Wow. I don't even know this can combined with algebra. Why is $A$ an ideal? How does multiply a function in $C(X \to \mathbb R )$ and a function in $A$ result a function in $A$? – user112564 Apr 16 '14 at 03:59
  • @T.A.E.: ... which more or less reduces the problem to http://math.stackexchange.com/questions/375400/maximal-ideals-in-the-ring-of-real-functions-on-0-1 – Nate Eldredge Apr 16 '14 at 03:59
  • @NateEldredge Thanks for linking! Reading now. – user112564 Apr 16 '14 at 04:01
  • @user112564: To amplify, let $A \oplus \mathbb{R} = {f + c : f \in A, c \in \mathbb{R}}$. Verify that $A$ is a closed algebra that contains the unit and separates points, hence $A = C(X)$. So every function $g \in C(X)$ can be written (uniquely) as $g = f + c$. This will make it easy to see that $A$ is an ideal. – Nate Eldredge Apr 16 '14 at 04:03
  • Missed the edit window. I mean, verify that $A \oplus \mathbb{R}$ is a closed algebra, etc, and hence $A \oplus \mathbb{R} = C(X)$. (For the "closed" part you may need the Hahn-Banach theorem.) And this also makes it clear that $A$ is maximal. – Nate Eldredge Apr 16 '14 at 04:15
  • @NateEldredge Thanks a lot! But I feel it's still hard for me to establish that $A \oplus \mathbb R$ is closed... – user112564 Apr 16 '14 at 06:27
  • Hint: use Hahn-Banach to find a continuous linear functional that vanishes on A but not on 1. Suppose $f_n + c_n$ converges. Apply your linear functional. – Nate Eldredge Apr 16 '14 at 07:02
  • @NateEldredge Thanks again. But I know Hahn-Banach only allows me to extend a functional from a subspace and maintain its norm. How do I find one that vanishes on $A$ but not on $\mathbb R$? – user112564 Apr 16 '14 at 16:17

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