Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem is

If $\mathscr{A}$ is an algebra of continuous real-valued functions on a compact set $K$ that separates points, then either
1. $\overline{\mathscr{A}}=\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous}\right\}$
2. $\overline{\mathscr{A}}=\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous and }f(p)=0\right\}$ for some $p\in K$.

I think I have part of the proof—I can show that under the assumption that (1) is not true then $\overline{\mathscr{A}}\subseteq\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous & }f(p)=0\right\}$. Unfortunately this is not very far. Here's what I figure so far.

Suppose $\overline{\mathscr{A}}\ne\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous}\right\} $. Then $\mathscr{A}$ does not vanish nowhere, so there exists some $p\in K$ such that $f(p)=0$ for all $f\in\mathscr{A}$. Let $\left\{ f_{n}\right\} $ be a sequence in $\mathscr{A}$ such that $f_{n}\rightarrow f$ uniformly. Then $f_{n}(p)\rightarrow0$ so $f(p)=0$, and since the convergence is uniform, $f$ is continuous. Hence $f\in\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous & }f(p)=0\right\} $ and so $\overline{\mathscr{A}}\subseteq\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous & }f(p)=0\right\} $.

Now I need to show that $\overline{\mathscr{A}}\supseteq\left\{ f:K\rightarrow\mathbb{R}|f\text{ is continuous & }f(p)=0\right\}$, so I figure I should let $f \in \left\{f:K\rightarrow\mathbb{R}|f\text{ is continuous & }f(p)=0\right\}$ and see what happens but I can't figure out where to go next. And I can't figure out how to that if (2) doesn't hold then (1) does.

share|cite|improve this question
I suggest restructuring your proof. You don't have to blindly choose between (1) & (2); in this case, you know exactly what property of $\mathscr{A}$ determines (1) vs (2), it's whether or not it vanishes anywhere. Prove both cases similarly, one is "if $\mathscr{A}$ vanishes at $p$ ... choose any continuous function $f$ with $f(p) = 0$ ... show $f \in \overline{\mathscr{A}}$", the other is "if $\mathscr{A}$ does not vanish at any point (i.e., for every $p$, there is a $g \in \mathscr{A}$ so that $g(p) \neq 0$) ... choose any continuous function $f$ ... show $f \in \overline{\mathscr{A}}$" – BaronVT Mar 13 '13 at 6:28
@BaronVT Thanks that already helps immensely! I'm still stuck on one part though. So we have either $\mathscr{A}$ vanishes or it does not. If it does not then Stone-Weierstrass gives us $\mathscr{A}=C(K)$. Otherwise, here I'm still stuck. I can show that if $\mathscr{A}$ vanishes at $p$ then a sequence $f_n$ in $\mathscr{A}$ which converges to $f$ vanishes at $p$ and is continuous, but I can't show the inclusion in the other direction. – crf Mar 13 '13 at 7:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.