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If $X$ is compact Hausdorff and $A$ a closed subalgebra ( a vector space and closed under multiplication) of $C( X \to \mathbb R)$ the set of continuous real valued functions which separates points, and $A$ does not contain the unit. How can I prove that that there must exists a $x_0 \in X$ such that $f(x_0) = 0$ for all $f \in A$.

This is a part of my bigger problem. I got stuck here. My approach is if for any $x \in X$, there exists a $f_x \in A$ such that $ f_x (x) \neq 0$, then we have an open neighbourhood of $x$ that $f$ does not equal to $0$ on it. Since $X$ is compact we can have a finite cover of $X$ of this open sets. Thus we have $f_1,...f_n$ each not equal to $0$ on an open set which the union covers $X$. And let $ f = f_1^2+...+f^2$ we have $ f > 0$ for all $X$, and $f \in A$.

This is as far as I can get. I want to take $ f * \frac{1}{f} = 1$ to get a contradiction but I don't know how $ \frac{1}{f}$ is in $A$.

Any help is appreciated.

user112564
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1 Answers1

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(I'm assuming that your $A$ is closed with respect to the supremum norm)

Take your $f$. As $f$ is continuous, its image is a compact subset of $(0,\infty)$. In particular, $e>f(x)>d>0$ for some fixed $e,d$ and all $x\in X.

Let $\{p_n\}$ be a sequence of polynomials on $[d,e]$ that converge uniformly to $t\mapsto1/t$. with the additional property that $p_n(0)=0$ for all $n$ (this can be achieved by approximating a continuous function on $[0,e]$ that his $0$ at $0$ and that agrees with $1/t$ on $[d,e]$). As $A$ is an algebra, $p_n(f)\in A$ for all $n$. As the convergence is uniform, $\{p_n(f)\}$ is Cauchy in $A$ and it converges to $1/f$.

Then $1\in A$, and that is your contradiction. Thus there exists $x_0\in X$ such that $f(x_0)=0$ for all $f\in A$.

Martin Argerami
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