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I know that this question has been asked before (for example here), but I am looking for a different answer.

Let $X$ be a compact Hausdorff space and consider the commutative Banach algebra $C(X):=\{\text{continuous functions $f\colon X\to\mathbb{C}$}\}$ endowed with the supremum norm. I want to prove that every closed ideal $I\subset C(X)$ is of the form $I_{A}:=\{f\in C(X) \ | \ A\subset f^{-1}\{0\}\}$ for some closed subset $A\subset Y$. I proved that $I_{A}$ is a closed ideal for any subset $A\subset X$. The exercise gives a hint that I should prove and use the following:

Given $f\in I$ and $\varepsilon\in(0,1)$, there exists a continuous map $u\colon X\to\mathbb{C}$ such that

  • $u\in I$,
  • $u(x)\in[0,1]$ for all $x\in X$,
  • $u(x)=0$ whenever $|f(x)|\leq\varepsilon$,
  • $u(x)=1$ whenever $|f(x)|\geq1$.

I managed to prove this hint, but still I'm having trouble with the exercise. As the link above suggests, I tried to prove that $I=I_{A}$ where $$A:=\bigcap_{g\in I}g^{-1}\{0\}.$$ Any suggestions on how to prove this using the hint above would be greatly appreciated.

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Calculix
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  • That hint seems a little strange; for instance, $\epsilon$ is pretty pointless because you can actually achieve such a $u$ with $\epsilon=0$. Maybe it was meant to instead say $u(x)=0$ whenever $f(x)=0$ and $u(x)=1$ whenever $|f(x)|\geq\epsilon$. – Eric Wofsey Nov 25 '19 at 23:12
  • @EricWofsey I think that you need $\varepsilon>0$ in order to prove that $u$ (which exists because of Urysohn’s lemma) lies in the ideal $I$. – Calculix Nov 25 '19 at 23:40
  • Actually I guess I misread it to say $u(x)$ iff $|f(x)|\leq \epsilon$. As written (with only one half of the iff), making $\epsilon$ smaller makes the conclusion weaker, not stronger! But it still makes no sense to me why it would be helpful to introduce such an $\epsilon$ instead of just asking for $u(x)=0$ when $f(x)=0$. – Eric Wofsey Nov 25 '19 at 23:47

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