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The following result are true if we assume full axiom of choice:

  • A. If $X$ is a compact Hausdorff space, then every maximal ideal of the ring $C(X)$ has the form $A_p=\{f\in C(X); f(p)=0\}$.

  • B. If $X$ is a compact Hausdorff space, then every ideal of the ring $C(X)$ is contained in an ideal of the form $A_p=\{f\in C(X); f(p)=0\}$.

I wonder how much choice is needed. To be precise, do we get a statement equivalent to some known form of AC if we assume validity of A/B for every compact space, for every compact metric space, for every complete totally bounded metric space or for the case $X=[0,1]$?

I've tried to answer A at least partially in my answer here. If I did not make a mistake there, I've shown that in ZF the claim A holds for complete totally bounded metric space, B holds for compact spaces. I've also gathered a few relevant references in that answer. According to those references validity of A of every compact regular space is equivalent to ultrafilter theorem. According to the same book, in ZF it can be shown that A holds if and only if $X$ the form $[0,1]^I$ and B holds if and only of $X$ is compact.

As I am not experienced with working in ZF (without AC), I'll be glad if you check my work there and point out any mistakes and add any additional references/proofs/insights.

This question is also related, but not identical: https://math.stackexchange.com/questions/97603/realizing-a-homomorphism-mathcalcx-to-mathbbr-as-an-evaluation

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    This doesn't answer your question, but for B I would think of it like this. If $I$ is a proper ideal, then for all $f\in I$, $\|f-1\|\geq 1$, and therefore $\overline I\neq C(X)$. Then $\overline I$ is a closed ideal and there exists a nonempty closed set $K\subseteq X$ such that $\overline I=\{f\in C(X):f|_K=0\}$. Therefore $I\subseteq \overline I =\bigcap\limits_{p\in K} A_p$.2012-01-09

2 Answers 2