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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: Realizing a homomorphism $\mathcal{C}(X) \to \mathbb{R}$ 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 $I\subseteq \overline I =\bigcap\limits_{p\in K} A_p$. –  Jonas Meyer Jan 9 '12 at 20:04

2 Answers 2

EDIT : The following does not answer the question, because it uses Urysohn's lemma, as pointed out in the comments. However, the only choice needed is, I think, countable dependant choice. It also gives a proof in ZF of both A and B simultaneously, in the case of compact metric space.


First, let's show that $A_p$ is a maximal ideal. Let $J$ by an ideal of $C(X)$ such that $A_p \subsetneq J$. Let's show that necessarily $J=C(X)$. Let $g \in J$, $g \notin A_p$. So $g(p) \neq 0$, and we can write $$1 = \frac{g(p)-g}{g(p)} + \frac{g}{g(p)}$$ The first term vanishes at $p$, so is in $A_p$, and the second term belongs to $J$. It follows that the function $1 \in J$, and thus $J=C(X)$.

Now, let $M$ be a maximal ideal of $C(X)$. Suppose that for all $p \in X$, there exists $g_p \in M$ such that $g_p(p) \neq 0$. By continuity, $g_p \neq 0$ on some neighborhood $V_p$ of $p$. By compacity of $X$, there exists $p_1, p_2, \dots, p_n$ such that $$X=V_{p_1} \cup V_{p_2} \cup \dots \cup V_{p_n}.$$ Let $h_1, h_2, \dots, h_n$ be a partition of unity subordinate to $V_{p_1}, \dots, V_{p_n}$ (This follows from Urysohn's lemma). Now write $$1= h_1 + \dots h_n = \frac{h_1}{g_{p_1}}g_{p_1} + \frac{h_2}{g_{p_2}}g_{p_2} + \dots + \frac{h_n}{g_{p_n}}g_{p_n}.$$ The functions $h_j/g_{p_j}$ are continuous in $X$, since the support of each $h_j$ is contained in $V_{p_j}$, and $g_{p_j} \neq 0$ in $V_{p_j}$. It follows that $1 \in M$, a contradiction.

We have shown that there exists a $p \in X$ such that for all $g \in M$, $g(p)=0$, i.e. $M \subseteq A_p$. By maximality of $M$, $M=A_p$.

The second part of the proof also shows that $B$ holds.

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The usual "onion shell" proof of Urysohn's lemma uses countable dependent choice for choosing the shells inductively, see also here. In Läuchli's model of ZF Urysohn's lemma breaks. Good and Tree in Example 2.4 on p.81 of Continuing horrors of topology without choice show that it is consistent with ZF that there is a compact Suslin line with the property that all real-valued functions are constant. –  t.b. Jan 10 '12 at 3:18
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Malik, as t.b. says Urysohn's lemma does need some choice. Just a little bit, but still some choice is needed. Not all uses of choice come as Zorn's lemma or well orders and such. It can get very subtle. –  Asaf Karagila Jan 10 '12 at 5:08
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@t.b. : ah I see now that I had missed something. Thanks for the comment and the references. However, if I'm not mistaken again, Urysohn's lemma is the only choice needed in the above proof, so this gives a proof of both A and B in ZF, in the case of compact metric spaces, right? –  Malik Younsi Jan 10 '12 at 13:43
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@t.b.: Note that C. Good has made Continuing horrors of topology without choice freely available here. –  Brian M. Scott Jan 12 '12 at 7:35
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@Malik Thanks a lot for your answer; +1. It was interesting to see another approach and I've learned about AC in Urysohn's lemma, which is good to know. I might be mistaken, but I still think there is one more usage of AC apart from Urysohn, namely, you have to choose for each $p\in X$ a function $g_p\in M$ such that $g_p(p)\ne 0$. (I also apologize for reacting very late to your post, but I've been busy in the last few days and I also have some health problems at the moment.) –  Martin Sleziak Jan 19 '12 at 18:34

Let me mention here at least some case, which I was able to solve. (Hopefully, I have not missed some hidden use of AC.)


$\newcommand{\mc}[1]{\mathcal{#1}}\newcommand{\inv}[1]{{#1}^{-1}}\newcommand{\emps}{\emptyset}\newcommand{\Invobr}[2]{\inv{#1}(#2)}$ We say that an ideal $A\subseteq C(X)$ is fixed, if there is a point $p$ such that $f(p)=0$ for each $f\in A$, i.e., there is a point $p$ such that $A\subseteq A_p$.

Claim 1 (ZF) If $X$ is compact, then every proper ideal in $C(X)$ is fixed.

Proof. Let $A$ be a proper ideal in $C(X)$. Then $\{\Invobr f0; f\in A\}$ is a system, which has finite intersection property. (Suppose that $\Invobr f0\cap\Invobr g0=\emps$ for some $f,g\in A$. This implies that $h=f^2+g^2$ is a non-zero function such that $h\in A$. Consequently $1\in A$ and $A=C(X)$, a contradiction.)

By compactness, the above system has a non-empty intersection, i.e., there exists a point $p\in\bigcup\limits_{f\in A} \Invobr f0$. This means that $A\subseteq A_p$. $\square$

The basic idea used in the following proof is that in a complete regular space every closed set is an intersection of zero-sets.

Claim 2 (ZF) If $X$ is completely regular, and every proper ideal in $C(X)$ is fixed, then $X$ is compact.

Proof. Let $\mc K$ be a system of closed subsets of $X$ which has finite intersection property.

Let $$A=\{f\in C(X); \text{ there exists a finite set }\mc F\subseteq\mc K\text{ such that }\Invobr f0\supseteq \bigcap\mc F\}.$$

Then $A$ is an ideal in $C(X)$. (It is easy to see that $f\in A$ implies $gf\in A$ for any $g\in C(X)$. If $f,g\in A$ then $\Invobr f0\supseteq\bigcap\mc F_1$ and $\Invobr g0\supseteq\bigcap\mc F_2$ for some finite sets $\mc F_1$ and $\mc F_2$. Thus for $h=f+g$ we get $\Invobr h0\supseteq\bigcap (\mc F_1\cup\mc F_2)$.) Clearly $1\notin A$ and thus $A$ is proper.

Using complete regularity we can show that $\bigcap\limits_{f\in A} \Invobr f0 = \bigcap \mc K$. (It is clear from the definition of $A$ that $\bigcap\mc K\subseteq\bigcap\limits_{f\in A} \Invobr f0$. On the other hand if $p\notin\bigcap\mc K$, then we have a closed set $K\in\mc K$ such that $p\notin K$. Since $X$ is complete regular, there exists a function $f\in C(X)$ such that $f(p)=1$ and $f|_K=0$. This function belongs to $A$. Thus we get that $p\notin \bigcap\limits_{f\in A} \Invobr f0 $.)

By the assumption, the intersection $\bigcap\limits_{f\in A} \Invobr f0$ is non-empty. So we see that $\bigcap\mc K$ is non-empty. $\square$

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