$X$ compact Hausdorff space, characterize the maximal ideals of $C(X)$

Question

I know this question has been asked before, but I think I'm very close to a new solution and wanted to know if it is a viable approach.

Let $C(X)$ be the ring of continuous functions $X \rightarrow \mathbb{C}$ where $X$ is compact Hausdorff, and let $\mathfrak M$ be a maximal ideal. We want to show that $\mathfrak M = I_{x_0} = \{ f \in C(X) : f(x_0) = 0\}$ for some $x_0$.

For each nonempty closed set $A$, we define the ideal $I_A = \{ f \in C(X) : f(x) = 0, \forall x \in A\}$. Clearly $A \subset B$ implies $I_B \subset I_A$ (I also need to show that $I_B \subseteq I_A$ implies $A \subseteq B$, which I haven't done yet). We let $\mathcal S$ be the set of ideals $I_A$ which are contained in $\mathfrak M$.

An ascending chain $I_{A_i} \in \mathcal S$ corresponds to a descending chain of closed sets $A_i$, whose intersection is nonempty because $X$ is compact. Thus the union $J$ of the chain $I_{A_i}$ is an ideal which is contained in $I_{\bigcap A_i}$ (now we have to show that $I_{\bigcap A_i} \subseteq \mathfrak M$ to show that it is an upper bound for the chain, I haven't been able to do this).

So by Zorn's Lemma, $\mathcal S$ has a maximal element $I_D$ for some nonempty closed set $D$. Now, I just want to show that $D$ is a singleton set $\{x_0\}$; then $I_D$ is a maximal ideal contained in $\mathfrak M$, whence $I_D = \mathfrak M$. That will finish the proof.

Edit: Assuming I can fix the holes, I finished the problem (answered below).

Current problems with the proof:

(i) Need to show that $I_B \subseteq I_A$ implies $A \subseteq B$.

(ii) Need to finish the Zorn's lemma argument by showing that $I_{\bigcap\limits_i A_i} \subseteq \mathfrak M$.

Second edit: (i) follows from the Urysohn lemma. A compact Hausdorff space is normal. If $x \in A$, but not in $B$, then $\{x\}$ and $B$ are disjoint closed sets, so by Urysohn there exists a continuous function $f: X \rightarrow [0,1] \subseteq \mathbb{C}$ for which $f(x) = 1$ and $f(b) = 0$ for all $b \in B$. Thus $f \in I_B$, but not in $I_A$. So there is only one hole in the proof left.

Third edit: (ii) also follows from the Urysohn lemma. See the edited answer for details.

Answer 1

Proof of the Zorn's lemma argument:

Suppose $I_{A_i}$ is an ascending chain of ideals (set $i < j$ if and only if $I_{A_i} \subseteq I_{A_j}$). By order reversing correspondence, $A_i$ is a descending chain of closed subsets whose intersection $A$ is nonempty, as $X$ is compact. Now $I_A$ is an ideal which clearly contains each $I_{A_i}$. So $I_A$ will be an upper bound once we show that $I_A$ is contained in $\mathfrak m$.

So let $f \in I_A$. Since $\mathfrak M$ is closed, it suffices to show that $f$ is a limit point of $\mathfrak M$. So given $\epsilon > 0$, we will produce a $g \in \mathfrak M$ for which $|f(x) - g(x)| \leq \epsilon$ for all $x \in X$.

Being continuous on a compact set, $f$ is uniformly continuous and, since $f$ takes the value $0$ on $A$, we see that there exists an open set $U$ containing $A$ for which $|f(x)| < \epsilon$ whenever $x \in U$. It should also be possible to find a sufficiently large index $n$ for which $A_n$ is contained in $U$: since $$A = \bigcap\limits_i A_i \subseteq U$$ we have an open cover of the closed, hence compact, set $X \setminus U$: $$X \setminus U \subseteq \bigcup\limits_i X \setminus A_i $$ The cover being linearly ordered, a finite subcover entails a single index $n$ for which $X \setminus U \subseteq X \setminus A_n$, so $A_n \subseteq U$, as asserted. Now, $A_n$ and $X \setminus U$ are then disjoint closed sets, so by Uryson we can produce a continuous function $h: X \rightarrow [0,1]$ taking the values $0$ and $1$ on these sets respectively. Then $hf \in I_{A_n} \subseteq \mathfrak M$, and one can easily see that $|f(x) - h(x)f(x)|$ is $\leq \epsilon$ on all of $X$ by checking on the sets $A_n, U \setminus A_n$, and $X \setminus U$.

Remainder of the proof, once we have shown that $\mathcal S$ admits a maximal element $I_D$ with $D$ closed and $I_D \subseteq \mathfrak M$.

If $E$ is a proper closed subset of $D$, then $I_E$ cannot be contained in $\mathfrak M$; this implies by maximality that $I_D = I_E$, which implies $E = D$.

Now, I claim that the set $D$ cannot be the union of any of its proper closed subsets (in the language of algebraic geometry, $D$ is an irreducible space). For if $D = E_1 \cup E_2$, with $E_1$ and $E_2$ closed and proper, then by maximality we cannot have neither $I_{E_1}$ nor $I_{E_2}$ contained in $\mathfrak M$. Pick $f, g$ in $I_{E_1}, I_{E_2}$ respectively, neither of which are in $\mathfrak M$. Then $fg \in I_D \subseteq \mathfrak M$, which implies $f$ or $g$ is in $\mathfrak M$, contradiction.

But an irreducible Hausdorff space is exactly a singleton set. Done.

Answer 2

It's even easier. Take a strict ideal $I$ and show, as you intended to, that $I\subseteq I_x$ for some $x\in X$. For this, suppose the contrary : then for all $x\in X$ you can find an $f_x \in I$ such that $f_x \not\in I_x$ i.e. $f_x (x) \not= 0$. Since $f_x$ is continuous there is a open neighbour hood of $x$ in $X$ such that $f_x(y)\not=0$ for all $y\in U_x$. Now $(U_x)_{x\in X}$ is an open cover of the compact space $X$, and has therefore a finite subcover : we can find $x_1,\ldots,x_n \in X$ such that $X = \cup_{i=1}^n U_{x_i}$. Consider the function $f = \sum_{i=1}^n |f_{x_i}|^2 = \sum_{i=1}^n \overline{f_{x_i}} f_{x_i}$. This $f$ is obviously in $I$ as each $f_{x_i}$ is an as $I$ is an ideal. But by construction $f > 0$. Indeed : let $x\in X$. That $x$ is in some $U_{x_i}$ and then $f(x)\geq |f_i (x)|^2 > 0$ by definition of the $U_x$'s. Now, this function is invertible, so that $I$ contains an invertible element, so that $I$ is not strict. Absurd.