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What properties hold on a compact infinite Hausdorff space?

I encountered this example in Atiyah-Macdonald on chain conditions:

Let $X$ be a compact infinite Hausdorff space, $C(X)$ the ring of real-valued continuous functions on $X$. Take a strictly decreasing sequence $F_1 \supseteq F_2 \supseteq \cdots$ of closed sets in $X$, and let $\mathfrak{a_n} = \{f \in C(X): f(F_n) = 0\}$. Then the $\mathfrak{a}_n$ form a strictly increasing sequence of ideals in $C(X)$ : so $C(X)$ is not Noetherian ring.

The example itself I understand, but I wasn't sure why we'd need the condition "$X$ be a compact infinite Hausdorff space". I don't have much analysis background so might be missing out some basics. Thanks!

nekodesu
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    You need some hypotheses to show that there exists a strictly decreasing sequence of closed sets, and then to show that $a_{n+1}\ne a_n$. – David C. Ullrich Mar 18 '18 at 15:13

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Maybe to help show the strictly part of strictly increasing? The normality and infiniteness of $X$ ensures that we indeed can find such a strictly decreasing family of sets (this need not hold in general spaces $X$, and certainly never for finite $X$).

If $F_{n+1} \subsetneq F_n$, we can pick $p \in F_n, p \notin F_{n+1}$ and then the fact that $X$ is Tychonoff allows to find us a function $f \in C(X)$ such that $f(p) =1$ and $f[F_{n+1}] = \{0\}$, so that $f$ witnesses that $\mathfrak{a}_{n+1} \setminus \mathfrak{a}_{n} \neq \emptyset$.

But this only needs the Tychonoff-ness of $X$ (which is often assumed anyway, to ensure that $C(X)$ is interesting/large enough.) The compactness also ensures that $F:= \cap_n F_n \neq \emptyset$, so that the union of the $\mathfrak{a}_n$ is not $C(X)$?

Maybe he does use the compactness of $X$ later (this does ensure that all maximal ideals are of the form $I_p = \{f \in C(X): f(p) = 0\}$ for some $p \in X$, e.g. so the ring has a different structure for compact $X$.

But for this particular case it's not needed, really.

Henno Brandsma
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