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The title question says it all: if $X$ is a topological space, then a subset $Z$ of $X$ is a zero set if there is a continuous function $f: X \rightarrow \mathbb{R}$ with $Z = f^{-1}(0)$.

Now I know the following:

1) Every zero set is a closed subset.
2) Every closed subset is a zero set iff $X$ is perfectly normal, e.g. if $X$ is metrizable.
3) Every closed subset is an intersection of zero sets iff $X$ is Tychonoff.

What I want to know is whether there is a similarly clean characterization of topological spaces $X$ such that for every point $x \in X$, there is a continuous function $f: X \rightarrow \mathbb{R}$ vanishing only at $x$. In particular, is there a compact (Hausdorff!) space that does not have this property?

Added: Having gotten some nice answers, maybe I should say a little more about my ulterior motive (which is sort of a motive in a teapot). I was mulling over a recent note of B. Sury in which he shows that in the ring $C([0,1])$ of continuous functions $f: X \rightarrow [0,1]$, for any $c \in [0,1]$, the maximal ideal $\mathfrak{m}_c$ of all functions vanishing at $c$ is not only infinitely generated (as is standard: I think this was a question on a qualifying exam I took as a graduate student!) but uncountably generated. I was thinking of generalizations to rings of continuous functions on other spaces $X$.

There is, it seems to me, a very small gap in his proof: about the function $f$ he constructs, he writes "since $f$ vanishes only at $c$". He hasn't argued for this, and depending on the choices of the sequence $\{f_n\}$, $f$ might vanish at other points. But no problem: if $\mathfrak{m}_c = \langle f_1,\ldots,f_n,\ldots \rangle$, since there is obviously some continuous function on $[0,1]$ which vanishes only on $c$ (e.g. $I(x) = |x-c|$), if $\bigcap_{n=1}^{\infty} f_n^{-1}(0) \supsetneq \{c\}$ then these functions cannot generate $\mathfrak{m}_c$.

If I am not mistaken, the following is a straightforward generalization of Sury's result.

Theorem: Let $X$ be a compact (Hausdorff!) space, and let $c \in X$. Suppose that there is a continuous function $I: X \rightarrow \mathbb{R}$ such that $I^{-1}(0) = \{c\}$. Then the following are equivalent:
(i) The point $c$ is isolated in $X$.
(ii) The ideal $\mathfrak{m}_c$ is principal.
(iii) The ideal $\mathfrak{m}_c$ is finitely generated.
(iv) The ideal $\mathfrak{m}_c$ is countably generated.

Well, this would be a better result without the weird hypothesis about the existence of $I$. Hence the question. (Maybe someone can see a better way to get around this hypothesis or replace it with something more natural...)

By the way, compactness also feels a little too strong here. This is being used to ensure that $C(X)$ is a Banach space under the supremum norm, but maybe there's a way around this as well.

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I believe the singleton $\{ \omega_1 \}$ is not a zero set in $\omega_1 + 1$, as $\omega_1$ has uncountable character in this space. (In normal spaces, the zero-sets are exactly the closed G$_\delta$-sets.) – arjafi Aug 9 '12 at 21:05
And slightly more generally, if $\alpha$ is a limit ordinal and $\{\alpha\}$ is a zero set in some ordinal $\beta > \alpha$, then $\alpha$ must be the limit of an increasing sequence (namely $\alpha_n = \min \{x: x < \alpha, |f(x)| < 1/n\}$). – Robert Israel Aug 9 '12 at 21:15
Two extreme non-examples: compact Hausdorff spaces in which no point is $G_\delta$: the Stone-Čech corona $\beta\mathbb N \smallsetminus \mathbb N$ and $\{0,1\}^\kappa$ with $\kappa \geq \aleph_1$. The former assertion is proved in detail here. – t.b. Aug 9 '12 at 21:40
I knew I saw a similar question here before: Brian mentions in this thread that singletons are zero sets in submetrizable spaces (definition and links there). In particular there are non-regular examples. – t.b. Aug 9 '12 at 23:40
I should have mentioned this before: Did you have a look at Gillman-Jerison, Rings of continuous functions? Chapter 4 and in particular the exercises therein contain a number of results on generation and ideals in completely regular spaces. – t.b. Aug 10 '12 at 0:01
up vote 14 down vote accepted

Note that if $f^{-1}(0)=\{x\}$ then there exists a sequence of neighbourhoods $U_n $ of $x$ so that $\{x\}=\bigcap_n U_n$. In the particular case that $X$ is compact Hausdorff this implies that $X$ is first countable. So any non-first countable compact topological space does not have that property. Edit:

We can have a similar characterization as 2. If $X$ is a Tychonoff space then $x$ is a zero set if and only if is a $G_\delta$

proof: Suppose $\{x\}=\bigcap_n U_n$ where $U_n$ is a decreasing sequence of open neighbourhoods of $x$. For each $n$ there is a function $f_n: X\to [0,1]$ so that $f_n(x)=0$ and $f_n(y)=1 $ for all $y\in X\setminus U_n$ ($f_n$ exists since $X$ is Tychonoff). Now, the function $g=\sum_n \frac{1}{2^n} f_n$ satisfies that $g^{-1}(0)=\{x\}$. The other implication is trivial.

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Thanks, this is a key observation. Could you say a little more about deducing that $X$ is first countable? I tried arguing this way: we may assume $U_n \supset U_{n+1}$ for all $n$. Let $V$ be a neighborhood of $x$. Suppose that for all $n$, there is $x_n \in U_n \setminus (V \cup U_{n-1})$. By compactness, the set $\{x_n\}$ has a limit point $L$. If we can show $L = x$, then $V$ is an open subset of $L$ not meeting any term of the sequence, contradiction. But I am having a little trouble showing this... – Pete L. Clark Aug 9 '12 at 21:39
@PeteL.Clark Since the space is regular you may assume that $U_{n+1}\subseteq \overline{U_{n+1}}\subseteq U_n$. It follows that $X\setminus \overline{U_{n+1}}$ is an open cover of $X\setminus V$. The rest follows from the compactness of $X\setminus V$. – azarel Aug 9 '12 at 21:53
Great, thanks. I believe my mistake was insisting on working with the original family $U_n$ (well, I was willing to replace it with a nested family) rather than taking interesctions to get your condition. – Pete L. Clark Aug 9 '12 at 22:00

This is a partial answer; I will provide an example of a compact Hausdorff space with a singleton that isn't a zero set.

Consider an uncoutable discrete space $X$ and let $X^+$ be its one-point compactification. Clearly $X^+$ is compact Hausdorff.

We claim that the singleton $\{\infty\}$ isn't $G_\delta$ and therefore can't be a zero set. If it were a countable intersection of open sets $U_n$, each of these must have a finite complement. But since we took $X$ to be uncountable, we cannot hope to get $X\setminus\{\infty\}$ as the countable union of finite sets.

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