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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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    Is there something unsatisfactory in the statement of the question?2014-01-22

2 Answers 2

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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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    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.2012-08-09
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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.