30

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.

Community
  • 1
Pete L. Clark
  • 97,892
  • 3
    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.) – user642796 Aug 09 '12 at 21:05
  • 1
    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 09 '12 at 21:15
  • 2
    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 09 '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 09 '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 00:01
  • @t.b.: Thanks for the suggestion. I have looked briefly at Rings of continuous functions at various points in the past (in particular, while writing the section of my commutative algebra notes with that name), but not recently. Just now I did an automated search of that book and found that the terms "uncountable" and "uncountably" are never applied to generators of ideals. But it definitely has a lot of material on zero sets: I should take a closer look. – Pete L. Clark Aug 10 '12 at 01:41
  • Is there something unsatisfactory in the statement of the question? – Pete L. Clark Jan 22 '14 at 15:52

2 Answers2

14

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.

azarel
  • 13,120
  • 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 09 '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 09 '12 at 21:53
  • 1
    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 09 '12 at 22:00
13

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.

Miha Habič
  • 7,164