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I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?

Martin
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Thomas Martin
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3 Answers3

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It is not true.

John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:X\to\Bbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, $\{p\}$ is a closed $G_\delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.

A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:X\to\Bbb R$ and $x\in F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again $\{p\}$ is a closed $G_\delta$ that cannot be a zero-set.

Brian M. Scott
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  • Why is it clear that ${p}$ is a $G_{\delta}$-set? For a topology which is sufficently rich (e.g. if it is metrizable), we can find a sequence of open sets $(U_n)$ with $x=\bigcap_n U_n$, and so ${x}$ is $G_{\delta}$. But how do we construct the sequence here? (I guess I'm missing something obvious...) – saz Oct 26 '19 at 13:19
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Not even in completely regular Hausdorff spaces. In general we have $$ \text{compact $G_\delta$}\qquad\Longrightarrow\qquad \text{zero-set}\qquad\Longrightarrow\qquad \text{closed $G_\delta$} $$ but none reversible.

GEdgar
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  • A bit late too the party, but could you point me to a reference to the proof of compact $G_\delta \implies $ zero-set? – John Aug 29 '17 at 11:45
  • what can be example for completely regular space? – Sushil Mar 09 '18 at 18:49
  • @John The implication compact $G_\delta \Rightarrow$ zero-set holds for a completely regular Hausdorff space, see e.g. [Fremlin, "Measure Theory", 4A2Fh(v)] for a proof. In a completely regular Hausdorff space you can separate compact sets from open sets by continuous functions. – yada Feb 27 '20 at 08:48
  • @yada I think you meant, you can separate compact sets from closed sets by continuous function in a Tychonoff (which I prefer to call them) space. – Jakobian Jun 22 '23 at 20:28
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Brian's answer covers the question fully. For fun, here's another example:

Bing's irrational slope space is a countable and connected Hausdorff space.

Now observe:

  1. If $f \colon X \to \mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $\mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f \colon X \to \mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.

  2. Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_\delta$-set: $F = \bigcap_{x \in X \setminus F} X \setminus \{x\}$, in particular, there is an abundance of closed $G_\delta$ sets that are not zero sets.

Martin
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