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Is there a completely regular (Hausdorff) space in which all singleton subsets are Gδ but which has a singleton subset which is not a zero-set? (Better yet, a first-countable such space.)

Recall that a subset $A$ of a topological space $X$ is called a zero-set if there is a continuous function $f : X \to [0,1]$ such that $A = f^{-1} (0)$. Clearly zero-sets are always closed Gδ. However a closed Gδ-set need not be a zero set: the Moore plane has the property that all closed subsets are Gδ, and furthermore every subset of the $x$-axis is closed. However the closed Gδ-set $A = \{ (x,0) : x \in \mathbb{Q} \}$ is not a zero-set.

A relatively simple consequence of Urysohn's Lemma is that in a normal space the zero-sets are exactly the closed Gδ-sets. Therefore, in a normal (Hausdorff) space with Gδ singletons, all singletons are zero-sets. I have been trying to think of a completely regular (Hausdorff) space with Gδ singletons which has a singleton which is not a zero-set.

Note that the Moore plane is not an example. Though it is completely regular (Hausdorff) and first-countable (hence all singletons are Gδ), every singleton can be shown to be a zero-set.

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There is not.

Let $X$ be Tikhonov and have countable pseudocharacter (i.e., singletons are $G_\delta$ sets). Fix $x\in X$, and let $\mathscr{U}=\{U_n:n\in\Bbb N\}$ be a family of open nbhds of $x$ such that $\bigcap\mathscr{U}=\{x\}$. For $n\in\Bbb N$ there is a continuous $f_n:X\to[0,2^{-n}]$ such that $f_n(x)=0$, and $f_n(y)=2^{-n}$ for $y\in X\setminus U_n$. Let

$$f:X\to[0,1]:y\mapsto\sum_{n\in\Bbb N}f_n(y)\;;$$

Then $f$ is continuous, and $f^{-1}[\{0\}]=\{x\}$, so $\{x\}$ is a zero-set of $X$.

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  • $\begingroup$ you've done it again! $\endgroup$ – Forever Mozart Nov 22 '14 at 20:51
  • $\begingroup$ @TomCruise It's the standard proof that in a normal space a closed $G_\delta$ is a zero-set. But here we only use that singletons and closed sets are functionally separated, so we need complete regularity instead of normality for singleton sets. $\endgroup$ – Henno Brandsma Nov 23 '14 at 6:11

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