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.


1 Answer 1


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 $X$.

  • $\begingroup$ you've done it again! $\endgroup$ Nov 22, 2014 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. $\endgroup$ Nov 23, 2014 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.