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Question is in the title.

Zero-dimensional means "has a basis of clopen sets".

Hausdorff is not enough to guarantee a countable space has dimension zero (in fact, a countable Hausdorff space can be connected).

Is regular enough?

Note 1: I assume that singletons are closed, so that regular is stronger than Hausdorff. I'm not sure what would happen here if we allow non-closed singletons...

Note 2: (countable + regular) implies normal, if that helps (?)

Forever Mozart
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    "In fact, a connected Hausdorff space can be connected"? – Sahiba Arora Jul 08 '17 at 19:12
  • I meant a countable Hausdorff space... – Forever Mozart Jul 08 '17 at 19:19
  • @ForeverMozart I read your brief profile statement that a couple users found distasteful. I googled the expression, and suspect I know to what you are referring, but you need to be aware that the sentence, even if you meant no harm, can be legitimately seen as offensive. Some matters are not worth joking about. – amWhy Jul 08 '17 at 19:45
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    @amWhy it's a movie quote from the movie "To Be and Not To Be", a comedy from the Hollywood's 'golden age'. highly recommended! – Forever Mozart Jul 08 '17 at 19:47
  • but I guess it's not that big of a deal to me. I'll replace with something else soon – Forever Mozart Jul 08 '17 at 19:49
  • That's what I found upon googling the phrase. But your use of the phrase lacks any context to say as much, and even if it comes from a comedy, doesn't justify its use apart from that context. I simply conclude, in your brief short profile statement, with no context, even knowing to what you are referring (of which very many users are unaware of), it reflects bad taste, at best, and offensive, at worst. – amWhy Jul 08 '17 at 19:52
  • @ForeverMozart That would be great, I think, for you to do. I wasn't aware of the phrase in connection to the movie, until googling. After googling the phrase, I knew you meant no harm. But in the end, I think it'd be best if you can come up with a different, clever reference you like in your profile. – amWhy Jul 08 '17 at 19:54
  • Henno Brandsma, Still true even with strong assumptions. If $X$ is countable completely regular space and using only the definition without using the Urysohn function. Right? – 00GB Oct 18 '21 at 16:16

1 Answers1

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Yes:

  1. a countable regular space is normal. (A regular Lindelöf space will even do). See my note e.g.

  2. A countable normal $T_2$ space is zero-dimensional:if $O$ is open, and $x \in O$, then $X\setminus O$ and $\{x\}$ are disjoint closed sets. Take a continuous Urysohn function $f: X \to [0,1]$ with $f(x)= 0$ and $f[X\setminus O] = \{1\}$. Note that $f[X] \subset [0,1]$ is at most countable, so pick $c \in [0,1]\setminus f[X]$. Then $V = f^{-1}[[0,c]] = f^{-1}[[0,c)]$ is clopen and $x \in V \subset O$, so clopen subsets form a base for $X$.

The $T_2$ (or equivalently $T_1$) hypothesis is needed here. For example, the Sierpinski space is $T_0$ (but not $T_1$) countable normal, but not zero-dimensional.

The comment by 00GB shows the more general fact by a similar argument:

  1. A completely normal space of size less than the continuum is zero-dimensional.
PatrickR
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Henno Brandsma
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  • You beat me by a few seconds! (when I noticed normality, I saw the light) – Forever Mozart Jul 08 '17 at 19:45
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    This can be taken further to covering dimension zero as the notions coincide for countable spaces: for an open cover, select a clopen subset of an open set from the cover that contains the least $x_n$ from $X={x_n:n<\omega}$ not already covered by previous selections and disjoint from the previous selections. – Steven Clontz Aug 23 '21 at 02:28