2
$\begingroup$

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 (?)

$\endgroup$
  • 1
    $\begingroup$ "In fact, a connected Hausdorff space can be connected"? $\endgroup$ – Sahiba Arora Jul 8 '17 at 19:12
  • $\begingroup$ I meant a countable Hausdorff space... $\endgroup$ – Forever Mozart Jul 8 '17 at 19:19
  • $\begingroup$ @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. $\endgroup$ – amWhy Jul 8 '17 at 19:45
  • 1
    $\begingroup$ @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! $\endgroup$ – Forever Mozart Jul 8 '17 at 19:47
  • $\begingroup$ but I guess it's not that big of a deal to me. I'll replace with something else soon $\endgroup$ – Forever Mozart Jul 8 '17 at 19:49
2
$\begingroup$

Yes:

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

  2. A countable normal space is zero-dimensional:if $O$ is open, and $x \in O$, then 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 $X$.

$\endgroup$
  • $\begingroup$ You beat me by a few seconds! (when I noticed normality, I saw the light) $\endgroup$ – Forever Mozart Jul 8 '17 at 19:45

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.