Let $X$ be a topological space. (All of the spaces I'm considering are $T_0$, but in general they are not $T_1$. To be even more concrete, one can even consider $X={\rm Spec}(R)$ to be the space of prime ideals of a [not necessarily commutative] ring $R$, under the Zariski topology.)

Let us say that $X$ has clopen separation if for any two closed, disjoint sets $A,B\subseteq X$, then there is a clopen set $U$ containing $A$, whose complement contains $B$.

I'm reading a paper which states that this is the definition of a strongly zero-dimensional space; but other sources give a (seemingly) different definition. Is this equivalence true? If not, how close to true is it?

  • $\begingroup$ what other sources give what other seemingly different definition? Engelking's book defines strongly zero-dimensional as a non-empty Tychonoff (he includes T$_1$) space such that every functionally (i.e. co-zero) open cover has a finite disjoint open refinement. Different authors need not agree on the terminology. $\endgroup$
    – Mirko
    May 5, 2015 at 22:40
  • $\begingroup$ The definition's I've seen are: (1) a space $X$ is zero-dimensional if it is $T_1$ with a base of clopen sets, and (2) a space $X$ is strongly zero-dimensional if the Stone-Cech compactification is zero-dimensional (which I believe is equivalent to the definition you gave). The source that gives this alternate definition is arxiv.org/pdf/1406.1236v1.pdf (see the beginning of Section 3). $\endgroup$ May 5, 2015 at 22:53
  • 1
    $\begingroup$ Mirko. I think you meant to say that a strongly zero-dimensional space is a Tychonoff space where every $\textit{finite}$ cozero open cover has a finite disjoint open refinement which I can believe is equivalent to having zero-dimensional Stone-Cech compactification. $\endgroup$ May 5, 2015 at 23:01
  • $\begingroup$ What´s the Stone-Cech compactification of a space that is not $T_1$? (or even a space that is not $T_{3\frac{1}{2}}$?). $\endgroup$ May 6, 2015 at 11:26

1 Answer 1


$\textbf{All the spaces mentioned in this answer are completely regular}$

I usually call a space $T_{1}$-space ultranormal if whenever there are two disjoint closed sets $A,B\subseteq X$ there is a clopen set with $A\subseteq C,B\subseteq C^{c}$. A strongly zero-dimensional space is usually defined to be a completely regular space $X$ such that the Stone-Cech compactification $\beta X$ is zero-dimensional. Equivalently, a completely regular space $X$ is strongly zero-dimensional if and only if whenever $f:X\rightarrow[0,1]$ is continuous, then there is a clopen set $C$ with $f^{-1}[\{0\}]\subseteq C,f^{-1}[\{1\}]\subseteq C^{c}$. The ultranormal spaces are precisely the normal strongly zero-dimensional spaces, but not every strongly zero-dimensional space is ultranormal. For example, the Tychonoff plank $(\omega+1)\times(\omega_{1}+1)\setminus\{(\omega,\omega_{1})\}$ is strongly zero-dimensional since $\beta((\omega+1)\times(\omega_{1}+1)\setminus\{(\omega,\omega_{1})\})=(\omega+1)\times(\omega_{1}+1)$ but not ultranormal. I gave a long answer here giving (probably too much) information about the distinction between various notions related to zero-dimensionality.

Let me now prove that every normal strongly zero-dimensional space is ultranormal. Suppose that $X$ is strongly zero-dimensional and normal. Let $C,D\subseteq X$ be disjoint closed sets. Then by normality, there is some continuous $f:X\rightarrow[0,1]$ with $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$. However, since $X$ is strongly zero-dimensional, there is a clopen set $R$ with $f^{-1}[\{0\}]\subseteq R,f^{-1}[\{1\}]\subseteq R^{c}$. Therefore $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$.

  • $\begingroup$ This is interesting, but I'm still a bit confused. Consider the following example. Let $R=\mathbb{Z}_{(2)}$ be the ring of integers localized at 2 (so the elements are just rational numbers with odd denominators). Then ${\rm Spec}(R)$ has two elements: the zero ideal $0$ and the unique maximal ideal $M=2R$. The closed sets are just $\emptyset, \{M\}, \{0,M\}$. This space has the separation property (which I called clopen separation), since the only way two closed sets are disjoint is if one of them is empty. But it isn't completely regular since it has non-closed points. $\endgroup$ May 5, 2015 at 23:11
  • $\begingroup$ Pace Nielsen. You are correct. In my answer, I should have mentioned that ultranormality is the clopen separation property + $T_{1}$ (when dealing with normality and similar ideas I always assume the spaces are completely regular so that I do not have to deal with these issues). I edited my answer to address this issue. $\endgroup$ May 5, 2015 at 23:42
  • $\begingroup$ I have two further questions and a comment. (1) Is there a nice source for the equivalence between $T_1 +$ clopen separation and normal strongly zero-dimensional? (2) Can much more be said about these properties in $T_0$ spaces which are not $T_1$? (3) The spaces I care about are quasi-compact, so it seems to me that merely under the $T_1$ hypothesis (along with quasi-compactness), we get an equivalence between strong zero-dimensional and clopen separation. (This was pointed out to me by Manny Reyes.) $\endgroup$ May 6, 2015 at 2:26
  • $\begingroup$ For question 1, I am not immediately aware of any sources (besides my own) of this fact since the notion of an ultranormal space is not a well known notion. However, the proof that every normal strongly zero-dimensional space is ultranormal is easy and I have edited my answer to include such a proof. $\endgroup$ May 6, 2015 at 13:04

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