Assume all spaces are Hausdorff.


$X$ is totally disconnected if the only nonempty connected subsets of $X$ are singletons.

$X$ is zero dimensional if it has a base of clopen sets.

Now consider the following:

A) $X$ is zero dimensional,

B) for any $p\neq q\in X$ there exists a clopen set containing $p$ missing $q$ (the quasicomponents of $X$ are singletons),

C) $X$ is totally disconnected (the components of $X$ are singletons)

It is clear to me that $A\Rightarrow B\Rightarrow C$, and I know that $A$ and $C$ are not equivalent.

Question: Is $B$ equivalent to either $A$ or $C$ ? I think that $B\Leftrightarrow C$ if $X$ is compact since in this case quasicomponents are connected, but I'm not sure in general.

Edit: I am really interested in knowing if $B$ is strictly weaker than $A$, even for normal spaces.

  • $\begingroup$ Property (B) is called totally separated in Steen & Seebach. Note that (A) zero dimensional implies (B) totally separated only for $T_0$ spaces. For example, a space with the indiscrete topology is zero-dimensional, but not totally separated (since not even $T_1$). $\endgroup$
    – PatrickR
    Sep 6, 2021 at 3:32

1 Answer 1


No two of the three properties are equivalent for Hausdorff spaces in general. In this answer Martin Sleziak gives a proof that the Erdős space $\ell^2\cap\Bbb Q^\omega$, the space of square-summable sequence of rational numbers, has (B) but not (A). The space obtained by removing the dispersion point $p=\left\langle\frac12,\frac12\right\rangle$ from the Knaster-Kuratowski fan has (C) but not (B): its quasicomponents are the segments $L_c\setminus\{p\}$ for $c\in C$, the middle-thirds Cantor set.

  • $\begingroup$ Thanks. I should have considered your second example. I am not at all familiar with the Erdos space. $\endgroup$ Dec 22, 2014 at 23:45
  • $\begingroup$ @ForeverMozart: You’re welcome. I confess that I’d completely forgotten about the Erdős space until I went digging a bit. $\endgroup$ Dec 22, 2014 at 23:46

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