A topological space is zero dimensional if, and only if, it has a basis consisting of sets which are both open and closed (that is, "clopen").

This definition is according to "Counterexamples in Topology" by Steen and Seebach, 2nd ed. 1978.

I see that "zero dimensional" is tied in with various levels of being "disconnected", e.g. extremally disconnected, totally separated, totally, disconnected, scattered, etc. (although the property of being "zero dimensional" is itself strictly independent of all disconnectivity properties).

I understand (vaguely) the concept of a "dimension" in the context of manifolds: a space of $n$ dimensions has a boundary of $n - 1$ dimensions. I also note that a set is clopen if, and only if, it has no boundary. So would that be the basis of this terminology? Is there a source which states this definitively?

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    $\begingroup$ Yes, the dimension is defined in an inductive manner by Brouwer, and zero dimensional object is a natural concept since a geometric line is "clearly" a one dimensional object. $\endgroup$
    – Hulkster
    Commented May 23, 2020 at 6:52
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    $\begingroup$ The inductive definition of dimension can be initialized by defining the dimension of the empty set to be $-1$. In a zero-dimensional space there is a base for the topology consisting of sets whose boundary is empty, i.e., has dimension $-1$. $\endgroup$
    – bof
    Commented May 23, 2020 at 7:25

2 Answers 2


This notion of zero-dimensionality is based on the small inductive dimension $\operatorname{ind}(X)$ which can be defined for lots of spaces. It's defined inductively, based on the intuition that the boundary of open sets in two-dimensional spaces (which are boundaries disks, so circles in the plane, e.g.) have a dimension that is one lower than that of the space itself (this works nicely, at least intuitively, for the Euclidean spaces), setting up a recursion: We define $\operatorname{ind}(X) = -1$ iff $X=\emptyset$ (!) and a space has $\operatorname{ind}(X) \le n$ whenever $X$ has a base $\mathcal{B}$ of open sets such that $\operatorname{ind}(\partial O) \le n-1$ for all $O \in \mathcal{B}$, where $\partial A$ denotes the boundary of a set $A$. Finally, $\operatorname{ind}(X) = n$ holds if $\operatorname{ind}(X) \le n$ holds and $\operatorname{ind}(X)\le n-1$ does not hold. Also, $\operatorname{ind}(X)=\infty$ if for no $n$ we have $\operatorname{ind}(X) \le n$. It's clear that this is a topological invariant (homeomorphic spaces have the same dimension w.r.t. $\operatorname{ind}$) and zero-dimensional (i.e. $\operatorname{ind}(X)=0$) exactly means that there is a base of open sets with empty boundary (from the $-1$ clause!) and $\partial O=\emptyset$ means that $O$ is clopen.

Note that $\Bbb Q$ and $\Bbb P = \Bbb R\setminus \Bbb Q$ are both zero-dimensional in $\Bbb R$ and that $\Bbb R$ is one-dimensional (i.e. $\operatorname{ind}(\Bbb R)=1$) as boundaries of $(a,b)$ are $\{a,b\}$, which is zero-dimensional (discrete), etc.

In dimension theory more dimension functions have been defined as well, e.g. large inductive dimension $\operatorname{Ind}(X)$, which is a variant of $\operatorname{ind}(X)$, and the (Lebesgue) covering dimension $\dim(X)$, which has a different flavour and is about refinements of open covers and the order of those covers. For separable metric spaces however it can be shown that all $3$ that were mentioned are the same (i.e. give the same (integer) value). There are also metric-based definitions (fractal dimensions) which have more possible values, but are not topological, but metric invariants. Outside of metric spaces, we can have gaps between the dimension functions and stuff gets hairy quickly. See Engelking's book "Theory of dimensions, finite and infinite" for a taste of this field.

So in summary: the name "comes" (can be justified) from the small inductive dimension definition $\operatorname{ind}(X)$, but the name itself for that special class (clopen base) is older I think, and other names (like Boolean space etc) have been used too. It's a nice way to be very disconnected, giving a lot of structure.

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    $\begingroup$ Two comments: (1) I'd reserve the name "Boolean" for compact Hausdorff 0-dimensional spaces, i.e., Stone spaces of Boolean algebras. (2) Although it's been a long time since I read it, I remember the book "Dimension Theory" by Hurewicz and Wallman as being readable and short. $\endgroup$ Commented May 24, 2020 at 0:44
  • $\begingroup$ @AndreasBlass I agree that the compact ones area best called Boolean ((Stone duality etc). H&W book mostly covers the separable metric case, while Engelking strives for the largest generality, treats more examples. It’s less accessible. Van Mill’s book on infinite- dimensional topology has a few chapters covering H&W in a more modern terminology. $\endgroup$ Commented May 24, 2020 at 6:56
  • $\begingroup$ To the proposer. Re: The last 2 lines of the 1st paragraph. Note that an open set with empty boundary is also a closed set. So a non-empty $X$ has $ind(X)=0$ iff $X$ has a base of open-and-closed sets. $\endgroup$ Commented May 24, 2020 at 15:23

There are various notions of dimension in topology. One of the "most common" is Lebesgue covering dimension, which relies on open covers. More precisely, let $(X_p)_p$ be an open cover of the topological space $X$, that is, a family of open subsets of $X$ whose union is $X$. The order of $(X_p)_p$ is defined to be the smallest non negative integer $n$ (if it exists) such that each point $x \in X$ belongs at most to $n$ sets of the cover. A refinenent of $(X_p)_p$ is another open cover such that each of its sets is contained in some $X_t \in (X_p)_p$. Lebesgue covering dimension, denoted by $\dim X$, is the smallest value of $n$ such that every cover of $X$ has an open refinement with order $\leq n+1$. If there is no such $n$, $\dim X := \infty$. A topological space $X$ which is zero dimensional is a space having $\dim X=0$.

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    $\begingroup$ The small inductive dimension $\operatorname{ind}(X)$ is a better fit for the clopen base definition. It's not always true that $\dim(X)=0 \iff \operatorname{ind}(X)=0$, but it's a theorem in separable metric spaces, e.g. $\endgroup$ Commented May 23, 2020 at 8:05
  • $\begingroup$ Nice point, thanks! $\endgroup$ Commented May 23, 2020 at 8:11
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    $\begingroup$ While I agree with @HennoBrandsma's comment that the small inductive dimension is a better fit I think it's worth pointing out that for remotely reasonable spaces ($T_1$ is already enough) $\dim X=0$ implies that $X$ has a basis of clopen sets. I know that this implication can be reversed in compact Hausdorff spaces, I'm not sure whether this is sharp for the reverse implication though. $\endgroup$ Commented May 23, 2020 at 15:19

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