# Minimal zero-dimensional Hausdorff spaces

A topological space $(X,\tau)$ is said to be zero-dimensional Hausdorff (zdH) if for $x\neq y\in X$ there is $C\subseteq X$ clopen (closed and open) such that $x\in C$, but $y\notin C$.

We say a zdH space $(X,\tau)$ is minimal zdH if for each topology $\tau' \subseteq \tau$ on $X$ with $\tau' \neq \tau$ the space $(X,\tau')$ is not zdH any more.

Clearly every compact zdH, such as $\omega+1$ with the interval topology, is minimal zdH as it is minimal Hausdorff. (I don't know whether there are non-compact minimal zdH spaces.)

Does every zdH topology contain a minimal zdH topology?

• In General Topology zdH-spaces are usually called totally disconnected spaces, see e.g., en.wikipedia.org/wiki/Totally_disconnected_space May 2 '16 at 15:25
• @TarasBanakh Isn't the space of all square-summable sequences of rational numbers under the usual $L^2$-metric a totally disconnected Hausdorff space which is not zero-dimensional? (Engelking, p.364 seems to think so.) May 2 '16 at 16:10
• @arjafi: Exactly! May 2 '16 at 16:24
• @TarasBanakh, maybe I am just being silly, but it seems that, in your response to arjafi above, you agree that zdH is not equivalent to total disconnectedness; but, in your answer below, you assert that it is. May 2 '16 at 18:33
• @L_Spice: The problem arises from a double use of the adverb "zero-dimensional". Its standard meaning is the existence of a base consisting of clopen sets and in this meaning the Erdos space mentined by arjafi is not zero-dimensional. On the other hand, "zero-dimensional Hausdorff" in the sense of Dominic van der Zypen is the same as "totally disconnected". So, the Erdos space mentiened by arjafi is zero-dimensional Hausdorff (i.e. totally disconnected) but not zero-dimensional (in the standard sense). May 2 '16 at 20:33

Theorem 1. Each minimal totally disconnected topological space $X$ is compact.
Proof: The total disconnectedness implies that $X$ admits an injective continuous map $f:X\to K$ to the Cantor cube $K$. The minimality of $X$ implies that the map $f$ is a topological embedding. Assuming that $Y=f(X)$ is not compact, take any point $y\in Y$, and any cluster point $z\in \bar Y\setminus Y$. Consider the quotient space $K/\{y,z\}$ and the quotient map $q:K\to K/\{y,z\}$. It can be shown that the space $K/\{y,z\}$ is zero-dimensional and the map $q\circ f:X\to K/\{y,z\}$ is not a topological embedding, which contradicts the minimality of $X$. This contradiction completes the proof of the compactness of $X$.
Corollary. Let $X$ be a countable zero-dimensional space without isolated points. The topology of $X$ is totally disconnected but contains no minimal totally disconnected topology.
Proof. Assuming that the topology of $X$ contains a minimal totally disconnected topology, we conclude that $X$ admits a continuous bijective map $f:X\to Y$ onto a minimal totally disconnected space $Y$. By the continuity of $f$, the countable space $Y$ has no isolated points and hence cannot be compact (by the Baire Theorem). By Theorem 1, $Y$ cannot be minimal totally disconnected.