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?



Minimal zero-dimensional Hausdorff spaces
First note that a topological space is zdH if and only if it is totally disconnected.


**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.


