A zero-dimensional Hausdorff space is totally disconnected

Question

The full question: A space is zero-dimensional if the clopen subsets form a basis for the topology. Show that a zero-dimensional Hausdorff space is totally disconnected. Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets).

Let X = {X1, X2, ...} be the set of clopen subsets of the space. We know that X is a basis for the topology T, so any open set in T can be written as a union or finite intersection of elements in X.

In a topological space, we know the union/finite intersection of open sets are also open (by definition) and the union/finite intersection of closed sets are closed, so any union/finite intersection of clopen sets is also clopen.

Since X is a basis, then any open set in T is also closed, since it will be the union/finite intersection of clopen sets. Does this mean our space is discrete? If it is discrete, then the only connected subsets are singletons, and then our space is totally disconnected.

I have a strong feeling I've gone in circles and my argument is incorrect (especially the discrete part...) Any help would be appreciated.

Answer 1

I'm not quite sure where you were trying to get with your suggested argument. But here's a possible outline for a solution.

Let $X$ be a zero-dimensional Hausdorff space, and let $B$ be a clopen basis. Now suppose that $C\subseteq X$ is a connected set, let us show that $C$ is a singleton.

Suppose that $x\in C$, then there is a clopen environment $U$ of $x$. Conclude that $U\cap C$ is both open and closed relative to $C$, and therefore either $C$ is a singleton, or that $C$ is not connected which is a contradiction.

Answer 2

Suppose $X$ is a zero dimensional Hausdorff space, and $\mathcal{B}$ a clopen basis for it.

Suppose $A$ is a set containing points $x$ and $y$. If they are distinct we can find an open set containing $x$ not containing $y$, hence we can find a basis element $B \in \mathcal{B}$ containing $x$ but not $y$. The sets $A\cap B$ and $A\cap B^c$ are both open in $A$ and disjoint, and their union is $A$, hence $A$ is disconnected.

Thus the space doesn't even need to be Hausdorff - the $T_1$ axiom is sufficient.