A zero-dimensional Hausdorff space is totally disconnected

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.

• You can't conclude that all open sets are closed : An open set is a (possibly uncountable) union of basic open sets, and so may not be closed even if the basic open sets are closed. – Prahlad Vaidyanathan Apr 20 '14 at 4:03
• An arbitrary union of clopen sets need not be clopen since it might not be closed. – Cheerful Parsnip Apr 20 '14 at 6:42

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.

• So could I say this: Suppose that C is not a singleton (i.e., there are distinct x and y in C). Since our space is Hausdorff, there exist disjoint open neighborhoods around x and y, say U and V respectively. So then C can be written as the union of nonempty sets U, V and C complement, which is a contradiction, so C must be singleton. The only problem with this argument is I fail to see the significance of a clopen basis... – Sultan of Swing Apr 20 '14 at 4:34
• Well, yes, but it suffices to note that $U\cap C$ is relatively clopen and is not $C$ itself. The significance of a clopen basis is that we can have $U$ which is clopen. – Asaf Karagila Apr 20 '14 at 4:36
• I'm just not exactly sure what relatively clopen means. My argument would suggest that that zero-dimensional condition isn't actually required; any Hausdorff space is totally disconnected. I need to find the significance of the space being zero-dimensional. So what does it matter that our environment U is clopen? – Sultan of Swing Apr 20 '14 at 4:39
• No no no. Being disconnected means that you can partition the space into two open nonempty parts. Not two open parts and one closed part. The significance of a clopen basis is that you can assume that the open environment for $x$ is also closed, i.e. has an open complement. The Hausdorff part comes into play when you assume $C$ has more than one point to allow $U$ to be taken such that $U\cap C\neq C$. – Asaf Karagila Apr 20 '14 at 4:44
• Alright, so my argument (original comment) is saying U is open, V is open, but C complement is actually closed, so C could be connected or disonnected at that point..we don't know...but since our space is zero-dimensional, then our neighborhoods U and V are open AND closed, so then the complement of C is open AND closed, and the union of these three clopen sets U, V, and C complement, form C, which would mean C is not connected, a contradiction, so C must be singleton. Right? – Sultan of Swing Apr 20 '14 at 4:49

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.

• But if the space isn't Hausdorff, then we might not be able to find the set and basis element B containing x but not y. – Sultan of Swing Apr 20 '14 at 5:33
• Yes, it needs to have something more than just zero dimensionality. An axiom called $T_1$, satisfied by what are called Frechet spaces is weaker than Hausdorff. In this type of space any two points have neighborhoods not containing the other points, those neighborhoods aren't necessarily disjoint. – guest196883 Apr 20 '14 at 5:37