# Is $\beta \mathbb Q$ totally disconnected?

Using the filter description of $\beta \mathbb N$ one can easily prove $\beta \mathbb N$ is totally disconnected; as $\mathbb N$ is a discrete space, $\beta \mathbb N$ is clearly a stone space and thus it is totally disconnected.

Now a question comes up, if $X$ is totally disconnected and Tychonoff, does this imply $\beta X$ is totally disconnected?

I tried to make a counterexample; namely, I tried to show $\beta\mathbb Q$ is not totally disconnected, but I showed that $\beta \mathbb Q$ is totally disconnected if whenever $Z_1$, $Z_2$ are disjoint closed subsets of $\mathbb Q$ there is some clopen subset $A$ of $\mathbb Q$ such that $Z_1\subseteq A$ and $A\cap Z_2=\emptyset$, but I don't know whether this last property is true or not. So, is this a counterexample?

Thanks for any help.

• YOU can take $A= Z_1$. Since $\Bbb N$ is discrete, then $A$ is clopen. – Paul Jun 14 '13 at 3:54
• A topological space X is totally disconnected if the connected components in X are the one-point sets. – Paul Jun 14 '13 at 3:55

## 1 Answer

That last property is true for all zero-dimensional separable metric spaces, I believe. As every separable metric space $X$ with a clopen base has $\dim(X) = 0$, it has $\operatorname{Ind}(X) = 0$ as well, and the latter means that $\operatorname{Ind(\beta X)} = 0$ too. And this implies $\beta X$ is indeed totally disconnected.

Proofs can be found in Engelking's book "General Topology", or his book on dimension theory.