2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ YOU can take $A= Z_1$. Since $\Bbb N$ is discrete, then $A$ is clopen. $\endgroup$ – Paul Jun 14 '13 at 3:54
  • 1
    $\begingroup$ A topological space X is totally disconnected if the connected components in X are the one-point sets. $\endgroup$ – Paul Jun 14 '13 at 3:55
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.