0
$\begingroup$

Is a 0-dimensional, locally compact and pseudocompact space $X$ necessarily strongly 0-dimensional? I.e., must $\beta X$ be 0-dimensional?

It is known that a 0-dimensional locally compact space which is also paracompact must be strongly 0-dimensional (Engelking, 1989, p. 362). But the answer to a recent question posted here points out that $\omega_1$ is locally compact and pseudocompact but not paracompact, so an approach attempting to use that fact will not answer the present question.

$\endgroup$
1
$\begingroup$

An infinite collection $\mathcal{A}$ of infinite subsets of $\mathbb{N}$ is said to be almost disjoint (AD) if $A\cap B$ is finite whenever $A,B \in \mathcal{A}$ with $A \neq B$. If the family is maximal with respect to this property, then it is called a MAD family.

Given an AD family $\mathcal{A}$, there is a well known way (introduced by Mrowka in "On completely regular spaces", Fund.Math.,1954) to construct a topological space $\Psi(\mathcal{A})$. This space has the following properties:

1) For any AD family, $\Psi(\mathcal{A})$ is 0-dimensional, locally compact and first countable.

2) $\Psi(\mathcal{A})$ is pseudocompact if and only if $\mathcal{A}$ is a MAD family.

Teresawa (in "Spaces N∪R need not be strongly 0-dimensional", Bull. Acad. Polon. Sci. Sér. Sci. Math. Astonom. Phys., 1977) proved that there is a MAD family $\mathcal{A}$ for which $\Psi(\mathcal{A})$ is not strongly 0-dimensional. This provides a counterexample to your question.

$\endgroup$
  • $\begingroup$ Thanks folks. I even know this example from a recent post here. $\endgroup$ – Fred Dashiell May 3 '12 at 17:13
1
$\begingroup$

The spaces in this answer are pseudocompact.

$\endgroup$

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.