3
$\begingroup$

For a maximal almost disjoint family $\mathcal A$ of subsets of $\omega$ we choose a set $\{x_A:A\in\mathcal A\}$ of distinct points not in $\omega$ and define $\Psi (\mathcal A)=\omega\cup \{x_A:A\in \mathcal A\}$.

A basis for a topology on $\Psi(\mathcal{A}$) is given by $$\{ \{n \}:n\in\omega\}\cup \{\{x_A\}\cup (A\setminus F):A\in\mathcal A, F\subseteq \omega\text{ is finite}\}.$$

Then $\Psi (\mathcal A)$ is

  • locally compact Hausdorff
  • completely regular
  • first countable
  • separable
  • zero dimensional
  • pseudocompact

It is not countably compact as $\{x_A:A\in\mathcal A\}$ is an infinite closed discrete subspace. As normal plus pseudocompact implies countably compact, $\Psi (\mathcal A)$ also fails to be normal.

There is a theorem which says any compact metric space without isolated points is homeomorphic to the Stone-Čech remainder of some $\Psi (\mathcal A)$. In particular the Stone-Čech remainder of $\Psi (\mathcal A)$ can be metric. This means in some cases $\{x_A:A\in\mathcal A\}$, if it is a zero set, is not C$^*$-embedded in $\Psi (\mathcal A)$ since otherwise the Stone-Čech remainder would contain a copy of $\omega ^*$.

For an arbitrary m.a.d. family $\mathcal A$ is there a nice way to characterize the zero sets of $\Psi (\mathcal A)$?

Can you give an example of a closed subset of $\Psi (\mathcal A)$ which is not a zero set?

Can you prove $\{x_A:A\in\mathcal A\}$ is a zero set?

$\endgroup$
5
$\begingroup$

Actually, $\{ x_A : A \in \mathcal{A} \}$ is always a zero-set in $\Psi(\mathcal{A})$: define the function $f : \Psi(\mathcal{A}) \to [0,1]$ by $$f(n) = \tfrac{1}{n+1} \\ f(x_A) = 0.$$ This is clearly a continuous function (any neighbourhood of $0$ contains all but finitely many of the $\frac{1}{n+1}$, and so it's inverse image under $f$ includes a cofinite subset of $A$ for each $A \in \mathcal{A}$).

I am really unaware of any characterisation of the zero-sets in $\Psi(\mathcal{A})$. The following are either easy or can be found in the literature:

  • All finite subsets of $\{ x_A : A \in \mathcal{A} \}$ are zero-sets in $\Psi(\mathcal{A})$. (For $A \in \mathcal{A}$ define $f_A : \Psi(A) \to [0,1]$ by $$ f_A(x) = \begin{cases} \tfrac{1}{n+1}, &\text{if }x \in A; \\ 0, &\text{if }x=x_A; \\ 1, &\text{otherwise.}\end{cases}$$ The almost-disjointness of $\mathcal{A}$ helps showing that each $f_A$ is continuous.)

  • No countably infinite subset of $\{ x_A : A \in \mathcal{A} \}$ can be a zero-set in $\Psi(\mathcal{A})$. (Follows from the pseudocompactness of $\Psi(\mathcal{A})$ and the fact that no countably infinite subset of $\{ x_A : A \in \mathcal{A} \}$ is compact.) (This is problem 5I(6) (p.79) in Gillman & Jerison, Rings of Continuous Functions.)

  • There is a mad family $\mathcal{A}$ such that an infinite $E \subseteq \{ x_A : A \in \mathcal{A} \}$ is a zero-set in $\Psi(\mathcal{A})$ iff it is co-countable. (Theorem 3.11 in Mrówka, Some set-theoretic constructions in topology, Fund. Math., vol.94 (1977) pp.83-92, link.)

$\endgroup$
  • $\begingroup$ Can you explain where I went wrong in my second to last paragraph? $\endgroup$ – Forever Mozart Nov 22 '14 at 6:36
  • $\begingroup$ Am I confusing "zero set" with "$C^*$-embedded"? $\endgroup$ – Forever Mozart Nov 22 '14 at 6:39
  • $\begingroup$ @TomCruise: That might be it. $\{ x_A : A \in \mathcal{A} \}$ is never C$^*$-embedded in $\Psi(\mathcal{A})$. $\endgroup$ – user642796 Nov 22 '14 at 6:43
  • $\begingroup$ Ok I see why that is true. Thanks $\endgroup$ – Forever Mozart Nov 22 '14 at 7:05

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.