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?


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.)

  • $\begingroup$ 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

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