# Ψ-Spaces – spaces from almost disjoint families

As the title suggests the $\Psi-$spaces are defined using almost disjoint families, in our case, of subsets of $\omega$. This is a classic example of a pseudocompact space that is not countably compact. This example is due to Mrowka ([3]) and Isbell (credited in [2]), It is sometimes called the Mrowka space in the literature. This is another example that is a useful counterexample and set-theoretic construction. This being an introduction, I prove that the $\Psi-$space, when it is defined using a maximal almost disjoint family of subsets of $\omega$, is pseudocompact and not countably compact. On the other hand, I show that if a normal space is pseudocompact space, then it is countably compact. All spaces in this note is at least Hausdorff.

A space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. According to Theorem 3.10.3 in [1], a space $X$ is countably compact if and only if every infinite subset of $X$ has an accumulation point. A space $X$ is pseudocompact if every real-valued continuous function defined on $X$ is bounded. It is clear that if $X$ is a countably compact space, then it is pseudocompact. We show that the converse does not hold by using the example of $\Psi-$space. We also show that the converse does hold for normal spaces.

Let $\mathcal{A}$ be a family of infinite subsets of $\omega$. The family $\mathcal{A}$ is said to be an almost disjoint family if for each two distinct $A,B \in \mathcal{A}$, $A \cap B$ is finite. The almost disjoint family $\mathcal{A}$ is said to be a maximal almost disjoint family if $B$ is an infinite subset of $\omega$ such that $B \notin \mathcal{A}$, then $B \cap A$ is infinite for some $A \in \mathcal{A}$.

There is an almost disjoint family $\mathcal{A}$ of subsets of $\omega$ such that $\lvert \mathcal{A} \lvert=\text{continuum}$. To see this, identify $\omega$ (the set of all natural numbers) with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). For each real number $x$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$. Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of $\omega$. See comment below Theorem 2.

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. Let’s define the space $\Psi(\mathcal{A})$. The underlying set is $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$. Points in $\omega$ are isolated. For $A \in \mathcal{A}$, a basic open set of of the form $\lbrace{A}\rbrace \cup (A-F)$ where $F \subset \omega$ is finite. It is straightforward to verify that $\Psi(\mathcal{A})$ is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that $\mathcal{A}$ is an infinite discrete and closed set in the space $\Psi(\mathcal{A})$. Thus $\Psi(\mathcal{A})$ is not countably compact. We have the following theorems.

Theorem 1. Let $\mathcal{A}$ be an infinite maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ is pseudocompact.

Proof. Suppose we have a continuous $f:\Psi(\mathcal{A}) \rightarrow \mathbb{R}$ that is unbounded. We can find an infinite $B \subset \omega$ such that $f$ is unbounded on $B$. If $B \in \mathcal{A}$, then we have a contradiction since $\lbrace{f(n):n \in B}\rbrace$ is a sequence that does not converge to $f(B)$. So we have $B \notin \mathcal{A}$. By the maximality of $\mathcal{A}$, $C=B \cap A$ is infinite for some $A \in \mathcal{A}$. Then $\lbrace{f(n):n \in C}\rbrace$ is a sequence that does not converge to $f(A)$, another contradiction. So $\Psi(\mathcal{A})$ is pseudocompact.

Theorem 2. For a normal space $X$, if $X$ is pseudocompact, then $X$ is countably compact.

Proof. Suppose $X$ is not countably compact. Then we have an infinite closed and discrete set $A=\lbrace{a_0,a_1,a_2,...}\rbrace$ in $X$. Define $f:A \rightarrow \mathbb{R}$ by $f(a_n)=n$. According to the Tietze-Urysohn Theorem, in a normal space, any continuous function defined on a closed subset of $X$ can be extended to a continuous function defined on all of $X$. Then $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $f^*:X \rightarrow \mathbb{R}$, making $X$ not pseudocompact.

Comment
If there is a countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$, then $\Psi(\mathcal{B})$ is a countable first countable space and is thus has a countable base (hence is normal). By Theorem 1, it is pseudocompact. By Theorem 2, it is countably compact. Yet $\mathcal{B}$ is an infinite closed and discrete subset of $\Psi(\mathcal{B})$, contradicting that it is countably compact. Thus there is no countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$. In fact, we have the following corollary.

Corollary. If $\mathcal{A}$ is an infinite maximal almost disjoint family of subsets of $\omega$, then $\Psi(\mathcal{A})$ cannot be normal.

Reference

1. Engelking, R., General Topology, Revised and Completed Edition, 1988, Heldermann Verlag Berlin.
2. Gillman, L., and Jerison, M., Rings of Continuous Functions, 1960, Van Nostrand, Princeton, NJ.
3. Mrowka, S., On completely regular spaces, Fund. Math., 41, (1954) 105-106.

## 2 thoughts on “Ψ-Spaces – spaces from almost disjoint families”

1. There is a more concrete construction of a continuum-sized family of almost disjoint subsets of $\mathbb N$ (not quite the same as $\omega$, but close enough):

For each irrational $x \in (0,1)$ we can form its continued fraction expansion $[0;x_1,x_2,\dotsc]$ and associate to that a set $A_x = \{p_1^{x_1}, p_1^{x_1}p_2^{x_2},\dotsc\} \subset \mathbb N$ where $p_n$ denotes the $n$th prime number. The fundamental theorem of arithmetic does the rest of the work.