# Stone-Cech Compactification of the Integers – Basic Facts

This is another post Stone-Cech compactification. The links for other posts on Stone-Cech compactification can be found below. In this post, we prove a few basic facts about $\beta \omega$, the Stone-Cech compactification of the discrete space of the non-negative integers, $\omega=\left\{0,1,2,3,\cdots \right\}$. We use several characterizations of Stone-Cech compactification to find out what $\beta \omega$ is like. These characterizations are proved in the blog posts listed below. Let $c$ denote the cardinality of the real line $\mathbb{R}$. We prove the following facts.

1. The cardinality of $\beta \omega$ is $2^c$.
2. The weight of $\beta \omega$ is $c$.
3. The space $\beta \omega$ is zero-dimensional.
4. Every infinite closed subset of $\beta \omega$ contains a topological copy of $\beta \omega$.
5. The space $\beta \omega$ contains no non-trivial convergent sequence.
6. No point of $\beta \omega-\omega$ is an isolated point.
7. The space $\beta \omega$ fails to have many properties involving the existence of non-trivial convergent sequence. For example:
$\text{ }$

1. The space $\beta \omega$ is not first countable at each point of the remainder $\beta \omega-\omega$.
2. The space $\beta \omega$ is not a Frechet space.
3. The space $\beta \omega$ is not a sequential space.
4. The space $\beta \omega$ is not sequentially compact.

$\text{ }$

8. No point of the remainder $\beta \omega-\omega$ is a $G_\delta$-point.
9. The remainder $\beta \omega-\omega$ does not have the countable chain condition. In fact, it has a disjoint open collection of cardinality $c$.

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Characterization Theorems

For any completely regular space $X$, let $C(X,I)$ be the set of all continuous functions from $X$ into $I=[0,1]$. The Stone-Cech compactification $\beta X$ is the subspace of the product space $[0,1]^{C(X,I)}$ which is the closure of the image of $X$ under the evaluation map $\beta:X \rightarrow [0,1]^{C(X,I)}$ (for the details, see Embedding Completely Regular Spaces into a Cube).

The brief sketch of $\beta \omega$ we present here is not based on the definition using the evaluation map. Instead we reply on some characterization theorems that are stated here (especially Theorem U3.1). These theorems uniquely describe the Stone-Cech compactification $\beta X$ of a given completely regular space $X$. For example, $\beta X$ satisfies the function extension property in Theorem C3 below. Furthermore any compactification $\alpha X$ of $X$ that satisfies the same property must be $\beta X$ (Theorem U3.1). So a “C” theorem tells us a property possessed by $\beta X$. The corresponding “U” theorem tells us that there is only one compactification (up to equivalence) that has this property.

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Theorem C1
Let $X$ be a completely regular space. Let $f:X \rightarrow Y$ be a continuous function from $X$ into a compact Hausdorff space $Y$. Then there is a continuous $F: \beta X \rightarrow Y$ such that $F \circ \beta=f$.

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Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.

See Two Characterizations of Stone-Cech Compactification.
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Theorem C2
Let $X$ be a completely regular space. Among all compactifications of the space $X$, the Stone-Cech compactification $\beta X$ of the space $X$ is maximal with respect to the partial order $\le$.

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Theorem U2
The property in Theorem C2 is unique to $\beta X$. That is, if, among all compactifications of the space $X$, $\alpha X$ is maximal with respect to the partial order $\le$, then $\alpha X \approx \beta X$.

See Two Characterizations of Stone-Cech Compactification.
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Theorem C3
Let $X$ be a completely regular space. The space $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$.

$\text{ }$

Theorem U3.1
Let $X$ be a completely regular space. Let $I=[0,1]$. Let $\alpha X$ be a compactification of $X$ such that each continuous $f:X \rightarrow I$ can be extended to a continuous $\hat{f}:\alpha X \rightarrow I$. Then $\alpha X$ must be $\beta X$.

$\text{ }$

Theorem U3.2
If $\alpha X$ is any compactification of $X$ that satisfies the property in Theorem C3 (i.e., $X$ is $C^*$-embedded in $\alpha X$), then $\alpha X$ must be $\beta X$.

See C*-Embedding Property and Stone-Cech Compactification.
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$\text{ }$

The following discussion illustrates how we can use some of these characterizations theorem to obtain information about $\beta X$ and $\beta \omega$ in particular.

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Result 1 and Result 2

According to the previous post (Stone-Cech Compactification is Maximal), we have for any completely regular space $X$, $\lvert \beta X \lvert \le 2^{2^{d(X)}}$ where $d(X)$ is the density (the smallest cardinality of a dense set in $X$). With $\omega$ being a countable space, $\lvert \beta \omega \lvert \le 2^{2^{\omega}}=2^c$.

Result 1 is established if we have $2^c \le \lvert \beta \omega \lvert$. Consider the cube $I^I$ where $I$ is the unit interval $I=[0,1]$. Since the product space of $c$ many separable space is separable (see Product of Separable Spaces), $I^I$ is separable. Let $S \subset I^I$ be a countable dense set. Let $f:\omega \rightarrow S$ be a bijection. Clearly $f$ is a continuous function from the discrete space $\omega$ into $I^I$. By Theorem C1, $f$ is extended by a continuous $F:\beta \omega \rightarrow I^I$. Note that the image $F(\beta \omega)$ is dense in $I^I$ since $F(\beta \omega)$ contains the dense set $S$. On the other hand, $F(\beta \omega)$ is compact. So $F(\beta \omega)=I^I$. Thus $F$ is a surjection. The cardinality of $I^I$ is $2^c$. Thus we have $2^c \le \lvert \beta \omega \lvert$.

From the same previous post (Stone-Cech Compactification is Maximal), it is shown that $w(\beta X) \le 2^{d(X)}$. Thus $w(\beta \omega) \le 2^{\omega}=c$. The same function $F:\beta \omega \rightarrow I^I$ in the above paragraph shows that $c \le w(\beta \omega)$ (see Lemma 2 in Stone-Cech Compactification is Maximal). Thus we have $w(\beta \omega)=c$ $\blacksquare$

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Result 3

A space is said to be zero-dimensional whenever it has a base consisting of open and closed sets. The proof that $\beta X$ is zero-dimensional comes after the following lemmas and theorems.

Theorem 1
Let $X$ be a normal space. If $H$ and $K$ are disjoint closed subsets of $X$, then $H$ and $K$ have disjoint closures in $\beta X$.

Proof of Theorem 1
Let $H$ and $K$ be disjoint closed subsets of $X$. By the normality of $X$ and by the Urysohn’s lemma, there is a continuous function $g:X \rightarrow [0,1]$ such that $g(H) \subset \left\{0 \right\}$ and $g(K) \subset \left\{1 \right\}$. By Theorem C3.1, $g$ can be extended by $G:\beta X \rightarrow [0,1]$. Note that $\overline{H} \subset G^{-1}(0)$ and $\overline{K} \subset G^{-1}(1)$. Thus $\overline{H} \cap \overline{K} = \varnothing$. $\blacksquare$

Theorem 2
Let $X$ be a completely regular space. Let $H$ be a closed and open subset of $X$. Then $\overline{H}$ (the closure of $H$ in $\beta X$) is also a closed and open set in $\beta X$.

Proof of Theorem 2
Let $H$ be a closed and open subset of $X$. Let $K=X-H$. Define $\gamma:X \rightarrow [0,1]$ by letting $\gamma(x)=0$ for all $x \in H$ and $\gamma(x)=1$ for all $x \in K$. Since both $H$ and $K$ are closed and open, the map $\gamma$ is continuous. By Theorem C3, $\gamma$ is extended by some continuous $\Gamma:\beta X \rightarrow [0,1]$. Note that $\overline{H} \subset \Gamma^{-1}(0)$ and $\overline{K} \subset \Gamma^{-1}(1)$. Thus $H$ and $K$ have disjoint closures in $\beta X$, i.e. $\overline{H} \cap \overline{K} = \varnothing$. Both $H$ and $K$ are closed and open in $\beta X$ since $\beta X=\overline{H} \cup \overline{K}$. $\blacksquare$

Lemma 3
For every $A \subset \omega$, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is both closed and open in $\beta \omega$.

Note that Lemma 3 is a corollary of Theorem 2.

Lemma 4
Let $O \subset \beta \omega$ be a set that is both closed and open in $\beta \omega$. Then $O=\overline{A}$ where $A= O \cap \omega$.

Proof of Lemma 4
Let $A=O \cap \omega$. Either $O \subset \omega$ or $O \cap (\beta \omega-\omega) \ne \varnothing$. Thus $A \ne \varnothing$. We claim that $O=\overline{A}$. Since $A \subset O$, it follows that $\overline{A} \subset \overline{O}=O$. To show $O \subset \overline{A}$, pick $x \in O$. If $x \in \omega$, then $x \in A$. So focus on the case that $x \notin \omega$. It is clear that $x \notin \overline{B}$ where $B=\omega -A$. But every open set containing $x$ must contain some points of $\omega$. These points of $\omega$ must be points of $A$. Thus we have $x \in \overline{A}$. $\blacksquare$

Proof of Result 3
Let $\mathcal{A}$ be the set of all closed and open sets in $\beta \omega$. Let $\mathcal{B}=\left\{\overline{A}: A \subset \omega \right\}$. Lemma 3 shows that $\mathcal{B} \subset \mathcal{A}$. Lemma 4 shows that $\mathcal{A} \subset \mathcal{B}$. Thus $\mathcal{A}= \mathcal{B}$. We claim that $\mathcal{B}$ is a base for $\beta \omega$. To this end, we show that for each open $O \subset \beta \omega$ and for each $x \in O$, we can find $\overline{A} \in \mathcal{B}$ with $x \in \overline{A} \subset O$. Let $O$ be open and let $x \in O$. Since $\beta \omega$ is a regular space, we can find open set $V \subset \beta \omega$ with $x \in V \subset \overline{V} \subset O$. Let $A=V \cap \omega$.

We claim that $x \in \overline{A}$. Suppose $x \notin \overline{A}$. There exists open $U \subset V$ such that $x \in U$ and $U$ misses $\overline{A}$. But $U$ must meets some points of $\omega$, say, $y \in U \cap \omega$. Then $y \in V \cap \omega=A$, which is a contradiction. So we have $x \in \overline{A}$.

It is now clear that $x \in \overline{A} \subset \overline{V} \subset O$. Thus $\beta \omega$ is zero-dimensional since $\mathcal{B}$ is a base consisting of closed and open sets. $\blacksquare$

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Result 4 and Result 5

Result 5 is a corollary of Result 4. We first prove two lemmas before proving Result 4.

Lemma 5
For each infinite $A \subset \omega$, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is a homeomorphic copy of $\beta \omega$ and thus has cardinality $2^c$.

Proof of Lemma 5
Let $A \subset \omega$. Let $g:A \rightarrow [0,1]$ be any function (necessarily continuous). Let $f:\omega \rightarrow [0,1]$ be defined by $f(x)=g(x)$ for all $x \in A$ and $f(x)=0$ for all $x \in \omega-A$. By Theorem C3, $f$ can be extended by $F:\beta \omega \rightarrow [0,1]$. Let $G=F \upharpoonright \overline{A}$.

Note that the function $G: \overline{A} \rightarrow [0,1]$ extends $g:A \rightarrow [0,1]$. Thus by Theorem U3.1, $\overline{A}$ must be $\beta A$. Since $A$ is a countably infinite discrete space, $\beta A$ must be equivalent to $\beta \omega$. $\blacksquare$

Lemma 6
For each countably infinite $A \subset \beta \omega-\omega$ such that $A$ is relatively discrete, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is a homeomorphic copy of $\beta \omega$ and thus has cardinality $2^c$.

Proof of Lemma 6
Let $A=\left\{t_1,t_2,t_3,\cdots \right\} \subset \beta \omega -\omega$ such that $A$ is discrete in the relative topology inherited from $\beta \omega$. There exist disjoint open sets $G_1,G_2,G_3,\cdots$ (open in $\beta \omega$) such that for each $j$, $t_j \in G_j$. Since $\beta \omega$ is zero-dimensional (Result 3), $G_1,G_2,G_3,\cdots$ can be made closed and open.

Let $f:A \rightarrow [0,1]$ be a continuous function. We show that $f$ can be extended by $F:\overline{A} \rightarrow [0,1]$. Once this is shown, by Theorem U3.1, $\overline{A}$ must be $\beta A$. Since $A$ is a countable discrete space, $\beta A$ must be equivalent to $\beta \omega$.

We first define $w:\omega \rightarrow [0,1]$ by:

$\displaystyle w(n)=\left\{\begin{matrix}f(t_j)& \exists \ j \text{ such that } n \in \omega \cap G_j\\{0}&\text{otherwise} \end{matrix}\right.$

The function $w$ is well defined since each $n \in \omega$ is in at most one $G_j$. By Theorem C3, the function $w$ is extended by some continuous $W:\beta \omega \rightarrow [0,1]$. By Lemma 4, for each $j$, $G_j=\overline{\omega \cap G_j}$. Thus, for each $j$, $t_j \in \overline{\omega \cap G_j}$. Note that $W$ is a constant function on the set $\omega \cap G_j$ (mapping to the constant value of $f(t_j)$). Thus $W(t_j)=f(t_j)$ for each $j$. So let $F=W \upharpoonright \overline{A}$. Thus $F$ is the desired function that extends $f$. $\blacksquare$

Proof of Result 4
Let $C \subset \beta \omega$ be an infinite closed set. Either $C \cap \omega$ is infinite or $C \cap (\beta \omega-\omega)$ is infinite. If $C \cap \omega$ is infinite, then by Lemma 5, $\overline{C \cap \omega}$ is a homeomorphic copy of $\beta \omega$. Now focus on the case that $C_0=C \cap (\beta \omega-\omega)$ is infinite. We can choose inductively a countably infinite set $A \subset C_0$ such that $A$ is relatively discrete. Then by Lemma 6 $\overline{A}$ is a copy of $\beta \omega$ that is a subset of $C$. $\blacksquare$

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Result 6

We prove that no point in the remainder $\beta \omega-\omega$ is an isolated point. To see this, pick $x \in \beta \omega-\omega$ and pick an arbitrary closed and open set $O \subset \beta \omega$ with $x \in O$. Let $V=O \cap (\beta \omega-\omega)$ (thus an arbitrary open set in the remainder containing $x$). By Lemma 4, $O=\overline{A}$ where $A=O \cap \omega$. According to Lemma 5, $O=\overline{A}$ is a copy of $\beta \omega$ and thus has cardinality $2^c$. The set $V$ is $O$ minus a subset of $\omega$. Thus $V$ must contains $2^c$ many points. This means that $\left\{ x \right\}$ can never be open in the remainder $\beta \omega-\omega$. In fact, we just prove that any open and closed subset of $\beta \omega-\omega$ (thus any open subset) must have cardinality at least $2^c$. $\blacksquare$

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Result 7

The results under Result 7 are corollary of Result 5 (there is no non-trivial convergent sequence in $\beta \omega$). To see Result 7.1, note that every point $x$ in the remainder is not an isolated point and hence cannot have a countable local base (otherwise there would be a non-trivial convergent sequence converging to $x$).

A space $Y$ is said to be a Frechet space if $A \subset Y$ and for each $x \in \overline{A}$, there is a sequence $\left\{ x_n \right\}$ of points of $A$ such that $x_n \rightarrow x$. A set $A \subset Y$ is said to be sequentially closed in $Y$ if for any sequence $\left\{ x_n \right\}$ of points of $A$, $x_n \rightarrow x$ implies $x \in A$. A space $Y$ is said to be a sequential space if $A \subset Y$ is a closed set if and only if $A$ is a sequentially closed set. If a space is Frechet, then it is sequential. It is clear that $\beta \omega$ is not a sequential space.

A space is said to be sequentially compact if every sequence of points in this space has a convergent subsequence. Even though $\beta \omega$ is compact, it cannot be sequentially compact.

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Result 8

Result 7.1 indicates that no point of remainder $\beta \omega-\omega$ can have a countable local base. In fact, no point of the remainder can be a $G_\delta$-point (a point that is the intersection of countably many open sets). The remainder $\beta \omega-\omega$ is a compact space (being a closed subset of $\beta \omega$). In a compact space, if a point is a $G_\delta$-point, then there is a countable local base at that point (see 3.1.F (a) on page 135 of [1] or 17F.7 on page 125 of [2]). $\blacksquare$

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Result 9

The space $\beta \omega$ is a separable space since $\omega$ is a dense set. Thus $\beta \omega$ has the countable chain condition. However, the remainder $\beta \omega-\omega$ does not have the countable chain condition. We show that there is a disjoint collection of $c$ many open sets in $\beta \omega-\omega$.

There is a family $\mathcal{A}$ of infinite subsets of $\omega$ such that for every $A,B \in \mathcal{A}$ with $A \ne B$, $A \cap B$ is finite. Such a collection of sets is said to be an almost disjoint family. There is even an almost disjoint family of cardinality $c$ (see A Space with G-delta Diagonal that is not Submetrizable). Let $\mathcal{A}$ be such a almost disjoint family.

For each $A \in \mathcal{A}$, let $U_A=\overline{A}$ and $V_A=\overline{A} \cap (\beta \omega -\omega)$. By Lemma 3, each $U_A$ is a closed and open set in $\beta \omega$. Thus each $V_A$ is a closed and open set in the remainder $\beta \omega-\omega$. Note that $\left\{V_A: A \in \mathcal{A} \right\}$ is a disjoint collection of open sets in $\beta \omega-\omega$. $\blacksquare$

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Blog Posts on Stone-Cech Compactification

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a $G_\delta$-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The diagonal of a space $X$ is the set $\Delta=\left\{(x,x): x \in X \right\}$, a subset of the square $X \times X$. An interesting property is when the diagonal of a space is a $G_\delta$-set in $X \times X$ (the space is said to have a $G_\delta$-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space $(X,\tau)$ is said to be submetrizable if there is a topology $\tau^*$ that can be defined on $X$ such that $(X,\tau^*)$ is a metrizable space and $\tau^* \subset \tau$. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a $G_\delta$-diagonal. Note that when $X$ has a weaker metric topology, the diagonal $\Delta$ is always a $G_\delta$-set in the metric square $X \times X$ (hence in the square in the original topology). The property of having a $G_\delta$-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let $\omega$ be the first infinite ordinal (or more conveniently the set of all nonnegative integers). 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. An 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 $A \in \mathcal{A}$. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of $\omega$. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify $\omega$ with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). Let $\mathbb{P}=\mathbb{R}-\mathbb{Q}$ be the set of all irrational numbers. For each $x \in \mathbb{P}$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ (in the Euclidean topology). 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 this post).

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. We now define a Mrowka space (or $\Psi$-space), denoted by $\Psi(\mathcal{A})$. 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 is 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 would like to point out that the definition of a Mrowka space $\Psi(\mathcal{A})$ only requires that the family $\mathcal{A}$ is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, $\mathcal{A}$ needs to be a maximal almost disjoint family of subsets of $\omega$.

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Example
Let $\mathcal{A}$ be a maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ as defined above is a space in which there is a $G_\delta$-diagonal that is not submetrizable.

Note that $\Psi(\mathcal{A})$ is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of $\omega$, $\mathcal{A}$ must be an uncountable in addition to being a closed and discrete subspace of $\Psi(\mathcal{A})$ (thus the space is not Lindelof). Since $\Psi(\mathcal{A})$ is separable and is not Lindelof, $\Psi(\mathcal{A})$ is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus $\Psi(\mathcal{A})$ must not be submetrizable.

On the other hand, any $\Psi$-space $\Psi(\mathcal{A})$ (even if $\mathcal{A}$ is not maximal) is a Moore space. It is well known that any Moore space has a $G_\delta$-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

Suppose that $X$ is a Moore space. We show that $X$ has a $G_\delta$-diagonal. That is, we wish to show that $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$.

Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace$. Clearly $\Delta \subset \bigcap_{n<\omega} U_n$. Let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{D}_n$. We claim that $x=y$. Suppose that $x \ne y$. By the definition of development, there exists some $m$ such that every open set in $\mathcal{D}_m$ containing the point $x$ has to be a subset of $X-\left\{y \right\}$. Then $V_m \subset X-\left\{y \right\}$, which contradicts $y \in V_m$. Thus we have $\Delta = \bigcap_{n<\omega} U_n$.

The remaining thing to show is that $\Psi(\mathcal{A})$ is a Moore space. For each positive integer $n$, let $F_n=\left\{0,1,\cdots,n-1 \right\}$ and let $F_0=\varnothing$. The development is defined by $\lbrace{\mathcal{E}_n}\rbrace_{n<\omega}$, where for each $n$, $\mathcal{E}_n$ consists of open sets of the form $\lbrace{A}\rbrace \cup (A-F_n)$ where $A \in \mathcal{A}$ plus any singleton $\left\{j \right\}$ ($j \in \omega$) that has not been covered by the sets $\lbrace{A}\rbrace \cup (A-F_n)$.

Reference

1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# 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.