# The normality of the product of the first uncountable ordinal with a compact factor

The product of a normal space with a compact space needs not be normal. For example, the product space $\omega_1 \times (\omega_1+1)$ is not normal where $\omega_1$ is the first uncountable ordinal with the order topology and $\omega_1+1$ is the immediate successor of $\omega_1$ (see this post). However, $\omega_1 \times I$ is normal where $I=[0,1]$ is the unit interval with the usual topology. The topological story here is that $I$ has countable tightness while the compact space $\omega_1+1$ does not. In this post, we prove the following theorem:

Theorem 1

Let $Y$ be an infinite compact space. Then the following conditions are equivalent:

1. The product space $\omega_1 \times Y$ is normal.
2. $Y$ has countable tightness, i.e., $t(Y)=\omega$.

Theorem 1 is a special case of the theorem found in [4]. The proof for the direction of countable tightness of $Y$ implies $\omega_1 \times Y$ is normal given in [4] relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction $2 \Longrightarrow 1$, we give a complete proof. For the direction $1 \Longrightarrow 2$, we essentially give the same proof as in [4], proving it by using a series of lemmas (stated below).

The authors in [2] studied the normality of $X \times \omega_1$ where $X$ is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.

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Definitions and Lemmas

Let $X$ be a topological space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal number $\kappa$ such that for any $A \subset X$ and for any $x \in \overline{A}$, there exists a $B \subset A$ such that $x \in \overline{B}$ and $\lvert B \lvert \le \kappa$. When $t(X)=\omega$, we say $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say $X$ has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is $\omega_1+1=\omega_1 \cup \left\{\omega_1 \right\}$. This space has uncountable tightness at the point $\omega_1$. All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.

A sequence $\left\{x_\alpha: \alpha<\tau \right\}$ of points of a space $X$ is said to be a free sequence if for each $\alpha<\tau$, $\overline{\left\{x_\beta: \beta<\alpha \right\}} \cap \overline{\left\{x_\beta: \beta \ge \alpha \right\}}=\varnothing$. When a free sequence is indexed by the cardinal number $\tau$, the free sequence is said to have length $\tau$. The cardinal function $F(X)$ is the least infinite cardinal $\kappa$ such that if $\left\{x_\alpha \in X: \alpha<\tau \right\}$ is a free sequence of length $\tau$, then $\tau \le \kappa$. The concept of tightness was introduced by Arkhangelskii and he proved that $t(X)=F(X)$ (see p. 15 of [3]). This fact implies the following lemma.

Lemma 2

Let $X$ be compact. If $t(X) \ge \tau$, then there exists a free sequence $\left\{x_\alpha \in X: \alpha<\tau \right\}$ of length $\tau$.

A proof of Lemma 2 can be found here.

The proof of the direction $1 \Longrightarrow 2$ also uses the following lemmas.

Lemma 3

For any compact space $Y$, $\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$.

Lemma 4

Let $X$ be a normal space. For every pair $H$ and $K$ of disjoint closed subsets of $X$, $H$ and $K$ have disjoint closures in $\beta X$.

For Lemma 3, see 3.12.20(c) on p. 237 of [1]. For Lemma 4, see 3.6.4 on p. 173 of [1].

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Proof of Theorem 1

$1 \Longrightarrow 2$
Let $X=\omega_1 \times Y$. Suppose that $X$ is normal. Suppose that $Y$ has uncountable tightness, i.e., $t(Y) \ge \omega_1$. By Lemma 2, there exists a free sequence $\left\{y_\alpha \in Y: \alpha<\omega_1 \right\}$. For each $\beta<\omega_1$, let $C_\beta=\left\{y_\alpha: \alpha>\beta \right\}$. Then the collection $\left\{\overline{C_\beta}: \beta<\omega_1 \right\}$ has the finite intersection property. Since $Y$ is compact, $\bigcap_{\beta<\omega_1} \overline{C_\beta} \ne \varnothing$. Let $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$. Consider the following closed subsets of $X=\omega_1 \times Y$.

$H=\overline{\left\{(\alpha,y_\alpha): \alpha<\omega_1 \right\}}$
$K=\left\{(\alpha,p): \alpha<\omega_1 \right\}$

We claim that $H \cap K=\varnothing$. Suppose that $(\alpha,p) \in H \cap K$. Either $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ or $p \in \overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$. The latter case is not possible. Note that $[0,\alpha] \times Y$ is an open set containing $(\alpha,p)$. This open set cannot contain points of the form $(\delta,p)$ where $\delta \ge \alpha+1$. So the first case $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ must hold. Since $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$, $p \in \overline{C_\alpha}=\overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$, a contradiction. So $H$ and $K$ are disjoint closed subsets of $X=\omega_1 \times Y$.

Now consider $\beta X$, the Stone-Cech compactification of $X=\omega_1 \times Y$. By Lemma 3, $\beta X=\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$. Let $H^*=\overline{H}$ and $K^*=\overline{K}$ (closures in $\beta X$). We claim that $(\omega_1,p) \in H^* \cap K^*$. Let $O=(\theta,\omega_1] \times V$ be an open set in $\beta X$ with $(\omega_1,p) \in O$. Note that $p \in \overline{C_\theta}=\left\{y_\delta: \delta>\theta \right\}$. Thus $V \cap \overline{C_\theta} \ne \varnothing$. Choose $\delta>\theta$ such that $y_\delta \in V$. We have $(\delta,y_\delta) \in (\theta,\omega_1] \times V$ and $(\delta,y_\delta) \in H^*$. On the other hand, $(\delta,p) \in K^*$. Thus $(\omega_1,p) \in H^* \cap K^*$, a contradiction. Since $X=\omega_1 \times Y$ is normal, Lemma 4 indicates that $H$ and $K$ should have disjoint closures in $\beta X=(\omega_1+1) \times Y$. Thus $Y$ has countable tightness.

$2 \Longrightarrow 1$
Suppose $t(Y)=\omega$. Let $H$ and $K$ be disjoint closed subsets of $\omega_1 \times Y$. The following series of claims will complete the proof:

Claim 1
For each $y \in Y$, there exists an $\alpha<\omega_1$ such that either $W_{H,y} \subset \alpha+1$ or $W_{K,y} \subset \alpha+1$ where

$W_{H,y}=\left\{\delta<\omega_1: (\delta,y) \in H \right\}$
$W_{K,y}=\left\{\delta<\omega_1: (\delta,y) \in K \right\}$

Proof of Claim 1
Let $y \in Y$. The set $V=\omega_1 \times \left\{y \right\}$ is a copy of $\omega_1$. It is a known fact that in $\omega_1$, there cannot be two disjoint closed and unbounded sets. Let $V_H=V \cap H$ and $V_K=V \cap K$. If $V_H \ne \varnothing$ and $V_K \ne \varnothing$, they cannot be both unbounded in $V$. Thus the claim follows if both $V_H \ne \varnothing$ and $V_K \ne \varnothing$. Now suppose only one of $V_H$ and $V_K$ is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say $V_K$. Then $W_{H,y}=\varnothing$ and the claim follows.

Claim 2
For each $y \in Y$, there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that one and only one of the following holds:

$H \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (1)$
$K \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (2)$

Proof of Claim 2
Let $y \in Y$. Let $\alpha<\omega_1$ be as in Claim 1. Assume that $W_{H,y} \subset \alpha+1$. We want to show that there exists an open set $O_y \subset Y$ with $y \in O_y$ such that (1) holds. Suppose that for each open $O \subset Y$ with $y \in O$, there is a $q \in \overline{O}$ and there exists $\delta_q>\alpha$ such that $(\delta_q,q) \in H$. Let $S$ be the set of all such points $q$. Then $y \in \overline{S}$. Since $Y$ has countable tightness, there exists countable $T \subset S$ such that $y \in \overline{T}$. Since $T$ is countable, choose $\gamma >\omega_1$ such that $\alpha<\delta_q<\gamma$ for all $q \in T$. Note that $[\alpha,\gamma] \times \left\{y \right\}$ does not contain points of $H$ since $W_{H,y} \subset \alpha+1$. For each $\theta \in [\alpha,\gamma]$, the point $(\theta,y)$ has an open neighborhood that contains no point of $H$. Since $[\alpha,\gamma] \times \left\{y \right\}$ is compact, finitely many of these neighborhoods cover $[\alpha,\gamma] \times \left\{y \right\}$. Let these finitely many open neighborhoods be $M_i \times N_i$ where $i=1,\cdots,m$. Let $N=\bigcap_{i=1}^m N_i$. Then $y \in N$ and $N$ would contain a point of $T$, say $q$. Then $(\delta_q,q) \in M_i \times N_i$ for some $i$, a contradiction. Note that $(\delta_q,q)$ is a point of $H$. Thus there exists an open $O_y \subset Y$ with $y \in O_y$ such that (1) holds. This completes the proof of Claim 2.

Claim 3
For each $y \in Y$, there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that there are disjoint open subsets $Q_H$ and $Q_K$ of $\omega_1 \times \overline{O_y}$ with $H \cap (\omega_1 \times \overline{O_y}) \subset Q_H$ and $K \cap (\omega_1 \times \overline{O_y}) \subset Q_K$.

Proof of Claim 3
Let $y \in Y$. Let $\alpha$ and $O_y$ be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that $(\alpha+1) \times \overline{O_y}$ is a product of two compact spaces and is thus compact (and normal). Let $R_{H,y}$ and $R_{K,y}$ be disjoint open sets in $(\alpha+1) \times \overline{O_y}$ such that $H \cap (\alpha+1) \times \overline{O_y} \subset R_{H,y}$ and $K \cap (\alpha+1) \times \overline{O_y} \subset R_{K,y}$. Note that $[\alpha+1,\omega_1) \times \overline{O_y}$ contains no points of $H$. Then $Q_{H,y}=R_{H,y}$ and $Q_{K,y}=R_{K,y} \cup [\alpha+1,\omega_1) \times \overline{O_y}$ are the desired open sets. This completes the proof of Claim 3.

To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.

Claim 4
Let $Z$ be a space. Let $C$ and $D$ be disjoint closed subsets of $Z$. Suppose that $\left\{U_1,U_2,\cdots,U_m \right\}$ is a collection of open subsets of $Z$ covering $C \cup D$ such that for each $i=1,2,\cdots,m$, only one of the following holds:

$C \cap \overline{U_i} \ne \varnothing \text{ and } D \cap \overline{U_i}=\varnothing$
$C \cap \overline{U_i} = \varnothing \text{ and } D \cap \overline{U_i} \ne \varnothing$

Then there exist disjoint open subsets of $Z$ separating $C$ and $D$.

Proof of Claim 4
Let $U_C=\cup \left\{U_i: \overline{U_i} \cap C \ne \varnothing \right\}$ and $U_D=\cup \left\{U_i: \overline{U_i} \cap D \ne \varnothing \right\}$. Note that $\overline{U_C}=\cup \left\{\overline{U_i}: \overline{U_i} \cap C \ne \varnothing \right\}$. Likewise, $\overline{U_D}=\cup \left\{\overline{U_i}: \overline{U_i} \cap D \ne \varnothing \right\}$. Let $V_C=U_C-\overline{U_D}$ and $V_D=U_D-\overline{U_C}$. Then $V_C$ and $V_D$ are disjoint open sets. Furthermore, $C \subset V_C$ and $D \subset V_D$. This completes the proof of Claim 4.

Now back to the proof of Theorem 1. For each $y \in Y$, let $O_y$, $Q_{H,y}$ and $Q_{K,y}$ be as in Claim 3. Since $Y$ is compact, there exists $\left\{y_1,y_2,\cdots,y_n \right\} \subset Y$ such that $\left\{O_{y_1},O_{y_2},\cdots,O_{y_n} \right\}$ is a cover of $Y$. For each $i=1,\cdots,n$, let $L_i=Q_{H,y_i} \cap (\omega_1 \times O_y)$ and $M_i=Q_{K,y_i} \cap (\omega_1 \times O_y)$. Note that both $L_i$ and $M_i$ are open in $\omega_1 \times Y$. To apply Claim 4, rearrange the open sets $L_i$ and $M_i$ and re-label them as $U_1,U_2,\cdots,U_m$. By letting $Z=\omega_1 \times Y$, $C=H$ and $D=K$, the open sets $U_i$ satisfy Claim 4. Tracing the $U_i$ to $L_j$ or $M_j$ and then to $Q_{H,y_j}$ and $Q_{K,y_j}$, it is clear that the two conditions in Claim 4 are satisfied:

$H \cap \overline{U_i} \ne \varnothing \text{ and } K \cap \overline{U_i}=\varnothing$
$H \cap \overline{U_i} = \varnothing \text{ and } K \cap \overline{U_i} \ne \varnothing$

Then by Claim 4, the disjoint closed sets $H$ and $K$ can be separated by two disjoint open subsets of $\omega_1 \times Y$. $\blacksquare$

The theorem proved in [4] is essentially the statement that for any compact space $Y$, the product $\kappa^+ \times Y$ is normal if and only $t(Y) \le \kappa$. Here $\kappa^+$ is the first ordinal of the next cardinal that is greater than $\kappa$.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Gruenhage, G., Nogura, T., Purisch, S., Normality of $X \times \omega_1$, Topology and its Appl., 39, 263-275, 1991.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

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$\copyright \ 2014 - 2015 \text{ by Dan Ma}$

# Cartesian Products of Two Paracompact Spaces

In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking two-factor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

With Results 1 and 2, compact spaces and $\sigma$-compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.

Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.

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Paracompact Spaces

First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a cover of $X$ if $X=\bigcup \mathcal{A}$ (in words every point of the space belongs to one set in the collection). Furthermore, $\mathcal{A}$ is an open cover of $X$ is it is a cover of $X$ consisting of open subsets of $X$.

Let $\mathcal{A}$ and $\mathcal{B}$ be covers of the space $X$. The cover $\mathcal{B}$ is said to be a refinement of $\mathcal{A}$ ($\mathcal{B}$ is said to refine $\mathcal{A}$) if for every $B \in \mathcal{B}$, there is some $A \in \mathcal{A}$ such that $B \subset A$. The cover $\mathcal{B}$ is said to be an open refinement of $\mathcal{A}$ if $\mathcal{B}$ refines $\mathcal{A}$ and $\mathcal{B}$ is an open cover.

A collection $\mathcal{A}$ of subsets of $X$ is said to be a locally finite collection if for each point $x \in X$, there is a non-empty open subset $V$ of $X$ such that $x \in V$ and $V$ has non-empty intersection with at most finitely many sets in $\mathcal{A}$. An open cover $\mathcal{A}$ of $X$ is said to have a locally finite open refinement if there exists an open cover $\mathcal{C}$ of $X$ such that $\mathcal{C}$ refines $\mathcal{A}$ and $\mathcal{C}$ is a locally finite collection. We have the following definition.

Definition

The space $X$ is said to be paracompact if every open cover of $X$ has a locally finite open refinement.

A collection $\mathcal{U}$ of subsets of the space $X$ is said to be a $\sigma$-locally finite collection if $\mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i$ such that each $\mathcal{U}_i$ is a locally finite collection of subsets of $X$. Consider the property that every open cover of $X$ has a $\sigma$-locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following holds:

For each open cover $\left\{U_t: t \in T \right\}$ of $X$, there exists a locally finite open cover $\left\{V_t: t \in T \right\}$ such that $\overline{V_t} \subset U_t$ for each $t \in T$.

Theorem 3 below shows that paracompactness is hereditary with respect to $F_\sigma$-subsets.

Theorem 3
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proof of Theorem 3
Let $X$ be paracompact. Let $Y \subset X$ such that $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is a closed subset of $X$. Let $\mathcal{U}$ be an open cover of $Y$. For each $U \in \mathcal{U}$, let $U^*$ be open in $X$ such that $U^* \cap Y=U$.

For each $i$, let $\mathcal{U}_i^*$ be the set of all $U^*$ such that $U \cap Y_i \ne \varnothing$. Let $\mathcal{V}_i^*$ be a locally finite refinement of $\mathcal{U}_i^* \cup \left\{X-Y_i \right\}$. Let $\mathcal{V}_i$ be the following:

$\mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}$

It is clear that each $\mathcal{V}_i$ is a locally finite collection of open set in $Y$ covering $Y_i$. All the $\mathcal{V}_i$ together form a refinement of $\mathcal{U}$. Thus $\mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. By Theorem 1, the $F_\sigma$-set $Y$ is paracompact. $\blacksquare$
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Result 1

Result 1 is the statement that:

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let $X$ be any space and $Y$ be compact. For each $x \in X$ and for each open set $U \subset X \times Y$ such that $\left\{x \right\} \times Y \subset U$, there is an open set $O \subset X$ such that $\left\{x \right\} \times Y \subset O \times Y \subset U$.

Proof of Result 1
Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, choose a finite $\mathcal{U}_x \subset \mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x \right\} \times Y$. By the Tube Lemma, for each $x \in X$, there is an open set $O_x \subset X$ such that $\left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is paracompact, by Theorem 2, let $\left\{W_x: x \in X \right\}$ be a locally finite open refinement of $\left\{O_x: x \in X \right\}$ such that $W_x \subset O_x$ for each $x \in X$.

Let $\mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}$. We claim that $\mathcal{W}$ is a locally finite open refinement of $\mathcal{U}$. First, this is an open cover of $X \times Y$. To see this, let $(a,b) \in X \times Y$. Then $a \in W_x$ for some $x \in X$. Furthermore, $a \in O_x$ and $(a,b) \in \cup \mathcal{U}_x$. Thus, $(a,b) \in (W_x \times Y) \cap U$ for some $U \in \mathcal{U}_x$. Secondly, it is clear that $\mathcal{W}$ is a refinement of the original cover $\mathcal{U}$.

It remains to show that $\mathcal{W}$ is locally finite. To see this, let $(a,b) \in X \times Y$. Then there is an open $V$ in $X$ such that $x \in V$ and $V$ can meets only finitely many $W_x$. Then $V \times Y$ can meet only finitely many sets in $\mathcal{W}$. $\blacksquare$

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Result 2

Result 2 is the statement that:

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Proof of Result 2
Note that the $\sigma$-compact space $Y$ is Lindelof. Since regular Lindelof are normal, $Y$ is normal and is thus completely regular. So we can embed $Y$ into a compact space $K$. For example, we can let $K=\beta Y$, which is the Stone-Cech compactification of $Y$ (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing $Y$ will do. By Result 1, $X \times K$ is paracompact. Note that $X \times Y$ can be regarded as a subspace of $X \times K$.

Let $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is compact in $Y$. Note that $X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i$ and each $X \times Y_i$ is a closed subset of $X \times K$. Thus the product $X \times Y$ is an $F_\sigma$-subset of $X \times K$. According to Theorem 3, $F_\sigma$-subsets of any paracompact space is paracompact space. Thus $X \times Y$ is paracompact. $\blacksquare$

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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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$\copyright \ \ 2012$

# Stone-Cech Compactifications – Another Two Characterizations

Let $X$ be a completely regular space. Let $\beta X$ be the Stone-Cech compactification of $X$. We present two characterizations of $\beta X$ in addition to three others that are discussed previously. In all, these five characterizations can help us derive many of the basic properties of $\beta X$. We prove the following theorems.

Theorem C4
Let $X$ be a completely regular space. Every two completely separated subsets of $X$ have disjoint closures in $\beta X$.

Theorem U4
The property described in Theorem C4 is unique to $\beta X$. That is, if $\alpha X$ is a compactification of $X$ satisfying the condition that every two completely separated subsets of $X$ have disjoint closures in $\alpha X$, then $\alpha X$ must be $\beta X$.

Theorem C5
Let $X$ be a normal space. Then every two disjoint closed subsets of $X$ have disjoint closures in $\beta X$.

Theorem U5
If $\alpha X$ is a compactification of $X$ satisfying the property that every two disjoint closed subsets of $X$ have disjoint closures in $\alpha X$, then $X$ is normal and $\alpha X$ must be $\beta X$.

The C theorem and U theorem with the same number work as a pair. The C theorem asserts that $\beta X$ has a certain property. The corresponding U theorem asserts that of all the compactifications of $X$, $\beta X$ is the only one with the property in question. Whenever we can show a given compactification does not possess the property described in the C-U theorem pair, we know that that compactification is not $\beta X$ (consequence of the C theorem). Whenever we can show that a given compactification has the property described in the C-U theorem pair, we know that that compactification must be $\beta X$ (a consequence of the U theorem).

Three other sets of characterizations (Theorems C1, U1, C2, U2, C3 and U3) have been established previously. See the links found below.
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Completely Separated Sets

Let $Y$ be a completely regular space. Let $H \subset Y$ and $K \subset Y$. The sets $H$ and $K$ are said to be completely separated in $Y$ if there is a continuous function $f:Y \rightarrow [0,1]$ such that for each $y \in H$, $f(y)=0$ and for each $y \in K$, $f(y)=1$ (this can also be expressed as $f(H) \subset \left\{0 \right\}$ and $f(K) \subset \left\{1 \right\}$). If $H$ and $K$ are completely separated, $\overline{H}$ and $\overline{K}$ are necessarily disjoint closed sets, since $\overline{H} \subset f^{-1}(0)$ and $\overline{K} \subset f^{-1}(1)$.

The Urysohn’s lemma can be stated as: a space is a normal space if and only if every two disjoint closed sets are completely separated. Thus disjoint closed sets are not necessarily completely separated (such sets can be found in non-normal spaces).

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To prove Theorem U4, we need a lemma and a theorem. Most of the work in proving Theorem U4 is carried out in Theorem 2 below.

Lemma 1
Let $Y$ be a compact space. Let $U$ be an open subset of $Y$. Let $\mathcal{C}$ be a collection of compact subsets of $Y$ such that $\cap \mathcal{C} \subset U$. Then there exists a finite collection $\left\{C_1,C_2,\cdots,C_n \right\} \subset \mathcal{C}$ such that $\bigcap \limits_{i=1}^n C_i \subset U$.

Proof of Lemma 1
Let $D=Y-U$, which is compact. Let $\mathcal{O}$ be the collection of all $Y-C$ where $C \in \mathcal{C}$. Note that $\cap \mathcal{C} \subset U$ implies that $D \subset \cup \mathcal{O}$. Thus $\mathcal{O}$ is a collection of open sets covering the compact set $D$. We have $\left\{O_1,O_2,\cdots,O_n \right\} \subset \mathcal{O}$ such that $D \subset \bigcup \limits_{i=1}^n O_i$. Each $O_i=Y-C_i$ for some $C_i \in \mathcal{C}$. Now $\left\{C_1,C_2,\cdots,C_n \right\}$ is the desired finite collection. $\blacksquare$

Theorem 2
Let $T$ be a completely regular space. Let $S$ be a dense subspace of $T$. Let $f:S \rightarrow K$ be a continuous function from $S$ into a compact space $K$. Suppose that every two completely separated subsets of $S$ have disjoint closures in $T$. Then $f$ can be extended to a continuous $F:T \rightarrow K$.

Proof
For each $t \in T$, let $\mathcal{O}(t)$ be the set of all open subsets of $T$ containing $t$. For each $t \in T$, let $\mathcal{W}(t)$ be the set of all $\overline{f(S \cap O)}$ where $O \in \mathcal{O}(t)$. Note that each $\mathcal{W}(t)$ consists of compact subsets of $K$. The theorem is established by proving the following claims.

Claim 1
For each $t \in T$, the collection $\mathcal{W}(t)$ has non-empty intersection.

For any $O_1, O_2, \cdots, O_n \in \mathcal{O}(t)$, we have the following:

$\overline{f(S \cap O_1 \cap O_2 \cap \cdots \cap O_n)} \subset \overline{f(S \cap O_1)} \cap \overline{f(S \cap O_2)} \cap \cdots \cap \overline{f(S \cap O_n)}$

The above shows that $\mathcal{W}(t)$ has the finite intersection property (f. i. p.). It is a well known fact that in a compact space, any collection of sets with f. i. p. has non-empty intersection (see [1] or [2] or see The Finite Intersection Property in Compact Spaces and Countably Compact Spaces in this blog).

Claim 2
For each $t \in T$, $\cap \mathcal{W}(t)$ has only one point.

Let $t \in T$. Suppose that

$\left\{k_1,k_2 \right\} \subset \cap \mathcal{W}(t)$ where $k_1 \ne k_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Then there exist open subsets $U_1$ and $U_2$ of $K$ such that $k_1 \in U_1$, $k_2 \in U_2$ and $\overline{U_1} \cap \overline{U_2} = \varnothing$. Since $K$ is compact, it is a normal space. By the Urysohn’s lemma, there exists a continuous $g:K \rightarrow [0,1]$ such that for each $k \in \overline{U_1}$, $g(k)=0$ and for each $k \in \overline{U_2}$, $g(k)=1$. Then because of the function $g \circ f:S \rightarrow [0,1]$, the sets $f^{-1}(\overline{U_1})$ and $f^{-1}(\overline{U_2})$ are completely separated sets in $S$. By assumption, these two sets have disjoint closures in $T$, i.e.,

$\text{ }$
$\overline{f^{-1}(\overline{U_1})} \cap \overline{f^{-1}(\overline{U_2})} = \varnothing \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
$\text{ }$

The point $t$ cannot be in both of the sets in $(2)$. Assume the following:

$\text{ }$
$t \notin \overline{f^{-1}(\overline{U_1})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$
$\text{ }$

Then $H=T- \overline{f^{-1}(\overline{U_1})} \in \mathcal{O}(t)$. Note that $S \cap H=S-\overline{f^{-1}(\overline{U_1})}$. Furthermore, $\overline{f(S-\overline{f^{-1}(\overline{U_1})})} \in \mathcal{W}(t)$. Thus we have:

$\text{ }$
$k_1 \in \cap \mathcal{W}(t) \subset \overline{f(S-\overline{f^{-1}(\overline{U_1})})}=W$
$\text{ }$

Since $k_1 \in W$ and $U_1$ is an open set containing $k_1$, $U_1$ contains at least one point of $f(S-\overline{f^{-1}(\overline{U_1})})$. Choose $z \in U_1$ such that $z \in f(S-\overline{f^{-1}(\overline{U_1})})$. Now choose $a \in S-\overline{f^{-1}(\overline{U_1})}$ such that $f(a)=z$. First we have $a \notin \overline{f^{-1}(\overline{U_1})}$ and thus $a \notin f^{-1}(\overline{U_1})$. Secondly since $f(a)=z \in U_1$, we have $a \in f^{-1}(U_1) \subset f^{-1}(\overline{U_1})$. We now have $a \notin f^{-1}(\overline{U_1})$ and $a \in f^{-1}(\overline{U_1})$, a contradiction. If we assume $t \notin \overline{f^{-1}(\overline{U_2})}$, we can also derive a contradiction in a similar derivation. Thus the assumption in $(1)$ above is faulty. The intersection $\cap \mathcal{W}(t)$ can only have one point.

Claim 3
For each $t \in S$, $\cap \mathcal{W}(t) =\left\{f(t) \right\}$.

Let $t \in S$. Suppose that $\cap \mathcal{W}(t) =\left\{p \right\}$ where $p \ne f(t)$. the rest of the proof for Claim 3 is similar to that of Claim 2. For the sake of completeness, we give a sketch.

There exist open subsets $U_1$ and $U_2$ of $K$ such that $p \in U_1$, $f(t) \in U_2$ and $\overline{U_1} \cap \overline{U_2} = \varnothing$. By the same argument as in Claim 2, we have the condition $(2)$, i.e., $\overline{f^{-1}(\overline{U_1})} \cap \overline{f^{-1}(\overline{U_2})} = \varnothing$. Since $t \in f^{-1}(U_2)$, $t \notin \overline{f^{-1}(\overline{U_1})}$. The remainder of the proof of Claim 3 is the same as above starting with condition $(3)$ with $p=k_1$. A contradiction will be obtained. We can conclude that the assumption that $\cap \mathcal{W}(t) =\left\{p \right\}$ where $p \ne f(t)$ must be faulty. Thus Claim 3 is established.

Claim 4
For each $t \in T$, define $F:T \rightarrow K$ by letting $F(t)$ be the point in $\cap \mathcal{W}(t)$. Note that this function extends $f$. Furthermore, the map $F:T \rightarrow K$ is continuous.

To show $F$ is continuous, let $t \in T$ and let $F(t) \in E$ where $E$ is open in $K$. The collection $\mathcal{W}(t)$ is a collection of compact subsets of $K$ such that $\left\{F(t) \right\} =\cap \mathcal{W}(t) \subset E$. By Lemma 1, there exists $\left\{C_1,\cdots,C_n \right\} \subset \mathcal{W}(t)$ such that $\bigcap \limits_{i=1}^n C_i \subset E$. By the definition of $\mathcal{W}(t)$, there exists $\left\{O_1,O_2,\cdots,O_n \right\} \subset \mathcal{O}(t)$ such that each $C_i=\overline{f(S \cap O_i)}$. Let $O=O_1 \cap O_2 \cap \cdots \cap O_n$. We have:

$\text{ }$
$\overline{f(S \cap O)} \subset \bigcap \limits_{i=1}^n \overline{f(S \cap O_i)} \subset E \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$
$\text{ }$

Note that $O$ is an open subset of $T$ and $t \in O$. We show that $F(O) \subset E$. Pick $a \in O$. According to the definition of $\mathcal{W}(a)$, we have $\left\{F(a) \right\}=\bigcap \limits_{U \in \mathcal{O}(a)} \overline{f(S \cap U)}$. Since $O \in \mathcal{O}(a)$, we have $F(a) \in \overline{f(S \cap O)}$. Thus by $(4)$, we have $F(a) \in E$. Thus Claim 4 is established.

With all the above claims established, we completed the proof of Theorem 2. $\blacksquare$

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Theorem C4 and Theorem U4

Proof of Theorem C4
In proving C4, we use Theorem C3, which is found in C*-Embedding Property and Stone-Cech Compactification.

Let $E$ and $F$ be two completely separated sets in $X$. Then there exists some continuous $g:X \rightarrow [0,1]$ such that for each $x \in E$, $g(x)=0$ and for each $x \in F$, $g(x)=1$. By Theorem C3, $g$ is extended by some continuous $G:\beta X \rightarrow [0,1]$. The sets $G^{-1}(0)$ and $G^{-1}(1)$ are disjoint closed sets in $\beta X$. Furthermore, $E \subset G^{-1}(0)$ and $F \subset G^{-1}(1)$. Thus $E$ and $F$ have disjoint closures in $\beta X$. $\blacksquare$

Proof of Theorem U4
In proving U4, we use Theorem U1, which is stated and proved in Two Characterizations of Stone-Cech Compactification.

Suppose that $\alpha X$ is a compactification of $X$ satisfying the condition that every two completely separated subsets of $X$ have disjoint closures in $\alpha X$. Let $g:X \rightarrow Y$ be a continuous function from $X$ into a compact space $Y$. By Theorem 2, $g$ can be extended by a continuous $G:\alpha X \rightarrow Y$. By Theorem U1, $\alpha X$ must be $\beta X$. $\blacksquare$

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Theorem C5 and Theorem U5

Proof of Theorem C5
Let $X$ be a normal space. According to the Urysohn’s lemma, every two disjoint closed sets are completely separated. Thus by Theorem C4, every two disjoint closed subsets of $X$ have disjoint closures in $\beta X$. $\blacksquare$

Proof of Theorem U5
Suppose that $\alpha X$ is a compactification of $X$ satisfying the property that every two disjoint closed subsets of $X$ have disjoint closures in $\alpha X$. To show that $X$ is normal, let $H$ and $K$ be disjoint closed subsets of $X$. By assumption about $\alpha X$, $\overline{H}$ and $\overline{K}$ (closures in $\alpha X$) are disjoint. Since $\alpha X$ are compact and Hausdorff, $\alpha X$ is normal. Then $\overline{H}$ and $\overline{K}$ can be separated by disjoint open subsets $U$ and $V$ of $\alpha X$. Thus $U \cap X$ and $V \cap X$ are disjoint open subsets of $X$ separating $H$ and $K$.

We use Theorem U4 to prove Theorem U5. We show that $\alpha X$ satisfies Theorem U4. To this end, let $E$ and $F$ be two completely separated sets in $X$. We show that $E$ and $F$ have disjoint closures in $\alpha X$. There exists some continuous $f:X \rightarrow [0,1]$ such that for each $x \in E$, $f(x)=0$ and for each $x \in F$, $f(x)=1$. Then $f^{-1}(0)$ and $f^{-1}(1)$ are disjoint closed sets in $X$ such that $E \subset f^{-1}(0)$ and $F \subset f^{-1}(1)$. By assumption about $\alpha X$, $f^{-1}(0)$ and $f^{-1}(1)$ have disjoint closures in $\alpha X$. This implies that $E$ and $F$ have disjoint closures in $\alpha X$. Then by Theorem U4, $\alpha X$ must be $\beta X$. $\blacksquare$

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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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$\copyright \ \ 2012$

# 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.

_______________________________________________________________________________________
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 $F \circ \beta=f$.

$\text{ }$

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.
_______________________________________________________________________________________
$\text{ }$

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 $\le$.

$\text{ }$

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.
_______________________________________________________________________________________
$\text{ }$

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.
_______________________________________________________________________________________
$\text{ }$

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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$\copyright \ \ 2012$

# Stone-Cech Compactification is Maximal

Let $X$ be a completely regular space. Let $\beta X$ be the Stone-Cech compactification of $X$. In a previous post, we show that among all compactifcations of $X$, the Stone-Cech compactification $\beta X$ is maximal with respect to a partial order $\le$ (see Theorem C2 in Two Characterizations of Stone-Cech Compactification). As a result of the maximality, $\beta X$ is the largest among all compactifications of $X$ both in terms of cardinality and weight. We also establish an upper bound for the cardinality of $\beta X$ and an upper bound for the weight of $\beta X$. As a result, we have upper bounds for cardinalities and weights for all compactifications of $X$. We prove the following points.

Upper Bounds for Stone-Cech Compactification

1. $\lvert \beta X \lvert \le 2^{2^{d(X)}}$.
2. $w(\beta X) \le 2^{d(X)}$.
3. Stone-Cech Compactification is Maximal

4. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le \lvert \beta X \lvert$.
5. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le w(\beta X)$.
6. Upper Bounds for all Compactifications

7. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le 2^{d(X)}$.
8. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le 2^{2^{d(X)}}$.

It is clear that Results 5 and 6 follow from the preceding results. The links for other posts on Stone-Cech compactification can be found toward the end of this post

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Some Cardinal Functions

Let $X$ be a space. The density of $X$ is denoted by $d(X)$ and is defined to be the smallest cardinality of a dense set in $X$. For example, if $X$ is separable, then $d(X)=\omega$. The weight of the space $X$ is denoted by $w(X)$ and is defined to be the smallest cardinality of a base of the space $X$. For example, if $X$ is second countable (i.e. having a countable space), then $w(X)=\omega$. Both $d(X)$ and $w(X)$ are cardinal functions that are commonly used in topological discussion. Most authors require that cardinal functions only take on infinite cardinals. We also adopt this convention here. We use $c$ to denote the cardinality of the continuum (the cardinality of the real line $\mathbb{R}$).

If $\mathcal{K}$ is a cardinal number, then $2^{\mathcal{K}}$ refers to the cardinal number that is the cardinallity of the set of all functions from $\mathcal{K}$ to $2=\left\{0,1 \right\}$. Equivalently, $2^{\mathcal{K}}$ is also the cardinality of the power set of $\mathcal{K}$ (i.e. the set of all subsets of $\mathcal{K}$). If $\mathcal{K}=\omega$ (the first infinite ordinal), then $2^\omega=c$ is the cardinality of the continuum.

If $X$ is separable, then $d(X)=\omega$ (as noted above) and we have $2^{d(X)}=c$ and $2^{2^{d(X)}}=2^c$. Result 5 and Result 6 imply that $2^c$ is an upper bound for the cardinality of all compactifications of any separable space $X$ and $c$ is an upper bound of the weight of all compactifications of any separable space $X$.

In general, Result 5 and Result 6 indicate that the density of $X$ bounds the cardinality of any compactification of $X$ by two exponents and the density of $X$ bounds the weight of any compactification of $X$ by one exponent.

Another cardinal function related to weight is that of the network weight. A collection $\mathcal{N}$ of subsets of the space $X$ is said to be a network for $X$ if for each point $x \in X$ and for each open subset $U$ of $X$ with $x \in U$, there is some set $A \in \mathcal{N}$ with $x \in A \subset U$. Note that sets in a network do not have to be open. However, any base for a topology is a network. The network weight of the space $X$ is denoted by $nw(X)$ and is defined to be the least cardinality of a network for $X$. Since any base is a network, we have $nw(X) \le w(X)$. It is also clear that $nw(X) \le \lvert X \lvert$ for any space $X$. Our interest in network and network weight is to facilitate the discussion of Lemma 2 below. It is a well known fact that in a compact space, the weight and the network weight are the same (see Result 5 in Spaces With Countable Network).
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Some Basic Facts

We need the following two basic results.

Lemma 1
Let $X$ be a space. Let $C(X)$ be the set of all continuous functions $f:X \rightarrow \mathbb{R}$. Then $\lvert C(X) \lvert \le 2^{d(X)}$.

Lemma 2
Let $S$ be a space and let $T$ be a compact space. Suppose that $T$ is the continuous image of $S$. Then $w(T) \le w(S)$.

Proof of Lemma 1
Let $A \subset X$ be a dense set with $\lvert A \lvert=2^{d(X)}$. Let $\mathbb{R}^A$ be the set of all functions from $A$ to $\mathbb{R}$. Consider the map $W:C(X) \rightarrow \mathbb{R}^A$ by $W(f)= f \upharpoonright A$. This is a one-to-one map since $f=g$ whenever $f$ and $g$ agree on a dense set. Thus we have $\lvert C(X) \lvert \le \lvert \mathbb{R}^A \lvert$. Upon doing some cardinal arithmetic, we have $\lvert \mathbb{R}^A \lvert=2^{d(X)}$. Thus Lemma 1 is established. $\blacksquare$

Proof of Lemma 2
Let $g:S \rightarrow T$ be a continuous function from $S$ onto $T$. Let $\mathcal{B}$ be a base for $S$ such that $\lvert \mathcal{B} \lvert=w(S)$. Let $\mathcal{N}$ be the set of all $g(B)$ where $B \in \mathcal{B}$. Note that $\mathcal{N}$ is a network for $T$ (since $g$ is a continuous function). So we have $nw(T) \le \lvert \mathcal{N} \lvert \le \lvert \mathcal{B} \lvert = w(S)$. Since $T$ is compact, $w(T)=nw(T)$ (see Result 5 in Spaces With Countable Network). Thus we have $nw(T)=w(T) \lvert \le w(S)$. $\blacksquare$

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Results 1 and 2

Let $X$ be a completely regular space. Let $I$ be the unit interval $[0,1]$. We show that the Stone-Cech compactification $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ (the product of $2^{d(X)}$ many copies of $I$). The cardinality of $I^{\mathcal{K}}$ is $2^{2^{d(X)}}$, thus leading to Result 1.

Let $C(X,I)$ be the set of all continuous functions $f:X \rightarrow I$. The Stone-Cech compactification $\beta X$ is constructed by embedding $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$ (see Embedding Completely Regular Spaces into a Cube or A Beginning Look at Stone-Cech Compactification). Thus $\beta X$ is a subspace of $I^{\mathcal{K}_1}$ where $\mathcal{K}_1=\lvert C(X,I) \lvert$.

Note that $C(X,I) \subset C(X)$. Thus $\beta X$ can be regarded as a subspace of $I^{\mathcal{K}_2}$ where $\mathcal{K}_2=\lvert C(X) \lvert$. By Lemma 1, $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$.

To see Result 2, note that the weight of $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ is $2^{d(X)}$. Then $\beta X$, as a subspace of the product space, must have weight $\le 2^{d(X)}$. $\blacksquare$

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Results 3 and 4

What drives Result 3 and Result 4 is the following theorem (established in Two Characterizations of Stone-Cech Compactification).

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 $\le$.

$\text{ }$

To define the partial order, for $\alpha_1 X$ and $\alpha_2 X$, both compactifications of $X$, we say that $\alpha_2 X \le \alpha_1 X$ if there is a continuous function $f:\alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1=\alpha_2$. See the following figure.

Figure 1

In this post, we use $\le$ to denote this partial order as well as the order for cardinal numbers. Thus we need to rely on context to distinguish this partial order from the order for cardinal numbers.

Let $\alpha X$ be a compactification of $X$. Theorem C2 indicates that $\alpha X \le \beta X$ (partial order), which means that there is a continuous $f:\beta X \rightarrow \alpha X$ such that $f \circ \beta=\alpha$ (the same point in $X$ is mapped to itself by $f$). Note that $\alpha X$ is the image of $\beta X$ under the function $f:\beta X \rightarrow \alpha X$. Thus we have $\lvert \alpha X \lvert \le \lvert \beta X \lvert$ (cardinal number order). Thus Result 3 is established.

By Lemma 2, the existence of the continuous function $f:\beta X \rightarrow \alpha X$ implies that $w(\alpha X) \le w(\beta X)$ (cardinal number order). Thus Result 4 is established.

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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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$\copyright \ \ 2012$

# C*-Embedding Property and Stone-Cech Compactification

This is a continuation of an introduction of Stone-Cech compactification started in two previous posts (first post: A Beginning Look at Stone-Cech Compactification; second post: Two Characterizations of Stone-Cech Compactification). In this post, we present another characterization of the Stone-Cech compactification, that is, for any completely regular space $X$, $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$ and that any compactification of $X$ in which $X$ is $C^*$-embedded must be $\beta X$. In other words, this property of $C^*$-embedding is unique to Stone-Cech compactification. We prove the following two theorems (U3 has two versions).

The links for other posts on Stone-Cech compactification can be found toward the end of this post.

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Definition. Let $Y$ be a space. Let $A \subset Y$. The subspace $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ is extendable to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

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 $\beta X$.
$\text{ }$

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Other Characterizations

Two other characterizations of $\beta X$ are proved in the previous post (Two Characterizations of Stone-Cech Compactification).

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 $F \circ \beta=f$.

$\text{ }$

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

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 the largest compactification.

$\text{ }$

Theorem U2
The property in Theorem C2 is unique to $\beta X$. That is, if $\alpha X$ is a compactification of $X$, then $\alpha X$ must be equivalent to $\beta X$.

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Remark

The C theorems and the U theorems are a great tool to determine whether a given compactification is $\beta X$. Whenever a compactification $\alpha X$ of a space $X$ satisfies the property belonging to a C theorem, based on the corresponding U theorem, we know that this compactification $\alpha X$ must be $\beta X$. For example, any compactification $\alpha X$ that satisfies the function extension property in Theorem C1 must be $\beta X$. Th $C^*$-embedding property in Theorem C3 and Theorem U3 (both versions) is also a function extension property much like that in Theorems C1 and U1, but is easier to use. The reason being that we only need to extend a smaller class of continuous functions (i.e., to check whether functions from $X$ into $I=[0,1]$ can be extended), rather than checking all continuous functions from $X$ to arbitrary compact spaces. As the following example below about $\beta \omega_1$ illustrates that the $C^*$-embedding in Theorem C3 and U3.1 can be used to describe $\beta X$ explicitly.

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Proving Theorem U3.1 and Theorem U3.2

Let $Y$ be a space. Let $A$ be a subspace of $X$. Recall that $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

Any bounded continuous function $f: X \rightarrow \mathbb{R}$ can be regarded as $f: X \rightarrow I_f$ where $I_f$ is some closed and bounded interval. The $C^*$-embedding property in Theorem C3 is a function extension property like the one in Theorem C1, except that it deals with function from $X$ into a specific type of compact spaces $Y$, namely the closed and bounded intervals in $\mathbb{R}$. Theorem C3 is a corollary of Theorem C1 (see below). So we only need to prove Theorem U3.1 and Theorem U3.2. Theorem U3.2 is a corollary of Theorem U3.1.

Proof of Theorem U3.1
By Theorem C2, we have $\alpha X \le \beta X$. So we only need to show $\beta X \le \alpha X$. To this end, we need to produce a continuous function $H: \alpha X \rightarrow \beta X$ such that $H \circ \alpha=\beta$.

Let $C(X,I)$ be the set of all continuous functions from $X$ into $I$. For each $f \in C(X,I)$, let $I_f=I$. Recall that $\beta X$ is embedded in the cube $\prod \limits_{f \in C(X,I)} I_f$ by the mapping $\beta$. For each $f \in C(X,I)$, let $\pi_f$ be the projection map from this cube into $I_f$.

Each $f \in C(X,I)$ can be expressed as $f=\pi_f \circ \beta$. Thus by assumption, each $f$ can be extended by $\hat{f}: \alpha X \rightarrow I$. Now define $H: \alpha X \rightarrow \prod \limits_{f \in C(X,I)} I_f$ by the following:

For each $t \in \alpha X$, $H(t)=a=< a_f >_{f \in C(X,I)}$ such that $a_f=\hat{f}(t)$

For each $x \in \alpha(X)$, we have $H(\alpha(x))=\beta(x)$. Note that $\hat{f}$ agrees with $f$ on $\alpha(X)$ since $\hat{f}$ extends $f$. So we have $H(\alpha(x))=a$ where $a_f=\hat{f}(\alpha(x))=f(x)$ for each $f \in C(X,I)$. On the other hand, by definition of $\beta$, we have $\beta(x)=a$ where $a_f=f(x)$ for each $f \in C(X,I)$. Thus we have $H \circ \alpha=\beta$ and the following:

$H(\alpha(X)) \subset \beta(X) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

It is straightforward to verify that $H$ is continuous. Note that $\alpha(X)$ is dense in $\alpha X$. Since $H$ is continuous, $H(\alpha(X))$ is dense in $H(\alpha X)$. Thus we have:

$H(\alpha X)=\overline{H(\alpha(X))} \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Putting $(1)$ and $(2)$ together, we have the following:

$H(\alpha X)=\overline{H(\alpha(X))} \subset \overline{\beta(X)}=\beta X$

Thus we can describe the map $H$ as $H: \alpha X \rightarrow \beta X$. As noted before, we have $H \circ \alpha=\beta$. Thus $\beta X \le \alpha X$. $\blacksquare$

Proof of Theorem U3.2
Suppose $\alpha X$ is a compactification of $X$ such that $X$ is $C^*$-embedded in $\alpha X$. Then every bounded continuous $f:X \rightarrow I_f$ can be extended to $\hat{f}:\alpha X \rightarrow I_f$ where $I_f$ is some closed and bounded interval containing the range. In particular, this means every continuous $f:X \rightarrow I$ can be extended. By Theorem U3.1, we have $\alpha X \approx \beta X$. $\blacksquare$

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Example

This is one example where we can use $C^*$-embedding to describe $\beta X$ explicitly.

Let $\omega_1$ be the first uncountable ordinal. Let $\omega_1+1$ be the successor ordinal of $\omega_1$ (i.e. $\omega_1$ with one additional point at the end). Consider $X=\omega_1$ and $Y=\omega_1+1$ as topological spaces with the order topology derived from the well ordering of the ordinals. The space $Y$ is a compactification of $X$. In fact $Y$ is the one-point compactification of $X$.

It is well known that every continuous real-valued function on $X$ is bounded (note that $X$ here is countably compact and hence pseudocompact). Furthermore, every continuous real-valued function on $X$ is eventually constant. This means that if $f:X \rightarrow \mathbb{R}$ is continuous, for some $\alpha < \omega_1$, $f$ is constant on the final segment $X_\alpha=\left\{\rho < \omega_1: \rho>\alpha \right\}$ (see result B in The First Uncountable Ordinal). As a result, every continuous bounded real-valued function $f:X \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$. Then according to Theorem U3.2, $\beta X=\beta \omega_1=Y=\beta \omega_1+1$.

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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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$\copyright \ \ 2012$

# Two Characterizations of Stone-Cech Compactification

This is the second post on Stone-Cech compactification (continuing from A Beginning Look at Stone-Cech Compactification). In this post, we establish two characterizations of Stone-Cech compactification. The first one is represented in the following diagram. The second one is that Stone-Cech compactification is maximal with respect to a certain partial order.

The first characterization is a central characteristic of Stone-Cech compactification. It is a function extension property that uniquely characterizes the Stone-Cech compactification of a completely regularly space. Here’s the diagram.

Figure 1

In this diagram, $X$ is a completely regular space and $\beta X$ is the Stone-Cech compactification of $X$ where $\beta$ is the homeomorphism mapping $X$ onto $\beta(X)$, which is dense in $\beta X$. The function $f: X \rightarrow Y$ is an arbitrary continuous function where $Y$ is compact. Then there exists a continuous function $F:\beta X \rightarrow Y$ such that $F$ restricted to $\beta(X)$ is identical to the function $f$. In other words, if we think of $X$ as a subset of $\beta X$, any continuous function from $X$ to a compact space can be extended to all of $\beta X$. This function extension property is stated in Theorem C1 below.

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 $F \circ \beta=f$. See Figure 1 above.
$\text{ }$
Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.

Theorem C1 is the statement of the extension property described at the beginning. Theorem U1 states that this property is unique to $\beta X$. That is, of all the possible compactifications of $X$, only $\beta X$ can satisfy Theorem C1.

For the other characterization, see Theorem C2 and Theorem U2 below.

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Defining Stone-Cech Compactification

The definition of $\beta X=\beta_X X$ is given in this previous post (A Beginning Look at Stone-Cech Compactification) and is repeated here again for the sake of completeness. Let $C(X,I)$ be the set of all continuous functions from $X$ into $I=[0,1]$. For each $g \in C(X,I)$, $I_g=[0,1]$. The map $\beta_X:X \rightarrow \prod \limits_{g \in C(X,I)} I_g$ is defined by:

For each $x \in X$, $\beta_X(x)=t=< t_g >_{g \in C(X,I)}$ is the point $t \in \prod \limits_{g \in C(X,I)} I_g$ such that $t_g=g(x)$ for each $g \in C(X,I)$ (i.e. the $g^{th}$ coordinate of the point $t$ is $g(x)$).

For the proof that $\beta_X$ is a homeomorphism, see A Beginning Look at Stone-Cech Compactification. We have the following definition.

Definition
Under the map $\beta_X$, $\beta_X(X)$ is the topological copy of $X$ within the cube $\prod \limits_{f \in C(X,I)} I_f$. The Stone-Cech compactification of $X$ is defined to be the closure of $\beta_X(X)$ in the cube $\prod \limits_{f \in C(X,I)} I_f$, i.e., set $\beta_X X=\overline{\beta_X(X)}$.

When there is no ambiguity as to what the space $X$ is, the embedding $\beta_X$ is written as $\beta$ and the compactification $\beta_X X$ is written as $\beta X$ (as in Figure 1 above). When more than one space is involved, we use subscripts to distinguish the embeddings, e.g., $\beta_X$ and $\beta_Y$.

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Proof of Theorem U1

Let $f:X \rightarrow Y$ be a continuous function from $X$ into a compact Hausdorff space $Y$. Let $\beta_X X$ be the Stone-Cech compactification of $X$ where $\beta_X$ is the homeomorphic embedding that defines $\beta_X X$. Since $Y$ is a completely regular space, it has a Stone-Cech compactification $\beta_Y Y$, where $\beta_Y$ is the homeomorphic embedding. We also define a map $W$ from $\prod \limits_{g \in C(X,I)} I_g$ into $\prod \limits_{g \in C(Y,I)} I_k$. We have the following diagram.

Figure 2

The desired function $F$ will be defined by $F=\beta_Y^{-1} \circ (W \upharpoonright \beta_X X)$. The rest of the proof is to define $W$ and to show that this definition of $F$ makes sense.

To define the function $W$, for each $t \in \prod \limits_{g \in C(X,I)} I_g$, let $W(t)=a$ such that $a_k=t_{k \circ f}$ (i.e. the $k^{th}$ coordinate of $W(t)=a$ is the $(k \circ f)^{th}$ coordinate of $t$). With the definition of $W$, the diagram in Figure 2 commutes, i.e.,

$W \circ \beta_X=\beta_Y \circ f \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Starting with a point $x \in X$ (the upper left corner of the diagram), we can reach the same point in the lower right corner regardless the path we take ($W \circ \beta_X$ or $\beta_Y \circ f$). The following shows the derivation.

One direction:
$x \in X$

$\downarrow$

$\beta_X(x)=t \text{ where } t_g=g(x) \ \forall \ g \in C(X,I)$

$\downarrow$

$W(t)=a \text{ where } a_k=t_{k \circ f}=(k \circ f)(x)=k(f(x)) \ \forall \ k \in C(Y,I)$

_________________________________
The other direction:
$x \in X$

$\downarrow$

$f(x) \in Y$

$\downarrow$

$\beta_Y(f(x))=a \text{ where } a_k=k(f(x)) \ \forall \ k \in C(Y,I)$

It is straightforward to verify that the map $W$ is continuous. Based on $(1)$ above, note that $W(\beta_X(X)) \subset \beta_Y(Y)$. The following derivation shows that $W(\beta_X X) \subset \beta_Y(Y)$.

\displaystyle \begin{aligned} W(\beta_X X)&=W(\overline{\beta_X(X)}) \\&\subset \overline{W(\beta_X(X))} \ \ \ \ \text{ based on the continuity of } W\\&\subset \overline{\beta_Y(Y)} \ \ \ \ \ \ \ \ \ \ \text{ based on (1)}\\&=\beta_Y Y \\&=\beta_Y(Y) \ \ \ \ \ \ \ \ \ \ \text{ based on the compactness of Y} \end{aligned}

With the above derivation, we now know that the function $W$ maps points of $\beta_X X$ to points of $\beta_Y(Y)$. So it makes sense to define $F=\beta_Y^{-1} \circ (W \upharpoonright \beta_X X)$. Note that for each $x \in X$, we have:

\displaystyle \begin{aligned} F(\beta_X(x))&=\beta_Y^{-1}(W(\beta_X(x)) \\&=\beta_Y^{-1}(\beta_Y(f(x))) \\&=f(x) \end{aligned}

Then we have $F \circ \beta_X=f$ and $F$ is the desired function. $\blacksquare$

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Compactifications

In order to prove Theorem U1, we first have a basic discussion on compactifications. Most importantly, we pin down what we mean when we say two compactifications of $X$ are equivalent. In the process, we produce another characterization of Stone-Cech compactification (see Theorem C2 and Theorem U2 below).

Let $X$ be a completely regular space. A pair $(T,\alpha)$ is said to be a compactification of the space $X$ if $T$ is a compact Hausdorff space and $\alpha:X \rightarrow T$ is a homeomorphism from $X$ into $T$ such that $\alpha(X)$ is dense in $T$. More informally, a compactification of the space $X$ can also be thought of as a compact space $T$ containing a topological copy of the space $X$ as a dense subspace.

Given a compactification $(T,\alpha)$, we use the notation $\alpha X$ rather than the pair $(T,\alpha)$. By saying that $\alpha X$ is a compactification of $X$, we mean $\alpha X$ is the compact space $T$ where $\alpha$ is the homeomorphism embedding $X$ onto $\alpha(X)$.

The Stone-Cech compactification construction above is an example of a compactification. There can be more than one compactification of a given space $X$. For example, for $X=\mathbb{R}$, we have the Stone-Cech compactification $\beta \mathbb{R}$, which is a subspace of the cube $\prod \limits_{f \in C(\mathbb{R},I)} I_f$. The circle $S^1=\left\{(x,y) \in \mathbb{R}^2: x^2+y^2=1 \right\}$ contains a copy of the real line $\mathbb{R}$ as a dense subspace, as does the unit interval $[0,1]$. Thus both $S^1$ and $I=[0,1]$ are also compactifications of $\mathbb{R}$. See A Beginning Look at Stone-Cech Compactification for a discussion of these examples.

We say that compactifications $\alpha_1 X$ and $\alpha_2 X$ are equivalent (we write $\alpha_1 X \approx \alpha_2 X$) if there exists a homeomorphism $f: \alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1= \alpha_2$. In other words, the following diagram commutes.

Figure 3

Essentially, two compactifications $\alpha_1 X$ and $\alpha_2 X$ of $X$ are equivalent if there is a homeomorphism $f$ between the two and if each $x \in X$ is mapped by $f$ to itself, i.e., $\alpha_1(x)$ is mapped to $\alpha_2(x)$.

For a given completely regular space $X$, let $\mathcal{C}(X)$ be the class of all compactifications of $X$. We define a partial order $\le$ on $\mathcal{C}(X)$. For $\alpha_1 X$ and $\alpha_2 X$, both in $\mathcal{C}(X)$, we say that $\alpha_2 X \le \alpha_1 X$ if there is a continuous function $f:\alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1=\alpha_2$. See Figure 4 below.

Figure 4

The following theorem ties the partial order $\le$ to the equivalence relation $\approx$ for compactifications.

Theorem 1
Let $\alpha_1 X$ and $\alpha_2 X$ be two compactifications of $X$. Then $\alpha_1 X \le \alpha_2 X$ and $\alpha_2 X \le \alpha_1 X$ if and only if $\alpha_1 X \approx \alpha_2 X$.

Proof of Theorem 1
$\Rightarrow$ With $\alpha_2 X \le \alpha_1 X$, there exists continuous $f_1:\alpha_1 X \rightarrow \alpha_2 X$ such that

$f_1 \circ \alpha_1=\alpha_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A1)$

With $\alpha_1 X \le \alpha_2 X$, there exists continuous $f_2:\alpha_2 X \rightarrow \alpha_1 X$ such that

$f_2 \circ \alpha_2=\alpha_1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A2)$

Applying $f_2$ to $(A1)$, we have $f_2 \circ f_1 \circ \alpha_1=f_2 \circ \alpha_2$. Applying $(A2)$ to this result, we have

$f_2 \circ f_1 \circ \alpha_1=\alpha_1 \ \ \ \ \ \ \ \ \ \ \ (A3)$

Note that $f_2 \circ f_1$ is a map from $\alpha_1 X$ into $\alpha_1 X$. The equation $(A3)$ indicates that when $f_2 \circ f_1$ is restricted to $\alpha_1(X)$, it is the identity map. Thus $f_2 \circ f_1$ agrees with the identity map on the dense set $\alpha_1(X)$. This implies that $\alpha_1(X)$ must agree with the identity map on all of $\alpha_1 X$.

Likewise we can see that $f_1 \circ f_2$ must equal to the identity map on $\alpha_2 X$. So $f_1:\alpha_1 X \rightarrow \alpha_2 X$ is a homeomorphism and it follows that $\alpha_1 X$ and $\alpha_2 X$ are equivalent compactifications of $X$.

$\Leftarrow$ This direction is straightforward. Let $f:\alpha_1 X \rightarrow \alpha_2 X$ a homeomorphism that makes $\alpha_1 X$ and $\alpha_2 X$ equivalent (as described by Figure 3). Then the map $f$ implies $\alpha_2 X \le \alpha_1 X$ and the map $f^{-1}$ implies $\alpha_1 X \le \alpha_2 X$. $\blacksquare$

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Another Characterization of the Stone-Cech Compactification

The next theorem says that the Stone-Cech compactification is the maximal compactification with respect to the partial order $\le$ defined here. Furthermore, this property is unique (there is only one maximal compactification up to equivalence). This result will simplify the work when we need to show that a given compactification is equivalent to $\beta X$.

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 $\le$.

$\text{ }$

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 $\alpha X \approx \beta X$.

Proof Theorem C2
Let $\alpha X$ be any compactification of $X$. Consider the continuous map $\alpha:X \rightarrow \alpha X$. By Theorem C1, $\alpha$ can be extended to $\beta X$. In other words, there exists a continuous $F: \beta X \rightarrow \alpha X$ such that $F \circ \beta = \alpha$. The existence of the map $F$ implies that $\alpha X \le \beta X$. $\blacksquare$

Proof Theorem U2
Let $\alpha X$ be another maximal compactification of $X$. This implies that $\beta X \le \alpha X$. By Theorem C2, we have $\alpha X \le \beta X$. By Theorem 1, $\alpha X$ must be equivalent to $\beta X$. $\blacksquare$

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Proof of Theorem U1

We are now ready to prove Theorem U1.

Proof of Theorem U1
Let $\alpha X$ be a compactification of $X$ that satisfies the extension property in Theorem C1. In light of Theorem C2, we have $\alpha X \le \beta X$. So we only need to show $\beta X \le \alpha X$. Consider the map $\beta: X \rightarrow \beta X$. By the assumption that $\alpha X$ satisfies the extension property in Theorem C1, there exists a continuous function $F:\alpha X \rightarrow \beta X$ such that $F \circ \alpha=\beta$. The existence of $F$ implies that $\beta X \le \alpha X$. By Theorem 1, $\alpha X$ must be equivalent to $\beta X$. $\blacksquare$

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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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