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