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}

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One thought on “The normality of the product of the first uncountable ordinal with a compact factor

  1. Pingback: The product of a normal countably compact space and a metric space is normal | Dan Ma's Topology Blog

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