The product of a perfectly normal space and a metric space is perfectly normal

The previous post gives a positive result for normality in product space. It shows that the product of a normal countably compact space and a metric space is always normal. In this post, we discuss another positive result, which is the following theorem.

Main Theorem
If X is a perfectly normal space and Y is a metric space, then X \times Y is a perfectly normal space.

As a result of this theorem, perfectly normal spaces belong to a special class of spaces called P-spaces. K. Morita defined the notion of P-space and he proved that a space Y is a Normal P-space if and only if X \times Y is normal for every metric space X (see the section below on P-spaces). Thus any perfectly normal space is a Normal P-space.

All spaces under consideration are Hausdorff. A subset A of the space X is a G_\delta-subset of the space X if A is the intersection of countably many open subsets of X. A subset B of the space X is an F_\sigma-subset of the space X if B is the union of countably many closed subsets of X. Clearly, a set A is a G_\delta-subset of the space X if and only if X-A is an F_\sigma-subset of the space X.

A space X is said to be a perfectly normal space if X is normal with the additional property that every closed subset of X is a G_\delta-subset of X (or equivalently every open subset of X is an F_\sigma-subset of X).

The perfect normality has a characterization in terms of zero-sets and cozero-sets. A subset A of the space X is said to be a zero-set if there exists a continuous function f: X \rightarrow [0,1] such that A=f^{-1}(0), where f^{-1}(0)=\left\{x \in X: f(x)=0 \right\}. A subset B of the space X is a cozero-set if X-B is a zero-set, or more explicitly if there is a continuous function f: X \rightarrow [0,1] such that B=\left\{x \in X: f(x)>0 \right\}.

It is well known that the space X is perfectly normal if and only if every closed subset of X is a zero-set, equivalently every open subset of X is a cozero-set. See here for a proof of this result. We use this result to show that X \times Y is perfectly normal.

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The Proof

Let X be a perfectly normal space and Y be a metric space. Since Y is a metric space, let \mathcal{B}=\bigcup_{j=1}^\infty \mathcal{B}_j be a base for Y such that each \mathcal{B}_j is locally finite. We show that X \times Y is perfectly normal. To that end, we show that every open subset of X \times Y is a cozero-set. Let U be an open subset of X \times Y.

For each (x,y) \in X \times Y, there exists open O_{x,y} \subset X and there exists B_{x,y} \in \mathcal{B} such that (x,y) \in O_{x,y} \times B_{x,y} \subset U. Then U is the union of all sets O_{x,y} \times B_{x,y}. Observe that B_{x,y} \in \mathcal{B}_{j} for some integer j. For each B \in \mathcal{B} such that B=B_{x,y} for some (x,y) \in X \times Y, let O(B) be the union of all corresponding open sets O_{x,y} for all applicable (x,y).

For each positive integer j, let \mathcal{W}_j be the collection of all open sets O(B) \times B such that B \in \mathcal{B}_j and B=B_{x,y} for some (x,y) \in X \times Y. Let \mathcal{V}_j=\cup \mathcal{W}_j. As a result, U=\bigcup_{j=1}^\infty \mathcal{V}_j.

Since both X and Y are perfectly normal, for each O(B) \times B \in \mathcal{W}_j, there exist continuous functions

    F_{O(B),j}: X \rightarrow [0,1]

    G_{B,j}: Y \rightarrow [0,1]

such that

    O(B)=\left\{x \in X: F_{O(B),j}(x) >0 \right\}

    B=\left\{y \in Y: G_{B,j}(y) >0 \right\}

Now define H_j: X \times Y \rightarrow [0,1] by the following:

    \displaystyle H_j(x,y)=\sum \limits_{O(B) \times B \in \mathcal{W}_j} F_{O(B),j}(x) \ G_{B,j}(y)

for all (x,y) \in X \times Y. Note that the function H_j is well defined. Since \mathcal{B}_j is locally finite in Y, \mathcal{W}_j is locally finite in X \times Y. Thus H_j(x,y) is obtained by summing a finite number of values of F_{O(B),j}(x) \ G_{B,j}(y). On the other hand, it can be shown that H_j is continuous for each j. Based on the definition of H_j, it can be readily verified that H_j(x,y)>0 for all (x,y) \in \cup \mathcal{W}_j and H_j(x,y)=0 for all (x,y) \notin \cup \mathcal{W}_j.

Define H: X \times Y \rightarrow [0,1] by the following:

    \displaystyle H(x,y)=\sum \limits_{j=1}^\infty \biggl[ \frac{1}{2^j} \ \frac{H_j(x,y)}{1+H_j(x,y)} \biggr]

It is clear that H is continuous. We claim that U=\left\{(x,y) \in X \times Y: H(x,y) >0 \right\}. Recall that the open set U is the union of all O(B) \times B \in \mathcal{W}_j for all j. Thus if (x,y) \in \cup \mathcal{W}_j for some j, then H(x,y)>0 since H_j(x,y)>0. If (x,y) \notin \cup \mathcal{W}_j for all j, H(x,y)=0 since H_j(x,y)=0 for all j. Thus the open set U is an F_\sigma-subset of X \times Y. This concludes the proof that X \times Y is perfectly normal. \square

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Remarks

The main theorem here is a classic result in general topology. An alternative proof is to show that any perfectly normal space is a P-space (definition given below). Then by Morita’s theorem, the product of any perfectly normal space and any metric space is normal (Theorem 1 below). For another proof that is elementary, see Lemma 7 in this previous post.

The notions of perfectly normal spaces and paracompact spaces are quite different. By the theorem discussed here, perfectly normal spaces are normally productive with metric spaces. It is possible for a paracompact space to have a non-normal product with a metric space. The classic example is the Michael line (discussed here).

On the other hand, there are perfectly normal spaces that are not paracompact. One example is Bing’s Example H, which is perfectly normal and not paracompact (see here).

Even though a perfectly normal space is normally productive with metric spaces, it cannot be normally productive in general. For each non-discrete perfectly normal space X, there exists a normal space Y such that X \times Y is not normal. This follows from Morita’s first conjecture (now a true statement). Morita’s first conjecture is discussed here.

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P-Space in the Sense of Morita

Morita defined the notion of P-spaces [1] and [2]. Let \kappa be a cardinal number such that \kappa \ge 1. Let \Gamma be the set of all finite ordered sequences (\alpha_1,\alpha_2,\cdots,\alpha_n) where n=1,2,\cdots and all \alpha_i < \kappa. Let X be a space. The collection \left\{F_\sigma \subset X: \sigma \in \Gamma \right\} is said to be decreasing if this condition holds: \sigma =(\alpha_1,\alpha_2,\cdots,\alpha_n) and \delta =(\alpha_1,\alpha_2,\cdots,\alpha_n, \cdots, \alpha_m) with n<m imply that F_{\delta} \subset F_{\sigma}. The space X is a P-space if for any cardinal \kappa \ge 1 and for any decreasing collection \left\{F_\sigma \subset X: \sigma \in \Gamma \right\} of closed subsets of X, there exists open set U_\sigma for each \sigma \in \Gamma such that the following conditions hold:

  • for all \sigma \in \Gamma, F_\sigma \subset U_\sigma,
  • for any infinite sequence (\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots) where each each finite subsequence \sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n) is an element of \Gamma, if \bigcap_{n=1}^\infty F_{\sigma_n}=\varnothing, then \bigcap_{n=1}^\infty U_{\sigma_n}=\varnothing.

If \kappa=1 where 1=\left\{0 \right\}. Then the index set \Gamma defined above can be viewed as the set of all positive integers. As a result, the definition of P-space with \kappa=1 implies the a condition in Dowker’s theorem (see condition 6 in Theorem 1 here). Thus any space X that is normal and a P-space is countably paracompact (or countably shrinking or that X \times Y is normal for every compact metric space or any other equivalent condition in Dowker’s theorem). The following is a theorem of Morita.

Theorem 1 (Morita)
Let X be a space. Then X is a normal P-space if and only if X \times Y is normal for every metric space Y.

In light of Theorem 1, both perfectly normal spaces and normal countably compact spaces are P-spaces (see here). According to Theorem 1 and Dowker’s theorem, it follows that any normal P-space is countably paracompact.

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Reference

  1. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
  2. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.

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\copyright \ 2017 \text{ by Dan Ma}

The product of a normal countably compact space and a metric space is normal

It is well known that normality is not preserved by taking products. When nothing is known about the spaces X and Y other than the facts that they are normal spaces, there is not enough to go on for determining whether X \times Y is normal. In fact even when one factor is a metric space and the other factor is a hereditarily paracompact space, the product can be non-normal (discussed here). This post discusses a productive scenario – the first factor is a normal space and second factor is a metric space with the first factor having the additional property that it is countably compact. In this scenario the product is always normal. This is a well known result in general topology. The goal here is to nail down a proof for use as future reference.

Main Theorem
Let X be a normal and countably compact space. Then X \times Y is a normal space for every metric space Y.

The proof of the main theorem uses the notion of shrinkable open covers.

Remarks
The main theorem is a classic result and is often used as motivation for more advanced results for products of normal spaces. Thus we would like to present a clear and complete proof of this classic result for anyone who would like to study the topics of normality (or the lack of) in product spaces. We found that some proofs of this result in the literature are hard to follow. In A. H. Stone’s paper [2], the result is stated in a footnote, stating that “it can be shown that the topological product of a metric space and a normal countably compact space is normal, though not necessarily paracompact”. We had seen several other papers citing [2] as a reference for the result. The Handbook [1] also has a proof (Corollary 4.10 in page 805), which we feel may not be the best proof to learn from. We found a good proof in [3] using the idea of shrinking of open covers.

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The Notion of Shrinking

The key to the proof is the notion of shrinkable open covers and shrinking spaces. Let X be a space. Let \mathcal{U} be an open cover of X. The open cover of \mathcal{U} is said to be shrinkable if there is an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} of X such that \overline{V(U)} \subset U for each U \in \mathcal{U}. When this is the case, the open cover \mathcal{V} is said to be a shrinking of \mathcal{U}. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed, e.g. \left\{U_\alpha: \alpha<\kappa \right\}, then a shrinking has the same indexing, e.g. \left\{V_\alpha: \alpha<\kappa \right\}.

A space X is a shrinking space if every open cover of X is shrinkable. Every open cover of a paracompact space has a locally finite open refinement. With a little bit of rearranging, the locally finite open refinement can be made to be a shrinking (see Theorem 2 here). Thus every paracompact space is a shrinking space. For other spaces, the shrinking phenomenon is limited to certain types of open covers. In a normal space, every finite open cover has a shrinking, as stated in the following theorem.

Theorem 1
The following conditions are equivalent.

  1. The space X is normal.
  2. Every point-finite open cover of X is shrinkable.
  3. Every locally finite open cover of X is shrinkable.
  4. Every finite open cover of X is shrinkable.
  5. Every two-element open cover of X is shrinkable.

The hardest direction in the proof is 1 \Longrightarrow 2, which is established in this previous post. The directions 2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5 are immediate. To see 5 \Longrightarrow 1, let H and K be two disjoint closed subsets of X. By condition 5, the two-element open cover \left\{X-H,X-K \right\} has a shrinking \left\{U,V \right\}. Then \overline{U} \subset X-H and \overline{V} \subset X-K. As a result, H \subset X-\overline{U} and K \subset X-\overline{V}. Since the open sets U and V cover the whole space, X-\overline{U} and X-\overline{V} are disjoint open sets. Thus X is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space may or may not be shrinkable. It turns out that finding a normal space with an infinite open cover that is not shrinkable is no trivial matter (see Dowker’s theorem in this previous post). However, if an open cover in a normal space point-finite or locally finite, then it is shrinkable.

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Key Idea

We now discuss the key idea to the proof of the main theorem. Consider the produce space X \times Y. Let \mathcal{U} be an open cover of X \times Y. Let M \subset X \times Y. The set M is stable with respect to the open cover \mathcal{U} if for each x \in X, there is an open set O_x containing x such that O_x \times M \subset U for some U \in \mathcal{U}.

Let \kappa be a cardinal number (either finite or infinite). A space X is a \kappa-shrinking space if for each open cover \mathcal{W} of X such that the cardinality of \mathcal{W} is \le \kappa, then \mathcal{W} is shrinkable. According to Theorem 1, any normal space is 2-shrinkable.

Theorem 2
Let \kappa be a cardinal number (either finite or infinite). Let X be a \kappa-shrinking space. Let Y be a paracompact space. Suppose that \mathcal{U} is an open cover of X \times Y such that the following two conditions are satisfied:

  • Each point y \in Y has an open set V_y containing y such that V_y is stable with respect to \mathcal{U}.
  • \lvert \mathcal{U} \lvert = \kappa.

Then \mathcal{U} is shrinkable.

Proof of Theorem 2
Let \mathcal{U} be any open cover of X \times Y satisfying the hypothesis. We show that \mathcal{U} has a shrinking.

For each y \in Y, obtain the open covers \left\{G(U,y): U \in \mathcal{U} \right\} and \left\{H(U,y): U \in \mathcal{U} \right\} of X as follows. For each U \in \mathcal{U}, define the following:

    G(U,y)=\cup \left\{O: O \text{ is open in } X \text{ such that } O \times V_y \subset U \right\}

Then \left\{G(U,y): U \in \mathcal{U} \right\} is an open cover of X. Since X is \kappa-shrinkable, there is an open cover \left\{H(U,y): U \in \mathcal{U} \right\} of X such that \overline{H(U,y)} \subset G(U,y) for each U \in \mathcal{U}.

Now \left\{V_y: y \in Y \right\} is an open cover of Y. By the paracompactness of Y, let \left\{W_y: y \in Y \right\} be a locally finite open cover of Y such that \overline{W_y} \subset V_y for each y \in Y. For each U \in \mathcal{U}, define the following:

    W_U=\cup \left\{H(U,y) \times W_y: y \in Y \text{ such that } \overline{H(U,y) \times W_y} \subset U \right\}

We claim that \mathcal{W}=\left\{ W_U: U \in \mathcal{U} \right\} is a shrinking of \mathcal{U}. First it is a cover of X \times Y. Let (x,t) \in X \times Y. Then t \in W_y for some y \in Y. There exists U \in \mathcal{U} such that x \in H(U,y). Note the following.

    \overline{H(U,y) \times W_y} \subset \overline{H(U,y)} \times \overline{W_y} \subset G(U,y) \times V_y \subset U

This means that H(U,y) \times W_y \subset W_U. Since (x,t) \in H(U,y) \times W_y, (x,t) \in W_U. Thus \mathcal{W} is an open cover of X \times Y.

Now we show that \mathcal{W} is a shrinking of \mathcal{U}. Let U \in \mathcal{U}. To show that \overline{W_U} \subset U, let (x,t) \in \overline{W_U}. Let L be open in Y such that t \in L and that L meets only finitely many W_y, say for y=y_1,y_2,\cdots,y_n. Immediately we have the following relations.

    \forall \ i=1,\cdots,n, \ \overline{W_{y_i}} \subset V_{y_i}

    \forall \ i=1,\cdots,n, \ \overline{H(U,y_i)} \subset G(U,y_i)

    \forall \ i=1,\cdots,n, \ \overline{H(U,y_i) \times W_{y_i}} \subset \overline{H(U,y_i)} \times \overline{W_{y_i}} \subset G(U,y_i) \times V_{y_i} \subset U

Then it follows that

    \displaystyle (x,t) \in \overline{\bigcup \limits_{j=1}^n H(U,y_j) \times W_{y_j}}=\bigcup \limits_{j=1}^n \overline{H(U,y_j) \times W_{y_j}} \subset U

Thus U \in \mathcal{U}. This shows that \mathcal{W} is a shrinking of \mathcal{U}. \square

Remark
Theorem 2 is the Theorem 3.2 in [3]. Theorem 2 is a formulation of Theorem 3.2 [3] for the purpose of proving Theorem 3 below.

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Main Theorem

Theorem 3 (Main Theorem)
Let X be a normal and countably compact space. Let Y be a metric space. Then X \times Y is a normal space.

Proof of Theorem 3
Let \mathcal{U} be a 2-element open cover of X \times Y. We show that \mathcal{U} is shrinkable. This would mean that X \times Y is normal (according to Theorem 1). To show that \mathcal{U} is shrinkable, we show that the open cover \mathcal{U} satisfies the two bullet points in Theorem 2.

Fix y \in Y. Let \left\{B_n: n=1,2,3,\cdots \right\} be a base at the point y. Define G_n as follows:

    G_n=\cup \left\{O \subset X: O \text{ is open such that } O \times B_n \subset U \text{ for some } U \in \mathcal{U} \right\}

It is clear that \mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\} is an open cover of X. Since X is countably compact, choose m such that \left\{G_1,G_2,\cdots,G_m \right\} is a cover of X. Let E_y=\bigcap_{j=1}^m B_j. We claim that E_y is stable with respect to \mathcal{U}. To see this, let x \in X. Then x \in G_j for some j \le m. By the definition of G_j, there is some open set O_x \subset X such that x \in O_x and O_x \times B_j \subset U for some U \in \mathcal{U}. Furthermore, O_x \times E_y \subset O_x \times B_j \subset U.

To summarize: for each y \in Y, there is an open set E_y such that y \in E_y and E_y is stable with respect to the open cover \mathcal{U}. Thus the first bullet point of Theorem 2 is satisfied. The open cover \mathcal{U} is a 2-element open cover. Thus the second bullet point of Theorem 2 is satisfied. By Theorem 2, the open cover \mathcal{U} is shrinkable. Thus X \times Y is normal. \square

Corollary 4
Let X be a normal and pseudocompact space. Let Y be a metric space. Then X \times Y is a normal space.

The corollary follows from the fact that any normal and pseudocompact space is countably compact (see here).

Remarks
The proof of Theorem 3 actually gives a more general result. Note that the second factor only needs to be paracompact and that every point has a countable base (i.e. first countable). The first factor X has to be countably compact. The shrinking requirement for X is flexible – if open covers of a certain size for X are shrinkable, then open covers of that size for the product are shrinkable. We have the following corollaries.

Corollary 5
Let X be a \kappa-shrinking and countably compact space and let Y be a paracompact first countable space. Then X \times Y is a \kappa-shrinking space.

Corollary 6
Let X be a shrinking and countably compact space and let Y be a paracompact first countable space. Then X \times Y is a shrinking space.

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Remarks

The main theorem (Theorem 3) says that any normal and countably compact space is productively normal with one class of spaces, namely the metric spaces. Thus if one wishes to find a non-normal product space with one factor being countably compact, the other factor must not be a metric space. For example, if W=\omega_1, the first uncountable ordinal with the ordered topology, then W \times X is always normal for every metric X. For non-normal example, W \times C is not normal for any compact space C with uncountable tightness (see Theorem 1 in this previous post). Another example, W \times L_{\omega_1} is not normal where L_{\omega_1} is the one-point Lindelofication of a discrete space of cardinality \omega_1 (follows from Example 1 and Theorem 7 in this previous post).

Another comment is that normal countably paracompact spaces are examples of Normal P-spaces. K. Morita defined the notion of P-space and he proved that a space Y is a Normal P-space if and only if X \times Y is normal for every metric space X.

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Reference

  1. Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  2. Stone A. H., Paracompactness and Product Spaces, Bull. Amer. Math. Soc., Vol. 54, 977-982, 1948. (paper)
  3. Yang L., The Normality in Products with a Countably Compact Factor, Canad. Math. Bull., Vol. 41 (2), 245-251, 1998. (abstract, paper)

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\copyright \ 2017 \text{ by Dan Ma}

kappa-Dowker space and the first conjecture of Morita

Recall the product space of the Michael line and the space of the irrational numbers. Even though the first factor is a normal space (in fact a paracompact space) and the second factor is a metric space, their product space is not normal. This is one of the classic examples demonstrating that normality is not well behaved with respect to product space. This post presents an even more striking result, i.e., for any non-discrete normal space Y, there exists another normal space X such that X \times Y is not normal. The example of the non-normal product of the Michael line and the irrationals is not some isolated example. Rather it is part of a wide spread phenomenon. This result guarantees that no matter how nice a space Y is, a counter part X can always be found that the product of the two spaces is not normal. This result is known as Morita’s first conjecture and was proved by Atsuji and Rudin. The solution is based on a generalization of Dowker’s theorem and a construction done by Rudin. This post demonstrates how the solution is put together.

All spaces under consideration are Hausdorff.

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Morita’s First Conjecture

In 1976, K. Morita posed the following conjecture.

    Morita’s First Conjecture
    If Y is a normal space such that X \times Y is a normal space for every normal space X, then Y is a discrete space.

The proof given in this post is for proving the contrapositive of the above statement.

    Morita’s First Conjecture
    If Y is a non-discrete normal space, then there exists some normal space X such that X \times Y is not a normal space.

Though the two forms are logically equivalent, the contrapositive form seems to have a bigger punch. The contrapositive form gives an association. Each non-discrete normal space is paired with a normal space to form a non-normal product. Examples of such pairings are readily available. Michael line is paired with the space of the irrational numbers (as discussed above). The Sogenfrey line is paired with itself. The first uncountable ordinal \omega_1 is paired with \omega_1+1 (see here) or paired with the cube I^I where I=[0,1] with the usual topology (see here). There are plenty of other individual examples that can be cited. In this post, we focus on a constructive proof of finding such a pairing.

Since the conjecture had been affirmed positively, it should no longer be called a conjecture. Calling it Morita’s first theorem is not appropriate since there are other results that are identified with Morita. In this discussion, we continue to call it a conjecture. Just know that it had been proven.

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Dowker’s Theorem

Next, we examine Dowker’s theorem, which characterizes normal countably paracompact spaces. The following is the statement.

Theorem 1 (Dowker’s Theorem)
Let X be a normal space. The following conditions are equivalent.

  1. The space X is countably paracompact.
  2. Every countable open cover of X has a point-finite open refinement.
  3. If \left\{U_n: n=1,2,3,\cdots \right\} is an open cover of X, there exists an open refinement \left\{V_n: n=1,2,3,\cdots \right\} such that \overline{V_n} \subset U_n for each n.
  4. The product space X \times Y is normal for any compact metric space Y.
  5. The product space X \times [0,1] is normal where [0,1] is the closed unit interval with the usual Euclidean topology.
  6. For each sequence \left\{A_n \subset X: n=1,2,3,\cdots \right\} of closed subsets of X such that A_1 \supset A_2 \supset A_3 \supset \cdots and \cap_n A_n=\varnothing, there exist open sets B_1,B_2,B_3,\cdots such that A_n \subset B_n for each n such that \cap_n B_n=\varnothing.

The theorem is discussed here and proved here. Any normal space that violates any one of the conditions in the theorem is said to be a Dowker space. One such space was constructed by Rudin in 1971 [2]. Any Dowker space would be one factor in a non-normal product space with the other factor being a compact metric space. Actually much more can be said.

The Dowker space constructed by Rudin is the solution of Morita’s conjecture for a large number of spaces. At minimum, the product of any infinite compact metric space and the Dowker space is not normal as indicated by Dowker’s theorem. Any nontrivial convergent sequence plus the limit point is a compact metric space since it is homeomorphic to S=\left\{0 \right\} \cup \left\{\frac{1}{n}: n=1,2,3,\cdots \right\} (as a subspace of the real line). Thus Rudin’s Dowker space has non-normal product with S. Furthermore, the product of Rudin’s Dowker space and any space containing a copy of S is not normal.

Spaces that contain a copy of S extend far beyond the compact metric spaces. Spaces that have lots of convergent sequences include first countable spaces, Frechet spaces and many sequential spaces (see here for an introduction for these spaces). Thus any Dowker space is an answer to Morita’s first conjecture for the non-discrete members of these classes of spaces. Actually, the range for the solution is wider than these spaces. It turns out that any space that has a countable non-discrete subspace would have a non-normal product with a Dowker space. These would include all the classes mentioned above (first countable, Frechet, sequential) as well as countably tight spaces and more.

Therefore, any Dowker space, a normal space that is not countably paracompact, is severely lacking in ability in forming normal product with another space. In order to obtain a complete solution to Morita’s first conjecture, we would need a generalized Dowker’s theorem.

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Shrinking Properties

The key is to come up with a generalized Dowker’s theorem, a theorem like Theorem 1 above, except that it is for arbitrary infinite cardinality. Then a \kappa-Dowker space is a space that violates one condition in the theorem. That space would be a candidate for the solution of Morita’s first conjecture. Note that Theorem 1 is for the infinite countable cardinal \omega only. Before stating the theorem, let’s gather all the notions that will go into the theorem.

Let X be a space. Let \mathcal{U} be an open cover of the space X. The open cover \mathcal{U} is said to be shrinkable if there is an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} such that \overline{V(U)} \subset U for each U \in \mathcal{U}. When this is the case, the open cover \mathcal{V} is said to be a shrinking of \mathcal{U}. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking).

Let \kappa be a cardinal. The space X is said to be a \kappa-shrinking space if every open cover of cardinality \le \kappa of the space X is shinkable. The space X is a shrinking space if it is a \kappa-shrinking space for every cardinal \kappa.

When a family of sets are indexed by ordinals, the notion of an increasing or decreasing family of sets is possible. For example, the family \left\{A_\alpha \subset X: \alpha<\kappa \right\} of subsets of the space X is said to be increasing if A_\beta \subset A_\gamma whenever \beta<\gamma. In other words, for an increasing family, the sets are getting larger whenever the index becomes larger. A decreasing family of sets is defined in the reverse way. These two notions are important for some shrinking properties discussed here – e.g. using an open cover that is increasing or using a family of closed sets that is decreasing.

In the previous discussion on shrinking spaces, two other shrinking properties are discussed – property \mathcal{D}(\kappa) and property \mathcal{B}(\kappa). A space X is said to have property \mathcal{D}(\kappa) if every increasing open cover of cardinality \le \kappa for the space X is shrinkable. A space X is said to have property \mathcal{B}(\kappa) if every increasing open cover of cardinality \le \kappa for the space X has a shrinking that is increasing. See this previous post for a discussion on property \mathcal{D}(\kappa) and property \mathcal{B}(\kappa).

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An Attempt for a Generalized Dowker’s Theorem

Let \kappa be an infinite cardinal. The space X is said to be a \kappa-paracompact space if every open cover \mathcal{U} of X with \lvert \mathcal{U} \lvert \le \kappa has a locally finite open refinement. Thus a space is paracompact if it is \kappa-paracompact for every infinite cardinal \kappa. Of course, an \omega-paracompact space is a countably paracompact space.

For any infinite \kappa, let D_\kappa be a discrete space of size \kappa. Let p be a point not in D_\kappa. Define the space Y_\kappa=D_\kappa \cup \left\{p \right\} as follows. The subspace D_\kappa is discrete as before. The open neighborhoods at p are of the form \left\{ p \right\} \cup B where B \subset D_\kappa and \lvert D_\kappa-B \lvert<\kappa. In other words, any open set containing p contains all but less than \kappa many discrete points.

Another concept that is needed is the cardinal function called minimal tightness. Let Y be any space. Define the minimal tightness mt(Y) as the least infinite cardinal \kappa such that there is a non-discrete subspace of Y of cardinality \kappa. If Y is a discrete space, then let mt(Y)=0. For any non-discrete space Y, mt(Y)=\kappa for some infinite \kappa. Note that for the space Y_\kappa defined above would have mt(Y_\kappa)=\kappa. For any space Y, mt(Y)=\omega if and only if Y has a countable non-discrete subspace.

The following theorem can be called a \kappa-Dowker’s Theorem.

Theorem 2
Let X be a normal space. Let \kappa be an infinite cardinal. Consider the following conditions.

  1. The space X is a \kappa-paracompact space.
  2. The space X is a \kappa-shrinking space.
    • For each open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.
  3. The space X has property \mathcal{D}(\kappa).
    • For each increasing open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.
  4. For each decreasing family \left\{F_\alpha: \alpha<\kappa \right\} of closed subsets of X such that \bigcap_{\alpha<\kappa} F_\alpha=\varnothing, there exists a family \left\{G_\alpha: \alpha<\kappa \right\} of open subsets of X such that \bigcap_{\alpha<\kappa} G_\alpha=\varnothing and F_\alpha \subset G_\alpha for each \alpha<\kappa.
  5. The space X has property \mathcal{B}(\kappa).
    • For each increasing open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an increasing open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.
  6. The product space X \times Y_\kappa is a normal space.
  7. The product space X \times Y is a normal space for some space Y with mt(Y)=\kappa.

The following diagram shows how these conditions are related.

Diagram 1
\displaystyle \begin{array}{ccccc}  1 &\text{ } & \Longrightarrow & \text{ } & 5 \\   \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\   \Downarrow & \text{ } & \text{ } & \text{ } & \Updownarrow \\   \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\   2 &\text{ } & \text{ } & \text{ } & 6 \\      \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\   \Downarrow & \text{ } & \text{ } & \text{ } & \Downarrow \\   \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\    3 & \text{ } & \Longleftarrow & \text{ } & 7 \\  \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\  \Updownarrow & \text{ } & \text{ } & \text{ } & \text{ } \\  \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\  4 & \text{ } & \text{ } & \text{ } & \text{ }  \end{array}

In addition to Diagram 1, we have the relations 5 \Longrightarrow 3 and 2 \not \Longrightarrow 5.

Remarks
At first glance, Diagram 1 might give the impression that the conditions in the theorem form a loop. It turns out the strongest property is \kappa-paracompactness (condition 1). Since condition 2 does not imply condition 5, condition 2 does not imply condition 1. Thus the conditions do not form a loop.

The implications 1 \Longrightarrow 2 \Longrightarrow 3 \Longleftarrow 5 and 6 \Longrightarrow 7 are immediate. The following implications are established in this previous post.

    3 \Longleftrightarrow 4 (Theorem 4)

    5 \Longleftrightarrow 6 (Theorem 7)

    2 \not \Longrightarrow 5 (Example 1)

The remaining implications to be shown are 1 \Longrightarrow 5 and 7 \Longrightarrow 3.

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Proof of Theorem 2

1 \Longrightarrow 5
Let \mathcal{U}=\left\{U_\alpha: \alpha<\kappa \right\} be an increasing open cover of X. By \kappa-paracompactness, let \mathcal{G} be a locally finite open refinement of \mathcal{U}. For each \alpha<\kappa, define W_\alpha as follows:

    W_\alpha=\cup \left\{G \in \mathcal{G}: G \subset U_\alpha \right\}

Then \mathcal{W}=\left\{W_\alpha: \alpha<\kappa \right\} is still a locally finite refinement of \mathcal{U}. Since the space X is normal, any locally finite open cover is shrinkable. Let \mathcal{E}=\left\{E_\alpha: \alpha<\kappa \right\} be a shrinking of \mathcal{W}. The open cover \mathcal{E} is also locally finite. For each \alpha, let V_\alpha=\bigcup_{\beta<\alpha} E_\beta. Then \mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\} is an increasing open cover of X. Note that

    \overline{V_\alpha}=\overline{\bigcup_{\beta<\alpha} E_\beta}=\bigcup_{\beta<\alpha} \overline{E_\beta}

since \mathcal{E} is locally finite and thus closure preserving. Since \mathcal{U} is increasing, \overline{E_\beta} \subset W_\beta \subset U_\beta \subset U_\alpha for all \beta<\alpha. This means that \overline{V_\alpha} \subset U_\alpha for all \alpha.

7 \Longrightarrow 3
Since condition 3 is equivalent to condition 4, we show 7 \Longrightarrow 4. Suppose that X \times Y is normal where Y is a space such that mt(Y)=\kappa. Let D=\left\{d_\alpha: \alpha<\kappa \right\} be a non-discrete subset of Y. Let p be a point such that p \ne d_\alpha for all \alpha and such that p is a limit point of D (this means that every open set containing p contains some d_\alpha). Let \mathcal{F}=\left\{F_\alpha: \alpha<\kappa \right\} be a decreasing family of closed subsets of X such that \bigcap_{\alpha<\kappa} F_\alpha=\varnothing. Define H and K as follows:

    H=\cup \left\{F_\alpha \times \left\{d_\alpha \right\}: \alpha<\kappa \right\}

    K=X \times \left\{p \right\}

The sets H and K are clearly disjoint. The set K is clearly a closed subset of X \times Y. To show that H is closed, let (x,y) \in (X \times Y)-H. Two cases to consider: x \in F_0 or x \notin F_0 where F_0 is the first closed set in the family \mathcal{F}.

The first case x \in F_0. Let \beta<\kappa be least such that x \notin F_\beta. Then y \ne d_\gamma for all \gamma<\beta since (x,y) \in (X \times Y)-H. In the space Y, any subset of cardinality <\kappa is a closed set. Let E=Y-\left\{d_\gamma: \gamma<\beta \right\}, which is open containing y. Let O \subset X be open such that x \in O and O \cap F_\beta=\varnothing. Then (x,y) \in O \times E and O \times E misses points of H.

Now consider the second case x \notin F_0. Let O \subset X be open such that x \in O and O misses F_0. Then O \times Y is an open set containing (x,y) such that O \times Y misses H. Thus H is a closed subset of X \times Y.

Since X \times Y is normal, choose open V \subset X \times Y such that H \subset V and \overline{V} \cap K=\varnothing. For each \alpha<\kappa, define G_\alpha as follows:

    G_\alpha=\left\{x \in X: (x,d_\alpha) \in V \right\}

Note that each G_\alpha is open in X and that F_\alpha \subset G_\alpha for each \alpha<\kappa. We claim that \bigcap_{\alpha<\kappa} G_\alpha=\varnothing. Let x \in X. The point (x,p) is in K. Thus (x,p) \notin \overline{V}. Choose an open set L \times M such that (x,p) \in L \times M and (L \times M) \cap \overline{V}=\varnothing. Since p \in M, there is some \gamma<\kappa such that d_\gamma \in M. Since (x,d_\gamma) \notin \overline{V}, (x,d_\gamma) \notin V. Thus x \notin G_\gamma. This establishes the claim that \bigcap_{\alpha<\kappa} G_\alpha=\varnothing.

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\kappa-Dowker Space

Analogous to the Dowker space, a \kappa-space is a normal space that violates one condition in Theorem 2. Since the seven conditions listed in Theorem 7 are not all equivalent, which condition to use? Condition 1 is the strongest condition since it implies all the other condition. At the lower left corner of Diagram 1 is condition 3, which follows from every other condition. Thus condition 3 (or 4) is the weakest property. An appropriate definition of a \kappa-Dowker space is through negating condition 3 or condition 4. Thus, given an infinite cardinal \kappa, a \kappa-Dowker space is a normal space X that satisfies the following condition:

    There exists a decreasing family \left\{F_\alpha: \alpha<\kappa \right\} of closed subsets of X with \bigcap_{\alpha<\kappa} F_\alpha=\varnothing such that for every family \left\{G_\alpha: \alpha<\kappa \right\} of open subsets of X with F_\alpha \subset G_\alpha for each \alpha, \bigcap_{\alpha<\kappa} G_\alpha \ne \varnothing.

The definition of \kappa-Dowker space is through negating condition 4. Of course, negating condition 3 would give an equivalent definition.

When \kappa is the countably infinite cardinal \omega, a \kappa-Dowker space is simply the ordinary Dowker space constructed by M. E. Rudin [2]. Rudin generalized the construction of the ordinary Dowker space to obtain a \kappa-Dowker space for every infinite cardinal \kappa [4]. The space that Rudin constructed in [4] would be a normal space X such that condition 4 of Theorem 2 is violated. This means that the space X would violate condition 7 in Theorem 2. Thus X \times Y is not normal for every space Y with mt(Y)=\kappa.

Here’s the solution of Morita’s first conjecture. Let Y be a normal and non-discrete space. Determine the least cardinality \kappa of a non-discrete subspace of Y. Obtain the \kappa-Dowker space X as in [4]. Then X \times Y is not normal according to the preceding paragraph.

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Remarks

Answering Morita’s first conjecture is a two-step approach. First, figure out what a generalized Dowker’s theorem should be. Then a \kappa-Dowker space is one that violates an appropriate condition in the generalized Dowker’s theorem. By violating the right condition in the theorem, we have a way to obtain non-normal product space needed in the answer. The second step is of course the proof of the existence of a space that violates the condition in the generalized Dowker’s theorem.

Figuring out the form of the generalized Dowker’s theorem took some work. It is more than just changing the countable infinite cardinal in Dowker’s theorem (Theorem 1 above) to an arbitrary infinite cardinal. This is because the conditions in Theorem 1 are unequal when the cardinality is changed to an uncountable one.

We take the cue from Rudin’s chapter on Dowker spaces [3]. In the last page of that chapter, Rudin pointed out the conditions that should go into a generalized Dowker’s theorem. However, the explanation of the relationship among the conditions is not clear. The previous post and this post are an attempt to sort out the conditions and fill in as much details as possible.

Rudin’s chapter did have the right condition for defining \kappa-Dowker space. It seems that prior to the writing of that chapter, there was some confusion on how to define a \kappa-Dowker space, i.e. a condition in the theorem the violation of which would give a \kappa-Dowker space. If the condition used is a stronger property, the violation may not yield enough information to get non-normal products. According to Diagram 1, condition 3 in Theorem 2 is the right one to use since it is the weakest condition and is down streamed from the conditions about normal product space. So the violation of condition 3 would answer Morita’s first conjecture.

We do not discuss the other step in the solution in any details. Any interested reader can review Rudin’s construction in [2] and [4]. The \kappa-Dowker space is an appropriate subspace of a product space with the box topology.

One interesting observation about the ordinary Dowker space (the one that violates a condition in Theorem 1) is that the product of any Dowker space and any space with a countable non-discrete subspace is not normal. This shows that Dowker space is badly non-productive with respect to normality. This fact is actually not obvious in the usual formulation of Dowker’s theorem (Theorem 1 above). What makes this more obvious in the direction 7 \Longrightarrow 3 in Theorem 2. For the countably infinite case, 7 \Longrightarrow 3 is essentially this: If X \times Y is normal where Y has a countable non-discrete subspace, then X is not a Dowker space. Thus if the goal is to find a non-normal product space, a Dowker space should be one space to check.

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Loose Ends

In the course of working on the contents in this post and the previous post, there are some questions that we do not know how to answer and have not spent time to verify one way or the other. Possibly there are some loose ends to tie. They for the most parts are not open questions, but they should be interesting questions to consider.

For the \kappa-Dowker’s theorem (Theorem 2), one natural question is on the relative strengths of the conditions. It will be interesting to find out the implications not shown in Diagram 1. For example, for the three shrinking properties (conditions 2, 3 and 5), it is straightforward from definition that 2 \Longrightarrow 3 and 5 \Longrightarrow 3. The example of X=\omega_1 (the first uncountable ordinal) shows that 2 \not \Longrightarrow 5 and hence 2 \not \Longrightarrow 1. What about 3 \Longrightarrow 2? In [5], Beslagic and Rudin showed that 3 \not \Longrightarrow 2 using \Diamond ^{++}. A natural question would be: can there be ZFC example? Perhaps searching on more recent papers can yield some answers.

Another question is 5 \Longrightarrow 1? The answer is no with the example being a Navy space – Example 7.6 in p. 194 [1]. The other two directions that have not been accounted for are: 7 \Longrightarrow 6 and 3 \Longrightarrow 7? We do not know the answer.

Another small question that we come across is about X=\omega_1 (the first uncountable ordinal). This is an example for showing 2 \not \Longrightarrow 5. Thus condition 6 is false. Thus X \times Y_{\omega_1} is not normal. Here Y_{\omega_1} is simply the one-point Lindelofication of a discrete space of cardinality \omega_1. The question is: is condition 7 true for X=\omega_1? The product of X=\omega_1 and Y_{\omega_1} (a space with minimal tightness \omega_1) is not normal. Is there a normal X \times Y where Y is another space with minimal tightness \omega_1?

Dowker’s theorem and \kappa-Dowker’s theorem show that finding a normal space that is not shrinking is not a simple matter. To find a normal space that is not countably shrinking took 20 years (1951 to 1971). For any uncountable \kappa, the \kappa-Dowker space that is based on the same construction of an ordinary Dowker space is also a space that is not \kappa-shrinking. With an uncountable \kappa, is the \kappa-Dowker space countably shrinking? This is not obvious one way or the other just from the definition of \kappa-Dowker space. Perhaps there is something obvious and we have not connected the dots. Perhaps we need to go into the definition of the \kappa-Dowker space in [4] to show that it is countably shrinking. The motivation is that we tried to find a normal space that is countably shrinking but not \kappa-shrinking for some uncountable \kappa. It seems that the \kappa-Dowker space in [4] is the natural candidate.

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Reference

  1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
  2. Rudin M. E., A Normal Space X for which X \times I is not Normal, Fund. Math., 73, 179-486, 1971. (link)
  3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
  4. Rudin M. E., \kappa-Dowker Spaces, Czechoslovak Mathematical Journal, 28, No.2, 324-326, 1978. (link)
  5. Rudin M. E., Beslagic A.,Set-Theoretic Constructions of Non-Shrinking Open Covers, Topology Appl., 20, 167-177, 1985. (link)
  6. Yasui Y., On the Characterization of the \mathcal{B}-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
  7. Yasui Y., Some Characterization of a \mathcal{B}-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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\copyright \ 2017 \text{ by Dan Ma}

Spaces with shrinking properties

Certain covering properties and separation properties allow open covers to shrink, e.g. paracompact spaces, normal spaces, and countably paracompact spaces. The shrinking property is also interesting on its own. This post gives a more in-depth discussion than the one in the previous post on countably paracompact spaces. After discussing shrinking spaces, we introduce three shrinking related properties. These properties show that there is a deep and delicate connection among shrinking properties and normality in products. This post is also a preparation for the next post on \kappa-Dowker space and Morita’s first conjecture.

All spaces under consideration are Hausdorff and normal or Hausdorff and regular (if not normal).

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Shrinking Spaces

Let X be a space. Let \mathcal{U} be an open cover of X. The open cover of \mathcal{U} is said to be shrinkable if there is an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} of X such that \overline{V(U)} \subset U for each U \in \mathcal{U}. When this is the case, the open cover \mathcal{V} is said to be a shrinking of \mathcal{U}. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed in a certain way, e.g. \left\{U_\alpha: \alpha<\kappa \right\}, then a shrinking has the same indexing, e.g. \left\{V_\alpha: \alpha<\kappa \right\}.

A space X is a shrinking space if every open cover of X is shrinkable. The property can also be broken up according to the cardinality of the open cover. Let \kappa be a cardinal. A space X is \kappa-shrinking if every open cover of cardinality \le \kappa for X is shrinkable. A space X is countably shrinking if it is \omega-shrinking.

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Examples of Shrinking

Let’s look at a few situations where open covers can be shrunk either all the time or on a limited basis. For a normal space, certain covers can be shrunk as indicated by the following theorem.

Theorem 1
The following conditions are equivalent.

  1. The space X is normal.
  2. Every point-finite open cover of X is shrinkable.
  3. Every locally finite open cover of X is shrinkable.
  4. Every finite open cover of X is shrinkable.
  5. Every two-element open cover of X is shrinkable.

The hardest direction in the proof is 1 \Longrightarrow 2, which is established in this previous post. The directions 2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5 are immediate. To see 5 \Longrightarrow 1, let H and K be two disjoint closed subsets of X. By condition 5, the two-element open cover \left\{X-H,X-K \right\} has a shrinking \left\{U,V \right\}. Then \overline{U} \subset X-H and \overline{V} \subset X-K. As a result, H \subset X-\overline{U} and K \subset X-\overline{V}. Since the open sets U and V cover the whole space, X-\overline{U} and X-\overline{V} are disjoint open sets. Thus X is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space does not have to be shrinkable unless it is a point-finite or locally finite open cover.

The theorem of C. H. Dowker states that a normal space X is countably paracompact if and only every countable open cover of X is shrinkable if and only if the product space X \times Y is normal for every compact metric space Y if and only if the product space X \times [0,1] is normal. The theorem is discussed here. A Dowker space is a normal space that violates the theorem. Thus any Dowker space has a countably infinite open cover that cannot be shrunk, or equivalently a normal space that forms a non-normal product with a compact metric space. Thus the notion of shrinking has a connection with normality in the product spaces. A Dowker space space was constructed by M. E. Rudin in ZFC [2]. So far Rudin’s example is essentially the only ZFC Dowker space. This goes to show that finding a normal space that is not countably shrinking is not a trivial matter.

Several facts can be derived easily from Theorem 1 and Dowker’s theorem. For clarity, they are called out as corollaries.

Corollary 2

  • All shrinking spaces are normal.
  • All shrinking spaces are normal and countably paracompact.
  • Any normal and metacompact space is a shrinking space.

For the first corollary, if every open cover of a space can be shrunk, then all finite open covers can be shrunk and thus the space must be normal. As indicated above, Dowker’s theorem states that in a normal space, countably paracompactness is equivalent to countably shrinking. Thus any shrinking space is normal and countably paracompact.

Though an infinite open cover of a normal space may not be shrinkable, adding an appropriate covering property to any normal space will make it into a shrinking space. An easy way is through point-finite open covers. If every open cover has a point-finite open refinement (i.e. a metacompact space), then the point-finite open refinement can be shrunk (if the space is also normal). Thus the third corollary is established. Note that the metacompact is not the best possible result. For example, it is known that any normal and submetacompact space is a shrinking space – see Theorem 6.2 of [1].

In paracompact spaces, all open covers can be shrunk. One way to see this is through Corollary 2. Any paracompact space is normal and metacompact. It is also informative to look at the following characterization of paracompact spaces.

Theorem 3
A space X is paracompact if and only if every open cover \left\{U_\alpha: \alpha<\kappa \right\} of X has a locally finite open refinement \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha.

A proof can be found here. Thus every open cover of a paracompact space can be shrunk by a locally finite shrinking. To summarize, we have discussed the following implications.

    Diagram 1

    \displaystyle \begin{aligned} \text{Paracompact} \Longrightarrow & \text{ Normal + Metacompact}  \\&\ \ \ \ \ \ \Big \Downarrow \\&\text{ Shrinking} \\&\ \ \ \ \ \ \Big \Downarrow  \\& \text{ Normal + Countably Paracompact} \\&\ \ \ \ \ \ \Big \Downarrow  \\& \text{ Normal} \end{aligned}

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Three Shrinking Related Properties

None of the implications in Diagram 1 can be reversed. The last implication in the diagram cannot be reversed due to Rudin’es Dowker space. One natural example to look for would be spaces that are normal and countably paracompact but fail in shrinking at some uncountable cardinal. As indicated by the the theorem of C. H, Dowker, the notion of shrinking is intimately connected to normality in product spaces X \times Y. To further investigate, consider the following three properties.

Let X be a space. Let \kappa be an infinite cardinal. Consider the following three properties.

The space X is \kappa-shrinking if and only if any open cover of cardinality \le \kappa for the space X is shrinkable, i.e. the following condition holds.

    For each open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.

The space X has Property \mathcal{D}(\kappa) if and only if every increasing open cover of cardinality \le \kappa for the space X is shrinkable, i.e. the following holds.

    For each increasing open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.

The space X has Property \mathcal{B}(\kappa) if and only if the following holds.

    For each increasing open cover \left\{U_\alpha: \alpha<\kappa \right\} of X, there exists an increasing open cover \left\{V_\alpha: \alpha<\kappa \right\} such that \overline{V_\alpha} \subset U_\alpha for each \alpha<\kappa.

A family \left\{A_\alpha: \alpha<\kappa \right\} is increasing if A_\alpha \subset A_\beta for any \alpha<\beta<\kappa. It is decreasing if A_\beta \subset A_\alpha for any \alpha<\beta<\kappa.

In general, any space that is \kappa-shrinking for all cardinals \kappa is a shrinking space as defined earlier. Any space that has property \mathcal{D}(\kappa) for all cardinals \kappa is said to have property \mathcal{D}. Any space that has property \mathcal{B}(\kappa) for all cardinals \kappa is said to have property \mathcal{B}.

The first property \kappa-shrinking is simply the shrinking property for open covers of cardinality \le \kappa. The property \mathcal{D}(\kappa) is \kappa-shrinking with the additional requirement that the open covers to be shrunk must be increasing. It is clear that \kappa-shrinking implies property \mathcal{D}(\kappa). The property \mathcal{B}(\kappa) appears to be similar to \mathcal{D}(\kappa) except that \mathcal{B}(\kappa) has the additional requirement that the shrinking is also increasing. As a result \mathcal{B}(\kappa) implies \mathcal{D}(\kappa). The following diagram shows the implications.

    Diagram 2

    \displaystyle \begin{array}{ccccc} \kappa \text{-Shrinking} &\text{ } & \not \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\kappa) \\  \text{ } & \searrow & \text{ } & \swarrow & \text{ } \\  \text{ } &\text{ } & \text{Property } \mathcal{D}(\kappa) & \text{ } & \text{ } \\     \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\    \end{array}

The implications in Diagram 2 are immediate. An example is given below showing that \omega_1-shrinking does not imply property \mathcal{B}(\omega_1). If \kappa=\omega, then all three properties are equivalent in normal spaces, as displayed in the following diagram. The proof is in Theorem 5.

    Diagram 3

    \displaystyle \begin{array}{ccccc} \omega \text{-Shrinking} &\text{ } & \longrightarrow & \text{ } & \text{Property } \mathcal{B}(\omega) \\  \text{ } & \nwarrow & \text{ } & \swarrow & \text{ } \\  \text{ } &\text{ } & \text{Property } \mathcal{D}(\omega) & \text{ } & \text{ } \\     \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\    \end{array}

The property \mathcal{D}(\kappa) has a dual statement in terms of decreasing closed sets. The following theorem gives the dual statement.

Theorem 4
Let X be a normal space. Let \kappa be an infinite cardinal. The following two properties are equivalent.

  • The space X has property \mathcal{D}(\kappa).
  • For each decreasing family \left\{F_\alpha: \alpha<\kappa \right\} of closed subsets of X such that \bigcap_{\alpha<\kappa} F_\alpha=\varnothing, there exists a family \left\{G_\alpha: \alpha<\kappa \right\} of open subsets of X such that \bigcap_{\alpha<\kappa} G_\alpha=\varnothing and F_\alpha \subset G_\alpha for each \alpha<\kappa.

First bullet implies second bullet
Let \left\{F_\alpha: \alpha<\kappa \right\} be a decreasing family of closed subsets of X with empty intersection. Then \left\{U_\alpha: \alpha<\kappa \right\} is an increasing family of open subsets of X where U_\alpha=X-F_\alpha. Let \left\{V_\alpha: \alpha<\kappa \right\} be an open cover of X such that \overline{V_\alpha} \subset U_\alpha for each \alpha. Then \left\{G_\alpha: \alpha<\kappa \right\} where G_\alpha=X-\overline{V_\alpha} is the needed open expansion.

Second bullet implies first bullet
Let \left\{U_\alpha: \alpha<\kappa \right\} be an increasing open cover of X. Then \left\{F_\alpha: \alpha<\kappa \right\} is a decreasing family of closed subsets of X where F_\alpha=X-U_\alpha. Note that \bigcap_{\alpha<\kappa} F_\alpha=\varnothing. Let \left\{G_\alpha: \alpha<\kappa \right\} be a family of open subsets of X such that \bigcap_{\alpha<\kappa} G_\alpha=\varnothing and F_\alpha \subset G_\alpha for each \alpha. For each \alpha, there is open set W_\alpha such that F_\alpha \subset W_\alpha \subset \overline{W_\alpha} \subset G_\alpha since X is normal. For each \alpha, let V_\alpha=X-\overline{W_\alpha}. Then \left\{V_\alpha: \alpha<\kappa \right\} is a family of open subsets of X required by the first bullet. It is a cover because \bigcap_{\alpha<\kappa} \overline{W_\alpha}=\varnothing. To show \overline{V_\alpha} \subset U_\alpha, let x \in \overline{V_\alpha} such that x \notin U_\alpha. Then x \in W_\alpha. Since x \in \overline{V_\alpha} and W_\alpha is open, W_\alpha \cap V_\alpha \ne \varnothing. Let y \in W_\alpha \cap V_\alpha. Since y \in V_\alpha, y \notin \overline{W_\alpha}, which means y \notin W_\alpha, a contradiction. Thus \overline{V_\alpha} \subset U_\alpha.

Now we show that the three properties in Diagram 3 are equivalent.

Theorem 5
Let X be a normal space. Then the following implications hold.
\omega-shrinking \Longrightarrow Property \mathcal{B}(\omega) \Longrightarrow Property \mathcal{D}(\omega) \Longrightarrow \omega-shrinking

Proof of Theorem 5
\omega-shrinking \Longrightarrow Property \mathcal{B}(\omega)
Suppose that X is \omega-shrinking. By Dowker’s theorem, X \times (\omega+1) is a normal space. We can think of \omega+1 as a convergent sequence with \omega as the limit point. Let \left\{U_n:n=0,1,2,\cdots \right\} be an increasing open cover of X. Define H and K as follows:

    H=\cup \left\{(X-U_n) \times \left\{n \right\}: n=0,1,2,\cdots \right\}

    K=X \times \left\{\omega \right\}

It is straightforward to verify that H and K are disjoint closed subsets of X \times (\omega+1). By normality, let V and W be disjoint open subsets of X \times (\omega+1) such that H \subset W and K \subset V. For each integer n=0,1,2,\cdots, define V_n as follows:

    V_n=\left\{x \in X: \exists \ \text{open } O \subset X \text{ such that } x \in O \text{ and } O \times [n, \omega] \subset V \right\}

The set [n, \omega] consists of all integers \ge n and the limit point \omega. From the way the sets V_n are defined, \left\{V_n:n=0,1,2,\cdots \right\} is an increasing open cover of X. The remaining thing to show is that \overline{V_n} \subset U_n for each n. Suppose that x \in \overline{V_n} and x \notin U_n. Then (x,n) \in H by definition of H. There exists an open set E \times \left\{n \right\} such that (x,n) \in E \times \left\{n \right\} and (E \times \left\{n \right\}) \cap V=\varnothing. Since E is an open set containing x, E \cap V_n \ne \varnothing. Let y \in E \cap V_n. By definition of V_n, there is some open set O such that y \in O and O \times [n, \omega] \subset V, a contradiction since (E \cap O) \times \left\{n \right\} is supposed to miss V. Thus \overline{V_n} \subset U_n for all integers n.

The direction Property \mathcal{B}(\omega) \Longrightarrow Property \mathcal{D}(\omega) is immediate.

Property \mathcal{D}(\omega) \Longrightarrow \omega-shrinking
Consider the dual condition of \mathcal{D}(\omega) in Theorem 4, which is equivalent to \omega-shrinking according to Dowker’s theorem. \square

Remarks
The direction \omega-shrinking \Longrightarrow Property \mathcal{B}(\omega) is true because \omega-shrinking is equivalent to the normality in the product X \times (\omega+1). The same is not true when \kappa becomes an uncountable cardinal. We now show that \kappa-shrinking does not imply \mathcal{B}(\kappa) in general.

Example 1
The space X=\omega_1 is the set of all ordinals less than \omega_1 with the ordered topology. Since it is a linearly ordered space, it is a shrinking space. Thus in particular it is \omega_1-shrinking. To show that X does not have property \mathcal{B}(\omega_1), consider the increasing open cover \left\{U_\alpha: \alpha<\omega_1 \right\} where U_\alpha=[0,\alpha) for each \alpha<\omega_1. Here [0,\alpha) consists of all ordinals less than \alpha. Suppose X has property \mathcal{B}(\omega_1). Then let \left\{V_\alpha: \alpha<\omega_1 \right\} be an increasing open cover of X such that \overline{V_\alpha} \subset U_\alpha for each \alpha.

Let L be the set of all limit ordinals in X. For each \alpha \in L, \alpha \notin U_\alpha and thus \alpha \notin \overline{V_\alpha}. Thus there exists a countable ordinal f(\alpha)<\alpha such that (f(\alpha),\alpha] misses points in \overline{V_\alpha}. Thus the map f: L \rightarrow \omega_1 is a pressing down map. By the pressing down lemma, there exists some \alpha<\omega_1 such that S=f^{-1}(\alpha) is a stationary set in \omega_1, which means that S intersects with every closed and unbounded subset of X=\omega_1. This means that for each \gamma>\alpha, (\alpha, \gamma] would miss \overline{V_\gamma}. This means that for each \gamma>\alpha, \overline{V_\gamma} \subset [0,\alpha]. As a result \left\{V_\alpha: \alpha<\omega_1 \right\} would not be a cover of X, a contradiction. So X does not have property \mathcal{B}(\omega_1). \square

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Property \mathcal{B}(\kappa)

Of the three properties discussed in the above section, we would like to single out property \mathcal{B}(\kappa). This property has a connection with normality in the product X \times Y (see Theorem 7). First, we prove a lemma that is used in proving Theorem 7.

Lemma 6
Show that the property \mathcal{B}(\kappa) is hereditary with respect to closed subsets.

Proof of Lemma 6
Let X be a space with property \mathcal{B}(\kappa). Let A be a closed subspace of X. Let \left\{U_\alpha \subset A: \alpha<\kappa \right\} be an increasing open cover of A. For each \alpha, let W_\alpha be an open subset of X such that U_\alpha=W_\alpha \cap A. Since the open sets U_\alpha are increasing, the open sets W_\alpha can be chosen inductively such that W_\alpha \supset W_\gamma for all \gamma<\alpha. This will ensure that W_\alpha will form an increasing cover.

Then \left\{W_\alpha^* \subset X: \alpha<\kappa \right\} is an increasing open cover of X where W_\alpha^*=W_\alpha \cup (X-A). By property \mathcal{B}(\kappa), let \left\{E_\alpha \subset X: \alpha<\kappa \right\} be an increasing open cover of X such that \overline{E_\alpha} \subset W_\alpha^*. For each \alpha, let V_\alpha=E_\alpha \cap A. It can be readily verified that \left\{V_\alpha \subset A: \alpha<\kappa \right\} is an increasing open cover of A. Furthermore, \overline{V_\alpha} \subset U_\alpha for each \alpha (closure taken in A). \square

Let \kappa be an infinite cardinal. Let D_\kappa=\left\{d_\alpha: \alpha<\kappa \right\} be a discrete space of cardinality \kappa. Let p be a point not in D_\kappa. Let Y_\kappa=D_\kappa \cup \left\{p \right\}. Define a topology on Y_\kappa by letting D_\kappa be discrete and by letting open neighborhood of p be of the form \left\{p \right\} \cup E where E \subset D_\kappa and D_\kappa-E has cardinality less than \kappa. Note the similarity between Y_\kappa and the convergent sequence \omega+1 in the proof of Theorem 5.

Theorem 7
Let X be a normal space. Then the product space X \times Y_\kappa is normal if and only if X has property \mathcal{B}(\kappa).

Remarks
The property \mathcal{B}(\kappa) involves the shrinking of any increasing open cover with the added property that the shrinking is also increasing. The increasing shrinking is just what is needed to show that disjoint closed subsets of the product space can be separated.

Notations
Let’s set some notations that are useful in proving Theorem 7.

  • The set [d_\alpha,p] is an open set in Y_\kappa containing the point p and is defined as follows.
    • [d_\alpha,p]=\left\{d_\beta: \alpha \le \beta<\kappa \right\} \cup \left\{p \right\}.
  • For any two disjoint closed subsets H and K of the product space X \times Y_\kappa, define the following sets.
    • For each \alpha<\kappa, let H_\alpha=H \cap (X \times \left\{d_\alpha \right\}) and K_\alpha=K \cap (X \times \left\{d_\alpha \right\}).
    • Let H_p=H \cap (X \times \left\{p \right\}) and K_p=K \cap (X \times \left\{p \right\}).
    • For each \alpha<\kappa, choose open O_\alpha \subset X such that G_\alpha=O_\alpha \times \left\{d_\alpha \right\}, H_\alpha \subset G_\alpha and \overline{G_\alpha} \cap K_\alpha=\varnothing (due to normality of X).
    • Choose open O_p \subset X such that G_p=O_p \times \left\{p \right\}, H_p \subset G_p and \overline{G_p} \cap K_p=\varnothing (due to normality of X).

Proof of Theorem 7
Suppose that X has property \mathcal{B}(\kappa). Let H and K be two disjoint closed sets of X \times Y_\kappa. Consider the following cases based on the locations of the closed sets H and K.

    Case 1. H \subset X \times D_\kappa and K \subset X \times D_\kappa.
    Case 2a. H=X \times \left\{p\right\}
    Case 2b. Exactly one of H and K intersect the set X \times \left\{p\right\}.
    Case 3. Both H and K intersect the set X \times \left\{p\right\}.

Remarks
Case 1 is easy. Case 2a is the pivotal case. Case 2b and Case 3 use a similar idea. The result in Theorem 7 is found in [1] (Theorem 6.9 in p. 189) and [4]. The authors in these two sources claimed that Case 2a is the only case that matters, citing a lemma in another source. The lemma was not stated in these two sources and the source for the lemma is a PhD dissertation that is not readily available. Case 3 essentially uses the same idea but it has enough differences. For the sake of completeness, we work out all the cases. Case 3 applies property \mathcal{B}(\kappa) twice. Despite the complicated notations, the essential idea is quite simple. If any reader finds the proof too long, just understand Case 2a and then get the gist of how the idea is applied in Case 2b and Case 3.

Case 1.
H \subset X \times D_\kappa and K \subset X \times D_\kappa.

Let M =\bigcup_{\alpha<\kappa} G_\alpha. It is clear that H \subset M and \overline{M} \cap K=\varnothing.

Case 2a.
Assume that H=X \times \left\{p\right\}. We now proceed to separate H and K with disjoint open sets. For each \alpha<\kappa, define U_\alpha as follows:

    U_\alpha=\cup \left\{O \subset X: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}

Then \left\{U_\alpha: \alpha<\kappa \right\} is an increasing open cover of X. By property \mathcal{B}(\kappa), there is an increasing open cover \mathcal{V}=\left\{V_\alpha: \alpha<\kappa \right\} of X such that \overline{V_\alpha} \subset U_\alpha for each \alpha. The shrinking \mathcal{V} allows us to define an open set G such that H \subset G and \overline{G} \cap K=\varnothing.

Let G=\cup \left\{V_\alpha \times [d_\alpha,p]: \alpha<\kappa \right\}. It is clear that H \subset G. Next, we show that \overline{G} \cap K=\varnothing. Suppose that (x,d_\alpha) \in K. Then (x,d_\alpha) \notin U_\alpha \times [d_\alpha,p]. As a result, (x,d_\alpha) \notin \overline{V_\alpha} \times [d_\alpha,p]. Let O \subset X be open such that x \in O and (O \times \left\{d_\alpha \right\}) \cap (\overline{V_\alpha} \times [d_\alpha,p])=\varnothing. Since V_\beta \subset V_\alpha for all \beta<\alpha, it follows that (O \times \left\{d_\alpha \right\}) \cap (V_\beta \times [d_\beta,p])=\varnothing for all \beta < \alpha. It is clear that (O \times \left\{d_\alpha \right\}) \cap (V_\gamma \times [d_\gamma,p])=\varnothing for all \gamma>\alpha. What has been shown is that there is an open set containing the point (x,d_\alpha) that contains no point of G. This means that (x,d_\alpha) \notin \overline{G}. We have established that \overline{G} \cap K=\varnothing.

Case 2b.
Exactly one of H and K intersect the set X \times \left\{p\right\}. We assume that H is the set that intersects the set X \times \left\{p\right\}. The only difference between Case 2b and Case 2a is that there can be points of H outside of X \times \left\{p\right\} in Case 2b.

Now proceed as in Case 2a. Obtain the open cover \left\{U_\alpha: \alpha<\kappa \right\}, the open cover \left\{V_\alpha: \alpha<\kappa \right\} and the open set G as in Case 2a. Let M=G \cup (\bigcup_{\alpha<\kappa} G_\alpha). It is clear that H \subset M. We claim that \overline{M} \cap K=\varnothing. Suppose that (x,d_\gamma) \in K. Since \overline{G} \cap K=\varnothing (as in Case 2a), there exists open set W=O \times \left\{ d_\gamma \right\} such that (x,d_\gamma) \in W and W \cap \overline{G}=\varnothing. There also exists open W_1 \subset W such that (x,d_\gamma) \in W_1 and W_1 \cap \overline{G_\gamma}=\varnothing. It is clear that W_1 \cap G_\beta=\varnothing for all \beta \ne \gamma. This means that W_1 is an open set containing the point (x,d_\gamma) such that W_1 misses the open set M. Thus \overline{M} \cap K=\varnothing.

Case 3.
Both H and K intersect the set X \times \left\{p\right\}.

Now project H_p and K_p onto the space X.

    H_p^*=\left\{x \in X: (x,p) \in H_p \right\}

    K_p^*=\left\{x \in X: (x,p) \in K_p \right\}

Note that H_p^* is simply the copy of H_p and K_p^* is the copy of K_p in X. Since X is normal, choose disjoint open sets E_1 and E_1 such that H_p^* \subset E_1 and K_p^* \subset E_2.

Let A_1=\overline{E_1} and B_1=X-K_p^*. Let A_2=\overline{E_2} and B_2=X-H_p^*. Note that A_1 is closed in X, B_1 is open in X and A_1 \subset B_1. Similarly A_2 is closed in X, B_2 is open in X and A_2 \subset B_2.

We now define two increasing open covers using property \mathcal{B}(\kappa). Define U_{\alpha,1} and T_{\alpha,1} and U_{\alpha,2} and T_{\alpha,2} as follows:

    U_{\alpha,1}=\cup \left\{O \subset B_1: O \text{ is open such that } (O \times [d_\alpha,p]) \cap K =\varnothing \right\}

    T_{\alpha,1}=U_{\alpha,1} \cap A_1

    U_{\alpha,2}=\cup \left\{O \subset B_2: O \text{ is open such that } (O \times [d_\alpha,p]) \cap H =\varnothing \right\}

    T_{\alpha,2}=U_{\alpha,2} \cap A_2

The open cover \mathcal{T}_1=\left\{T_{\alpha,1}: \alpha<\kappa \right\} is an increasing open cover of A_1. The open cover \mathcal{T}_2=\left\{T_{\alpha,2}: \alpha<\kappa \right\} is an increasing open cover of A_2.By property \mathcal{B}(\kappa) of A_1 and A_2, both covers have the following as shrinking (by Lemma 6). The two shrinkings are:

    \mathcal{V}_1=\left\{V_{\alpha,1} \subset A_1: \alpha<\kappa \right\}

    \mathcal{V}_2=\left\{V_{\alpha,2} \subset A_2: \alpha<\kappa \right\}

such that

    \overline{V_{\alpha,1}} \subset T_{\alpha,1}

    \overline{V_{\alpha,2}} \subset T_{\alpha,2}

for each \alpha<\kappa and such that both \mathcal{V}_1 and \mathcal{V}_2 are increasing open covers. Note that the closure \overline{V_{\alpha,1}} is taken in A_1 and the closure \overline{V_{\alpha,2}} is taken in A_2.

For each \alpha, let W_{\alpha,1} be the interior of V_{\alpha,1} and W_{\alpha,2} be the interior of V_{\alpha,2} (with respect to X). Note that W_{\alpha,1} is meaningful since V_{\alpha,1} is a subset of the closure of the open set E_1. Similar observation for W_{\alpha,2}. To make the rest of the argument easier to see, note the following fact about W_{\alpha,1} and W_{\alpha,2}.

    \overline{W_{\alpha,1}} \subset \overline{V_{\alpha,1}} \subset T_{\alpha,1} \subset U_{\alpha,1} (closure with respect to X)

    \overline{W_{\alpha,2}} \subset \overline{V_{\alpha,2}} \subset T_{\alpha,2} \subset U_{\alpha,2} (closure with respect to X)

For each \alpha<\kappa, choose open set O_\alpha \subset X such that

    L_\alpha=O_\alpha \times \left\{d_\alpha \right\}

    H_\alpha \subset L_\alpha

    \overline{L_\alpha} \cap K_\alpha=\varnothing

    L_\alpha \cap (\overline{W_{\alpha,2}} \times [d_\alpha,p])=\varnothing

The last point is possible because U_{\alpha,2} \times [d_\alpha,p] misses H and \overline{W_{\alpha,2}}  \subset U_{\alpha,2}. Define the open sets G and M as follows:

    G=\cup \left\{W_{\alpha,1} \times [d_\alpha,p]: \alpha<\kappa \right\}

    M=G \cup (\bigcup_{\alpha<\kappa} L_\alpha)

It is clear that H \subset M. We claim that \overline{M} \cap K=\varnothing. To this end, we show that if (x,y) \in K, then (x,y) \notin \overline{M}. If (x,y) \in K, then either (x,y)=(x,d_\gamma) for some \gamma or (x,y)=(x,p).

Let (x,d_\gamma) \in K. Note that (x,d_\gamma) \notin U_{\gamma,1} \times [d_\gamma,p]. Since \overline{W_{\gamma,1}} \subset \overline{V_{\gamma,1}} \subset T_{\gamma,1} \subset U_{\gamma,1}, (x,d_\gamma) \notin \overline{W_{\gamma,1}} \times [d_\gamma,p]. Choose an open set O \subset X such that x \in O and C=O \times \left\{d_\gamma \right\} misses \overline{W_{\gamma,1}} \times [d_\gamma,p]. Note that C misses W_{\beta,1} \times [d_\beta,p] for all \beta<\gamma since W_{\beta,1} \subset W_{\gamma,1} for all \beta<\gamma. It is clear that C misses W_{\beta,1} \times [d_\beta,p] for all \beta>\gamma.

We can also choose open C_1 \subset C such that (x,d_\gamma) \in C_1 and C_1 misses \overline{L_\gamma}. It is clear that C_1 misses L_\beta for all \beta \ne \gamma. Thus there is an open set C_1 containing the point (x,d_\gamma) such that C_1 contains no point of M.

Let (x,p) \in K. First we find an open set Q containing (x,p) such that Q misses G. From the way the open sets U_{\alpha,1} are defined, it follows that (x,p) \notin \overline{W_{\alpha,1}} \times [d_\alpha,p] for all \alpha. Furthermore W_{\alpha,1} \subset \overline{A_1}. Thus Q=(X-\overline{A_1}) \times Y_\kappa is the desired open set. On the other hand, there exists \alpha<\kappa such that x \in W_{\alpha,2}. Note that L_\gamma are chosen so that (W_{\gamma,2} \times [d_\gamma,p]) \cap L_\gamma=\varnothing for all \gamma. Since W_{\alpha,2} \subset W_{\beta,2} for all \beta \ge \alpha, (W_{\alpha,2} \times [d_\alpha,p]) \cap L_\beta=\varnothing for all \beta \ge \alpha. Thus the open set W_{\alpha,2} \times [d_\alpha,p] contains no points of L_\gamma for any \gamma. Then the open set Q \cap (W_{\alpha,2} \times [d_\alpha,p]) contains no point of M. This means that (x,p) \notin \overline{M}. Thus \overline{M} \cap K=\varnothing.

In each of the four cases (1, 2a, 2b and 3), there exists an open set M \subset X \times Y_\kappa such that H \subset M and \overline{M} \cap K=\varnothing. This completes the proof that X \times Y_\kappa is normal assuming that X has property \mathcal{B}(\kappa).

Now the other direction. Suppose that X \times Y_\kappa is normal. Then it can be shown that X has property \mathcal{B}(\kappa). The proof is similar to the proof for \omega-shrinking \Longrightarrow Property \mathcal{B}(\omega) in Theorem 5. \square

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Reference

  1. Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
  2. Rudin M. E., A Normal Space X for which X \times I is not Normal, Fund. Math., 73, 179-486, 1971. (link)
  3. Rudin M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
  4. Yasui Y., On the Characterization of the \mathcal{B}-Property by the Normality of Product Spaces, Topology and its Applications, 15, 323-326, 1983. (abstract and paper)
  5. Yasui Y., Some Characterization of a \mathcal{B}-Property, TSUKUBA J. MATH., 10, No. 2, 243-247, 1986.

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\copyright \ 2017 \text{ by Dan Ma}

Product Space – Exercise Set 1

This post presents several exercises concerning product spaces. All the concepts involved in the exercises have been discussed in the blog.

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Exercise 1

Exercise 1a
Prove or disprove:
If X and Y are both hereditarily separable, then X \times Y is hereditarily separable.

Exercise 1b
Show that if each X_\alpha is separable, then the product space \prod_{\alpha < \omega} \ X_\alpha is separable.

Exercise 1c
Prove or disprove:
If each X_\alpha is separable, then the product space \prod_{\alpha < \omega_1} \ X_\alpha is not separable.

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Exercise 2

Exercise 2a
Show that if the space X is normal, then every closed subspace of X is a normal space.

Exercise 2b
Prove or disprove:
If the space X is normal, then every dense open subspace of X is a normal space.

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Exercise 3

Consider the product space \prod_{\alpha \in W} \ X_\alpha.

Exercise 3a
Suppose that X_\alpha is compact for all but one \alpha \in W such that the non-compact factor is a Lindelof space. Show that the product space \prod_{\alpha \in W} \ X_\alpha is a normal space.

Exercise 3b
Prove or disprove:
Suppose that X_\alpha is compact for all but one \alpha \in W such that the non-compact factor is a normal space. Then the product space \prod_{\alpha \in W} \ X_\alpha is a normal space.

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Exercise 4

Exercise 4a
Let X be a compact space.
Show that if X^n is hereditarily Lindelof for all positive integer n, then X is metrizable.

Exercise 4b
Prove or disprove:
If X^n is hereditarily Lindelof for all positive integer n, then X is metrizable.

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Exercise 5

Let Y the product of uncountably many copies of the real line \mathbb{R}. If a specific example is desired, try Y=\mathbb{R}^{\omega_1} (\omega_1 many copies of \mathbb{R}) or Y=\mathbb{R}^{\mathbb{R}} (continuum many copies of \mathbb{R}). It is also OK to use a larger number of copies of the real line.

Note that the space Y is not normal (see here).

Exercise 5a
Since the product space Y is not normal, it is not Lindelof. As an exercise, find an open cover of Y that proves that Y is not Lindelof, i.e. an open cover \mathcal{U} of Y such that no countable subcollection of \mathcal{U} can cover Y.

Exercise 5b
Show that for every open cover \mathcal{U} of the space Y, there is a countable \mathcal{V} \subset \mathcal{U} of Y such that \overline{\mathcal{V}}=Y, i.e. \cup \mathcal{V} is dense in Y. Note that with this property, the space Y is said to be weakly Lindelof.

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Exercise 6

This exercise is about the product Y=\mathbb{R}^{\mathbb{R}} (continuum many copies of \mathbb{R}). Show the following.

  1. Show that Y is separable by exhibiting a countable dense set.
  2. Show that Y is not hereditarily separable by exhibiting a non-separable subspace.
  3. Show that the space Y has a closed and discrete subspace of cardinality continuum.
  4. Show that Y is not first countable.
  5. Show that Y is not a Frechet space.
  6. Show that Y is not a countably tight space.

See here for the definition of Frechet space.

See here for the definition of countably tight space.

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Exercise 7

Consider the product space Y=\mathbb{\omega}^{\omega_1}. It is not normal (see here).

Exercise 7a
Construct a dense normal subspace of Y.

Exercise 7b
Construct a dense Lindelof subspace of Y.

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\copyright \ 2016 \text{ by Dan Ma}

Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space X=\omega_1 \times I^I, which is the product of \omega_1 with the interval topology and the product space I^I=\prod_{t \in I} I where I is of course the unit interval [0,1]. The notation of \omega_1, the first uncountable ordinal, in Steen and Seebach is [0,\Omega).

Another way to notate the example X is the product space \prod_{t \in I} X_t where X_0 is \omega_1 and X_t is the unit interval I for all t>0. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of \omega_1 makes this product space an interesting example.

The following lists out the basic topological properties of the space that X=\omega_1 \times I^I are covered in [2].

  • The space X is Hausdorff and completely regular.
  • The space X is countably compact.
  • The space X is neither compact nor sequentially compact.
  • The space X is neither separable, Lindelof nor \sigma-compact.
  • The space X is not first countable.
  • The space X is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

  • The space X is not normal.
  • The space X has a dense subspace that is normal.

It follows from these bullet points that the space X is an example of a completely regular space that is not normal. Not being a normal space, X is then not metrizable. Of course there are other ways to show that X is not metrizable. One is that neither of the two factors \omega_1 or I^I is metrizable. Another is that X is not first countable.

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The space X is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example X=\omega_1 \times I^I is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example X=\omega_1 \times I^I is to show whether the second factor I^I is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space Y, the product \omega_1 \times Y is normal if and only if Y is countably tight. Thus the normality of the space X (or the lack of) hinges on whether the compact factor I^I=\prod_{t \in I} I is countably tight.

A space Y is countably tight (or has countable tightness) if for each S \subset Y and for each x \in \overline{S}, there exists some countable B \subset S such that x \in \overline{B}. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space I^I=\prod_{t \in I} I is not countably tight, we let S be the subspace of I^I consisting of points, each of which is non-zero on at most countably many coordinates. Specifically S is defined as follows:

    S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}

The set S just defined is also called the \Sigma-product of copies of unit interval I. Let g \in I^I be defined by g(t)=1 for all t \in I. It follows that g \in \overline{S}. It can also be verified that g \notin \overline{B} for any countable B \subset S. This shows that the product space I^I=\prod_{t \in I} I is not countably tight.

By Theorem 1 found in this link, the space X=\omega_1 \times I^I is not normal.

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The space X has a dense subspace that is normal

Now that we know X=\omega_1 \times I^I is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace \omega_1 \times S where S is the \Sigma-product S=\Sigma_{t \in I} I defined in the preceding section. The subspace S is dense in the product space I^I. Thus \omega_1 \times S is dense in X=\omega_1 \times I^I. The space S is normal since the \Sigma-product of separable metric spaces is normal. Furthermore, \omega_1 can be embedded as a closed subspace of S=\Sigma_{t \in I} I. Then \omega_1 \times S is homeomorphic to a closed subspace of S \times S. Since S \times S \cong S, the space \omega_1 \times S is normal.

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Reference

  1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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\copyright \ 2015 \text{ by Dan Ma}

Normality in the powers of countably compact spaces

Let \omega_1 be the first uncountable ordinal. The topology on \omega_1 we are interested in is the ordered topology, the topology induced by the well ordering. The space \omega_1 is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of \omega_1. We also make comments on normality in the powers of a countably compact non-compact space.

Let \omega be the first infinite ordinal. It is well known that \omega^{\omega_1}, the product space of \omega_1 many copies of \omega, is not normal (a proof can be found in this earlier post). This means that any product space \prod_{\alpha<\kappa} X_\alpha, with uncountably many factors, is not normal as long as each factor X_\alpha contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space \prod_{\alpha<\kappa} X_\alpha, the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of \omega). In other words, the interesting case is that each factor X_\alpha is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in X^{\kappa} where X is a countably non-compact space. In this post we start with the space X=\omega_1 of the countable ordinals. We examine \omega_1 power \omega_1^{\omega_1} as well as the countable power \omega_1^{\omega}. The former is not normal while the latter is normal. The proof that \omega_1^{\omega} is normal is an application of the normality of \Sigma-product of the real line.

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The uncountable product

Theorem 1
The product space \prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1} is not normal.

Theorem 1 follows from Theorem 2 below. For any space X, a collection \mathcal{C} of subsets of X is said to have the finite intersection property if for any finite \mathcal{F} \subset \mathcal{C}, the intersection \cap \mathcal{F} \ne \varnothing. Such a collection \mathcal{C} is called an f.i.p collection for short. It is well known that a space X is compact if and only collection \mathcal{C} of closed subsets of X satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space X=\omega_1, there is an f.i.p. collection of cardinality \omega_1 using its linear order. For each \alpha<\omega_1, let C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}. Let \mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}. It is a collection of closed subsets of X=\omega_1. It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space X with an f.i.p. collection of cardinality \omega_1 that has empty intersection, the product space X^{\omega_1} is not normal.

Theorem 2
Let X be a countably compact space. Suppose that there exists a collection \mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\} of closed subsets of X such that \mathcal{C} has the finite intersection property and that \mathcal{C} has empty intersection. Then the product space X^{\omega_1} is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space X^{\omega_1}=\prod_{\alpha<\omega_1} X, we mean a set of the form O=\prod_{\alpha<\omega_1} O_\alpha such that each O_\alpha is an open subset of X and that O_\alpha=X for all but finitely many \alpha<\omega_1. Given a standard basic open set O=\prod_{\alpha<\omega_1} O_\alpha, the notation \text{Supp}(O) refers to the finite set of \alpha for which O_\alpha \ne X. For any set M \subset \omega_1, the notation \pi_M refers to the projection map from \prod_{\alpha<\omega_1} X to the subproduct \prod_{\alpha \in M} X. Each element d \in X^{\omega_1} can be considered a function d: \omega_1 \rightarrow X. By (d)_\alpha, we mean (d)_\alpha=d(\alpha).

For each t \in X, let f_t: \omega_1 \rightarrow X be the constant function whose constant value is t. Consider the following subspaces of X^{\omega_1}.

    H=\prod_{\alpha<\omega_1} C_\alpha

    \displaystyle K=\left\{f_t: t \in X  \right\}

Both H and K are closed subsets of the product space X^{\omega_1}. Because the collection \mathcal{C} has empty intersection, H \cap K=\varnothing. We show that H and K cannot be separated by disjoint open sets. To this end, let U and V be open subsets of X^{\omega_1} such that H \subset U and K \subset V.

Let d_1 \in H. Choose a standard basic open set O_1 such that d_1 \in O_1 \subset U. Let S_1=\text{Supp}(O_1). Since S_1 is the support of O_1, it follows that \pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U. Since \mathcal{C} has the finite intersection property, there exists a_1 \in \bigcap_{\alpha \in S_1} C_\alpha.

Define d_2 \in H such that (d_2)_\alpha=a_1 for all \alpha \in S_1 and (d_2)_\alpha=(d_1)_\alpha for all \alpha \in \omega_1-S_1. Choose a standard basic open set O_2 such that d_2 \in O_2 \subset U. Let S_2=\text{Supp}(O_2). It is possible to ensure that S_1 \subset S_2 by making more factors of O_2 different from X. We have \pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U. Since \mathcal{C} has the finite intersection property, there exists a_2 \in \bigcap_{\alpha \in S_2} C_\alpha.

Now choose a point d_3 \in H such that (d_3)_\alpha=a_2 for all \alpha \in S_2 and (d_3)_\alpha=(d_2)_\alpha for all \alpha \in \omega_1-S_2. Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

  • a sequence d_1,d_2,d_3,\cdots of point of H=\prod_{\alpha<\omega_1} C_\alpha,
  • a sequence S_1 \subset S_2 \subset S_3 \subset \cdots of finite subsets of \omega_1,
  • a sequence a_1,a_2,a_3,\cdots of points of X

such that for all n \ge 2, (d_n)_\alpha=a_{n-1} for all \alpha \in S_{n-1} and \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U. Let A=\left\{a_1,a_2,a_3,\cdots \right\}. Either A is finite or A is infinite. Let’s examine the two cases.

Case 1
Suppose that A is infinite. Since X is countably compact, A has a limit point a. That means that every open set containing a contains some a_n \ne a. For each n \ge 2, define y_n \in \prod_{\alpha< \omega_1} X such that

  • (y_n)_\alpha=(d_n)_\alpha=a_{n-1} for all \alpha \in S_n,
  • (y_n)_\alpha=a for all \alpha \in \omega_1-S_n

From the induction step, we have y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U for all n. Let t=f_a \in K, the constant function whose constant value is a. It follows that t is a limit of \left\{y_1,y_2,y_3,\cdots \right\}. This means that t \in \overline{U}. Since t \in K \subset V, U \cap V \ne \varnothing.

Case 2
Suppose that A is finite. Then there is some m such that a_m=a_j for all j \ge m. For each n \ge 2, define y_n \in \prod_{\alpha< \omega_1} X such that

  • (y_n)_\alpha=(d_n)_\alpha=a_{n-1} for all \alpha \in S_n,
  • (y_n)_\alpha=a_m for all \alpha \in \omega_1-S_n

As in Case 1, we have y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U for all n. Let t=f_{a_m} \in K, the constant function whose constant value is a_m. It follows that t=y_n for all n \ge m+1. Thus U \cap V \ne \varnothing.

Both cases show that U \cap V \ne \varnothing. This completes the proof the product space X^{\omega_1} is not normal. \blacksquare

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The countable product

Theorem 3
The product space \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is collectionwise normal, which is stronger than normality. The proof makes use of the \Sigma-product of \kappa many copies of \mathbb{R}, which is the following subspace of the product space \mathbb{R}^{\kappa}.

    \Sigma(\kappa)=\left\{x \in \mathbb{R}^{\kappa}: x(\alpha) \ne 0 \text{ for at most countably many } \alpha<\kappa \right\}

It is well known that \Sigma(\kappa) is collectionwise normal (see this earlier post). We show that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega} is a closed subspace of \Sigma(\kappa) where \kappa=\omega_1. Thus \omega_1^{\omega} is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space \omega_1 is embedded as a closed subspace of \Sigma(\omega_1).

For each \beta<\omega_1, define f_\beta:\omega_1 \rightarrow \mathbb{R} such that f_\beta(\gamma)=1 for all \gamma<\beta and f_\beta(\gamma)=0 for all \beta \le \gamma <\omega_1. Let W=\left\{f_\beta: \beta<\omega_1 \right\}. We show that W is a closed subset of \Sigma(\omega_1) and W is homeomorphic to \omega_1 according to the mapping f_\beta \rightarrow W.

First, we show W is closed by showing that \Sigma(\omega_1)-W is open. Let y \in \Sigma(\omega_1)-W. We show that there is an open set containing y that contains no points of W.

Suppose that for some \gamma<\omega_1, y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}. Consider the open set Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1) where Q_\alpha=\mathbb{R} except that Q_\gamma=O. Then y \in Q and Q \cap W=\varnothing.

So we can assume that for all \gamma<\omega_1, y_\gamma \in \left\{0, 1 \right\}. There must be some \theta such that y_\theta=1. Otherwise, y=f_0 \in W. Since y \ne f_\theta, there must be some \delta<\gamma such that y_\delta=0. Now choose the open interval T_\theta=(0.9,1.1) and the open interval T_\delta=(-0.1,0.1). Consider the open set M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1) such that M_\alpha=\mathbb{R} except for M_\theta=T_\theta and M_\delta=T_\delta. Then y \in M and M \cap W=\varnothing. We have just established that W is closed in \Sigma(\omega_1).

Consider the mapping f_\beta \rightarrow W. Based on how it is defined, it is straightforward to show that it is a homeomorphism between \omega_1 and W.

Claim 2
The \Sigma-product \Sigma(\omega_1) has the interesting property it is homeomorphic to its countable power, i.e.

    \Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}.

Because each element of \Sigma(\omega_1) is nonzero only at countably many coordinates, concatenating countably many elements of \Sigma(\omega_1) produces an element of \Sigma(\omega_1). Thus Claim 2 can be easily verified. With above claims, we can see that

    \displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)

Thus \omega_1^{\omega} is a closed subspace of \Sigma(\omega_1). Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that \omega_1^{\omega} is normal. \blacksquare

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The normality in the powers of X

We have established that \prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1} is not normal. Hence any higher uncountable power of \omega_1 is not normal. We have also established that \prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}, the countable power of \omega_1 is normal (in fact collectionwise normal). Hence any finite power of \omega_1 is normal. However \omega_1^{\omega} is not hereditarily normal. One of the exercises below is to show that \omega_1 \times \omega_1 is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let X be a countably compact space has an f.i.p. collection \mathcal{C} of closed sets such that \bigcap \mathcal{C}=\varnothing. Then X^{\kappa} is not normal where \kappa=\lvert \mathcal{C} \lvert.

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let X be a countably compact space that is not compact. Then there exists a cardinal number \kappa such that X^{\kappa} is not normal and X^{\tau} is normal for all cardinal number \tau<\kappa.

By the non-compactness of X, there exists an f.i.p. collection \mathcal{C} of closed subsets of X such that \bigcap \mathcal{C}=\varnothing. Let \kappa be the least cardinality of such an f.i.p. collection. By Theorem 4, that X^{\kappa} is not normal. Because \kappa is least, any smaller power of X must be normal.

Theorem 6
Let X be a space that is not countably compact. Then X^{\kappa} is not normal for any cardinal number \kappa \ge \omega_1.

Since the space X in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, \omega^{\omega_1} is not normal. Then \omega^{\omega_1} is a closed subspace of X^{\omega_1}.

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space X, X^{\kappa} is not normal for some cardinal number \kappa. The \kappa from either Theorem 5 or Theorem 6 is at least \omega_1. Interestingly for some spaces, the \kappa can be much smaller. For example, for the Sorgenfrey line, \kappa=2. For some spaces (e.g. the Michael line), \kappa=\omega.

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space X is normal, then X is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of X is just another way of discussing Theorem 7. According to Theorem 7, if X is not compact, some power of X is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in X^{\kappa} for more examples of countably compact non-compact X. One particular interesting example would be a countably compact non-compact X such that the least power \kappa for non-normality in X^{\kappa} is more than \omega_1. A possible candidate could be the second uncountable ordinal \omega_2. By Theorem 2, \omega_2^{\omega_2} is not normal. The issue is whether the \omega_1 power \omega_2^{\omega_1} and countable power \omega_2^{\omega} are normal.

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Exercises

Exercise 1
Show that \omega_1 \times \omega_1 is not hereditarily normal.

Exercise 2
Show that the mapping f_\beta \rightarrow W in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space \omega_1 is a closed subspace of the \Sigma-product of the real line. Show that \omega_1 can be embedded in the \Sigma-product of arbitrary spaces.

For each \alpha<\omega_1, let X_\alpha be a space with at least two points. Let p \in \prod_{\alpha<\omega_1} X_\alpha. The \Sigma-product of the spaces X_\alpha is the following subspace of the product space \prod_{\alpha<\omega_1} X_\alpha.

    \Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}

The point p is the center of the \Sigma-product. Show that the space \Sigma(X_\alpha) contains \omega_1 as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that \omega_1^{\omega} is normal.

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\copyright \ 2015 \text{ by Dan Ma}