Weakly Lindelof spaces

The weakly Lindelof property is a natural weakening of the familiar Lindelof property. In this post, we discuss some of the basic properties of weakly Lindelof spaces.

We consider topological spaces that are at least T_1 (i.e. finite sets are closed) and regular. A space X is said to be Lindelof if for any open cover \mathcal{U} of X, there is a countable \mathcal{V} \subset \mathcal{U} such that X=\bigcup \mathcal{V}. A natural weakening of the Lindelof property is that we only require the countable \mathcal{V} to cover a dense subset of the space X. Specifically, a space X is said to be a weakly Lindelof space if for any open cover \mathcal{U} of X, there is a countable \mathcal{V} \subset \mathcal{U} such that \bigcup \mathcal{V} is dense in X.

The notion of weakly Lindelof has a brief mention in the Encyclopedia of General Topology (see page 183 in [4]), pointing out a connection to Banach space theory. Furthermore, assuming CH, the weakly Lindelof subspaces of \beta \mathbb{N} are precisely those subspaces which are C^*-embedded into \beta \mathbb{N}. In this post, we focus on the basic properties.

Clearly separable spaces and Lindelof spaces are weakly Lindelof. Another obvious property that implies weakly Lindelof is the existence of a dense Lindelof subspace. It is slightly less obvious that the countable chain condition implies the weakly Lindelof property. We have the following implications.

All the affirmative implications in the above diagram cannot be reversed (see Examples 1, 2 and 3 below).

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Some Cardinal Functions

Some of the properties discussed below can be described by cardinal functions, e.g., Lindelof number and weak Lindelof numbers. So we describe these before going into the basic properties. Let X be a space. The Lindelof number of the space X, denoted by L(X), is the least cardinal number \mathcal{K} such that every open cover \mathcal{U} of X has a subcollection \mathcal{V} \subset \mathcal{U} with \lvert \mathcal{V} \lvert \le \mathcal{K} such that \mathcal{V} is a cover of X. When L(X)=\omega, we say that the space is Lindelof.

The weak Lindelof number of the space X, denoted by wL(X), is the least cardinal number \mathcal{K} such that every open cover \mathcal{U} of X has a subcollection \mathcal{V} \subset \mathcal{U} with \lvert \mathcal{V} \lvert \le \mathcal{K} such that X=\overline{\bigcup \mathcal{V}}. When wL(X)=\omega, we say that the space is weakly Lindelof.

The character at x \in X, denoted by \chi(x,X), is the least cardinal number of a local base at the point x \in X. The character of the space X, denoted by \chi(X), is the supremum of all the cardinal numbers \chi(x,X) over all x \in X. When \chi(X)=\omega, we say that X is first countable.

The cellularity of the space X, denoted by c(X), is the least infinite cardinal number \mathcal{K} such that every collection of pairwise disjoint non-empty open subsets of X has cardinality \le \mathcal{K}. When c(X)=\omega, we say that X has the countable chain condition.

The extent of the space X, denoted by e(X), is the least infinite cardinal number \mathcal{K} such that if A is a closed and discrete subset of X, then \lvert A \lvert \le \mathcal{K}. If e(X)=\omega, then X is said to have countable extent (there are no uncountable closed and discrete subset). It is well known that Lindelof spaces have countable extent. The Lindelof number and the extent is related by the inequality: e(X) \le L(X).

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Basic Properties

Weakly Lindelof spaces behave differently from Lindelof spaces in some ways. For example, closed subsets of a weakly Lindelof space do not have to be weakly Lindelof. In other ways, weakly Lindelof spaces and Lindelof spaces behave similarly. For example, the product of weakly Lindelof spaces needs not be weakly Lindelof and that every continuous image of a weakly Lindelof space is weakly Lindelof. Any Lindelof, Hausdorff and first countable space has cardinality no more than continuum. There is a similar theorem for weakly Lindelof spaces. Despite all these similarities with Lindelof spaces, the weak Lindelof property is a very weak property. It is well known that every Lindelof space has countable extent. There is no bound on the extent of weakly Lindelof spaces. The extent of a weakly Lindelof space can be arbitrarily large (see Example 4 below).

We discuss the following properties of weakly Lindelof spaces.

  1. Any space with the countable chain condition is weakly Lindelof.
  2. Any paracompact weakly Lindelof space is Lindelof.
  3. Every continuous image of a weakly Lindelof space is weakly Lindelof.
  4. The product of a compact space and a weakly lindelof space is weakly Lindelof.
  5. The product of two Lindelof spaces needs not be weakly Lindelof.
  6. Any normal first countable weakly Lindelof space has cardinality \le 2^\omega.
  7. For any infinite cardinal \mathcal{K}, there exists a weakly Lindelof space X such that e(X) \ge \mathcal{K}, i.e., the extent is at least \mathcal{K}. See Example 4 below.

Proof of 1
A space X has the countable chain condition (has the CCC or is CCC for short) if there exists no uncountable collection of non-empty pairwise disjoint open subsets of X. “CCC \Longrightarrow weakly Lindelof” follows from the following theorem (proved in this previous post).

    Theorem
    A space X has the CCC if and only if for every collection \mathcal{U} of non-empty open subsets of X, there is a countable \mathcal{V} \subset \mathcal{U} such that \bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}.

To finish off, let \mathcal{U} be an open cover of X. By the theorem, there exists a countable \mathcal{V} \subset \mathcal{U} such that \bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}. This means that X=\overline{\bigcup \mathcal{V}}. \blacksquare

Comments
Even though CCC implies weakly Lindelof, CCC does not imply the stronger property of having a dense Lindelof subspace (see Example 3 below).

The proof of 1 can be generalized to show that wL(X) \le c(X) for any space X. However, the inequality cannot be made an equality. In fact, the inequality wL(X) \le c(X) can be made as wide as one wishes. Specifically, we can keep wL(X)=\omega while making c(X) as large as one wishes (see Example 2 below). Thus the notions of countable chain condition and the weakly Lindelof property are far apart.

Proof of 2
Let \mathcal{U} be an open cover of a paracompact weakly Lindelof space X. Using the regularity of the space, there is an open refinement \mathcal{V} of \mathcal{U} for each V \in \mathcal{V}, \overline{V} \subset U for some U \in \mathcal{U}. Using the paracompactness, let \mathcal{W} be a locally finite open refinement of \mathcal{V}. Using the weakly Lindelof property, choose a countable \mathcal{C} \subset \mathcal{W} such that X=\overline{\bigcup \mathcal{C}}. With the collection \mathcal{C} being locally finite, we have X=\overline{\bigcup \mathcal{C}}=\bigcup \left\{\overline{C}: C \in \mathcal{C} \right\}. Thus every point of X belongs to some \overline{C} for some C \in \mathcal{C}. Tracing from \mathcal{C} to \mathcal{W}, to \mathcal{V} and then to \mathcal{U}, we see that for every C \in \mathcal{C}, \overline{C} \subset U for some U \in \mathcal{U}. It follows that a countable subcollection of \mathcal{U} is a cover of X. This completes the proof of bullet point 2.

This result implies that in any metrizable space, the weakly Lindelof number coincides with the Lindelof number. So in metrizable spaces, the weak Lindelof number is just as good as an indicator of weight as the other cardinal functions such as density and Lindelof number.

Among CCC spaces, paracompactness and the Lindelof property coincide. This result shows that among weakly Lindelof spaces, paracompactness and the Lindelof property also coincide. \blacksquare

The proof of 3 is straightforward. It is very similar to the proof that continuous image of a Lindelof space is Lindelof.

Proof of 4
The proof that the product of a compact space and a weakly Lindelof space is weakly Lindelof makes use of the tube lemma, as in the proof that the product of a compact space and a Lindelof space is Lindelof.

Let X be weakly Lindelof. Let Y be compact. Let \mathcal{U} be an open cover of X \times Y. For each x \in X, let \mathcal{U}_x \subset \mathcal{U} be finite such that \mathcal{U}_x is a cover of \left\{ x \right\} \times Y. By the tube lemma, there exists some open set O_x \subset X such that \left\{ x \right\} \times Y \subset O_x \times Y \subset \bigcup \mathcal{U}_x.

Since X is weakly Lindelof, there exists a countable A \subset X such that X=\overline{\bigcup \limits_{x \in A} O_x}. Let \mathcal{U}_A=\bigcup \limits_{x \in A} \mathcal{U}_x. It is clear that \mathcal{U}_A is a countable subcollection of \mathcal{U}. Note that the set \bigcup \limits_{x \in A} (O_x \times Y) is dense in X \times Y. Thus the set \bigcup \bigcup \limits_{x \in A} \mathcal{U}_x is dense in X \times Y too. Thus X \times Y=\overline{\bigcup \bigcup \limits_{x \in A} \mathcal{U}_x}. This completes the proof that X \times Y is weakly Lindelof. \blacksquare

Proof of 5
An example of two Lindelof spaces whose product is not weakly Lindelof is provided in [3]. \blacksquare

Discussion of 6
Any Lindelof first countable Hausdorff space has cardinality no more than continuum (discussed in this previous post). This fact is a specific case of the general theorem that

    \lvert X \lvert \le 2^{\chi(X) \cdot L(X)}

for any Hausdorff space X. Hence, the cardinality of any first countable Lindelof space is bounded by 2^\omega. It is interesting that there is an analogous result for weakly Lindelof space. In [2], the following inequality was proved:

    \lvert X \lvert \le 2^{\chi(X) \cdot wL(X)}

for any normal space (Theorem 2.1 in [2]). Thus the cardinality of any normal weakly Lindelof space is bounded by 2^\omega.

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Examples

Example 1 and Example 2 below use Lindelof or compact spaces that do not have the CCC as starting point. Here’s several examples of Lindelof non-CCC spaces:

  • One-point Lindelofication of an uncountable set. The space is denoted by L(\mathcal{K}) and is the set \left\{p \right\} \cup D(\mathcal{K}) where D(\mathcal{K}) is the discrete space of cardinality \mathcal{K} and p is a point not in D(\mathcal{K}). The open neighborhoods at p have the form \left\{p \right\} \cup (D(\mathcal{K})-C) where C \subset D(\mathcal{K}) is countable.
  • The space \omega_1+1 with the order topology. Note that \omega_1+1 is the immediate successor of \omega_1, the first uncountable ordinal. See here.
  • The unit square [0,1] \times [0,1] with the lexicographic order. See here.
  • The Alexandroff Double Circle. See here.

In the above four spaces, the first one is Lindelof and the other three are compact. All four do not have the countable chain condition.

Example 1
A non-Lindelof space X_1 that has a dense Lindelof subspace. As a bonus, this space does not have the CCC.

The idea is to start with a space that has a countable dense set of isolated points and an uncountable closed and discrete subset. One such space is a so called psi-space, a space defined using an uncountable almost disjoint family of subsets of \omega. Then replace each of the countably many isolated points with a copy of one of the above examples of a Lindelof space without the CCC.

Let \omega the first infinite ordinal (or the set of all nonnegative integers). Let \mathcal{A} be an uncountable almost disjoint family of subsets of \omega (for the purpose of this example, it does not have to be an maximal almost disjoint family). Let \Psi(\mathcal{A})=\mathcal{A} \cup \omega, where each n \in \omega is isolated and each A \in \mathcal{A} has open neighborhoods of the form \left\{A \right\} \cup (A-F) where F \subset \omega is finite. For a more detailed discussion about Psi-space, see this previous post.

Let Y be any one of the above Lindelof space that is not CCC. For each n \in \omega, let Y_n=Y \times \left\{n \right\}. So the Y_n are distinct copies of the space Y. The underlying set of this example is the following set:

    X_1=\mathcal{A} \cup \bigcup \limits_{n \in \omega} Y_n

The topology on X_1 is defined in such a way that each Y_n is considered a copy of the space Y and each A \in \mathcal{A} has open neighborhoods of the form:

    \left\{A \right\} \cup \bigcup \limits_{n \in A-F} Y_n

where F \subset \omega is finite. The union of all Y_n is a dense Lindelof subspace of X_1. The set \mathcal{A} is an uncountable closed and discrete subset of X_1. Thus X_1 is not Lindelof. Each Y_n has uncountably many disjoint open sets. Thus X_1 does not have the CCC. This example shows that the existence of a dense Lindelof subspace implies neither the CCC nor the Lindelof property.

Example 2
A weakly Lindelof non-CCC space X_2.

Let X be any one of the above three non-CCC compact spaces. Let Y be any space with the CCC, hence is weakly Lindelof. Let X_2=X \times Y. Then X \times Y is weakly Lindelof. It is also clear that X \times Y does not have the CCC. This example shows that the weakly Lindelof property does not imply the countable chain condition.

This example shows that \omega=wL(X_2)<c(X_2). In fact, it is possible to make c(X_2) as large as possible. In the definition of X \times Y in this example, let X be the one-point Lindelofication L(\mathcal{K}) and Y be any CCC space. Then c(L(\mathcal{K})) can be made as large as possible. Hence c(X \times Y) can be made as large as possible.

Example 3
A CCC space X_3 that has no dense Lindelof subspace.

This example is found in a paper of Arhangel’skii (Theorem 1.1 in [1]). Let C(\omega_1+1) be the set of all continuous real-valued functions defined on \omega_1+1. The set C(\omega_1+1) endowed with the pointwise convergence topology is typically denoted by C_p(\omega_1+1). The space we want to use is X_3=C_p(\omega_1+1).

The space C_p(\omega_1+1) is a dense subspace of the product space \mathbb{R}^{\omega_1}. Thus C_p(\omega_1+1) has the CCC. In [1], it is shown that C_p(\omega_1+1) does not contain a dense normal subspace. Hence it does not contain a dense Lindelof subspace. The proof that C_p(\omega_1+1) does not contain a dense normal subspace is a deep and non-trivial result.

The example X_3=C_p(\omega_1+1) shows that even though CCC implies the weakly Lindelof property, it cannot give the stronger property of the existence of a dense Lindelof subspace. It is also an example showing that the implication “existence of a dense Lindelof subspace \Longrightarrow weakly Lindelof” cannot be reversed.

Example 4
An weakly Lindelof space X_4 such that the extent can be made arbitrarily large.

Let \mathcal{K} be any uncountable cardinal. Let W be a discrete space of cardinality \mathcal{K}. Let \beta W be the Stone-Cech compactification of W. Consider the ordinal S=\omega+1 with the order topology (can just think of it as a sequence of isolated points converging to the limit \omega). The space X_4 is defined as follows:

    X_4=\beta W \times S-(\beta W-W) \times \left\{\omega \right\}

Note that \beta W \times \omega is a \sigma-compact dense subspace of X_4. Hence X_4 is weakly Lindelof. On the other hand, the set W \times \left\{\omega \right\} is a closed and discrete subset of X_4. Since the cardinality of W can be made arbitrarily large, the extent of X_4 can be made arbitrarily large. Thus there is no upper bound on the extent of weakly Lindelof spaces (unlike Lindelof spaces).

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Reference

  1. Arhangel’skii A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
  2. Bell M., Ginsburg J., Woods G., Cardinal Inequalities for Topological Spaces Involving the Weak Lindelof Number, Pacific J. Math., 79, no. 1, 37-45, 1978.
  3. Hajnal A., Juhasz I., On the Products of Weakly Lindelof Spaces, Proc. Amer. Math. Soc., 130, no. 1, 454-456, 1975.
  4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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\copyright \ 2014 \text{ by Dan Ma}

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Another characterization about CCC spaces

In this post, we present another characterization about spaces with the countable chain condition (CCC spaces for short). The theorem presented here (Theorem 1 below) will provide more insight about CCC spaces and should be useful in proving theorems about CCC spaces. This characterization will make it easy to see that CCC spaces are weakly Lindelof.

This post can be considered a continuation of an earlier post, which discusses a different characterization of CCC spaces.

All spaces under consideration are at least T_1 and regular. A space X is said to have the countable chain condition (to have the CCC for short) if \mathcal{U} is a disjoint collection of non-empty open subsets of X (meaning that for any A,B \in \mathcal{U} with A \ne B, we have A \cap B=\varnothing), then \mathcal{U} is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if X has the CCC, we also say that X is a CCC space or X is CCC. We prove the following theorem.

    Theorem 1

      Let X be a space. Then the following conditions are equivalent.

      1. The space has the CCC.
      2. For any collection \mathcal{U} of non-empty open subsets of X, there exists a countable \mathcal{V} \subset \mathcal{U} such that \bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}.

Proof of Theorem 1
1 \Longrightarrow 2
Suppose that condition 2 does not hold. Then there exists a collection \mathcal{U} of non-empty open subsets of X such that for any countable \mathcal{V} \subset \mathcal{U}, there exists a point x \in \bigcup \mathcal{U}=Y such that x \notin \overline{\bigcup \mathcal{V}}. From the collection \mathcal{U}, performing a transfinite inductive process, we will generate an uncountable collection of pairwise disjoint non-empty open subsets of X.

Choose some U_0 \in \mathcal{U}. Choose some x_0 \in Y-\overline{U_0}. For \alpha < \omega_1, suppose that the following have been chosen

    \left\{x_\beta \in Y: \beta<\alpha \right\}

    \left\{U_\beta \in \mathcal{U}: \beta<\alpha  \right\}

such that for each \beta<\alpha, x_\beta \notin \overline{\bigcup \limits_{\gamma < \beta} U_\gamma} and x_\beta \in U_\beta. Then by the assumption about \mathcal{U}, there exists x_\alpha \in Y such that x_\alpha \notin \overline{\bigcup \limits_{\gamma < \alpha} U_\gamma}. Now choose some U_\alpha \in \mathcal{U} such that x_\alpha \in U_\alpha. The inductive process is completed.

For each \alpha<\omega_1, let O_\alpha=U_\alpha-\overline{\bigcup \limits_{\gamma < \alpha} U_\gamma}. Clearly O_\alpha \ne \varnothing since x_\alpha \in O_\alpha. For \beta<\alpha<\omega_1, we have O_\beta \cap O_\alpha=\varnothing. With the non-empty open sets O_\alpha being pairwise disjoint, we conclude that X does not have the CCC.

2 \Longrightarrow 1
This is the easier direction. Suppose X is not CCC. Let \mathcal{W}=\left\{W_\alpha: \alpha<\omega_1 \right\} be a collection of pairwise disjoint non-empty open subsets of X. It is clear that for any countable \mathcal{V} \subset \mathcal{W}, the closure \overline{\bigcup \mathcal{V}} has to miss some W_\alpha (e.g. choose some W_\alpha \notin \mathcal{V}). Thus condition 2 does not hold. \blacksquare

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Weakly Lindelof Spaces

With Theorem 1, the CCC property looks like a covering property. Let X be a CCC space. Let \mathcal{U} be an open cover of X. By Theorem 1, there is a countable \mathcal{V} \subset \mathcal{U} such that X=\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}} (in other words, \bigcup \mathcal{V} is dense in X). So any CCC space X satisfies the following covering property:

    For any open cover \mathcal{U} of X, there exists a countable \mathcal{V} \subset \mathcal{U} such that \bigcup \mathcal{V} is dense in X.

Any space satisfying the above property is called a weakly Lindelof space. Any CCC space is weakly Lindelof. On the other hand, the weakly Lindelof property is strictly weaker than CCC. For a further discussion, see the next post called Weakly Lindelof spaces.

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\copyright \ 2014 \text{ by Dan Ma}

An exercise involving non-normal spaces

A space is normal if any two disjoint closed subsets of the space can be separated by disjoint open sets. A space is pseudonormal if any two disjoint closed subsets of the space, one of which is countable, can be separated by disjoint open sets. In this post, we present an interesting exercise that deals with non-normal spaces:

    Take a space that is not normal. Then determine whether it is pseudonormal. You can supply your own examples or you can start with several non-normal spaces listed below. Once you have a list, determine which ones are psuedonormal and which ones are not.

To make the exercise more interesting, we propose that the focus is on spaces that are T_1 (i.e. singleton sets are closed) and regular. Since regular Lindelof spaces are normal, we will be certain that any non-normal (and regular) space is not Lindelof.

In the previous post called Pseudonormal spaces, we identify four spaces that are known to be non-normal. Three of these spaces are not normal because one countable closed set and another closed set cannot be separated, hence not pseudonormal (one is the Sorgenfrey plane and one is the Niemmytzkis’ plane). The fourth non-normal space is pseudonormal.

Here’s a list of several other non-normal spaces previously discussed in this blog.

  • The Tychonoff Plank.
  • The sigma-product of \omega_1 many copies of \omega_1+1.
  • The product space \omega^{\omega_1}.
  • The product of the Michael line and the space of irrationals.
  • The product of countably many copies of the Michael line.
  • The product of a Lindelof space and a Bernstein set.
  • The Pixley-Roy space \mathcal{F}[\mathbb{R}].
  • Mrowka space, defined on a maximal almost disjoint family of subsets of \omega.

Readers are welcome to submit other examples of non-normal spaces. Submit examples by entering a comment below. Submitted examples that are different from the ones listed above will be appended to this post.

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\copyright \ 2014 \text{ by Dan Ma}

A theorem about CCC spaces

It is a well known result in general topology that in any regular space with the countable chain condition, paracompactness and the Lindelof property are equivalent. The proof of this result hinges on one theorem about the spaces with the countable chain condition. In this post we are to put the spotlight on this theorem (Theorem 1 below) and then use it to prove a few results. These results indicate that in a space with the countable chain condition with some weaker covering property is either Lindelof or paracompact.

This post is centered on a theorem about the CCC property (Theorem 1 and Theorem 1a below). So it can be considered as a continuation of a previous post on CCC called Some basic properties of spaces with countable chain condition. The results that are derived from Theorem 1 are also found in [2]. But the theorem concerning CCC is only a small part of that paper among several other focuses. In this post, the exposition is to explain several interesting theorems that are derived from Theorem 1. One of the theorems is the statement that every locally compact metacompact perfectly normal space is paracompact, a theorem originally proved by Arhangelskii (see Theorem 11 below).

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CCC Spaces

All spaces under consideration are at least T_1 and regular. A space X is said to have the countable chain condition (to have the CCC for short) if \mathcal{U} is a disjoint collection of non-empty open subsets of X (meaning that for any A,B \in \mathcal{U} with A \ne B, we have A \cap B=\varnothing), then \mathcal{U} is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if X has the CCC, we also say that X is a CCC space or X is CCC.

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A Theorem about CCC Spaces

The theorem of CCC spaces we want to discuss has to do with collections of open sets that are “nice”. We first define what we mean by nice. Let \mathcal{A} be a collection of non-empty subsets of the space X. The collection \mathcal{A} is said to be point-finite (point-countable) if each point of X belongs to only finitely (countably) many sets in \mathcal{A}.

Now we define what we mean by “nice” collection of open sets. The collection \mathcal{A} is said to be locally finite (locally countable) at a point x \in X if there exists an open set O \subset X with x \in O such that O meets at most finitely (countably) many sets in \mathcal{A}. The collection \mathcal{A} is said to be locally finite (locally countable) if it is locally finite (locally countable) at each x \in X.

The property of being a separable space implies the CCC. The reverse is not true. However the CCC property is still a very strong property. The CCC property is equivalent to the property that if a collection of non-empty open sets is “nice” on a dense set of points, then the collection of open sets is a countable collection. The following is a precise statement.

    Theorem 1

      Let X be a CCC space. Then if \mathcal{U} is a collection of non-empty open subsets of X such that the following set

        D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}

      is dense in the open subspace \bigcup \mathcal{U}, then \mathcal{U} must be countable.

The collections of open sets in the above theorem do not have to be open covers. However, if they are open covers, the theorem can tie CCC spaces with some covering properties. As long as the space has the CCC, any open cover that is locally-countable on a dense set must be countable. Looking at it in the contrapositive angle, in a CCC space, any uncountable open cover is not locally-countable in some open set.

Proof of Theorem 1
Let \mathcal{U} be a collection of open subsets of X such that the set D(\mathcal{U}) as defined above is dense in the open subspace \bigcup \mathcal{U}. We show that \mathcal{U} is countable. Suppose not.

For each U \in \mathcal{U}, since U \cap D(\mathcal{U}) \ne \varnothing, we can choose a non-empty open set f(U) \subset U such that f(U) has non-empty intersection with only countably many sets in \mathcal{U}. Let \mathcal{U}_f be the following collection:

    \mathcal{U}_f=\left\{f(U): U \in \mathcal{U} \right\}

For H,K \in \mathcal{U}_f, by a chain from H to K, we mean a finite collection

    \left\{W_1,W_2,\cdots,W_n \right\} \subset \mathcal{U}_f

such that H=W_1, K=W_n and W_j \cap W_{j+1} \ne \varnothing for any 1 \le j <n. For each open set W \in \mathcal{U}_f, define \mathcal{C}(W) and \mathcal{E}(W) as follows:

    \mathcal{C}(W)=\left\{V \in \mathcal{U}_f: \text{there exists a chain from } W \text{ to } V \right\}

    \mathcal{E}(W)=\bigcup \mathcal{C}(W)

One observation we make is that for W_1,W_2 \in \mathcal{U}_f, if \mathcal{E}(W_1) \cap \mathcal{E}(W_2) \ne \varnothing, then \mathcal{C}(W_1)=\mathcal{C}(W_2) and \mathcal{E}(W_1)=\mathcal{E}(W_2). So the distinct \mathcal{E}(W) are pairwise disjoint. Because the space X has the CCC, there can be only countably many distinct open sets \mathcal{E}(W). Thus there can be only countably many distinct collections \mathcal{C}(W).

Note that each \mathcal{C}(W) is a countable collection of open sets. Each V \in \mathcal{U}_f meets only countably many open sets in \mathcal{U}. So each V \in \mathcal{U}_f can meet only countably many sets in \mathcal{U}_f, since for each V \in \mathcal{U}_f, V \subset U for some U \in \mathcal{U}. Thus for each W \in \mathcal{U}_f, in considering all finite-length chain starting from W, there can be only countably many open sets in \mathcal{U}_f that can be linked to W. Thus \mathcal{C}(W) must be countable. In taking the union of all \mathcal{C}(W), we get back the collection \mathcal{U}_f. Thus we have:

    \mathcal{U}_f=\bigcup \limits_{W \in \mathcal{U}_f} \mathcal{C}(W)

Because the space X is CCC, there are only countably many distinct collections \mathcal{C}(W) in the above union. Each \mathcal{C}(W) is countable. So \mathcal{U}_f is a countable collection of open sets.

Furthermore, each U \in \mathcal{U} contains at least one set in \mathcal{U}_f. From the way we choose sets in \mathcal{U}_f, we see that for each V \in \mathcal{U}_f, V=f(U) \subset U for at most countably many U \in \mathcal{U}. The argument indicates that we have a one-to-countable mapping from \mathcal{U}_f to \mathcal{U}. Thus the original collection \mathcal{U} must be countable. \blacksquare

The property in Theorem 1 is actually equivalent to the CCC property. Just that the proof of Theorem 1 represents the hard direction that needs to be proved. Theorem 1 can be expanded to be the following theorem.

    Theorem 1a

      Let X be a space. Then the following conditions are equivalent.

      1. The space X has the CCC.
      2. If \mathcal{U} is a collection of non-empty open subsets of X such that the following set

          D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}

        is dense in the open subspace \bigcup \mathcal{U}, then \mathcal{U} must be countable.

      3. If \mathcal{U} is a collection of non-empty open subsets of X such that \mathcal{U} is locally-countable at every point in the open subspace \bigcup \mathcal{U}, then \mathcal{U} must be countable.

The direction 1 \rightarrow 2 has been proved above. The directions 2 \rightarrow 3 and 3 \rightarrow 1 are straightforward.

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Tying Theorem 1 to “Nice” Open Covers

One easy application of Theorem 1 is to tie it to locally-finite and locally-countable open covers. We have the following theorem.

    Theorem 2

      In any CCC space, any locally-countable open cover must be countable. Thus any locally-finite open cover must also be countable.

Theorem 2 gives the well known result that any CCC paracompact space is Lindelof (see Theorem 5 below). In fact, Theorem 2 gives the result that any CCC para-Lindelof space is Lindelof (see Theorem 6 below). A space X is para-Lindelof if every open cover has a locally-countable open refinement.

Can Theorem 2 hold for point-finite covers (or point-countable covers)? The answer is no (see Example 1 below). With the additional property of having a Baire space, we have the following theorem.

    Theorem 3

      In any Baire space with the CCC, any point-finite open cover must be countable.

A Space X is a Baire space if U_1,U_2,U_3,\cdots are dense open subsets of X, then \bigcap \limits_{j=1}^\infty U_j \ne \varnothing. For more information about Baire spaces, see this previous post.
.

Proof of Theorem 3
Let X be a Baire space with the CCC. Let \mathcal{U} be a point-finite open cover of X. Suppose that \mathcal{U} is uncountable. We show that this assumption with lead to a contradiction. Thus \mathcal{U} must be countable.

By Theorem 1, there exists an open set V \subset X such that \mathcal{U} is not locally-countable at any point in V. For each positive integer n, let H_n be the following:

    H_n=\left\{x \in V: x \text{ is in at most } n \text{ sets in } \mathcal{U} \right\}

Note that V=\bigcup \limits_{j=1}^\infty H_j. Furthermore, each H_n is a closed set in the space V. Since X is a Baire space, every non-empty open subset of X is of second category (i.e. it cannot be a union of countably many closed and nowhere dense sets). Thus it cannot be that each H_n is nowhere dense in V. For some n, H_n is not nowhere dense. There must exist some open W \subset V such that H_n \cap W is dense in W. Because H_n is closed, W \subset H_n.

Choose y \in W. The point y is in at most n open sets in \mathcal{U}. Let U_1,U_2,\cdots,U_m \in \mathcal{U} such that y \in \bigcap \limits_{j=1}^m U_j. Clearly 1 \le m \le n. Let U=W \cap U_1 \cap \cdots \cap U_m. Note that y \in U \subset H_n \subset V.

Every point in U belongs to at most n many sets in \mathcal{U} and already belong to m sets in \mathcal{U}. So each point in U can belong to at most n-m additional open sets in \mathcal{U}. Consider the case n-m=0 and the case n-m>0. We show that each case leads to a contradiction.

Suppose that n-m=0. Then each point of U can only meet n open sets in \mathcal{U}, namely U_1,U_2,\cdots,U_m. This contradicts that \mathcal{U} is not locally-countable at points in U \subset V.

Suppose that k=n-m>0. Let \mathcal{U}^*=\mathcal{U}-\left\{U_1,\cdots,U_m \right\}. Let \mathcal{M} be the following collection:

    \mathcal{M}=\left\{U \cap \bigcap \limits_{O \in M} O \ne \varnothing: M \subset \mathcal{U}^* \text{ and } \lvert M \lvert=k \right\}

Each element of \mathcal{M} is an open subset of U that is the intersection of exactly n many open sets in \mathcal{U}. So \mathcal{M} is a collection of pairwise disjoint open sets. The open set U as a topological space has the CCC. So \mathcal{M} is at most countable. Thus the open set U meets at most countably many open sets in \mathcal{U}, contradicting that \mathcal{U} is not locally-countable at points in U \subset V.

Both cases n-m=0 and n-m>0 lead to contradiction. So \mathcal{U} must be countable. The proof to Theorem 3 is completed. \blacksquare

As a corollary to Theorem 3, we have the result that every Baire CCC metacompact space is Lindelof.

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Some Applications of Theorems 2 and 3

In proving paracompactness in some of the theorems, we need a theorem involving the concept of star-countable open cover. A collection \mathcal{A} of subsets of a space X is said to be star-finite (star-countable) if for each A \in \mathcal{A}, only finitely (countably) many sets in \mathcal{A} meets A, i.e., the following set

    \left\{B \in \mathcal{A}: B \cap A \ne \varnothing \right\}

is finite (countable). The proof of the following theorem can be found in Engleking (see the direction (iv) implies (i) in the proof of Theorem 5.3.10 on page 326 in [1]).

    Theorem 4

      If every open cover of a regular space X has a star-countable open refinement, then X is paracompact.

As indicated in the above section, Theorem 2 and Theorem 3 have some obvious applications. We have the following theorems.

    Theorem 5

      Let X be a CCC space. Then X is paracompact if and only of X is Lindelof.

Proof of Theorem 5
The direction \Longleftarrow follows from the fact that any regular Lindelof space is paracompact.

The direction \Longrightarrow follows from Theorem 2. \blacksquare

    Theorem 6

      Every CCC para-Lindelof space is Lindelof.

Proof of Theorem 6
This also follows from Theorem 2. \blacksquare

    Theorem 7

      Every Baire CCC metacompact space is Lindelof.

Proof of Theorem 7
Let X be a Baire CCC metacompact space. Let \mathcal{U} be an open cover of X. By metacompactness, let \mathcal{V} be a point-finite open refinement of \mathcal{U}. By Theorem 3, \mathcal{V} must be countable. \blacksquare

    Theorem 8

      Every Baire CCC hereditarily metacompact space is hereditarily Lindelof.

Proof of Theorem 8
Let X be a Baire CCC hereditarily metacompact space. To show that X is hereditarily Lindelof, it suffices to show that every non-empty open subset is Lindelof. Let Y \subset X be open. Then Y has the CCC and is also metacompact. Being a Baire space is hereditary with respect to open subspaces. So Y is a Baire space too. By Theorem 7, Y is Lindelof. \blacksquare

    Theorem 9

      Every locally CCC regular para-Lindelof space is paracompact.

Proof of Theorem 9
A space is locally CCC if every point has an open neighborhood that has the CCC. Let X be a regular space that is locally CCC and para-Lindelof. Let \mathcal{U} be an open cover of X. Using the locally CCC assumption and by taking a refinement of \mathcal{U} if necessary, we can assume that each open set in \mathcal{U} has the CCC. By the para-Lindelof assumption, let \mathcal{V} be a locally-countable open refinement of \mathcal{U}. So each open set in \mathcal{V} has the CCC too.

Now we show that \mathcal{V} is star-countable. Let V \in \mathcal{V}. Let \mathcal{G} be the following collection:

    \mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}

which is is open cover of V. Within the subspace V, \mathcal{G} is a locally-countable open cover. By Theorem 2, \mathcal{G} must be countable. The collection \mathcal{G} represents all the open sets in \mathcal{V} that have non-empty intersection with V. Thus only countably many open sets in \mathcal{V} can meet V. So \mathcal{V} is a star-countable open refinement of \mathcal{U}. By Theorem 4, X is paracompact. \blacksquare

    Theorem 10

      Every locally CCC regular metacompact Baire space is paracompact.

Proof of Theorem 10
Let X be a regular space that is locally CCC and is a metacompact Baire space. Let \mathcal{U} be an open cover of X. Using the locally CCC assumption and by taking a refinement of \mathcal{U} if necessary, we can assume that each open set in \mathcal{U} has the CCC. By the metacompact assumption, let \mathcal{V} be a point-finite open refinement of \mathcal{U}. So each open set in \mathcal{V} has the CCC too. Each open set in \mathcal{V} is also a Baire space.

Now we show that \mathcal{V} is star-countable. Let V \in \mathcal{V}. Let \mathcal{G} be the following collection:

    \mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}

which is is open cover of V. Within the subspace V, \mathcal{G} is a point-finite open cover. By Theorem 3, \mathcal{G} must be countable. The collection \mathcal{G} represents all the open sets in \mathcal{V} that have non-empty intersection with V. Thus only countably many open sets in \mathcal{V} can meet V. So \mathcal{V} is a star-countable open refinement of \mathcal{U}. By Theorem 4, X is paracompact. \blacksquare

    Theorem 11

      Every locally compact metacompact perfectly normal space is paracompact.

Proof of Theorem 11
This follows from Theorem 10 after we prove the following two points:

  • Any locally compact space is a Baire space.
  • Any perfect locally compact space is locally CCC.

To see the first point, let Y be a locally compact space. Let W_1,W_2,W_3,\cdots be dense open sets in Y. Let y \in Y and let W \subset Y be open such that y \in W and \overline{W} is compact. We show that W contains a point that belongs to all W_n. Let X_1=W \cap W_1, which is open and non-empty. Next choose non-empty open X_2 such that \overline{X_2} \subset X_1 and X_2 \subset W_2. Next choose non-empty open X_3 such that \overline{X_3} \subset X_2 and X_3 \subset W_3. Continue in this manner, we have a sequence of open sets X_1,X_2,X_3,\cdots such that for each n, \overline{X_{n+1}} \subset X_n and \overline{X_n} is compact. The intersection of all the X_n is non-empty. The points in the intersection must belong to each W_n.

To see the second point, let Y be a locally compact space such that every closed set is a G_\delta-set. Suppose that Y is not locally CCC at y \in Y. Let U \subset Y be open such that y \in U and \overline{U} is compact. Then U must not have the CCC. Let \left\{U_\alpha: \alpha<\omega_1 \right\} be a pairwise disjoint collection of open subsets of U. Let O=\bigcup \limits_{\alpha<\omega_1} U_\alpha and let C=Y-O.

Let C=\bigcap \limits_{n=1}^\infty V_n where each V_n is open in Y and V_{n+1} \subset V_n for each integer n. For each \alpha<\omega_1, pick y_\alpha \in U_\alpha. For each y_\alpha, there is some integer f(\alpha) such that y_\alpha \notin V_{f(\alpha)}. So there must exist some integer n such that A=\left\{y_\alpha: f(\alpha)=n \right\} is uncountable.

The set A is an infinite subset of the compact set \overline{U}. So A has a limit point, say p (also called cluster point). Clearly p \notin O. So p \in C. In particular, p \in V_n. Then V_n contains some points of A. But for any y_\alpha \in A, y_\alpha \notin V_n=V_{f(\alpha)}, a contradiction. So Y must be locally CCC at each y \in Y. \blacksquare

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Some Examples

Example 1
A CCC space X with an uncountable point-finite open covers. This example demonstrates that in Theorem 2, locally-finite or locally-countable cannot be replaced by point-finite. Consider the following product space:

    Y=\prod \limits_{\alpha < \omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}

i.e, the product space of \omega_1 many copies of the two-point discrete space \left\{0,1 \right\}. Let X be the set of all points h \in Y such that h(\alpha)=1 for only finitely many \alpha<\omega_1.

The product space Y is the product of separable spaces, hence has the CCC. The space X is dense in Y. Hence X has the CCC. For each \alpha<\omega_1, define U_\alpha as follows:

    U_\alpha=\left\{h \in X: h(\alpha)=1 \right\}

Then \left\{U_\alpha:\alpha<\omega_1  \right\} is a point-finite open cover of X. Of course, X in this example is not a Baire space. \blacksquare

The following three examples center around the four properties in Theorem 7 (Baire + CCC + metacompact imply Lindelof). These examples show that each property in the hypothesis is crucial.

Example 2
A separable non-Lindelof space that is a Baire space. This example shows that the metacompact assumption is crucial for Theorem 7.

The example is the Sorgenfrey plane S \times S where S is the real line with the Sorgenfrey topology (generated by the half-open intervals of the form [a,b)). It is well known that S \times S is not Lindelof. The Sorgenfrey plane is Baire and is separable (hence CCC). Furthermore, S \times S is not metacompact (if it were, it would be Lindelof by Theorem 7). \blacksquare

Example 3
A non-Lindelof metacompact Baire space M. This example shows that the CCC assumption in Theorem 7 is necessary.

This space M is the subspace of Bing’s Example G that has finite support (defined and discussed in the post A subspace of Bing’s example G. It is normal and not collectionwise normal (hence cannot be Lindelof) and metacompact. The space M does not have CCC since it has uncountably many isolated points. Any space with a dense set of isolated points is a Baire space. Thus the space M is also a Baire space. \blacksquare

Example 4
A non-Lindelof CCC metacompact non-Baire space W. This example shows that the Baire space assumption in Theorem 7 is necessary.

Let W be the set of all non-empty finite subsets of the real line with the Pixley-Roy topology. Note that W is non-Lindelof and has the CCC and is metacompact. Of course it is not Baire. For more information on Pixley-Roy spaces, see the post called Pixley-Roy hyperspaces. For the purpose of this example, the Pixley-Roy space can be built on any uncountable separable metrizable space.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Tall, F. D., The Countable Chain Condition Versus Separability – Applications of Martin’s Axiom, Gen. Top. Appl., 4, 315-339, 1974.

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\copyright \ 2014 \text{ by Dan Ma}

Pseudonormal spaces

When two disjoint closed sets in a topological space cannot be separated by disjoint open sets, the space fails to be a normal space. When one of the two closed sets is countable, the space fails to satisfy a weaker property than normality. A space X is said to be a pseudonormal space if H and K can always be separated by two disjoint open sets whenever H and K are disjoint closed subsets of X and one of them is countable. In this post, we discuss several non-normal spaces that actually fail to be pseudonormal. We also give an example of a pseudonormal space that is not normal.

We work with spaces that are at minimum T_1 spaces, i.e., spaces in which singleton sets are closed. Then any pseudonormal space is regular. To see this, let X be T_1 and pseudonormal. For any closed subset C of X and for any point x \in X-C, we can always separate the disjoint closed sets \left\{ x \right\} and C by disjoint open sets. This is one reason why we insist on having T_1 separation axiom as a starting point. We now show some examples of spaces that fail to be pseudonormal.

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Some Non-Pseudonormal Examples

All three examples in this section are spaces where the failure of normality is exhibited by the inability of separating a countable closed set and another disjoint closed set.

Example 1
This example of a non-normal space that fails to be pseudonormal is defined in the previous post called An Example of a Completely Regular Space that is not Normal. This is an example of a Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular space that is not normal. We state the definition of the space and present a proof that it is not pseudonormal.

Let E be the set of all points (x,y) \in \mathbb{R} \times \mathbb{R} such that y \ge 0. For each real number x, define the following sets:

    V_x=\left\{(x,y) \in E: 0 \le y \le 2 \right\}

    D_x=\left\{(s,s-x) \in E: x \le s \le x+2 \right\}

    O_x=V_x \cup D_x

The set V_x is the vertical line of height 2 at the point (x,0). The set D_x is the line originating at (x,0) and going in the Northeast direction reaching the same vertical height as V_x as shown in the following figure.

The topology on E is defined by the following:

  • Each point (x,y) \in E where y>0 is isolated.
  • For each point (x,0) \in E, a basic open set is of the form O_x - F where (x,0) \notin F and F is a finite subset of O_x.

The x-axis in this example is a closed and discrete set of cardinality continuum. Amy two disjoint subsets of the x-axis are disjoint closed sets. The two closed sets that cannot be separated are:

    H=\left\{(x,0) \in E: x \text{ is rational} \right\}

    K=\left\{(x,0) \in E: x \text{ is irrational} \right\}

For each (x,0), let W_x=O_x-F_x where F_x \subset O_x is finite and (x,0) \notin F_x. Furthermore, break up F_x by letting F_{x,d}=F_x \cap D_x and F_{x,v}=F_x \cap V_x. Let U and V be defined by:

    U_H=\bigcup \limits_{(x,0) \in H} W_x

    U_K=\bigcup \limits_{(x,0) \in K} W_x

The open sets U_H and U_K are essentially arbitrary open sets containing H and K respectively. We claims that U_H \cap U_K \ne \varnothing.

Define the projection map \tau_1:\mathbb{R}^2 \rightarrow \mathbb{R} by \tau_1(x,y)=x. Let A and B be defined by:

    A=\bigcup \left\{\tau_1(F_{x,d}): (x,0) \in H \right\}

    B=\left\{(x,0) \in K: (x,0) \notin A \right\}

The set A is countable. So the set B is uncountable. Choose (x,0) \in B. Choose (a,0) \in H on the left of (x,0) and close enough to (x,0) such that V_x \cap D_a=\left\{t \right\} and t \notin F_{x,v}. This means that

    t \in V_x \cup D_x -F_x=O_x-F_x=W_x

    t \in V_a \cup D_a -F_a=O_a-F_a=W_a.

Thus U_H \cap U_K \ne \varnothing. We have shown that the space E is not pseudonormal and thus not normal.

Example 2
The Sorgenfrey line is the real line \mathbb{R} topologized by the base consisting of half open and half closed intervals of the form [a,b)=\left\{x \in \mathbb{R}: a \le x < b \right\}. In this post, we use S to denote the real line \mathbb{R} with this topology.

The Sorgenfrey line S is a classic example of a normal space whose square S \times S is not normal. In the Sorgenfrey plane S \times S, the set \left\{(x,-x) \in S \times S: x \in \mathbb{R} \right\} is a closed and discrete set and is called the anti-diagonal. The proof presented in this previous post shows that the following two disjoint closed subsets of S \times S

    H=\left\{(x,-x) \in S \times S: x \text{ is rational} \right\}

    K=\left\{(x,-x) \in S \times S: x \text{ is irrational} \right\}

cannot be separated by disjoint open sets. The argument is based on the fact that the real line with the usual topology is of second category. The key point in the argument is that the set of the irrationals cannot be the union of countably many closed and nowhere dense sets (in the usual topology of the real line).

Thus S \times S fails to be pseudonormal. This example shows that normality can fail to be preserved by taking Cartesian product in such a way that even pseudonormality cannot be achieved in the Cartesian product!

Example 3
Another example of a non-normal space that fails to be pseudonormal is the Niemmytzkis’ plane (Example 2 in in this previous post). The underlying set is N=\left\{(x,y) \in \mathbb{R} \times \mathbb{R}: y \ge 0 \right\}. The points lying above the x-axis have the usual Euclidean open neighborhoods. A point (x,0) in the x-axis has as neighborhoods \left\{(x,0) \right\} together with the interior of a disc in the upper half plane that is tangent at the point (x,0). Consider the following the two disjoint closed sets on the x-axis:

    H=\left\{(x,0): x \text{ is rational} \right\}

    K=\left\{(x,0): x \text{ is irrational} \right\}

The disjoint closed sets H and K cannot be separated by disjoint open sets (see Niemytzki’s Tangent Disc Topology in [2], Example 82). Like Example 2 above, the argument that H and K cannot be separated is also a Baire category argument.

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An Example of Pseudonormal but not Normal

Example 4
One way to find such a space is to look for spaces that are non-normal and see which one is pseudonormal. On the other hand, in a pseudonormal space, countable closed sets are easily separated from other disjoint closed sets. One space in which “countable” is nice is the first uncountable ordinal \omega_1 with the order topology. But \omega_1 is normal. So we look at the Cartesian product \omega_1 \times (\omega_1 +1). The second factor is the successor ordinal to \omega_1 or as a space that is obtained by tagging one more point to \omega_1 that is considered greater than all the points in \omega_1. Let’s use X \times Y=\omega_1 \times (\omega_1 +1) to denote this space.

The space X \times Y is not normal (shown in this previous post). In the previous post, X \times Y is presented as an example showing that the product of a normal space with a compact space needs not be normal. However, in this case at least, the product is pseudonormal.

Let \alpha < \omega_1. Then the square \alpha \times \alpha as a subspace of X \times Y is a countable space and a first countable space. So it has a countable base (second countable) and thus metrizable, and in particular normal. Any countable subset of X \times Y is contained in one of these countable squares, making it easy to separate a countable closed set from another closed set.

Let H and K be disjoint closed sets in X \times Y such that H is countable. Then there is some successor ordinal \mu < \omega_1 (\mu=\alpha+1 for some ordinal \alpha<\omega_1) such that H \subset \mu \times \mu. Based on the discussion in the preceding paragraph, there are disjoint open sets O_H and O_K in \mu \times \mu such that H \subset O_H and (K \cap (\mu \times \mu)) \subset O_K. With \mu being a successor ordinal, the square \mu \times \mu is both closed and open in X \times Y. Then the following sets

    V_H=O_H

    V_K=O_K \cup (X \times Y-\mu \times \mu)

are disjoint open sets in X \times Y separating H and K.

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Some Comments about Examples 1 – 3

In each of Examples 1, 2 and 3 discussed above, there is a closed and discrete set of cardinality continuum (the x-axis in Examples 1 and 3 and the anti-diagonal in Example 2). So the extent of each of these three spaces is continuum. Note that the extent of a space is the maximum cardinality of a closed and discrete subset.

In each of these examples, it just so happens that it is not possible to separate the rationals from the irrationals in the x-axis or the anti-diagonal by disjoint open sets, making each example not only not normal but also not pseudonormal.

What if we consider a smaller subset of the x-axis or anti-diagonal? For example, consider an uncountable set of cardinality less than continuum. Then what can we say about the pseudonormality or normality of the resulting subspaces? For Example 1, the picture is clear cut.

In Example 1, the argument that H and K cannot be separated is a “countable vs. uncountable” argument. The argument will work as long as H is a countable dense set in the x-axis (dense in the usual topology) and K is any uncountable set.

For Example 2 and Example 3, the argument that H and K cannot be separated is not a “countable vs. uncountable” argument and instead is a Baire category argument. The fact that one of the closed sets is the irrationals is a crucial point. On the other hand, both Example 2 and Example 3 (especially Example 3) are set-theoretic sensitive examples. For Example 2 and Example 3, the normality of the resulting smaller subspaces is dependent on some extra axioms beyond ZFC. For pseudonormality, it could be set-theoretic sensitive too. We give some indication here why this is so.

Let S be the Sorgenfrey line as in Example 2 above. Assuming Martin’s Axiom and the negation of the continuum hypothesis (abbreviated by MA + not CH), for any uncountable X \subset S with \lvert X \lvert < c, X \times X is normal but not paracompact (see Example 6.3 in [1] and see [3]). Even though X \times X is not exactly a comparable example, this example shows that restricting to a smaller subset on the anti-diagonal seems to make the space normal.

Example 3 has an illustrious history with respect to the normal Moore space conjecture. There is not surprise that extra set-theory axioms are used. For any subset B of the x-axis, let N(B) be the space defined as in Example 3 above except that only points of B are used on the x-axis. Assuming MA + not CH, for any uncountable B that is of cardinality less than continuum, it can be shown that N(B) is normal non-metrizable Moore space (see Example F in [4]). So by assuming extra axiom of MA + not CH, we cannot get a non-pseudonormal example out of Example 3 by restricting to a smaller uncountable subset of the x-axis. Under other set-theoretic axioms, there exists no normal non-metrizable Moore space. Just because this is a set-theoretic sensitive example, it is conceivable that N(B) could be a space that is not pseudonormal under some other axioms.

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Reference

  1. Burke, D. K., Covering Properties, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 347-422, 1984.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc, Amsterdam, New York, 1995.
  3. Przymusinski, T. C., A Lindelof space X such that X \times X is normal but not paracompact, Fund. Math., 91, 161-165, 1973.
  4. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.

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\copyright \ 2014 \text{ by Dan Ma}

Pixley-Roy hyperspaces

In this post, we introduce a class of hyperspaces called Pixley-Roy spaces. This is a well-known and well studied set of topological spaces. Our goal here is not to be comprehensive but rather to present some selected basic results to give a sense of what Pixley-Roy spaces are like.

A hyperspace refers to a space in which the points are subsets of a given “ground” space. There are more than one way to define a hyperspace. Pixley-Roy spaces were first described by Carl Pixley and Prabir Roy in 1969 (see [5]). In such a space, the points are the non-empty finite subsets of a given ground space. More precisely, let X be a T_1 space (i.e. finite sets are closed). Let \mathcal{F}[X] be the set of all non-empty finite subsets of X. For each F \in \mathcal{F}[X] and for each open subset U of X with F \subset U, we define:

    [F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}

The sets [F,U] over all possible F and U form a base for a topology on \mathcal{F}[X]. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set \mathcal{F}[X] with this topology is called a Pixley-Roy space.

The hyperspace as defined above was first defined by Pixley and Roy on the real line (see [5]) and was later generalized by van Douwen (see [7]). These spaces are easy to define and is useful for constructing various kinds of counterexamples. Pixley-Roy played an important part in answering the normal Moore space conjecture. Pixley-Roy spaces have also been studied in their own right. Over the years, many authors have investigated when the Pixley-Roy spaces are metrizable, normal, collectionwise Hausdorff, CCC and homogeneous. For a small sample of such investigations, see the references listed at the end of the post. Our goal here is not to discuss the results in these references. Instead, we discuss some basic properties of Pixley-Roy to solidify the definition as well as to give a sense of what these spaces are like. Good survey articles of Pixley-Roy are [3] and [7].

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Basic Discussion

In this section, we focus on properties that are always possessed by a Pixley-Roy space given that the ground space is at least T_1. Let X be a T_1 space. We discuss the following points:

  1. The topology defined above is a legitimate one, i.e., the sets [F,U] indeed form a base for a topology on \mathcal{F}[X].
  2. \mathcal{F}[X] is a Hausdorff space.
  3. \mathcal{F}[X] is a zero-dimensional space.
  4. \mathcal{F}[X] is a completely regular space.
  5. \mathcal{F}[X] is a hereditarily metacompact space.

Let \mathcal{B}=\left\{[F,U]: F \in \mathcal{F}[X] \text{ and } U \text{ is open in } X \right\}. Note that every finite set F belongs to at least one set in \mathcal{B}, namely [F,X]. So \mathcal{B} is a cover of \mathcal{F}[X]. For A \in [F_1,U_1] \cap [F_2,U_2], we have A \in [A,U_1 \cap U_2] \subset   [F_1,U_1] \cap [F_2,U_2]. So \mathcal{B} is indeed a base for a topology on \mathcal{F}[X].

To show \mathcal{F}[X] is Hausdorff, let A and B be finite subsets of X where A \ne B. Then one of the two sets has a point that is not in the other one. Assume we have x \in A-B. Since X is T_1, we can find open sets U, V \subset X such that x \in U, x \notin V and A \cup B-\left\{ x \right\} \subset V. Then [A,U \cup V] and [B,V] are disjoint open sets containing A and B respectively.

To see that \mathcal{F}[X] is a zero-dimensional space, we show that \mathcal{B} is a base consisting of closed and open sets. To see that [F,U] is closed, let C \notin [F,U]. Either F \not \subset C or C \not \subset U. In either case, we can choose open V \subset X with C \subset V such that [C,V] \cap [F,U]=\varnothing.

The fact that \mathcal{F}[X] is completely regular follows from the fact that it is zero-dimensional.

To show that \mathcal{F}[X] is metacompact, let \mathcal{G} be an open cover of \mathcal{F}[X]. For each F \in \mathcal{F}[X], choose G_F \in \mathcal{G} such that F \in G_F and let V_F=[F,X] \cap G_F. Then \mathcal{V}=\left\{V_F: F \in \mathcal{F}[X] \right\} is a point-finite open refinement of \mathcal{G}. For each A \in \mathcal{F}[X], A can only possibly belong to V_F for the finitely many F \subset A.

A similar argument show that \mathcal{F}[X] is hereditarily metacompact. Let Y \subset \mathcal{F}[X]. Let \mathcal{H} be an open cover of Y. For each F \in Y, choose H_F \in \mathcal{H} such that F \in H_F and let W_F=([F,X] \cap Y) \cap H_F. Then \mathcal{W}=\left\{W_F: F \in Y \right\} is a point-finite open refinement of \mathcal{H}. For each A \in Y, A can only possibly belong to W_F for the finitely many F \subset A such that F \in Y.

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More Basic Results

We now discuss various basic topological properties of \mathcal{F}[X]. We first note that \mathcal{F}[X] is a discrete space if and only if the ground space X is discrete. Though we do not need to make this explicit, it makes sense to focus on non-discrete spaces X when we look at topological properties of \mathcal{F}[X]. We discuss the following points:

  1. If X is uncountable, then \mathcal{F}[X] is not separable.
  2. If X is uncountable, then every uncountable subspace of \mathcal{F}[X] is not separable.
  3. If \mathcal{F}[X] is Lindelof, then X is countable.
  4. If \mathcal{F}[X] is Baire space, then X is discrete.
  5. If \mathcal{F}[X] has the CCC, then X has the CCC.
  6. If \mathcal{F}[X] has the CCC, then X has no uncountable discrete subspaces,i.e., X has countable spread, which of course implies CCC.
  7. If \mathcal{F}[X] has the CCC, then X is hereditarily Lindelof.
  8. If \mathcal{F}[X] has the CCC, then X is hereditarily separable.
  9. If X has a countable network, then \mathcal{F}[X] has the CCC.
  10. The Pixley-Roy space of the Sorgenfrey line does not have the CCC.
  11. If X is a first countable space, then \mathcal{F}[X] is a Moore space.

Bullet points 6 to 9 refer to properties that are never possessed by Pixley-Roy spaces except in trivial cases. Bullet points 6 to 8 indicate that \mathcal{F}[X] can never be separable and Lindelof as long as the ground space X is uncountable. Note that \mathcal{F}[X] is discrete if and only if X is discrete. Bullet point 9 indicates that any non-discrete \mathcal{F}[X] can never be a Baire space. Bullet points 10 to 13 give some necessary conditions for \mathcal{F}[X] to be CCC. Bullet 14 gives a sufficient condition for \mathcal{F}[X] to have the CCC. Bullet 15 indicates that the hereditary separability and the hereditary Lindelof property are not sufficient conditions for the CCC of Pixley-Roy space (though they are necessary conditions). Bullet 16 indicates that the first countability of the ground space is a strong condition, making \mathcal{F}[X] a Moore space.

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To see bullet point 6, let X be an uncountable space. Let \left\{F_1,F_2,F_3,\cdots \right\} be any countable subset of \mathcal{F}[X]. Choose a point x \in X that is not in any F_n. Then none of the sets F_i belongs to the basic open set [\left\{x \right\} ,X]. Thus \mathcal{F}[X] can never be separable if X is uncountable.

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To see bullet point 7, let Y \subset \mathcal{F}[X] be uncountable. Let W=\cup \left\{F: F \in Y \right\}. Let \left\{F_1,F_2,F_3,\cdots \right\} be any countable subset of Y. We can choose a point x \in W that is not in any F_n. Choose some A \in Y such that x \in A. Then none of the sets F_n belongs to the open set [A ,X] \cap Y. So not only \mathcal{F}[X] is not separable, no uncountable subset of \mathcal{F}[X] is separable if X is uncountable.

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To see bullet point 8, note that \mathcal{F}[X] has no countable open cover consisting of basic open sets, assuming that X is uncountable. Consider the open collection \left\{[F_1,U_1],[F_2,U_2],[F_3,U_3],\cdots \right\}. Choose x \in X that is not in any of the sets F_n. Then \left\{ x \right\} cannot belong to [F_n,U_n] for any n. Thus \mathcal{F}[X] can never be Lindelof if X is uncountable.

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For an elementary discussion on Baire spaces, see this previous post.

To see bullet point 9, let X be a non-discrete space. To show \mathcal{F}[X] is not Baire, we produce an open subset that is of first category (i.e. the union of countably many closed nowhere dense sets). Let x \in X a limit point (i.e. an non-isolated point). We claim that the basic open set V=[\left\{ x \right\},X] is a desired open set. Note that V=\bigcup \limits_{n=1}^\infty H_n where

    H_n=\left\{F \in \mathcal{F}[X]: x \in F \text{ and } \lvert F \lvert \le n \right\}

We show that each H_n is closed and nowhere dense in the open subspace V. To see that it is closed, let A \notin H_n with x \in A. We have \lvert A \lvert>n. Then [A,X] is open and every point of [A,X] has more than n points of the space X. To see that H_n is nowhere dense in V, let [B,U] be open with [B,U] \subset V. It is clear that x \in B \subset U where U is open in the ground space X. Since the point x is not an isolated point in the space X, U contains infinitely many points of X. So choose an finite set C with at least 2 \times n points such that B \subset C \subset U. For the the open set [C,U], we have [C,U] \subset [B,U] and [C,U] contains no point of H_n. With the open set V being a union of countably many closed and nowhere dense sets in V, the open set V is not of second category. We complete the proof that \mathcal{F}[X] is not a Baire space.

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To see bullet point 10, let \mathcal{O} be an uncountable and pairwise disjoint collection of open subsets of X. For each O \in \mathcal{O}, choose a point x_O \in O. Then \left\{[\left\{ x_O \right\},O]: O \in \mathcal{O} \right\} is an uncountable and pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] is CCC then X must have the CCC.

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To see bullet point 11, let Y \subset X be uncountable such that Y as a space is discrete. This means that for each y \in Y, there exists an open O_y \subset X such that y \in O_y and O_y contains no point of Y other than y. Then \left\{[\left\{y \right\},O_y]: y \in Y \right\} is an uncountable and pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] has the CCC, then the ground space X has no uncountable discrete subspace (such a space is said to have countable spread).

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To see bullet point 12, let Y \subset X be uncountable such that Y is not Lindelof. Then there exists an open cover \mathcal{U} of Y such that no countable subcollection of \mathcal{U} can cover Y. We can assume that sets in \mathcal{U} are open subsets of X. Also by considering a subcollection of \mathcal{U} if necessary, we can assume that cardinality of \mathcal{U} is \aleph_1 or \omega_1. Now by doing a transfinite induction we can choose the following sequence of points and the following sequence of open sets:

    \left\{x_\alpha \in Y: \alpha < \omega_1 \right\}

    \left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}

such that x_\beta \ne x_\gamma if \beta \ne \gamma, x_\alpha \in U_\alpha and x_\alpha \notin \bigcup \limits_{\beta < \alpha} U_\beta for each \alpha < \omega_1. At each step \alpha, all the previously chosen open sets cannot cover Y. So we can always choose another point x_\alpha of Y and then choose an open set in \mathcal{U} that contains x_\alpha.

Then \left\{[\left\{x_\alpha \right\},U_\alpha]: \alpha < \omega_1 \right\} is a pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] has the CCC, then X must be hereditarily Lindelof.

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To see bullet point 13, let Y \subset X. Consider open sets [A,U] where A ranges over all finite subsets of Y and U ranges over all open subsets of X with A \subset U. Let \mathcal{G} be a collection of such [A,U] such that \mathcal{G} is pairwise disjoint and \mathcal{G} is maximal (i.e. by adding one more open set, the collection will no longer be pairwise disjoint). We can apply a Zorn lemma argument to obtain such a maximal collection. Let D be the following subset of Y.

    D=\bigcup \left\{A: [A,U] \in \mathcal{G} \text{ for some open } U  \right\}

We claim that the set D is dense in Y. Suppose that there is some open set W \subset X such that W \cap Y \ne \varnothing and W \cap D=\varnothing. Let y \in W \cap Y. Then [\left\{y \right\},W] \cap [A,U]=\varnothing for all [A,U] \in \mathcal{G}. So adding [\left\{y \right\},W] to \mathcal{G}, we still get a pairwise disjoint collection of open sets, contradicting that \mathcal{G} is maximal. So D is dense in Y.

If \mathcal{F}[X] has the CCC, then \mathcal{G} is countable and D is a countable dense subset of Y. Thus if \mathcal{F}[X] has the CCC, the ground space X is hereditarily separable.

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A collection \mathcal{N} of subsets of a space Y is said to be a network for the space Y if any non-empty open subset of Y is the union of elements of \mathcal{N}, equivalently, for each y \in Y and for each open U \subset Y with y \in U, there is some A \in \mathcal{N} with x \in A \subset U. Note that a network works like a base but the elements of a network do not have to be open. The concept of network and spaces with countable network are discussed in these previous posts Network Weight of Topological Spaces – I and Network Weight of Topological Spaces – II.

To see bullet point 14, let \mathcal{N} be a network for the ground space X such that \mathcal{N} is also countable. Assume that \mathcal{N} is closed under finite unions (for example, adding all the finite unions if necessary). Let \left\{[A_\alpha,U_\alpha]: \alpha < \omega_1 \right\} be a collection of basic open sets in \mathcal{F}[X]. Then for each \alpha, find B_\alpha \in \mathcal{N} such that A_\alpha \subset B_\alpha \subset U_\alpha. Since \mathcal{N} is countable, there is some B \in \mathcal{N} such that M=\left\{\alpha< \omega_1: B=B_\alpha \right\} is uncountable. It follows that for any finite E \subset M, \bigcap \limits_{\alpha \in E} [A_\alpha,U_\alpha] \ne \varnothing.

Thus if the ground space X has a countable network, then \mathcal{F}[X] has the CCC.

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The implications in bullet points 12 and 13 cannot be reversed. Hereditarily Lindelof property and hereditarily separability are not sufficient conditions for \mathcal{F}[X] to have the CCC. See [4] for a study of the CCC property of the Pixley-Roy spaces.

To see bullet point 15, let S be the Sorgenfrey line, i.e. the real line \mathbb{R} with the topology generated by the half closed intervals of the form [a,b). For each x \in S, let U_x=[x,x+1). Then \left\{[ \left\{ x \right\},U_x]: x \in S \right\} is a collection of pairwise disjoint open sets in \mathcal{F}[S].

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A Moore space is a space with a development. For the definition, see this previous post.

To see bullet point 16, for each x \in X, let \left\{B_n(x): n=1,2,3,\cdots \right\} be a decreasing local base at x. We define a development for the space \mathcal{F}[X].

For each finite F \subset X and for each n, let B_n(F)=\bigcup \limits_{x \in F} B_n(x). Clearly, the sets B_n(F) form a decreasing local base at the finite set F. For each n, let \mathcal{H}_n be the following collection:

    \mathcal{H}_n=\left\{[F,B_n(F)]: F \in \mathcal{F}[X] \right\}

We claim that \left\{\mathcal{H}_n: n=1,2,3,\cdots \right\} is a development for \mathcal{F}[X]. To this end, let V be open in \mathcal{F}[X] with F \in V. If we make n large enough, we have [F,B_n(F)] \subset V.

For each non-empty proper G \subset F, choose an integer f(G) such that [F,B_{f(G)}(F)] \subset V and F \not \subset B_{f(G)}(G). Let m be defined by:

    m=\text{max} \left\{f(G): G \ne \varnothing \text{ and } G \subset F \text{ and } G \text{ is proper} \right\}

We have F \not \subset B_{m}(G) for all non-empty proper G \subset F. Thus F \notin [G,B_m(G)] for all non-empty proper G \subset F. But in \mathcal{H}_m, the only sets that contain F are [F,B_m(F)] and [G,B_m(G)] for all non-empty proper G \subset F. So [F,B_m(F)] is the only set in \mathcal{H}_m that contains F, and clearly [F,B_m(F)] \subset V.

We have shown that for each open V in \mathcal{F}[X] with F \in V, there exists an m such that any open set in \mathcal{H}_m that contains F must be a subset of V. This shows that the \mathcal{H}_n defined above form a development for \mathcal{F}[X].

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Examples

In the original construction of Pixley and Roy, the example was \mathcal{F}[\mathbb{R}]. Based on the above discussion, \mathcal{F}[\mathbb{R}] is a non-separable CCC Moore space. Because the density (greater than \omega for not separable) and the cellularity (=\omega for CCC) do not agree, \mathcal{F}[\mathbb{R}] is not metrizable. In fact, it does not even have a dense metrizable subspace. Note that countable subspaces of \mathcal{F}[\mathbb{R}] are metrizable but are not dense. Any uncountable dense subspace of \mathcal{F}[\mathbb{R}] is not separable but has the CCC. Not only \mathcal{F}[\mathbb{R}] is not metrizable, it is not normal. The problem of finding X \subset \mathbb{R} for which \mathcal{F}[X] is normal requires extra set-theoretic axioms beyond ZFC (see [6]). In fact, Pixley-Roy spaces played a large role in the normal Moore space conjecture. Assuming some extra set theory beyond ZFC, there is a subset M \subset \mathbb{R} such that \mathcal{F}[M] is a CCC metacompact normal Moore space that is not metrizable (see Example I in [8]).

On the other hand, Pixley-Roy space of the Sorgenfrey line and the Pixley-Roy space of \omega_1 (the first uncountable ordinal with the order topology) are metrizable (see [3]).

The Sorgenfrey line and the first uncountable ordinal are classic examples of topological spaces that demonstrate that topological spaces in general are not as well behaved like metrizable spaces. Yet their Pixley-Roy spaces are nice. The real line and other separable metric spaces are nice spaces that behave well. Yet their Pixley-Roy spaces are very much unlike the ground spaces. This inverse relation between the ground space and the Pixley-Roy space was noted by van Douwen (see [3] and [7]) and is one reason that Pixley-Roy hyperspaces are a good source of counterexamples.

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Reference

  1. Bennett, H. R., Fleissner, W. G., Lutzer, D. J., Metrizability of certain Pixley-Roy spaces, Fund. Math. 110, 51-61, 1980.
  2. Daniels, P, Pixley-Roy Spaces Over Subsets of the Reals, Topology Appl. 29, 93-106, 1988.
  3. Lutzer, D. J., Pixley-Roy topology, Topology Proc. 3, 139-158, 1978.
  4. Hajnal, A., Juahasz, I., When is a Pixley-Roy Hyperspace CCC?, Topology Appl. 13, 33-41, 1982.
  5. Pixley, C., Roy, P., Uncompletable Moore spaces, Proc. Auburn Univ. Conf. Auburn, AL, 1969.
  6. Przymusinski, T., Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, 291-297, 1981.
  7. van Douwen, E. K., The Pixley-Roy topology on spaces of subsets, Set-theoretic Topology, Academic Press, New York, 111-134, 1977.
  8. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.
  9. Tanaka, H, Normality and hereditary countable paracompactness of Pixley-Roy hyperspaces, Fund. Math. 126, 201-208, 1986.

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\copyright \ 2014 \text{ by Dan Ma}

Bing’s Example H

In a previous post we introduced Bing’s Example G, a classic example of a normal but not collectionwise normal space. Other properties of Bing’s Example G include: completely normal, not perfectly normal and not metacompact. This is an influential example introduced in an influential paper of R. H. Bing in 1951 (see [1]). In the same paper, another example called Example H was introduced. This space has some of the same properties of Example G, except that it is perfectly normal. In this post, we define and discuss Example H.

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Defining Bing’s Example H

Throughout the discussion in this post, we use \omega to denote the first infinite ordinal, i.e., \omega =\left\{0,1,2,3,\cdots \right\}. Let P be any uncountable set. Let Q be the set of all subsets of the set P, i.e., it is the power set of P. Let H be the set of all functions f:Q \rightarrow \omega. In other words, the set H is the Cartesian product \prod \limits_{q \in Q} \omega. But the topology on H is not the product topology.

For each p \in P, consider the function f_p:Q \rightarrow 2=\left\{0,1 \right\} such that for each q \in Q:

    f_p(q) = \begin{cases} 1, & \mbox{if } p \in q \\ 0, & \mbox{if } p \notin q \end{cases}

Let H_P=\left\{f_p: p \in P \right\}. Now define a topology on the set H by the following:

  • Each point of H-H_P is an isolated point.
  • Each point f_p \in H_P has basic open sets of the form U(p,W,n) defined as follows:

      U(p,W,n)=\left\{f_p \right\} \cup D(p,W,n)

      D(p,W,n)=\left\{f \in H: \forall q \in Q, f(q) \ge n \text{ and } \forall q \in W, f(q) \equiv f_p(q) \ (\text{mod} \ 2) \right\}

    where p \in P, W \subset Q is finite, and n \in \omega.

If a and b are integers, the a \equiv b \ (\text{mod} \ 2) means that a-b is divisible by 2. The congruence equation f(q) \equiv f_p(q) \ (\text{mod} \ 2) means that f(q) is an even integer if f_p(q)=0. On the other hand, f(q) \equiv f_p(q) \ (\text{mod} \ 2) means that f(q) is an odd integer if f_p(q)=1.

The set D(p,W,n) seems to mimic a basic open set of the point f_p in the product topology: for each point in D(p,W,n), the value of each coordinate is an integer \ge n and the values for finitely many coordinates are fixed to agree with the function f_p modulo 2. Adding the point f_p to D(p,W,n), we have a basic open set U(p,W,n).

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Basic Discussion

The points in H-H_P are the isolated points in the space H. The points in H_P are the non-isolated points (limit points). The space H is a Hausdorff space. Another interesting point is that the set H_P is a closed and discrete set in the space H.

To see that H is Hausdorff, let h_1, h_2 \in H with h_1 \ne h_2. Consider the case that h_1 is an isolated point and h_2=f_p for some p \in P. Let n be the minimum of all h_1(q) over all q \in Q. Let O_1=\left\{h_1 \right\} and O_2=U(p,W,n+1) where W \subset Q is any finite set. Then O_1 and O_2 are disjoint open set containing h_1 and h_2, respectively.

Now consider the case that h_1=f_p and h_2=f_{p'} where p \ne p'. Let O_1=U(p,W,0) and O_2=U(p',W,0) where W=\left\{ \left\{ p \right\},\left\{ p' \right\} \right\}. Then O_1 and O_2 are disjoint open set containing h_1 and h_2, respectively.

The set H_P is a closed and discrete set in the space H. It is closed since H-H_P consists of isolated points. To see that H_P is discrete, note that U(p,W,0), where W=\left\{ \left\{ p \right\} \right\}, is an open set with f_p \in U(p,W,0) and f_{p'} \notin U(p,W,0) for all p' \ne p.

In the sections below, we show that the space H is normal, completely normal (thus hereditarily normal), and is perfectly normal. Furthermore, we show that it is not collectionwise Hausdorff (hence not collectionwise normal) and not meta-lindelof (hence not metacompact).

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Bing’s Example H is Normal

In the next section, we show that Bing’s Example H is completely normal (i.e. any two separated sets can be separated by disjoint open sets). Note that any two disjoint closed sets are separated sets.

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Bing’s Example H is Completely Normal

Let X be a space. Let A \subset X and B \subset X. The sets A and B are separated sets if A \cap \overline{B}=\varnothing=\overline{A} \cap B. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space X is said to be completely normal if for every two separated sets A and B in X, there exist disjoint open subsets U and V of X such that A \subset U and B \subset V. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space X, X is completely normal if and only if X is hereditarily normal. For more about completely normality, see [3] and [6].

Let S and T be separated sets in the space H, i.e.,

    S \cap \overline{T}=\varnothing=\overline{S} \cap T

We consider two cases. Case 1 is that one of the sets consists entirely of isolated points. Assume that S \subset H-H_P. Let O_1=S. For each x \in T, choose an open set V_x with x \in V_x and V_x \cap \overline{S}=\varnothing. Let O_2=\bigcup \limits_{x \in T} V_x. Then O_1 and O_2 are disjoint open sets containing S and T respectively.

Now consider Case 2 where S_1=S \cap H_P \ne \varnothing and T_1=T \cap H_P \ne \varnothing. Consider the sets q_1 and q_2 defined as follows:

    q_1=\left\{p \in P: f_p \in S_1 \right\}

    q_2=\left\{p \in P: f_p \in T_1 \right\}

Let W=\left\{q_1,q_2 \right\}. Let Y_1 and Y_2 be the following open sets:

    Y_1=\bigcup \limits_{p \in q_1} U(p,W,0)

    Y_2=\bigcup \limits_{p \in q_2} U(p,W,0)

Immediately, we know that S_1 \subset Y_1, T_1 \subset Y_2 and Y_1 \cap Y_2=\varnothing. Let S_2=S \cap (H-H_P) and T_2=T \cap (H-H_P) (both of which are open). Let O_1 and O_2 be the following open sets:

    O_1=(Y_1 \cup S_2)-\overline{T}

    O_2=(Y_2 \cup T_2)-\overline{S}

Then O_1 and O_2 are disjoint open sets containing S and T respectively.

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Bing’s Example H is Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is a G_\delta-set (i.e. the intersection of countably many open subsets). All we need to show here is that every closed subset is a G_\delta-set.

Let C \subset H be a closed set. Of course, if C consists entirely of isolated points, then we are done. So assume that C \cap H_P \ne \varnothing. Let q*=\left\{p \in P: f_p \in C \right\}. Let O=C \cap (H-H_P), which is open. For each positive integer n, define the open set Y_n as follows:

    Y_n=O \cup \biggl( \bigcup \limits_{p \in q*} U(p,\left\{q* \right\},n) \biggr)

Immediately we have C \subset Y_n for each n. Let g \in \bigcap \limits_{n=1}^\infty Y_n. We claim that g \in C. Suppose g \notin C. Then g \notin O. It follows that for each n g \in U(p_n,\left\{q* \right\},n) for some p_n \in q*. Recall that U(p_n,\left\{q* \right\},n)=\left\{f_{p_n} \right\} \cup D(p_n,\left\{q* \right\},n).

The assumption that g \notin C implies that g \ne f_{p_n} for all n. Then g \in D(p_n,\left\{q* \right\},n) for all n. By the definition of D(p_n,\left\{q* \right\},n), it follows that for all q \in Q, g(q) \ge n for all positive integer n. This is a contradiction. So it must be the case that g \in C. This completes the proof that Bing’s Example H is perfectly normal.

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Collectionwise Normal Spaces

Let X be a space. Let \mathcal{A} be a collection of subsets of X. We say \mathcal{A} is pairwise disjoint if A \cap B=\varnothing whenever A,B \in \mathcal{A} with A \ne B. We say \mathcal{A} is discrete if for each x \in X, there is an open set O containing x such that O intersects at most one set in \mathcal{A}.

The space X is said to be collectionwise normal if for every discrete collection \mathcal{D} of closed subsets fo X, there is a pairwise disjoint collection \left\{U_D: D \in \mathcal{D} \right\} of open subsets of X such that D \subset U_D for each D \in \mathcal{D}. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus both Bing’s Example G and Example H are not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. As shown below Bing’s Example H is actually not collectionwise Hausdorff.

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Bing’s Example H is not Collectionwise Hausdorff

To prove that Bing’s Example H is not collectionwise Hausdorff, we need an intermediate result (Lemma 1) that is based on an infinitary combinatorial result called the Delta-system lemma.

A family \mathcal{A} of sets is called a Delta-system (or \Delta-system) if there exists a set r, called the root of the \Delta-system, such that for any A,B \in \mathcal{A} with A \ne B, we have A \cap B=r. The following is a version of the Delta-system lemma (see Theorem 1.5 in p. 49 of [2]).

    Delta-System Lemma

      Let \mathcal{A} be an uncountable family of finite sets. Then there exists an uncountable \mathcal{B} \subset \mathcal{A} such that \mathcal{B} is a \Delta-system.
    Lemma 1

      Let P_0 \subset P be any uncountable subset. For each p \in P_0, let U(p,W_p,n_p) be a basic open subset containing f_p. Then there exists an uncountable P_1 \subset P_0 such that \bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing.

Proof of Lemma 1
Let \mathcal{A}=\left\{W_p: p \in P_0 \right\}. We need to break this up into two cases – \mathcal{A} is a countable family of finite sets or an uncountable family of finite sets. The first case is relatively easy to see. The second case requires using the Delta-system lemma.

Suppose that \mathcal{A} is countable. Then there exists an uncountable R \subset P_0 such that for all p,t \in R with p \ne t, we have W_p=W_t=W and n_p=n_t=n. Suppose that W=\left\{q_1,q_2,\cdots,q_m \right\}. By inductively working on the sets q_j, we can obtain an uncountable set P_1 \subset R such that for all p,t \in P_1 with p \ne t, we have f_p(q_j)=f_t(q_j) for each j=1,2,\cdots,m. Clearly, we have:

    \bigcap \limits_{p \in P_1} U(p,W,n) \ne \varnothing

To show the above, just define a function h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\} such that h(q_j)=f_p(q_j) for all j=1,2,\cdots,m for one particular p \in P_1. Then h belongs to the intersection.

Suppose that \mathcal{A} is uncountable. By the Delta-system lemma, there is an uncountable R \subset P_0 and there exists a finite set r \subset Q such that for all p,t \in R with p \ne t, we have W_p \cap W_t=r. Suppose that r=\left\{q_1,q_2,\cdots,q_m \right\}. As in the previous case, work inductively on the sets q_j, we can obtain an uncountable S \subset R such that for all p,t \in S with p \ne t, we have f_p(q_j)=f_t(q_j) for each j=1,2,\cdots,m. Now narrow down to an uncountable P_1 \subset S such that n_p=n_t=n for all p,t \in P_1 with p \ne t. We now show that

    \bigcap \limits_{p \in P_1} U(p,W_p,n) \ne \varnothing

To define a function h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\} that belongs to the above intersection, we define h so that h matches f_t (mod 2) with one particular t \in P_1 on the set r=\left\{q_1,q_2,\cdots,q_m \right\}. Note that W_p-r are disjoint over all p \in P_1. So h can be defined on W_p-r to match f_p (mod 2). For any remaining values in the domain, define h freely to be at least the integer n. Then the function h belongs to the intersection.

With the two cases established, the proof of Lemma 1 is completed. \blacksquare

The fact that Example H is not collectionwise Hausdorff is a corollary of Lemma 1. The set H_P is a discrete collection of points in the space H. It follows that H_P cannot be separated by disjoint open sets. For each p \in P, let U(p,W_p,n_p) be a basic open set containing the point f_p. By Lemma 1, there is an uncountable P_1 \subset P such that \bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing. Thus there can be no disjoint collection of open sets in H that separate the points in H_P.

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Bing’s Example H is not Metacompact

Let X be a space. A collection \mathcal{A} of subsets of X is said to be a point-finite (point-countable) collection if every point of X belongs to only finitely (countably) many sets in \mathcal{A}. A space X is said to be a metacompact space if every open cover \mathcal{U} of X has a point-finite open refinement \mathcal{V}. A space X is said to be a meta-Lindelof space if every open cover \mathcal{U} of X has a point-countable open refinement \mathcal{V}. Clearly, every metacompact space is meta-Lindelof.

It follows from Lemma 1 that Example H is not meta-Lindelof. Thus Example H is not metacompact. To see that it is not meta-Lindelof, for each f_p \in H_P, let U_{f_p}=U(p,\left\{\left\{p \right\} \right\},0), and for each x \in H-H_P, let U_x=\left\{x \right\}. Let \mathcal{U} be the following open cover of H:

    \mathcal{U}=\left\{U_x: x \in H \right\}

Each f_p \in H_P belongs to only one set in \mathcal{U}, namely U_{f_p}. So for any open refinement \mathcal{V} of \mathcal{U} (consisting of basic open sets), we have uncountably many open sets of the form U(p,W_p,n_p). By Lemma 1, we can find uncountably many such open sets with non-empty intersection. So no open refinement of \mathcal{U} can be point-countable.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

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\copyright \ 2014 \text{ by Dan Ma}