Countably paracompact spaces

This post is a basic discussion on countably paracompact space. A space is a paracompact space if every open cover has a locally finite open refinement. The definition can be tweaked by saying that only open covers of size not more than a certain cardinal number \tau can have a locally finite open refinement (any space with this property is called a \tau-paracompact space). The focus here is that the open covers of interest are countable in size. Specifically, a space is a countably paracompact space if every countable open cover has a locally finite open refinement. Even though the property appears to be weaker than paracompact spaces, the notion of countably paracompactness is important in general topology. This post discusses basic properties of such spaces. All spaces under consideration are Hausdorff.

Basic discussion of paracompact spaces and their Cartesian products are discussed in these two posts (here and here).

A related notion is that of metacompactness. A space is a metacompact space if every open cover has a point-finite open refinement. For a given open cover, any locally finite refinement is a point-finite refinement. Thus paracompactness implies metacompactness. The countable version of metacompactness is also interesting. A space is countably metacompact if every countable open cover has a point-finite open refinement. In fact, for any normal space, the space is countably paracompact if and only of it is countably metacompact (see Corollary 2 below).

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Normal Countably Paracompact Spaces

A good place to begin is to look at countably paracompactness along with normality. In 1951, C. H. Dowker characterized countably paracompactness in the class of normal spaces.

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.

Dowker’s Theorem is proved in this previous post. Condition 2 in the above formulation of the Dowker’s theorem is not in the Dowker’s theorem in the previous post. In the proof for 1 \rightarrow 2 in the previous post is essentially 1 \rightarrow 2 \rightarrow 3 for Theorem 1 above. As a result, we have the following.

Corollary 2
Let X be a normal space. Then X is countably paracompact if and only of X is countably metacompact.

Theorem 1 indicates that normal countably paracompact spaces are important for the discussion of normality in product spaces. As a result of this theorem, we know that normal countably paracompact spaces are productively normal with compact metric spaces. The Cartesian product of normal spaces with compact spaces can be non-normal (an example is found here). When the normal factor is countably paracompact and the compact factor is upgraded to a metric space, the product is always normal. The connection with normality in products is further demonstrated by the following corollary of Theorem 1.

Corollary 3
Let X be a normal space. Let Y be a non-discrete metric space. If X \times Y is normal, then X is countably paracompact.

Since Y is non-discrete, there is a non-trivial convergent sequence (i.e. the sequence represents infinitely many points). Then the sequence along with the limit point is a compact metric subspace of Y. Let’s call this subspace S. Then X \times S is a closed subspace of the normal X \times Y. As a result, X \times S is normal. By Theorem 1, X is countably paracompact.

C. H. Dowker in 1951 raised the question: is every normal space countably paracompact? Put it in another way, is the product of a normal space and the unit interval always a normal space? As a result of Theorem 1, any normal space that is not countably paracompact is called a Dowker space. The search for a Dowker space took about 20 years. In 1955, M. E. Rudin showed that a Dowker space can be constructed from assuming a Souslin line. In the mid 1960s, the existence of a Souslin line was shown to be independent of the usual axioms of set theorey (ZFC). Thus the existence of a Dowker space was known to be consistent with ZFC. In 1971, Rudin constructed a Dowker space in ZFC. Rudin’s Dowker space has large cardinality and is pathological in many ways. Zoltan Balogh constructed a small Dowker space (cardinality continuum) in 1996. Various Dowker space with nicer properties have also been constructed using extra set theory axioms. The first ZFC Dowker space constructed by Rudin is found in [2]. An in-depth discussion of Dowker spaces is found in [3]. Other references on Dowker spaces is found in [4].

Since Dowker spaces are rare and are difficult to come by, we can employ a “probabilistic” argument. For example, any “concrete” normal space (i.e. normality can be shown without using extra set theory axioms) is likely to be countably paracompact. Thus any space that is normal and not paracompact is likely countably paracompact (if the fact of being normal and not paracompact is established in ZFC). Indeed, any well known ZFC example of normal and not paracompact must be countably paracompact. In the long search for Dowker spaces, researchers must have checked all the well known examples! This probability thinking is not meant to be a proof that a given normal space is countably paracompact. It is just a way to suggest a possible answer. In fact, a good exercise is to pick a normal and non-paracompact space and show that it is countably paracompact.

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Some Examples

The following lists out a few classes of spaces that are always countably paracompact.

  • Metric spaces are countably paracompact.
  • Paracompact spaces are countably paracompact.
  • Compact spaces are countably paracompact.
  • Countably compact spaces are countably paracompact.
  • Perfectly normal spaces are countably paracompact.
  • Normal Moore spaces are countably paracompact.
  • Linearly ordered spaces are countably paracompact.
  • Shrinking spaces are countably paracompact.

The first four bullet points are clear. Metric spaces are paracompact. It is clear from definition that paracompact spaces, compact and countably compact spaces are countably paracompact. One way to show perfect normal spaces are countably paracompact is to show that they satisfy condition 6 in Theorem 1 (shown here). Any Moore space is perfect (closed sets are G_\delta). Thus normal Moore space are perfectly normal and hence countably paracompact. The proof of the countably paracompactness of linearly ordered spaces can be found in [1]. See Theorem 5 and Corollary 6 below for the proof of the last bullet point.

As suggested by the probability thinking in the last section, we now look at examples of countably paracompact spaces among spaces that are “normal and not paracompact”. The first uncountable ordinal \omega_1 is normal and not paracompact. But it is countably compact and is thus countably paracompact.

Example 1
Any \Sigma-product of uncountably many metric spaces is normal and countably paracompact.

For each \alpha<\omega_1, let X_\alpha be a metric space that has at least two points. Assume that each X_\alpha has a point that is labeled 0. Consider the following subspace of the product space \prod_{\alpha<\omega_1} X_\alpha.

    \displaystyle \Sigma_{\alpha<\omega_1} X_\alpha =\left\{f \in \prod_{\alpha<\omega_1} X_\alpha: \ f(\alpha) \ne 0 \text{ for at most countably many } \alpha \right\}

The space \Sigma_{\alpha<\omega_1} X_\alpha is said to be the \Sigma-product of the spaces X_\alpha. It is well known that the \Sigma-product of metric spaces is normal, in fact collectionwise normal (this previous post has a proof that \Sigma-product of separable metric spaces is collectionwise normal). On the other hand, any \Sigma-product always contains \omega_1 as a closed subset as long as there are uncountably many factors and each factor has at least two points (see the lemma in this previous post). Thus any such \Sigma-product, including the one being discussed, cannot be paracompact.

Next we show that T=(\Sigma_{\alpha<\omega_1} X_\alpha) \times [0,1] is normal. The space T can be reformulated as a \Sigma-product of metric spaces and is thus normal. Note that T=\Sigma_{\alpha<\omega_1} Y_\alpha where Y_0=[0,1], for any n with 1 \le n<\omega, Y_n=X_{n-1} and for any \alpha with \alpha>\omega, Y_\alpha=X_\alpha. Thus T is normal since it is the \Sigma-product of metric spaces. By Theorem 1, the space \Sigma_{\alpha<\omega_1} X_\alpha is countably paracompact. \square

Example 2
Let \tau be any uncountable cardinal number. Let D_\tau be the discrete space of cardinality \tau. Let L_\tau be the one-point Lindelofication of D_\tau. This means that L_\tau=D_\tau \cup \left\{\infty \right\} where \infty is a point not in D_\tau. In the topology for L_\tau, points in D_\tau are isolated as before and open neighborhoods at \infty are of the form L_\tau - C where C is any countable subset of D_\tau. Now consider C_p(L_\tau), the space of real-valued continuous functions defined on L_\tau endowed with the pointwise convergence topology. The space C_p(L_\tau) is normal and not Lindelof, hence not paracompact (discussed here). The space C_p(L_\tau) is also homeomorphic to a \Sigma-product of \tau many copies of the real lines. By the same discussion in Example 1, C_p(L_\tau) is countably paracompact. For the purpose at hand, Example 2 is similar to Example 1. \square

Example 3
Consider R. H. Bing’s example G, which is a classic example of a normal and not collectionwise normal space. It is also countably paracompact. This previous post shows that Bing’s Example G is countably metacompact. By Corollary 2, it is countably paracompact. \square

Based on the “probabilistic” reasoning discussed at the end of the last section (based on the idea that Dowker spaces are rare), “normal countably paracompact and not paracompact” should be in plentiful supply. The above three examples are a small demonstration of this phenomenon.

Existence of Dowker spaces shows that normality by itself does not imply countably paracompactness. On the other hand, paracompact implies countably paracompact. Is there some intermediate property that always implies countably paracompactness? We point that even though collectionwise normality is intermediate between paracompactness and normality, it is not sufficiently strong to imply countably paracompactness. In fact, the Dowker space constructed by Rudin in 1971 is collectionwise normal.

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More on Countably Paracompactness

Without assuming normality, the following is a characterization of countably paracompact spaces.

Theorem 4
Let X be a topological space. Then the space X is countably paracompact if and only of the following condition holds.

  • For any decreasing sequence \left\{A_n: n=1,2,3,\cdots \right\} of closed subsets of X such that \cap_n A_n=\varnothing, there exists a decreasing sequence \left\{B_n: n=1,2,3,\cdots \right\} of open subsets of X such that A_n \subset B_n for each n and \cap_n \overline{B_n}=\varnothing.

Proof of Theorem 4
Suppose that X is countably paracompact. Suppose that \left\{A_n: n=1,2,3,\cdots \right\} is a decreasing sequence of closed subsets of X as in the condition in the theorem. Then \mathcal{U}=\left\{X-A_n: n=1,2,3,\cdots \right\} is an open cover of X. Let \mathcal{V} be a locally finite open refinement of \mathcal{U}. For each n=1,2,3,\cdots, define the following:

    B_n=\cup \left\{V \in \mathcal{V}: V \cap A_n \ne \varnothing  \right\}

It is clear that A_n \subset B_n for each n. The open sets B_n are decreasing, i.e. B_1 \supset B_2 \supset \cdots since the closed sets A_n are decreasing. To show that \cap_n \overline{B_n}=\varnothing, let x \in X. The goal is to find B_j such that x \notin \overline{B_j}. Once B_j is found, we will obtain an open set V such that x \in V and V contains no points of B_j.

Since \mathcal{V} is locally finite, there exists an open set V such that x \in V and V meets only finitely many sets in \mathcal{V}. Suppose that these finitely many open sets in \mathcal{V} are V_1,V_2,\cdots,V_m. Observe that for each i=1,2,\cdots,m, there is some j(i) such that V_i \cap A_{j(i)}=\varnothing (i.e. V_i \subset X-A_{j(i)}). This follows from the fact that \mathcal{V} is a refinement \mathcal{U}. Let j be the maximum of all j(i) where i=1,2,\cdots,m. Then V_i \cap A_{j}=\varnothing for all i=1,2,\cdots,m. It follows that the open set V contains no points of B_j. Thus x \notin \overline{B_j}.

For the other direction, suppose that the space X satisfies the condition given in the theorem. Let \mathcal{U}=\left\{U_n: n=1,2,3,\cdots \right\} be an open cover of X. For each n, define A_n as follows:

    A_n=X-U_1 \cup U_2 \cup \cdots \cup U_n

Then the closed sets A_n form a decreasing sequence of closed sets with empty intersection. Let B_n be decreasing open sets such that \bigcap_{i=1}^\infty \overline{B_i}=\varnothing and A_n \subset B_n for each n. Let C_n=X-B_n for each n. Then C_n \subset \cup_{j=1}^n U_j. Define V_1=U_1. For each n \ge 2, define V_n=U_n-\bigcup_{j=1}^{n-1}C_{j}. Clearly each V_n is open and V_n \subset U_n. It is straightforward to verify that \mathcal{V}=\left\{V_n: n=1,2,3,\cdots \right\} is a cover of X.

We claim that \mathcal{V} is locally finite in X. Let x \in X. Choose the least n such that x \notin \overline{B_n}. Choose an open set O such that x \in O and O \cap \overline{B_n}=\varnothing. Then O \cap B_n=\varnothing and O \subset C_n. This means that O \cap V_k=\varnothing for all k \ge n+1. Thus the open cover \mathcal{V} is a locally finite refinement of \mathcal{U}. \square

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We present another characterization of countably paracompact spaces that involves the notion of shrinkable open covers. An open cover \mathcal{U} of a space X is said to be shrinkable if there exists an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} of the space X such that for each U \in \mathcal{U}, \overline{V(U)} \subset U. If \mathcal{U} is shrinkable by \mathcal{V}, then we also say that \mathcal{V} is a shrinking of \mathcal{U}. Note that Theorem 1 involves a shrinking. Condition 3 in Theorem 1 (Dowker’s Theorem) can rephrased as: every countable open cover of X has a shrinking. This for any normal countably paracompact space, every countable open cover has a shrinking (or is shrinkable).

A space X is a shrinking space if every open cover of X is shrinkable. Every shrinking space is a normal space. This follows from this lemma: A space X is normal if and only if every point-finite open cover of X is shrinkable (see here for a proof). With this lemma, it follows that every shrinking space is normal. The converse is not true. To see this we first show that any shrinking space is countably paracompact. Since any Dowker space is a normal space that is not countably paracompact, any Dowker space is an example of a normal space that is not a shrinking space. To show that any shrinking space is countably paracompact, we first prove the following characterization of countably paracompactness.

Theorem 5
Let X be a space. Then X is countably paracompact if and only of every countable increasing open cover of X is shrinkable.

Proof of Theorem 5
Suppose that X is countably paracompact. Let \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an increasing open cover of X. Then there exists a locally open refinement \mathcal{V}_0 of \mathcal{U}. For each n, define V_n=\cup \left\{O \in \mathcal{V}_0: O \subset U_n \right\}. Then \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is also a locally finite refinement of \mathcal{U}. For each n, define

    G_n=\cup \left\{O \subset X: O \text{ is open and } \forall \ m > n, O \cap V_m=\varnothing \right\}

Let \mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}. It follows that G_n \subset G_m if n<m. Then \mathcal{G} is an increasing open cover of X. Observe that for each n, \overline{G_n} \cap V_m=\varnothing for all m > n. Then we have the following:

    \displaystyle \begin{aligned} \overline{G_n}&\subset X-\cup \left\{V_m: m > n \right\} \\&\subset \cup \left\{V_k: k=1,2,\cdots,n \right\} \\&\subset \cup \left\{U_k: k=1,2,\cdots,n \right\}=U_n  \end{aligned}

We have just established that \mathcal{G} is a shrinking of \mathcal{U}, or that \mathcal{U} is shrinkable.

For the other direction, to show that X is countably paracompact, we show that the condition in Theorem 4 is satisfied. Let \left\{A_1,A_2,A_3,\cdots \right\} be a decreasing sequence of closed subsets of X with empty intersection. Then \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an open cover of X where U_n=X-A_n for each n. By assumption, \mathcal{U} is shrinkable. Let \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} be a shrinking. We can assume that \mathcal{V} is an increasing sequence of open sets.

For each n, let B_n=X-\overline{V_n}. We claim that \left\{B_1,B_2,B_3,\cdots \right\} is a decreasing sequence of open sets that expand the closed sets A_n and that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. The expansion part follows from the following:

    A_n=X-U_n \subset X-\overline{V_n}=B_n

The part about decreasing follows from:

    B_{n+1}=X-\overline{V_{n+1}} \subset X-\overline{V_n}=B_n

We show that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. To this end, let x \in X. Then x \in V_n for some n. We claim that x \notin \overline{B_n}. Suppose x \in \overline{B_n}. Since V_n is an open set containing x, V_n must contain a point of B_n, say y. Since y \in B_n, y \notin \overline{V_n}. This in turns means that y \notin V_n, a contradiction. Thus we have x \notin \overline{B_n} as claimed. We have established that every point of X is not in \overline{B_n} for some n. Thus the intersection of all the \overline{B_n} must be empty. We have established the condition in Theorem 4 is satisfied. Thus X is countably paracompact. \square

Corollary 6
If X is a shrinking space, then X is countably paracompact.

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Reference

  1. Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
  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. Wikipedia Entry on Dowker Spaces (link)

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\copyright \ 2016 \text{ by Dan Ma}

Cp(X) where X is a separable metric space

Let \tau be an uncountable cardinal. Let \prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau} be the Cartesian product of \tau many copies of the real line. This product space is not normal since it contains \prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1} as a closed subspace. However, there are dense subspaces of \mathbb{R}^{\tau} are normal. For example, the \Sigma-product of \tau copies of the real line is normal, i.e., the subspace of \mathbb{R}^{\tau} consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of \mathbb{R}^{\tau} that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space C_p(X) where X is a separable metrizable space.

For definitions of basic open sets and other background information on the function space C_p(X), see this previous post.

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C_p(X) when X is a separable metric space

In the remainder of the post, X denotes a separable metrizable space. Then, C_p(X) is more than normal. The function space C_p(X) has the following properties:

  • normal,
  • Lindelof (hence paracompact and collectionwise normal),
  • hereditarily Lindelof (hence hereditarily normal),
  • hereditarily separable,
  • perfectly normal.

All such properties stem from the fact that C_p(X) has a countable network whenever X is a separable metrizable space.

Let L be a topological space. A collection \mathcal{N} of subsets of L is said to be a network for L if for each x \in L and for each open O \subset L with x \in O, there exists some A \in \mathcal{N} such that x \in A \subset O. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for C_p(X), let \mathcal{B} be a countable base for the domain space X. For each B \subset \mathcal{B} and for any open interval (a,b) in the real line with rational endpoints, consider the following set:

    [B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}

There are only countably many sets of the form [B,(a,b)]. Let \mathcal{N} be the collection of sets, each of which is the intersection of finitely many sets of the form [B,(a,b)]. Then \mathcal{N} is a network for the function space C_p(X). To see this, let f \in O where O=\bigcap_{x \in F} [x,O_x] is a basic open set in C_p(X) where F \subset X is finite and each O_x is an open interval with rational endpoints. For each point x \in F, choose B_x \in \mathcal{B} with x \in B_x such that f(B_x) \subset O_x. Clearly f \in \bigcap_{x \in F} \ [B_x,O_x]. It follows that \bigcap_{x \in F} \ [B_x,O_x] \subset O.

Examples include C_p(\mathbb{R}), C_p([0,1]) and C_p(\mathbb{R}^\omega). All three can be considered subspaces of the product space \mathbb{R}^c where c is the cardinality of the continuum. This is true for any separable metrizable X. Note that any separable metrizable X can be embedded in the product space \mathbb{R}^\omega. The product space \mathbb{R}^\omega has cardinality c. Thus the cardinality of any separable metrizable space X is at most continuum. So C_p(X) is the subspace of a product space of \le continuum many copies of the real lines, hence can be regarded as a subspace of \mathbb{R}^c.

A space L has countable extent if every closed and discrete subset of L is countable. The \Sigma-product \Sigma_{\alpha \in A} X_\alpha of the separable metric spaces \left\{X_\alpha: \alpha \in A \right\} is a dense and normal subspace of the product space \prod_{\alpha \in A} X_\alpha. The normal space \Sigma_{\alpha \in A} X_\alpha has countable extent (hence collectionwise normal). The examples of C_p(X) discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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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}

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}

Some subspaces of Bing’s Example G

In the previous post called Compact Subspaces of Bing’s Example G, we give a characterization of the compact subspaces of Bing’s Example G (the space is denoted by the letter F). In this post, we discuss how this characterization of compact subsets of Bing’s G can shed light on certain subspaces of Bing’s G. This post should be read (or studied) alongside the previous post on the characterization of compact sets of F. The only thing we repeat is the definition of Bing’s Example G.

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Defining Bing’s Example G

First we repeat the definition of Bing’s Example G. Let P be any uncountable set. Let Q be the power set of P, i.e., the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Obviously 2^Q is simply the Cartesian product of \lvert Q \lvert many copies of the two-point discrete space \left\{0,1 \right\}, i.e., \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. The following is another topology on 2^Q:

    \tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}

Bing’s Example G is the set F=2^Q with the topology \tau^*. In other words, each x \in F-F_P is made an isolated point and points in F_P retain the usual product open sets.

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Compact Subspaces of Bing’s Example G

We do not repeat the results given in the previous post. For the purposes in this post, the following observation is crucial (see Theorem 2 in the previous post).

    Any infinite compact subspace of F is the union of a finite set and finitely many other compact sets each of which is a compact set with only one limit point.

Any compact subset of F that has exactly one limit point is a member of the collection of sets \mathcal{K}_p for some p \in P (see Theorem 1 in the previous post). For p \in P, the set \mathcal{K}_p is defined by:

    \mathcal{K}_p=\left\{K \in \mathcal{C}_p: \forall q \in Q, \left\{f \in K: f(q) \ne f_p(q) \right\} \text{ is finite} \right\}

where \mathcal{C}_p is the collection of all closed subsets of F each of which has the point f_p as the only limit point. For the results shown below, it suffices to work with a member of some \mathcal{K}_p when we work with an infinite compact subset of F.

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Some Subspaces of Example G

For each f \in F, let supp(f) be the support of f, i.e., supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}. For any infinite cardinal number \theta \le \lvert Q \lvert, we consider the following subspace:

    M_{\theta}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\theta \right\}

The subspace M_{\theta} consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for less than \theta many q \in Q. So the support of these functions is small (in relation to the size of the domain Q). Among these subspaces, of particular interest are the following two subspaces:

    M_{\lvert Q \lvert}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\lvert Q \lvert \right\}
    M_{\omega}=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}

The subspace M_{\omega} was discussed by Michael in [1] and is discussed in our blog in the post called A subspace of Bing’s example G. Michael in [1] used the letter G to denote the space M_{\omega}. We choose another letter to distinguish it from Example G. The subspace M_{\omega} consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for only finitely many q \in Q. Just like Example G, the space M is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points f \in F that have values of 1 for infinitely many q \in Q, we obtain a subspace that is metacompact.

We show the following claim about the subspace M_{\lvert Q \lvert}:

    Proposition 1
    All compact subsets of the space M_{\lvert Q \lvert} are finite.

Proof of Proposition 1
In light of the comment in the preceding section, we only need to show for any compact subset K of F such that K \in \mathcal{K}_p for some p \in P, K \cap M_{\lvert Q \lvert} is finite.

Suppose K \cap M_{\lvert Q \lvert} is infinite for K \in \mathcal{K}_p. Choose \left\{g_1,g_2,g_3,\cdots \right\} \subset K \cap M_{\lvert Q \lvert} such that g_i \ne g_j for i \ne j. Note that K_0=\left\{g_1,g_2,g_3,\cdots \right\} \cup \left\{f_p \right\} is a closed subset of K and is thus compact.

For each j let Q_j=\left\{q \in Q: g_j(q)=1 \right\}. Since each g_j \in M_{\lvert Q \lvert}, each Q_j has cardinality less than \lvert Q \lvert. Thus Q_1 \cup Q_2 \cup \cdots has cardinality less than \lvert Q \lvert too. On the other hand, let Q_\omega=\left\{q \in Q: p \in q \right\}. Since Q_\omega has cardinality equal to \lvert Q \lvert, we can pick r \in Q_\omega-(Q_1 \cup Q_2 \cup \cdots).

Right away we know that f_p(r)=1 and g_j(r)=0 for all j. Let V=\left\{f \in 2^Q: f(r)=1 \right\} which is an open set that contains f_p. But g_j \notin V for all j. Thus the following collection

    \left\{V \right\} \cup \left\{ \left\{g_j \right\}:j=1,2,3,\cdots \right\}

is an open cover of K_0 that has no finite subcover, contradicting the fact that K_0 is compact. Thus for any p \in P, for any compact set K \in \mathcal{K}_p, K \cap M_{\lvert Q \lvert} is finite. In other words, for any compact subset K of F with only one limit point, K \cap M_{\lvert Q \lvert} must be finite. It follows that in the space M_{\lvert Q \lvert}, all compact sets are finite. \blacksquare

Note that M_\theta \subset M_{\lvert Q \lvert} for all infinite cardinal numbers \theta < \lvert Q \lvert. Thus the compact subsets of all such subspaces M_\theta are finite. In particular, for the subspace M_{\omega}, there are no infinite compact subsets. Thus we have the following two easy propositions.

    Proposition 2
    For any infinite cardinal number \theta < \lvert Q \lvert, all compact subsets of the space M_{\theta} are finite.
    Proposition 3
    In particular, all compact subsets of the space M_{\omega} are finite.

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Another Proposition

The proof of Proposition 1 can be modified to show that every “small” subspace of the space M_{\lvert Q \lvert} has no limit point. We have the following proposition.

    Proposition 4
    For any Y \subset M_{\lvert Q \lvert} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\lvert Q \lvert}.

Once Proposition 4 is established, we have the following two propositions.

    Proposition 5
    Let \theta be any infinite cardinal number \theta < \lvert Q \lvert. Then for any set Y \subset M_{\theta} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\theta}.
    Proposition 6
    In particular, for any Y \subset M_{\omega} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\omega}.

The limit points in Bing’s Example G have large character. In the subspaces of Bing’s Example G discussed here, the characters at the points of F_P are still large. In these subspaces, the closure of any “small” subset cannot reach the limit points in the set F_P. So even by narrowing the focus on just the subspaces of points with “small” support, we still obtain subspaces that have large characters. For example, M_{\omega} (the subspace with finite support on the isolated points) is not only not first countable; it cannot even have any convergent sequence. In fact, any long as a subset is small (cardinality less than the cardinality of Q), the closure cannot reach any limit points at all.

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Reference

  1. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.

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\copyright \ 2014 \text{ by Dan Ma}

A subspace of Bing’s example G

Bing’s Example G is the first example of a topological space that is normal but not collectionwise normal (see [1]). Example G was an influential example from an influential paper. The Example G and its subspaces had been extensively studied. In addition to being normal and not collectionwise normal, Example G is not perfectly normal and not metacompact. See the previous post “Bing’s Example G” for a basic discussion of Example G. In this post we focus on one subspace of Example G examined by Michael in [3]. This subspace is normal, not collectionwise normal and not perfectly normal just like Example G. However it is metacompact. In [3], Michael proved that any metacompact collectionwise normal space is paracompact (metacompact was called pointwise paracompact in that paper). This subspace of Example G demonstrates that collectionwise normality in Michael’s theorem cannot be replaced by normality.

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Bing’s Example G

For a more detailed discussion of Bing’s Example G in this blog, see the blog post “Bing’s Example G”. For the sake of completeness, we repeat the definition of Example G. Let P be any uncountable set. Let Q be the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Obviously 2^Q is simply the Cartesian product of \lvert Q \lvert many copies of the two-point discrete space \left\{0,1 \right\}, i.e., \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. The following is another topology on 2^Q:

    \tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}

Bing’s Example G is the set F=2^Q with the topology \tau^*. In other words, each x \in F-F_P is made an isolated point and points in F_P retain the usual product open sets.

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Michael’s Subspace of Example G

For each f \in F, let supp(f) be the support of f, i.e., supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}. Michael in [3] considered the following subspace of F.

    M=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}

Michael in [3] used the letter G to denote the space M. We choose another letter to distinguish it from Example G. The subspace M consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for only finitely many q \in Q. The space M is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points f \in F that have values of 1 for infinitely many q \in Q, we obtain a subspace that is metacompact. We discuss the following points:

  • The space M is normal.
  • The space M is not collectionwise Hausdorff and hence not collectionwise normal.
  • The space M is metacompact.
  • The space M is not perfectly normal.

The space M is normal since the space F that is Example G is hereditarily normal (see the section called Bing’s Example G is Completely Normal in the post “Bing’s Example G”).

To show that the space M is not collectionwise Hausdorff, it is helpful to first look at M as a subspace of the product space 2^Q. The product space 2^Q has the countable chain condition (CCC) since it is a product of separable spaces. Note that M is dense in the product space 2^Q. Thus M as a subspace of the product space has the CCC.

In the space M, the set F_P is still a closed and discrete set. In the space M, open sets containing points of F_P are the same as product open sets in 2^Q relative to the set M. Since M has CCC (as a subspace of the product space 2^Q), M cannot have uncountably many pairwise disjoint open sets containing points of F_P (in either the product topology or the Example G subspace topology). It follows that M is not collectionwise Hausdorff. If it were, there would be uncountably many pairwise disjoint product open sets separating points in F_P, which is not possible.

To see that M is metacompact, let \mathcal{U} be an open cover of M. For each p \in P, choose U_p \in \mathcal{U} such that f_p \in U_p. For each p \in P, let W_p=\left\{f \in M: f(\left\{p \right\})=1 \right\}. Let \mathcal{V} be the following:

    \mathcal{V}=\left\{U_p \cap W_p: p \in P \right\} \cup \left\{\left\{x \right\}: x \in M-F_P \right\}

Note that \mathcal{V} is a point-finite open refinement of \mathcal{U}. Each U_p \cap W_p contains only one point of F_P, namely f_p. On the other hand, each f \in M with finite support can belong to at most finitely many U_p \cap W_p.

The space M is not perfectly normal. This point is alluded to in [3] by Michael and elsewhere in the literature, e.g. in Bing’s paper (see [1]) and in Engelking’s general topology text (see 5.53 on page 338 of [2]). In fact Michael indicated that one can obtain a perfectly normal example with the aforementioned properties using Example H defined in [1] instead of using the subspace M defined here in this post.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
  4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

An Example of a Completely Regular Space that is not Normal

We present an example of a space S that is Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular but not normal. This space S is a building block for defining other spaces. In a previous post, we show that the space S can be turned into a space that is regular but not completely regular by adding one point (see Regular but not Completely Regular). Even though S is discussed in the previous post, the space S is an interesting example by itself and deserves a separate post.

Let S 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 V_x to be the set V_x=\left\{(x,y) \in S: 0 \le y \le 2 \right\}, define D_x to be the set D_x=\left\{(s,s-x) \in S: x \le s \le x+2 \right\}, and define O_x=V_x \cup D_x. The topology on S is defined by the following:

  • Each point (x,y) \in S where y>0 is isolated.
  • For each point (x,0) \in S, 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.

It is straightforward to verify that the basic open sets defined above form a base for a topology on the set S and that the resulting topology is Hausdorff. One important observation to make is that this base consists of sets that are both closed and open. Whenever a space has a base consisting of closed and open sets, it is said to be a zero-dimensional space. It is straightforward to show that any zero-dimensional space is completely regular. Another interesting point about the space S is that it is metacompact. Recall that a space X is metacompact if every open cover of X has a point-finite open refinement.

Now we discuss why S is not normal. Note that the x-axis in S is a closed and discrete set of cardinality continuum. Jones’ lemma states that in a normal and separable space, the cardinality of any closed and discrete set must not equal to or exceed continuum. But Jones’ lemma is of no use here since S is not separable. However, the two disjoint closed sets that destroyed normality are from the x-axis, namely H=\left\{(x,0): x \in \mathbb{Q} \right\} and K=\left\{(x,0): x \in \mathbb{R}-\mathbb{Q} \right\}.

Note that H and K are disjoint closed sets in S. If S were normal, there would be a continuous f:S \rightarrow [0,1] such that f(a)=0 for all a \in H and f(a)=1 for all a \in K (using Urysohn lemma). But this function is not possible. It can be shown that any continuous function g:S \rightarrow [0,1] that maps H to zero would have to map the entire x-axis to zero except for a countable subset of x-axis. This fact follows from the Main Result presented in Regular but not Completely Regular.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012