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).

____________________________________________________________________

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 compact? 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.

____________________________________________________________________

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.

____________________________________________________________________

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

___________________________________

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.

____________________________________________________________________

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)

____________________________________________________________________
\copyright \ 2016 \text{ by Dan Ma}

A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

____________________________________________________________________

The countable chain conditon

A topological 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 discussion, if X has the CCC, we also say that X is a CCC space or X is CCC. A space X is separable if there exists a countable subset A of X such that A is dense in X (meaning that if U is a nonempty open subset of X, then U \cap A \ne \varnothing).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

____________________________________________________________________

The strategy

Here’s the strategy for finding CCC and not separable.

    The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

____________________________________________________________________

Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if c is the cardinality of continuum, then consider \left\{0,1 \right\}^{2^c}, the product of 2^c many copies of \left\{0,1 \right\}, or consider X^{2^c} where X is your favorite separable space.

____________________________________________________________________

Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the \Sigma-products. To make this clear, let’s focus on a specific example. Consider X=\left\{0,1 \right\}^{c} where c is the cardinality of continuum. Consider the following subspace.

    \Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}

The subspace \Sigma(\left\{0,1 \right\}^{c}) is dense in X, thus has CCC. It is straightforward to verify that \Sigma(\left\{0,1 \right\}^{c}) is not separable.

To implement this example, find any space X which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the \Sigma-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The \Sigma-product has the countable chain condition since it is a dense subspace of the CCC space X. The \Sigma-product is not separable since there are uncountably many factors in the product space X and that each factor has at least two points. This idea had been implemented in this previous post.

____________________________________________________________________

Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space X, let \mathcal{F}[X] be the set of all non-empty finite subsets of X. For F \in \mathcal{F}[X] and for any open subset U of X, let [F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}. The sets [F,U] over all 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 Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space X is uncountable, \mathcal{F}[X] is not a separable space. We need to identify the \mathcal{F}[X] that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space X with a countable network, \mathcal{F}[X] is CCC. Thus for any uncountable space X with a countable network, the Pixley-Roy space \mathcal{F}[X] is a CCC space that is not separable. The following gives a few such examples.

    \mathcal{F}[\mathbb{R}]

    \mathcal{F}[X] where X is any uncountable, separable and metrizable space.

    \mathcal{F}[X] where X is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space X that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

____________________________________________________________________

Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line \mathbb{R}. Let X be any uncountable completely regular space. The product space \mathbb{R}^X always has the CCC since it is a product of separable space. Now we single out a dense subspace of \mathbb{R}^X for which there is a characterization for separability, namely the subspace C(X), which is the set of all continuous functions from X into \mathbb{R}. The subspace C(X) as a topological space is usually denoted by C_p(X). For a basic discussion of C_p(X), see this previous post.

We know precisely when C_p(X) is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space C_p(X) is separable if and only if the domain space X has a weaker (or coarser) separable metric topology (in other words, X is submetrizable with a separable metric topology).

Based on the theorem, C_p(X) is separable for any separable metric space X. Other examples of separable C_p(X) include spaces X that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus C_p(X) would be separable if X is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces X that have no weaker separable metric topology.

  • Any compact space X that is not metrizable.
  • The space X=\omega_1, the first uncountable ordinal with the order topology.
  • Any space X=C_p(Y) where Y is not separable.

The function space C_p(X) for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact X are: the first uncountable successor ordinal \omega_1+1, the double arrow space, and the product space \left\{0,1 \right\}^{\omega_1}.

It can be shown that C_p(\omega_1) is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space C_p(Y) has a weaker (or coarser) separable metric topology if and only if the domain space Y is separable.

Thus picking a non-separable space Y would guarantee that C_p(Y) has a weaker separable metric topology. As a result, C_p(C_p(Y)) is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

____________________________________________________________________

Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space X that is CCC and not separable from any one of the avenues discussed here, the function space C_p(X) has no weaker separable metric topology (by Theorem 2) and the function space C_p(C_p(X)) is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

____________________________________________________________________

Exercises

  1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if X is metrizable and CCC, then X is separable.
  2. Show that any dense subspace of a CCC space is also CCC.
  3. Verify that the space \Sigma(\left\{0,1 \right\}^{c}) defined in Example 2 is dense in X and is not separable.
  4. Verify that the Pixley-Roy space \mathcal{F}[\mathbb{R}] defined in Example 3 is CCC and not separable.
  5. Verify that function space C_p(\omega_1) mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

____________________________________________________________________

Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

____________________________________________________________________
\copyright \ 2016 \text{ by Dan Ma}

Counterexample 106 from Steen and Seebach

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

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

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

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

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

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

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

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

____________________________________________________________________

The space X is not normal

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

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

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

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

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

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

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

____________________________________________________________________

The space X has a dense subspace that is normal

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

____________________________________________________________________

Reference

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

____________________________________________________________________
\copyright \ 2015 \text{ by Dan Ma}

An exercise gleaned from the proof of a theorem on pseudocompact space

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

____________________________________________________________________

Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space X is a pseudocompact space if every continuous real-valued function defined on X is bounded, i.e., if f:X \rightarrow \mathbb{R} is a continuous function, then f(X) is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

    \text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space \Psi(\mathcal{A}) where \mathcal{A} is a maximal almost disjoint family of subsets of \omega (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

____________________________________________________________________

The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let Y be a compact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto a space Z. Then g is a homeomorphism.

Theorem 2
Let Y be a pseudocompact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto Z where Z is a separable and metrizable space. Then g is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map g as described in Theorem 2, the domain Y must be compact. Then Theorem 1 will finish the job.

____________________________________________________________________

Reference

  1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

.

____________________________________________________________________
\copyright \ 2015 \text{ by Dan Ma}

A note on products of sequential fans

Two posts (the previous post and this post) are devoted to discussing the behavior of countable tightness in taking Cartesian products. The previous post shows that countable tightness behaves well in the product operation if the spaces are compact. In this post, we step away from the orderly setting of compact spaces. We examine the behavior of countable tightness in product of sequential fans. In this post, we show that countable tightness can easily be destroyed when taking products of sequential fans. Due to the combinatorial nature of sequential fans, the problem of determining the tightness of products of fans is often times a set-theoretic problem. In many instances, it is hard to determine the tightness of a product of two sequential fans without using extra set theory axioms beyond ZFC. The sequential fans is a class of spaces that have been studied extensively and are involved in the solutions of many problems that were seemingly unrelated. For one example, see [3].

For a basic discussion of countable tightness, see these previous post on the notion of tightness and its relation with free sequences. Also see chapter a-4 on page 15 of [4].

____________________________________________________________________

Sequential fans

Let S be a non-trivial convergent sequence along with its limit point. For convenience, let \displaystyle S=\left\{0 \right\} \cup \left\{1, 2^{-1}, 3^{-1}, 4^{-1}, \cdots \right\}, considered as a subspace of the Euclidean real line. Let \kappa be a cardinal number. The set \kappa is usually taken as the set of all the ordinals that precede \kappa. The set \omega is the first infinite ordinal, or equivalently the set of all non-negative integers. Let \omega^\kappa be the set of all functions from \kappa into \omega.

There are several ways to describe a sequential fan. One way is to describe it as a quotient space. The sequential fan S(\kappa) is the topological sum of \kappa many copies of the convergent sequence S with all non-isolated points identified as one point that is called \infty. To make the discussion easier to follow, we also use the following formulation of S(\kappa):

    \displaystyle S(\kappa)=\left\{\infty \right\} \cup (\kappa \times \omega)

In this formulation, every point is \kappa \times \omega is isolated and an open neighborhood of the point \infty is of the form:

    \displaystyle B_f=\left\{\infty \right\} \cup \left\{(\alpha,n) \in \kappa \times \omega: n \ge f(\alpha) \right\} where f \in \omega^\kappa.

According to the definition of the open neighborhood B_f, the sequence (\alpha,0), (\alpha,1), (\alpha,2),\cdots converges to the point \infty for each \alpha \in \kappa. Thus the set (\left\{\alpha \right\} \times \omega) \cup \left\{\infty \right\} is a homeomorphic copy of the convergent sequence S. The set \left\{\alpha \right\} \times \omega is sometimes called a spine. Thus the space S(\kappa) is said to be the sequential fan with \kappa many spines.

The point \infty is the only non-isolated point in the fan S(\kappa). The set \mathcal{B}=\left\{B_f: f \in \omega^\kappa \right\} is a local base at the point \infty. The base \mathcal{B} is never countable except when \kappa is finite. Thus if \kappa is infinite, the fan S(\kappa) can never be first countable. In particular, for the fan S(\omega), the character at the point \infty is the cardinal number \mathfrak{d}. See page 13 in chapter a-3 of [4]. This cardinal number is called the dominating number and is introduced below in the section “The bounding number”. This is one indication that the sequential fan is highly dependent on set theory. It is hard to pinpoint the character of S(\omega) at the point \infty. For example, it is consistent with ZFC that \mathfrak{d}=\omega_1. It is also consistent that \mathfrak{d}>\omega_1.

Even though the sequential fan is not first countable, it has a relatively strong convergent property. If \infty \in \overline{A} and \infty \notin A where A \subset S(\kappa), then infinitely many points of A are present in at least one of the spine \left\{\alpha \right\} \times \omega (if this is not true, then \infty \notin \overline{A}). This means that the sequential fan is always a Frechet space. Recall that the space Y is a Frechet space if for each A \subset Y and for each x \in \overline{A}, there exists a sequence \left\{x_n \right\} of points of A converging to x.

Some of the convergent properties weaker than being a first countable space are Frechet space, sequential space and countably tight space. Let's recall the definitions. A space X is a sequential space if A \subset X is a sequentially closed set in X, then A is a closed set in X. The set A is sequentially closed in X if this condition is satisfied: if the sequence \left\{x_n \in A: n \in \omega \right\} converges to x \in X, then x \in A. A space X is countably tight (have countable tightness) if for each A \subset X and for each x \in \overline{A}, there exists a countable B \subset A such that x \in \overline{B}. See here for more information about these convergent properties. The following shows the relative strength of these properties. None of the implications can be reversed.

    First countable space \Longrightarrow Frechet space \Longrightarrow Sequential space \Longrightarrow Countably tight space

____________________________________________________________________

Examples

The relatively strong convergent property of being a Frechet space is not preserved in products or squares of sequential fans. We now look at some examples.

Example 1
Consider the product space S(\omega) \times S where S is the convergent sequence defined above. The first factor is Frechet and the second factor is a compact metric space. We show that S(\omega) \times S is not sequential. To see this, consider the following subset A of S(\omega) \times S:

    \displaystyle A_f=\left\{(x,n^{-1}) \in S(\omega) \times S: n \in \omega \text{ and } x=(n,f(n)) \right\} \ \forall \ f \in \omega^\omega

    \displaystyle A=\bigcup_{f \in \omega^\omega} A_f

It follows that (\infty,0) \in \overline{A}. Furthermore, no sequence of points of A can converge to the point (\infty,0). To see this, let a_n \in A for each n. Consider two cases. One is that some spine \left\{m \right\} \times \omega contains the first coordinate of a_n for infinitely many n \in \omega. The second is the opposite of the first – each spine \left\{m \right\} \times \omega contains the first coordinate of a_n for at most finitely many n. Either case means that there is an open set containing (\infty,0) that misses infinitely many a_n. Thus the sequence a_n cannot converge to (\infty,0).

Let A_1 be the set of all sequential limits of convergent sequences of points of A. With A \subset A_1, we know that (\infty,0) \in \overline{A_1} but (\infty,0) \notin A_1. Thus A_1 is a sequentially closed subset of S(\omega) \times S that is not closed. This shows that S(\omega) \times S is not a sequential space.

The space S(\omega) \times S is an example of a space that is countably tight but not sequential. The example shows that the product of two Frechet spaces does not even have to be sequential even when one of the factors is a compact metric space. The next example shows that the product of two sequential fans does not even have to be countably tight.

Example 2
Consider the product space S(\omega) \times S(\omega^\omega). We show that it is not countably tight. To this end, consider the following subset A of S(\omega) \times S(\omega^\omega).

    \displaystyle S(\omega)=\left\{\infty \right\} \cup (\omega \times \omega)

    \displaystyle S(\omega^\omega)=\left\{\infty \right\} \cup (\omega^\omega \times \omega)

    \displaystyle A_f=\left\{(x,y) \in S(\omega) \times S(\omega^\omega): x=(n,f(n)) \text{ and } y=(f,j)  \right\} \ \forall \ f \in \omega^\omega

    \displaystyle A=\bigcup_{f \in \omega^\omega} A_f

It follows that (\infty,\infty) \in \overline{A}. We show that for any countable C \subset A, the point (\infty,\infty) \notin \overline{C}. Fix a countable C \subset A. We can assume that C=\bigcup_{i=1}^\infty A_{f_i}. Now define a function g \in \omega^\omega by a diagonal argument as follows.

Define g(0) such that g(0)>f_0(0). For each integer n>0, define g(n) such that g(n)>\text{max} \{ \ f_n(0),f_n(1),\cdots,f_n(n) \ \} and g(n)>g(n-1). Let O=B_g \times S(\omega^\omega). The diagonal definition of g ensures that O is an open set containing (\infty,\infty) such that O \cap C=\varnothing. This shows that the space S(\omega) \times S(\omega^\omega) is not countably tight.

Example 3
The space S(\omega_1) \times S(\omega_1) is not countably tight. In fact its tightness character is \omega_1. This fact follows from Theorem 1.1 in [2].

____________________________________________________________________

The set-theoretic angle

Example 2 shows that S(\omega) \times S(\omega^\omega) is not countably tight even though each factor has the strong property of a Frechet space with the first factor being a countable space. The example shows that Frechetness behaves very badly with respect to the product operation. Is there an example of \kappa>\omega such that S(\omega) \times S(\kappa) is countably tight? In particular, is S(\omega) \times S(\omega_1) countably tight?

First off, if Continuum Hypothesis (CH) holds, then Example 2 shows that S(\omega) \times S(\omega_1) is not countably tight since the cardinality of \omega^{\omega} is \omega_1 under CH. So for S(\omega) \times S(\omega_1) to be countably tight, extra set theory assumptions beyond ZFC will have to be used (in fact the extra axioms will have to be compatible with the negation of CH). In fact, it is consistent with ZFC for S(\omega) \times S(\omega_1) to be countably tight. It is also consistent with ZFC for t(S(\omega) \times S(\omega_1))=\omega_1. We point out some facts from the literature to support these observations.

Consider S(\omega) \times S(\kappa) where \kappa>\omega_1. For any regular cardinal \kappa>\omega_1, it is possible that S(\omega) \times S(\kappa) is countably tight. It is also possible for the tightness character of S(\omega) \times S(\kappa) to be \kappa (of course in a different model of set theory). Thus it is hard to pin down the tightness character of the product S(\omega) \times S(\kappa). It all depends on your set theory. In the next section, we point out some facts from the literature to support these observations.

Example 3 points out that the tightness character of S(\omega_1) \times S(\omega_1) is \omega_1, i.e. t(S(\omega_1) \times S(\omega_1))=\omega_1 (this is a fact on the basis of ZFC only). What is t(S(\omega_2) \times S(\omega_2)) or t(S(\kappa) \times S(\kappa)) for any \kappa>\omega_1? The tightness character of S(\kappa) \times S(\kappa) for \kappa>\omega_1 also depends on set theory. We also give a brief explanation by pointing out some basic information from the literature.

____________________________________________________________________

The bounding number

The tightness of the product S(\omega) \times S(\kappa) is related to the cardinal number called the bounding number denoted by \mathfrak{b}.

Recall that \omega^{\omega} is the set of all functions from \omega into \omega. For f,g \in \omega^{\omega}, define f \le^* g by the condition: f(n) \le g(n) for all but finitely many n \in \omega. A set F \subset \omega^{\omega} is said to be a bounded set if F has an upper bound according to \le^*, i.e. there exists some f \in \omega^{\omega} such that g \le^* f for all g \in F. Then F \subset \omega^{\omega} is an unbounded set if it is not bounded. To spell it out, F \subset \omega^{\omega} is an unbounded set if for each f \in \omega^{\omega}, there exists some g \in F such that g \not \le^* f.

Furthermore, F \subset \omega^{\omega} is a dominating set if for each f \in \omega^{\omega}, there exists some g \in F such that f \le^* g. Define the cardinal numbers \mathfrak{b} and \mathfrak{d} as follows:

    \displaystyle \mathfrak{b}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is an unbounded set} \right\}

    \displaystyle \mathfrak{d}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is a dominating set} \right\}

The cardinal number \mathfrak{b} is called the bounding number. The cardinal number \mathfrak{d} is called the dominating number. Note that continuum \mathfrak{c}, the cardinality of \omega^{\omega}, is an upper bound of both \mathfrak{b} and \mathfrak{d}, i.e. \mathfrak{b} \le \mathfrak{c} and \mathfrak{d} \le \mathfrak{c}. How do \mathfrak{b} and \mathfrak{d} relate? We have \mathfrak{b} \le \mathfrak{d} since any dominating set is also an unbounded set.

A diagonal argument (similar to the one in Example 2) shows that no countable F \subset \omega^{\omega} can be unbounded. Thus we have \omega < \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}. If CH holds, then we have \omega_1 = \mathfrak{b} = \mathfrak{d} = \mathfrak{c}. On the other hand, it is also consistent that \omega < \mathfrak{b} < \mathfrak{d} \le \mathfrak{c}.

We now relate the bounding number to the tightness of S(\omega) \times S(\kappa). The following theorem is from Theorem 1.3 in [3].

Theorem 1 – Theorem 1.3 in [3]
The following conditions hold:

  • For \omega \le \kappa <\mathfrak{b}, the space S(\omega) \times S(\kappa) is countably tight.
  • The tightness character of S(\omega) \times S(\mathfrak{b}) is \mathfrak{b}, i.e. t(S(\omega) \times S(\mathfrak{b}))=\mathfrak{b}.

Thus S(\omega) \times S(\kappa) is countably tight for any uncountable \kappa <\mathfrak{b}. In particular if \omega_1 <\mathfrak{b}, then S(\omega) \times S(\omega_1) is countably tight. According to Theorem 5.1 in [6], this is possible.

Theorem 2 – Theorem 5.1 in [6]
Let \tau and \lambda be regular cardinal numbers such that \omega_1 \le \tau \le \lambda. It is consistent with ZFC that \mathfrak{b}=\mathfrak{d}=\tau and \mathfrak{c}=\lambda.

Theorem 2 indicates that it is consistent with ZFC that the bounding number \mathfrak{b} can be made to equal any regular cardinal number. In the model of set theory in which \omega_1 <\mathfrak{b}, S(\omega) \times S(\omega_1) is countably tight. Likewise, in the model of set theory in which \omega_1 < \kappa <\mathfrak{b}, S(\omega) \times S(\kappa) is countably tight.

On the other hand, if the bounding number is made to equal an uncountable regular cardinal \kappa, then t(S(\omega) \times S(\kappa))=\kappa. In particular, t(S(\omega) \times S(\omega_1))=\omega_1 if \mathfrak{b}=\omega_1.

The above discussion shows that the tightness of S(\omega) \times S(\kappa) is set-theoretic sensitive. Theorem 2 indicates that it is hard to pin down the location of the bounding number \mathfrak{b}. Choose your favorite uncountable regular cardinal, there is always a model of set theory in which \mathfrak{b} is your favorite uncountable cardinal. Then Theorem 1 ties the bounding number to the tightness of S(\omega) \times S(\kappa). Thus the exact value of the tightness character of S(\omega) \times S(\kappa) depends on your set theory. If your favorite uncountable regular cardinal is \omega_1, then in one model of set theory consistent with ZFC, t(S(\omega) \times S(\omega_1))=\omega (when \omega_1 <\mathfrak{b}). In another model of set theory, t(S(\omega) \times S(\omega_1))=\omega_1 (when \omega_1 =\mathfrak{b}).

One comment about the character of the fan S(\omega) at the point \infty. As indicated earlier, the character at \infty is the dominating number \mathfrak{d}. Theorem 2 tells us that it is consistent that \mathfrak{d} can be any uncountable regular cardinal. So for the fan S(\omega), it is quite difficult to pinpoint the status of a basic topological property such as character of a space. This is another indication that the sequential fan is highly dependent on additional axioms beyond ZFC.

____________________________________________________________________

The collectionwise Hausdorff property

Now we briefly discuss the tightness of t(S(\kappa) \times S(\kappa)) for any \kappa>\omega_1. The following is Theorem 1.1 in [2].

Theorem 3 – Theorem 1.1 in [2]
Let \kappa be any infinite regular cardinal. The following conditions are equivalent.

  • There exists a first countable < \kappa-collectionwise Hausdorff space which fails to be a \kappa-collectionwise Hausdorff space.
  • t(S(\kappa) \times S(\kappa))=\kappa.

The existence of the space in the first condition, on the surface, does not seem to relate to the tightness character of the square of a sequential fan. Yet the two conditions were proved to be equivalent [2]. The existence of the space in the first condition is highly set-theory sensitive. Thus so is the tightness of the square of a sequential fan. It is consistent that a space in the first condition exists for \kappa=\omega_2. Thus in that model of set theory t(S(\omega_2) \times S(\omega_2))=\omega_2. It is also consistent that there does not exist a space in the first condition for \kappa=\omega_2. Thus in that model, t(S(\omega_2) \times S(\omega_2))<\omega_2. For more information, see [3].

____________________________________________________________________

Remarks

Sequential fans and their products are highly set-theoretic in nature and are objects that had been studied extensively. This is only meant to be a short introduction. Any interested readers can refer to the small list of articles listed in the reference section and other articles in the literature.

____________________________________________________________________

Exercise

Use Theorem 3 to show that t(S(\omega_1) \times S(\omega_1))=\omega_1 by finding a space X that is a first countable < \omega_1-collectionwise Hausdorff space which fails to be a \omega_1-collectionwise Hausdorff space.

For any cardinal \kappa, a space X is \kappa-collectionwise Hausdorff (respectively < \kappa-collectionwise Hausdorff) if for any closed and discrete set A \subset X with \lvert A \lvert \le \kappa (repectively \lvert A \lvert < \kappa), the points in A can be separated by a pairwise disjoint family of open sets.

____________________________________________________________________

Reference

  1. Bella A., van Mill J., Tight points and countable fan-tightness, Topology Appl., 76, (1997), 1-27.
  2. Eda K., Gruenhage G., Koszmider P., Tamano K., Todorčeviće S., Sequential fans in topology, Topology Appl., 67, (1995), 189-220.
  3. Eda K., Kada M., Yuasa Y., Tamano K., The tightness about sequential fans and combinatorial properties, J. Math. Soc. Japan, 49 (1), (1997), 181-187.
  4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  5. LaBerge T., Landver A., Tightness in products of fans and psuedo-fans, Topology Appl., 65, (1995), 237-255.
  6. Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 111-167.

.

____________________________________________________________________
\copyright \ 2015 \text{ by Dan Ma}

Products of compact spaces with countable tightness

In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.

The following theorems are the main results in this post.

Theorem 1
Let X and Y be countably tight spaces. If one of X and Y is compact, then X \times Y is countably tight.

Theorem 2
The product of finitely many compact countably tight spaces is countably tight.

Theorem 3
Suppose that X_1, X_2, X_3, \cdots are countably many compact spaces such that each X_i has at least two points. If each X_i is a countably tight space, then the product space \prod_{i=1}^\infty X_i is countably tight.

____________________________________________________________________

Finite products

Before proving Theorem 1 and Theorem 2, we prove the following results.

Theorem 4
Let f:Y_1 \rightarrow Y_2 be a continuous and closed map from the space Y_1 onto the space Y_2. Suppose that the space Y_2 is countably tight and that each fiber of the map f is countably tight. Then the space Y_1 is countably tight.

Proof of Theorem 4
Let x \in Y_1 and x \in \overline{A} where A \subset Y_1. We proceed to find a countable W \subset Y_1 such that x \in \overline{W}. Choose y \in Y_2 such that y=f(x).

Let M be the fiber of the map f at the point y, i.e. M=f^{-1}(y). By assumption, M is countably tight. Call a point w \in M countably reached by A if there is some countable C \subset A such that w \in \overline{C}. Let G be the set of all points in M that are countably reached by A. We claim that x \in \overline{G}.

Let U \subset Y_1 be open such that x \in U. Because the space Y_1 is regular, choose open V \subset U such that x \in V and \overline{V} \subset U. Then V \cap A \ne \varnothing. Furthermore, x \in \overline{V \cap A}. Let C=f(V \cap A). By the continuity of f, we have y \in \overline{C}. Since Y_2 is countably tight, there exists some countable D \subset C such that y \in \overline{D}. Choose a countable E \subset V \cap A such that f(E)=D. It follows that y \in \overline{f(E)}.

We show that that \overline{E} \cap M \ne \varnothing. Since E \subset \overline{E}, we have f(E) \subset f(\overline{E}). Note that f(\overline{E}) is a closed set since f is a closed map. Thus \overline{f(E)} \subset f(\overline{E}). As a result, y \in f(\overline{E}). Then y=f(t) for some t \in \overline{E}. We have t \in \overline{E} \cap M.

By the definition of the set G, we have \overline{E} \cap M \subset G. Furthermore, \overline{E} \cap M \subset \overline{V} \subset U. Note that the arbitrary open neighborhood U of x contains points of G. This establishes the claim that x \in \overline{G}.

Since M is a fiber of f, M is countably tight by assumption. Choose some countable T \subset G such that x \in \overline{T}. For each t \in T, choose a countable W_t \subset A with t \in \overline{W_t}. Let W=\bigcup_{t \in T} W_t. Note that W \subset A and W is countable with x \in \overline{W}. This establishes the space Y_1 is countably tight at x \in Y_1. \blacksquare

Lemma 5
Let f:X \times Y \rightarrow Y be the projection map. If X is a compact space, then f is a closed map.

Proof of Lemma 5
Let A be a closed subset of X \times Y. Suppose that f(A) is not closed. Let y \in \overline{f(A)}-f(A). It follows that no point of X \times \left\{y \right\} belongs to A. For each x \in X, choose open subset O_x of X \times Y such that (x,y) \in O_x and O_x \cap A=\varnothing. The set of all O_x is an open cover of the compact space X \times \left\{y \right\}. Then there exist finitely many O_x that cover X \times \left\{y \right\}, say O_{x_i} for i=1,2,\cdots,n.

Let W=\bigcup_{i=1}^n O_{x_i}. We have X \times \left\{y \right\} \subset W. Since X is compact, we can then use the Tube Lemma which implies that there exists open G \subset Y such that X \times \left\{y \right\} \subset X \times G \subset W. It follows that G \cap f(A) \ne \varnothing. Choose t \in G \cap f(A). Then for some x \in X, (x,t) \in A. Since t \in G, (x,t) \in W, implying that W \cap A \ne \varnothing, a contradiction. Thus f(A) must be a closed set in Y. This completes the proof of the lemma. \blacksquare

Proof of Theorem 1
Let X be the factor that is compact. let f: X \times Y \rightarrow Y be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space Y is countably tight by assumption. Each fiber of the projection map f is of the form X \times \left\{y \right\} where y \in Y, which is countably tight. Then use Theorem 4 to establish that X \times Y is countably tight. \blacksquare

Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. \blacksquare

____________________________________________________________________

Countable products

Our proof to establish that the product space \prod_{i=1}^\infty X_i is countably tight is an indirect one and makes use of two non-trivial results. We first show that \omega_1 \times \prod_{i=1}^\infty X_i is a closed subspace of a \Sigma-product that is normal. It follows from another result that the second factor \prod_{i=1}^\infty X_i is countably tight. We now present all the necessary definitions and theorems.

Consider a product space Y=\prod_{\alpha<\kappa} Y_\alpha where \kappa is an infinite cardinal number. Fix a point p \in Y. The \Sigma-product of the spaces Y_\alpha with p as the base point is the following subspace of the product space Y=\prod_{\alpha<\kappa} Y_\alpha:

    \displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}

The definition of the space \Sigma_{\alpha<\kappa} Y_\alpha depends on the base point p. The discussion here is on properties of \Sigma_{\alpha<\kappa} Y_\alpha that hold regardless of the choice of base point. If the factor spaces are indexed by a set A, the notation is \Sigma_{\alpha \in A} Y_\alpha.

If all factors Y_\alpha are identical, say Y_\alpha=Z for all \alpha, then we use the notation \Sigma_{\alpha<\kappa} Z to denote the \Sigma-product. Once useful fact is that if there are \omega_1 many factors and each factor has at least 2 points, then the space \omega_1 can be embedded as a closed subspace of the \Sigma-product.

Theorem 6
For each \alpha<\omega_1, let Y_\alpha be a space with at least two points. Then \Sigma_{\alpha<\omega_1} Y_\alpha contains \omega_1 as a closed subspace. See Exercise 3 in this previous post.

Now we discuss normality of \Sigma-products. This previous post shows that if each factor is a separable metric space, then the \Sigma-product is normal. It is also well known that if each factor is a metric space, the \Sigma-product is normal. The following theorem handles the case where each factor is a compact space.

Theorem 7
For each \alpha<\kappa, let Y_\alpha be a compact space. Then the \Sigma-product \Sigma_{\alpha<\kappa} Y_\alpha is normal if and only if each factor Y_\alpha is countably tight.

Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the \Sigma-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.

Theorem 8
Let Y be a compact space. Then the product space \omega_1 \times Y is normal if and only if Y is countably tight.

We now prove Theorem 3.

Proof of Theorem 3
Let \omega_1=\cup \left\{A_n: n \in \omega \right\}, where for each n, \lvert A_n \lvert=\omega_1 and that A_n \cap A_m=\varnothing if n \ne m. For each n=1,2,3,\cdots, let S_n=\Sigma_{\alpha \in A_n} X_n. By Theorem 7, each S_n is normal. Let S_0=\Sigma_{\alpha \in A_0} X_1, which is also normal. By Theorem 6, the space \omega_1 of countable ordinals is a closed subspace of S_0. Let T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots. We have the following derivation.

    \displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}

Recall that \omega_1=\cup \left\{A_n: n \in \omega \right\}. The space W=\Sigma_{\alpha<\omega_1} W_\alpha is defined such that for each n \ge 1 and for each \alpha \in A_n, W_\alpha=X_n. Furthermore, for n=0, for each \alpha \in A_0, let W_\alpha=X_1. Thus W is a \Sigma-product of compact countably tight spaces and is thus normal by Theorem 7. The space T=\omega_1 \times \prod_{n=1}^\infty X_n is a closed subspace of the normal space W. By Theorem 8, the product space \prod_{n=1}^\infty X_n must be countably tight. \blacksquare

____________________________________________________________________

Remarks

Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let X=X_1 \times X_2 \times \cdots \times X_n.

    \displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset  X_1 \times X_2 \times \cdots \times X_n \times X_{n+1}  \times X_{n+2} \times \cdots  \end{aligned}

In the above derivation, t_m is a point of X_m for all m >n. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.

____________________________________________________________________

Exercise

Exercise 1
Let f:X \times Y \rightarrow Y be the projection map. If X is a countably compact space and Y is a Frechet space, then f is a closed map.

Exercise 2
Let X and Y be countably tight spaces. If one of X and Y is a countably compact space and the other space is a Frechet space, then X \times Y is countably tight.

Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.

____________________________________________________________________

Reference

  1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

____________________________________________________________________
\copyright \ 2015 \text{ by Dan Ma}

Tightness and free sequences

The previous post discusses several definitions of the tightness of a topological space. In this post, we discuss another way of characterizing tightness using the notion of free sequences.

____________________________________________________________________

The main theorem

Let X be a space. The tightness of X, denoted by t(X), is the least infinite cardinal number \tau such that for each A \subset X and for each x \in \overline{A}, there is a set B \subset A with cardinality \le \tau such that x \in \overline{B}. There are various different statements that can be used to define t(X) (discussed in this previous post).

A sequence \left\{x_\alpha: \alpha<\tau \right\} of points of a space X is called a free sequence if for each \alpha<\tau, \overline{\left\{x_\gamma: \gamma<\alpha \right\}} \cap \overline{\left\{x_\gamma: \gamma \ge \alpha \right\}}=\varnothing. When the free sequence is indexed by a cardinal number \tau, the free sequence is said to be of length \tau.

The cardinal function F(X) is the least infinite cardinal number \kappa such that if \left\{x_\alpha \in X: \alpha<\tau \right\} is a free sequence of length \tau, then \tau \le \kappa. Thus F(X) is the least upper bound of all the free sequences of points of the space X. The cardinal function F(X) is another way to characterize tightness of a space. We prove the following theorem.

Theorem 1
Let X be a compact space. Then t(X)=F(X).

All spaces considered in this post are regular spaces.

____________________________________________________________________

One direction of the proof

We first show that F(X) \le t(X). Suppose that t(X)=\kappa. We show that F(X) \le \kappa. Suppose not. Then there is a free sequence of points of X of length greater than \kappa, say A=\left\{x_\alpha: \alpha<\tau \right\} where \tau>\kappa. For each \beta<\tau, let L_\beta=\left\{x_\alpha: \alpha<\beta \right\} and R_\beta=\left\{x_\alpha: \beta \le \alpha<\tau \right\}.

let x \in \overline{A}. By t(X)=\kappa, there is some \beta_x \le \kappa <\tau such that x \in \overline{L_{\beta_x}}. Furthermore, x \notin \overline{R_{\beta_x}} since A is a free sequence. Then choose some open O_x such that x \in O_x and O_x \cap \overline{R_{\beta_x}}=\varnothing. Note that O_x contains at most \kappa many points of the free sequence A.

Let \mathcal{O}=\left\{O_x: x \in \overline{A} \right\} \cup \left\{X-\overline{A} \right\}. The collection \mathcal{O} is an open cover of the compact space X. Thus some finite \mathcal{V} \subset \mathcal{O} is a cover of X. Then all the open sets O_x \in \mathcal{V} are supposed to cover all the elements of the free sequence A=\left\{x_\alpha: \alpha<\tau \right\}. But each O_x is supposed to cover at most \kappa many elements of A and there are only finitely many O_x in \mathcal{V}, a contradiction. Thus F(X) \le t(X)=\kappa.

____________________________________________________________________

Some lemmas

To show t(X) \le F(X), we need some basic results technical lemmas. Throughout the discussion below, \kappa is an infinite cardinal number.

A subset M of the space X is a G_\kappa set if M is the intersection of \le \kappa many open subsets of X. Clearly, the intersection of \le \kappa many G_\kappa sets is a G_\kappa set.

Lemma 2
Let X be any space. Let M be a G_\kappa subset of X. Then for each x \in M, there is a G_\kappa subset Z of X such that Z is closed and x \in Z \subset M.

Proof of Lemma 2
Let M=\bigcap_{\alpha<\lambda} O_\alpha where each O_\alpha is open and \lambda is an infnite cardinal number \le \kappa. Note that for each \alpha, x \in O_\alpha. We assume that the space X is regular. We can choose open sets U_{\alpha,0}=O_\alpha,U_{\alpha,1},U_{\alpha,2},\cdots such that for each integer n=0,1,2,\cdots, x \in U_{\alpha,n} and \overline{U_{\alpha,n+1}} \subset U_{\alpha,n}. Consider the following set Z.

    \displaystyle Z=\bigcap_{\alpha<\lambda} \biggl(\bigcap \limits_{n=0}^\infty U_{\alpha,n}  \biggr)

The set Z is a G_\kappa subset of X and x \in Z \subset M. To see that Z is closed, note that Z can be rearranged as follows:

    \displaystyle Z=\bigcap_{\alpha<\lambda} \biggl(\bigcap \limits_{n=0}^\infty U_{\alpha,n}  \biggr)=\bigcap_{\alpha<\lambda} \biggl(\bigcap \limits_{n=0}^\infty \overline{U_{\alpha,n+1}}  \biggr)

The right hand side is the intersection of closed sets, showing that Z a closed set. This concludes the proof of Lemma 2.

_________________________________

For any set A \subset X, define \text{CL}_\kappa(A) as follows:

    \text{CL}_\kappa(A)=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \kappa \right\}

In general \text{CL}_\kappa(A) is the part of \overline{A} that can be “reached” by the closure of a “small enough” subset of A. Note that t(X)=\kappa if and only if for each A \subset X, \text{CL}_\kappa(A)=\overline{A}.

For any set W \subset X, define the set W^* as follows:

    W^*=\left\{x \in X: \forall \ G_\kappa \text{ subset } M \text{ of } X \text{ with } x \in M, M \cap W \ne \varnothing  \right\}

A point y \in X is an accumulation point of the set W if O \cap W \ne \varnothing for all open set O with x \in O. As a contrast, \overline{W} is the set of all accumulation points of W. Any point x \in W^* is like an accumulation point of W except that G_\kappa sets are used instead of open sets. It is clear that W \subset W^*.

Lemma 3
Let X be a compact space as before. Let \kappa be any infinite cardinal number. Let A \subset X. Then \overline{A}=\text{CL}_\kappa(A)^*.

Proof of Lemma 3
It is clear that \text{CL}_\kappa(A)^* \subset \overline{A}. We only need to show \overline{A} \subset \text{CL}_\kappa(A)^*. Suppose that we have x \in \overline{A} and x \notin \text{CL}_\kappa(A)^*. This means there exists a G_\kappa subset M of X such that x \in M and M \cap \text{CL}_\kappa(A)=\varnothing. By Lemma 2, there is a closed G_\kappa subset Z of X such that x \in Z \subset M.

Since Z is a closed subset of a compact space and is a G_\kappa subset, there is a base \mathcal{U} for the set Z such that \mathcal{U} has cardinality \le \kappa (see the exercise below). For each U \in \mathcal{U}, U \cap A \ne \varnothing since U is an open set containing x. Choose t_U \in U \cap A. Let B=\left\{t_U: U \in \mathcal{U} \right\}. Note that B \subset A and \lvert B \lvert \le \kappa. Thus \overline{B} \subset \text{CL}_\kappa(A). On the other hand, Z \cap \text{CL}_\kappa(A)=\varnothing. Thus Z \cap \overline{B}=\varnothing.

Let’s look at what we have. The sets Z and \overline{B} are disjoint closed sets. We also know that \mathcal{U} is a base for Z. There exists U \in \mathcal{U} such that Z \subset U and U \cap \overline{B}=\varnothing. But t_U \in B and t_U \in U, a contradiction. Thus \overline{A} \subset \text{CL}_\kappa(A)^*. This concludes the proof of Lemma 3.

_________________________________

Let \kappa is an infinite cardinal number as before. Recall the concept of a \kappa-closed set from this previous post. A set A \subset X is a \kappa-closed set if for each B \subset A with \lvert B \lvert \le \kappa, we have \overline{B} \subset A. Theorem 2 in the previous post states that

    t(X)=\kappa if and only if every \kappa-closed set is closed.

This means that

    if t(X) > \kappa, then there is some \kappa-closed set that is not closed.

The above observation will be used in the proof below. Another observation that if A \subset X is a \kappa-closed set, we have A=\text{CL}_\kappa(A)=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \kappa \right\}.

____________________________________________________________________

The other direction of the proof

We now show that t(X) \le F(X). First we show the following:

    If t(X) > \kappa, then there exists a free sequence of length \kappa^+ where \kappa^+ is the next cardinal number larger than \kappa.

Suppose t(X) > \kappa. According to the observation on \kappa-closed set indicated above, there exists a set A \subset X such that A is a \kappa-closed set but A is not closed. By another observation on \kappa-closed set indicated above, we have A=\text{CL}_\kappa(A). By Lemma 3, \overline{A}=\text{CL}_\kappa(A)^*=A^*.

Since A is not closed, choose x \in \overline{A}-A. Then x \in A^*. This means the following:

    For each G_\kappa-subset M of X with x \in M, M \cap A \ne \varnothing \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

The fact indicated in (1) will make the construction of the free sequence feasible. To start the construction of the free sequence, choose x_0 \in A. Let F_0=X. Suppose that for \alpha<\kappa^+, we have obtained \left\{x_\gamma \in A: \gamma<\alpha \right\} and \left\{F_\gamma: \gamma<\alpha\right\} with the following properties:

  1. For each \gamma < \alpha, F_\gamma is a closed G_\kappa subset of X with x \in F_\gamma,
  2. For each \gamma < \alpha, x_\gamma \in F_\gamma,
  3. For all \gamma< \alpha, \overline{\left\{x_\theta: \theta<\gamma \right\}} \cap F_\gamma=\varnothing,
  4. For all \gamma < \delta < \alpha, F_\delta \subset F_\gamma.

We now proceed to choose define F_\alpha and choose x_\alpha \in A. Consider the set D=\left\{x_\gamma: \gamma<\alpha \right\}. Note that \lvert D \lvert \le \kappa and D \subset A. Thus \overline{D} \subset \text{CL}_\kappa(A)=A. Since x \notin A, x \notin \overline{D} and x \in X-\overline{D}. By Lemma 2, there exists some closed G_\kappa-subset M of X such that x \in M and M \cap \overline{D}=\varnothing. Let F_\alpha=M \cap (\cap \left\{F_\gamma: \gamma<\alpha \right\}), which is still a closed and G_\kappa-subset of the space X. By the observation (1), F_\alpha \cap A \ne \varnothing. Then choose x_\alpha \in F_\alpha \cap A.

The construction we describe can be done for any \alpha as long as \alpha \le \kappa. Thus the construction yields the sequence W=\left\{x_\alpha: \alpha < \kappa^+ \right\}. We now show that W is a free sequence. Let \alpha<\kappa^+. From the construction step for \alpha, we see that F_\alpha \cap \overline{\left\{x_\gamma: \gamma<\alpha \right\}}=\varnothing. From how F_\alpha is defined in step \alpha, we see that F_\rho \subset F_\alpha for any \alpha < \rho < \kappa^+. This means that \left\{x_\rho: \alpha \le \rho < \kappa^+\right\} \subset F_\alpha. Since F_\alpha is closed, \overline{\left\{x_\rho: \alpha \le \rho < \kappa^+\right\}} \subset F_\alpha. This shows that \overline{\left\{x_\gamma: \gamma<\alpha \right\}} \cap \overline{\left\{x_\rho: \alpha \le \rho < \kappa^+\right\}}=\varnothing. We have shown that W is a free sequence of points of X.

As a result of assuming t(X) > \kappa, a free sequence of length \kappa^+ is obtained. Thus if t(X) > \kappa, then F(X) > \kappa. Then it must be the case that t(X) \le F(X). This concludes the proof of Theorem 1. \blacksquare

____________________________________________________________________

Remarks

The easier direction of Theorem 1, the direction for showing F(X) \le t(X), does not require that the space X is compact. The proof will work as long as the Lindelof degree of X \le t(X).

The harder direction, the direction for showing t(X) \le F(X), does need the fact the compactness of the space X (see the exercise below). Proving t(X) = F(X) for a wider class of spaces than the compact spaces will probably require a different proof than the one given here. One generalization is found in [1]. It obtained theorem in the form of t(X) \le F(X) for pseudo-radial regular spaces as well as other theorems with various sufficient conditions that lead to t(X) = F(X).

Theorem 1 has been applied in this blog post to characterize the normality of X \times \omega_1 for any compact space X.

____________________________________________________________________

Exercise

Let X be a compact space. Let C be a closed subset of X such that C is the intersection of \le \kappa many open subsets of X. Show that there exists a base \mathcal{B} for the closed set C such that \lvert \mathcal{B} \lvert \le \kappa. To say that \mathcal{B} is a base for C, we mean that \mathcal{B} is a collection of open subsets of X such that each element of \mathcal{B} contains C and that if C \subset W with W open, then C \subset O \subset W for some O \in \mathcal{B}.

____________________________________________________________________

Reference

  1. Bella A., Free sequences in pseudo-radial spaces, Commentationes Mathematicae Universitatis Carolinae, Vol 27, No 1 (1986), 163-170

____________________________________________________________________
\copyright \ 2015 \text{ by Dan Ma}