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].

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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

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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].

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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.

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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.

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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].

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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.

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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.

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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.

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\copyright \ 2015 \text{ by Dan Ma}

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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.

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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

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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

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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.

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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”.

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Reference

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

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\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.

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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.

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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.

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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.

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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.

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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\}.

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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

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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.

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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}.

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Reference

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

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\copyright \ 2015 \text{ by Dan Ma}