# 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

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

# Several ways to define countably tight spaces

This post is an introduction to countable tight and countably generated spaces. A space being a countably tight space is a convergence property. The article [1] lists out 8 convergence properties. The common ones on that list include Frechet space, sequential space, k-space and countably tight space, all of which are weaker than the property of being a first countable space. In this post we discuss several ways to define countably tight spaces and to discuss its generalizations.

____________________________________________________________________

Several definitions

A space $X$ is countably tight (or has countable tightness) if for each $A \subset X$ and for each $x \in \overline{A}$, there is a countable $B \subset A$ such that $x \in \overline{B}$. According to this Wikipedia entry, a space being a countably generated space is the property that its topology is generated by countable sets and is equivalent to the property of being countably tight. The equivalence of the two definitions is not immediately clear. In this post, we examine these definitions more closely. Theorem 1 below has three statements that are equivalent. Any one of the three statements can be the definition of countably tight or countably generated.

Theorem 1
Let $X$ be a space. The following statements are equivalent.

1. For each $A \subset X$, the set equality (a) holds.$\text{ }$
• $\displaystyle \overline{A}=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \omega \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a)$

2. For each $A \subset X$, if condition (b) holds,
For all countable $C \subset X$, $C \cap A$ is closed in $C \ \ \ \ \ \ \ \ (b)$

then $A$ is closed.

3. For each $A \subset X$, if condition (c) holds,
For all countable $B \subset A$, $\overline{B} \subset A \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (c)$

then $A$ is closed.

Statement 1 is the definition of a countably tight space. The set inclusion $\supset$ in (a) is always true. We only need to be concerned with $\subset$, which is the definition of countable tightness given earlier.

Statement 2 is the definition of a countably generated space according to this Wikipedia entry. This definition is in the same vein as that of k-space (or compactly generated space). Note that a space $X$ is a k-space if Statement 2 holds when “countable” is replaced with “compact”.

Statement 3 is in the same vein as that of a sequential space. Recall that a space $X$ is a sequential space if $A \subset X$ is a sequentially closed set then $A$ is closed. The set $A$ is a sequentially closed set if the sequence $x_n \in A$ converges to $x \in X$, then $x \in A$ (in other words, for any sequence of points of $A$ that converges, the limit must be in $A$). If the sequential limit in the definition of sequential space is relaxed to be just topological limit (i.e. accumulation point), then the resulting definition is Statement 3. Thus Statement 3 says that for any countable subset $B$ of $A$, any limit point (i.e. accumulation point) of $B$ must be in $A$. Thus any sequential space is countably tight. In a sequential space, the closed sets are generated by taking sequential limit. In a space defined by Statement 3, the closed sets are generated by taking closures of countable sets.

All three statements are based on the countable cardinality and have obvious generalizations by going up in cardinality. For any set $A \subset X$ that satisfies condition (c) in Statement 3 is said to be an $\omega$-closed set. Thus for any cardinal number $\tau$, the set $A \subset X$ is a $\tau$-closed set if for any $B \subset A$ with $\lvert B \lvert \le \tau$, $\overline{B} \subset A$. Condition (c) in Statement 3 can then be generalized to say that if $A \subset X$ is a $\tau$-closed set, then $A$ is closed.

The proof of Theorem 1 is handled in the next section where we look at the generalizations of all three statements and prove their equivalence.

____________________________________________________________________

Generalizations

The definition in Statement 1 in Theorem 1 above can be generalized as a cardinal function called tightness. Let $X$ be a space. By $t(X)$ we mean the least infinite cardinal number $\tau$ such that the following holds:

For all $A \subset X$, and for each $x \in \overline{A}$, there exists $B \subset A$ with $\lvert B \lvert \le \tau$ such that $x \in \overline{B}$.

When $t(X)=\omega$, the space $X$ is countably tight (or has countable tightness). In keeping with the set equality (a) above, the tightness $t(X)$ can also be defined as the least infinite cardinal $\tau$ such that for any $A \subset X$, the following set equality holds:

$\displaystyle \overline{A}=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \tau \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\alpha)$

Let $\tau$ be an infinite cardinal number. To generalize Statement 2, we say that a space $X$ is $\tau$-generated if the following holds:

For each $A \subset X$, if the following condition holds:

For all $C \subset X$ with $\lvert C \lvert \le \tau$, the set $C \cap A$ is closed in $C \ \ \ \ \ \ \ \ \ \ \ (\beta)$

then $A$ is closed.

To generalize Statement 3, we say that a set $A \subset X$ is $\tau$-closed if for any $B \subset A$ with $\lvert B \lvert \le \tau$, $\overline{B} \subset A$. A generalization of Statement 3 is that

For any $A \subset X$, if $A \subset X$ is a $\tau$-closed set, then $A$ is closed $.\ \ \ \ \ \ \ \ \ \ \ (\chi)$

Theorem 2
Let $X$ be a space. Let $\tau$ be an infinite cardinal. The following statements are equivalent.

1. $t(X)=\tau$.
2. The space $X$ is $\tau$-generated.
3. For each $A \subset X$, if $A \subset X$ is a $\tau$-closed set, then $A$ is closed.

Proof of Theorem 2
$1 \rightarrow 2$
Suppose that (2) does not hold. Let $A \subset X$ be such that the set $A$ satisfies condition $(\beta)$ and $A$ is not closed. Let $x \in \overline{A}-A$. By (1), the point $x$ belongs to the right hand side of the set equality $(\alpha)$. Choose $B \subset A$ with $\lvert B \lvert \le \tau$ such that $x \in \overline{B}$. Let $C=B \cup \left\{x \right\}$. By condition $(\beta)$, $C \cap A=B$ is closed in $C$. This would mean that $x \in B$ and hence $x \in A$, a contradiction. Thus if (1) holds, (2) must holds.

$2 \rightarrow 3$
Suppose (3) does not hold. Let $A \subset X$ be a $\tau$-closed set that is not a closed set in $X$. Since (2) holds and $A$ is not closed, condition $(\beta)$ must not hold. Choose $C \subset X$ with $\lvert C \lvert \le \tau$ such that $B=C \cap A$ is not closed in $C$. Choose $x \in C$ that is in the closure of $C \cap A$ but is not in $C \cap A$. Since $A$ is $\tau$-closed, $\overline{B}=\overline{C \cap A} \subset A$, which implies that $x \in A$, a contradiction. Thus if (2) holds, (3) must hold.

$3 \rightarrow 1$
Suppose (1) does not hold. Let $A \subset X$ be such that the set equality $(\alpha)$ does not hold. Let $x \in \overline{A}$ be such that $x$ does not belong to the right hand side of $(\alpha)$. Let $A_0=\overline{A}-\left\{x \right\}$. Note that the set $A_0$ is $\tau$-closed. By (3), $A_0$ is closed. Furthermore $x \in \overline{A_0}$, leading to $x \in A_0=\overline{A}-\left\{x \right\}$, a contradiction. So if (3) holds, (1) must hold. $\blacksquare$

Theorem 1 obviously follows from Theorem 2 by letting $\tau=\omega$. There is another way to characterize the notion of tightness using the concept of free sequence. See the next post.

____________________________________________________________________

Examples

Several elementary convergence properties have been discussed in a series of blog posts (the first post and links to the other are found in the first one). We have the following implications and none is reversible.

First countable $\Longrightarrow$ Frechet $\Longrightarrow$ Sequential $\Longrightarrow$ k-space

Where does countable tightness place in the above implications? We discuss above that

Sequential $\Longrightarrow$ countably tight.

How do countably tight space and k-space compare? It turns out that none implies the other. We present some supporting examples.

Example 1
The Arens’ space is a canonical example of a sequential space that is not a Frechet space. A subspace of the Arens’ space is countably tight and not sequential. The same subspace is also not a k-space. There are several ways to represent the Arens’ space, we present the version found here.

Let $\mathbb{N}$ be the set of all positive integers. Define the following:

$\displaystyle V_{i,j}=\left\{\biggl(\frac{1}{i},\frac{1}{k} \biggr): k \ge j \right\}$ for all $i,j \in \mathbb{N}$

$V=\bigcup_{i \in \mathbb{N}} V_{i,j}$

$\displaystyle H=\left\{\biggl(\frac{1}{i},0 \biggr): i \in \mathbb{N} \right\}$

$V_i=V_{i,1} \cup \left\{ x \right\}$ for all $i \in \mathbb{N}$

Let $Y=\left\{(0,0) \right\} \cup H \cup V$. Each point in $V$ is an isolated point. Open neighborhoods at $(\frac{1}{i},0) \in H$ are of the form:

$\displaystyle \left\{\biggl(\frac{1}{i},0 \biggr) \right\} \cup V_{i,j}$ for some $j \in \mathbb{N}$

The open neighborhoods at $(0,0)$ are obtained by removing finitely many $V_i$ from $Y$ and by removing finitely many isolated points in the $V_i$ that remain. The open neighborhoods just defined form a base for a topology on the set $Y$, i.e. by taking unions of these open neighborhoods, we obtain all the open sets for this space. The space $Y$ can also be viewed as a quotient space (discussed here).

The space $Y$ is a sequential space that is not Frechet. The subspace $Z=\left\{(0,0) \right\} \cup V$ is not sequential. Since $Y$ is a countable space, the space $Z$ is by default a countably tight space. The space $Z$ is also not an k-space. These facts are left as exercises below.

Example 2
Consider the product space $X=\left\{0,1 \right\}^{\omega_1}$. The space $X$ is compact since it is a product of compact spaces. Any compact space is a k-space. Thus $X$ is a k-space (or compactly generated space). On the other hand, $X$ is not countably tight. Thus the notion of k-space and the notion of countably tight space do not relate.

____________________________________________________________________

Remarks

There is another way to characterize the notion of tightness using the concept of free sequence. See the next post.

The notion of tightness had been discussed in previous posts. One post shows that the function space $C_p(X)$ is countably tight when $X$ is compact (see here). Another post characterizes normality of $X \times \omega_1$ when $X$ is compact (see here)

____________________________________________________________________

Exercises

Exercise 1
This is to verify Example 1. Verify that

• The space $Y$ is a sequential space that is not Frechet.
• $Z=\left\{(0,0) \right\} \cup V$ is not sequential.
• The space $Z$ is not an k-space.

Exercise 2
Verify that any compact space is a k-space. Show that the space $X$ in Example 2 is not countably tight.

____________________________________________________________________

Reference

1. Gerlits J., Nagy Z., Products of convergence properties, Commentationes Mathematicae Universitatis Carolinae, Vol 23, No 4 (1982), 747–756

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

# Sequential spaces, V

In the previous post Sequential spaces, IV, we show that the uncountable product of sequential spaces is not sequential (e.g. the product $2^{\omega_1}$ is not sequential). What is more remarkable is that the product of two sequential spaces needs not be sequential. We present an example of a first countable space and a Frechet space whose product is not a k-space (thus not sequential). For the previous discussion on this blog on sequential spaces and k-spaces, see the links at the end of this post.

Let $\mathbb{R}$ be the real line and let $\mathbb{N}$ be the set of all positive integers. Let $X$ be the space $\mathbb{R}-\left\{1,\frac{1}{2},\frac{1}{3},\cdots\right\}$ with the topology inherited from the usual topology on the real line. Let $Y=\mathbb{R}$ with the positive integers identified as one point (call this point $p$). We claim that $X \times Y$ is not a k-space and thus not a sequential space. To this end, we define a non-closed $A \subset X \times Y$ such that $K \cap A$ is closed in $K$ for all compact $K \subset X \times Y$.

Let $A=\bigcup \limits_{i=1}^\infty A_i$ where for each $i \in \mathbb{N}$, the set $A_i$ is defined by the following:

$\displaystyle A_i =\left\{\biggl(\frac{1}{i}+\frac{a_i}{j},i+\frac{0.5}{j} \biggr) \in X \times Y:j \in \mathbb{N}\right\}$

where $\displaystyle a_i=\biggl(\frac{1}{i}-\frac{1}{i+1} \biggr) 10^{-i}$.

Clearly $A$ is not closed as $(0,p) \in \overline{A}-A$. In fact in the product space $X \times Y$, the point $(0,p)$ is the only limit point of the set $A$. Another observation is that for each $n \in \mathbb{N}$, $(0,p)$ is not a limit point of $\bigcup \limits_{i=1}^n A_i$. Furthermore, if $z_i \in A_i$ for each $i \in S$ where $S$ is an infinite subset of $\mathbb{N}$, then $(0,p)$ is not a limit point of $\left\{z_i:i \in S\right\}$. It follows that no infinite subset of $A$ is compact. Consequently, $K \cap A$ is finite for each compact $K \subset X \times Y$. Thus $X \times Y$ is not a k-space. To see that $X \times Y$ is not sequential directly, observe that $A$ is sequentially closed.

Previous posts on sequential spaces and k-spaces