# 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. $t(X) \le \tau$.
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 Arensâ€™ 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

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