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

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