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



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.



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}.



  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}


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