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}

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.

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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.

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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.

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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.

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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)

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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.

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Reference

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

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\copyright \ 2015 \text{ by Dan Ma}

Cp(X) is countably tight when X is compact

Let X be a completely regular space (also called Tychonoff space). If X is a compact space, what can we say about the function space C_p(X), the space of all continuous real-valued functions with the pointwise convergence topology? When X is an uncountable space, C_p(X) is not first countable at every point. This follows from the fact that C_p(X) is a dense subspace of the product space \mathbb{R}^X and that no dense subspace of \mathbb{R}^X can be first countable when X is uncountable. However, when X is compact, C_p(X) does have a convergence property, namely C_p(X) is countably tight.

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Tightness

Let X be a completely regular space. The tightness of X, denoted by t(X), is the least infinite cardinal \kappa such that for any A \subset X and for any x \in X with x \in \overline{A}, there exists B \subset A for which \lvert B \lvert \le \kappa and x \in \overline{B}. When t(X)=\omega, we say that Y has countable tightness or is countably tight. When t(X)>\omega, we say that X has uncountable tightness or is uncountably tight. Clearly any first countable space is countably tight. There are other convergence properties in between first countability and countable tightness, e.g., the Frechet-Urysohn property. The notion of countable tightness and tightness in general is discussed in further details here.

The fact that C_p(X) is countably tight for any compact X follows from the following theorem.

Theorem 1
Let X be a completely regular space. Then the function space C_p(X) is countably tight if and only if X^n is Lindelof for each n=1,2,3,\cdots.

Theorem 1 is the countable case of Theorem I.4.1 on page 33 of [1]. We prove one direction of Theorem 1, the direction that will give us the desired result for C_p(X) where X is compact.

Proof of Theorem 1
The direction \Longleftarrow
Suppose that X^n is Lindelof for each positive integer. Let f \in C_p(X) and f \in \overline{H} where H \subset C_p(X). For each positive integer n, we define an open cover \mathcal{U}_n of X^n.

Let n be a positive integer. Let t=(x_1,\cdots,x_n) \in X^n. Since f \in \overline{H}, there is an h_t \in H such that \lvert h_t(x_j)-f(x_j) \lvert <\frac{1}{n} for all j=1,\cdots,n. Because both h_t and f are continuous, for each j=1,\cdots,n, there is an open set W(x_j) \subset X with x_j \in W(x_j) such that \lvert h_t(y)-f(y) \lvert < \frac{1}{n} for all y \in W(x_j). Let the open set U_t be defined by U_t=W(x_1) \times W(x_2) \times \cdots \times W(x_n). Let \mathcal{U}_n=\left\{U_t: t=(x_1,\cdots,x_n) \in X^n \right\}.

For each n, choose \mathcal{V}_n \subset \mathcal{U}_n be countable such that \mathcal{V}_n is a cover of X^n. Let K_n=\left\{h_t: t \in X^n \text{ such that } U_t \in \mathcal{V}_n \right\}. Let K=\bigcup_{n=1}^\infty K_n. Note that K is countable and K \subset H.

We now show that f \in \overline{K}. Choose an arbitrary positive integer n. Choose arbitrary points y_1,y_2,\cdots,y_n \in X. Consider the open set U defined by

    U=\left\{g \in C_p(X): \forall \ j=1,\cdots,n, \lvert g(y_j)-f(y_j) \lvert <\frac{1}{n} \right\}.

We wish to show that U \cap K \ne \varnothing. Choose U_t \in \mathcal{V}_n such that (y_1,\cdots,y_n) \in U_t where t=(x_1,\cdots,x_n) \in X^n. Consider the function h_t that goes with t. It is clear from the way h_t is chosen that \lvert h_t(y_j)-f(x_j) \lvert<\frac{1}{n} for all j=1,\cdots,n. Thus h_t \in K_n \cap U, leading to the conclusion that f \in \overline{K}. The proof that C_p(X) is countably tight is completed.

The direction \Longrightarrow
See Theorem I.4.1 of [1].

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Remarks

As shown above, countably tightness is one convergence property of C_p(X) that is guaranteed when X is compact. In general, it is difficult for C_p(X) to have stronger convergence properties such as the Frechet-Urysohn property. It is well known C_p(\omega_1+1) is Frechet-Urysohn. According to Theorem II.1.2 in [1], for any compact space X, C_p(X) is a Frechet-Urysohn space if and only if the compact space X is a scattered space.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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\copyright \ 2014 - 2015 \text{ by Dan Ma}