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}

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