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) \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.

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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}

The product of uncountably many factors is never hereditarily normal

The space Y=\prod_{\alpha<\omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1} is the product of \omega_1 many copies of the two-element set \left\{0,1 \right\} where \omega_1 is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space Y hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space \prod X_\alpha cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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The connection with a theorem of Katetov

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space X, knowing whether X, X^2 and X^3 are hereditarily normal can tell us whether X is metrizable. If all three are hereditarily normal, then X is metrizable. If one of the three self products is not hereditarily normal, then X is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space Y=\left\{0,1 \right\}^{\omega_1} is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space X, determine where the hereditary normality fails. Does it fail in X, X^2 or X^3? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in X or X^2. The space Y=\left\{0,1 \right\}^{\omega_1} can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space X such that the hereditary normality holds in X^2 and fails in X^3? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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Exercises

Problem 1
Let X be a compact space. Show that X is normal.

Problem 2
For each \alpha<\omega_1, let A_\alpha be a set with cardinality \le \omega_1. Show that \lvert \bigcup_{\alpha<\omega_1} A_\alpha \lvert \le \omega_1.

Problem 2 holds for any infinite cardinal, not just \omega_1. One reference for Problem 2 is Lemma 10.21 on page 30 of Set Theorey, An Introduction to Independence Proofs by Kenneth Kunen.

Problem 3
For each \alpha<\omega_1, let X_\alpha be a space with at least two points. Show that for every point p \in \prod_{\alpha<\omega_1} X_\alpha, there does not exist a countable base at the point p. In other words, the product space \prod_{\alpha<\omega_1} X_\alpha is not first countable at every point. It follows that product space \prod_{\alpha<\omega_1} X_\alpha is not metrizable.

Problem 4
In any space, a G_\delta-set is a set that is the intersection of countably many open sets. When a singleton set \left\{ x \right\} is a G_\delta-set, we say the point x is a G_\delta-point. For each \alpha<\omega_1, let X_\alpha be a space with at least two points. Show that every point p in the product space \prod_{\alpha<\omega_1} X_\alpha is not a G_\delta-point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form \prod_{\alpha<\omega_1} O_\alpha and that it has only finitely many coordinates at which O_\alpha \ne X_\alpha.

Problem 5
For each \alpha<\omega_1, let X_\alpha=\left\{0,1,2,\cdots \right\} be the set of non-negative integers with the discrete topology. Show that the product space \prod_{\alpha<\omega_1} X_\alpha is not normal.

See here for a discussion of Problem 5.

Problem 6
Let \displaystyle Y=\left\{0,1 \right\}^{\omega_1}. Show that Y has a countably infinite subspace

    W=\left\{y_0,y_1,y_2,y_3\cdots \right\}

such that W is relatively discrete. In other words, W is discrete in the subspace topology of W. However W is not discrete in the product space Y since Y is compact.

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Proof #1

Let \displaystyle Y=\left\{0,1 \right\}^{\omega_1}. We show that Y is not hereditarily normal.

Note that the product space \displaystyle Y=\left\{0,1 \right\}^{\omega_1} can be written as the product of \omega_1 many copies of itself:

    \displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

The fact (1) follows from the fact that the union of \omega_1 many pairwise disjoint sets, each of which has cardinality \omega_1, has cardinality \omega_1 (see Problem 2). The space \left\{0,1 \right\}^{\omega_1} has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to \omega=\left\{0,1,2,\cdots \right\} where \omega has the discrete topology. Thus the following is homeomorphic to a subspace of \displaystyle Y=\left\{0,1 \right\}^{\omega_1}.

    \displaystyle \omega^{\omega_1} = \omega \times \omega \times \omega \times \cdots \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

By Problem 5, the space \omega^{\omega_1} is not normal. Hence the compact space \displaystyle Y=\left\{0,1 \right\}^{\omega_1} contains the non-normal space \omega^{\omega_1} and is thus not hereditarily normal. \blacksquare

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Proof #2

Let \displaystyle Y=\left\{0,1 \right\}^{\omega_1}. We show that Y is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

Theorem 1
If X_1 \times X_2 is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

  • The factor X_1 is perfectly normal.
  • Every countable and infinite subset of the factor X_2 is closed.

First, Y can be written as the product of two copies of itself:

    \displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

This is because the union of two disjoints sets, each of which has cardinality \omega_1, has carinality \omega_1. Note that the countably infinite subset W from Problem 6 is not a closed subset of Y. If it were, the compact space Y would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If Y \cong Y \times Y were to be hereditarily normal, then the first condition must be satisfied, i.e. Y is perfectly normal (meaning that Y is normal and that every closed subset of it is a G_\delta-set). However, Problem 4 indicates that no point in Y can be a G_\delta point. Therefore Y cannot be hereditarily normal. \blacksquare

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Corollary

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space \displaystyle Y=\left\{0,1 \right\}^{\omega_1}. Thus such a product space can never be hereditarily normal. We state this more formally below.

Theorem 2
Let \kappa be any uncountable cardinal. For each \alpha<\kappa, let X_\alpha be a space with at least two points. Then \prod_{\alpha<\kappa} X_\alpha is not hereditarily normal.

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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}

Every Corson compact space has a dense first countable subspace

In any topological space X, a point x \in X is a G_\delta point if the one-point set \left\{ x \right\} is the intersection of countably many open subsets of X. It is well known that any compact Hausdorff space is first countable at every G_\delta point, i.e., if a point of a compact space is a G_\delta point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space [0,1]^{\omega_1} can be a G_\delta point. In this post, we show that any Corson compact space has a dense set of G_\delta point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.

The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.

For any infinite cardinal number \kappa, the \Sigma-product of \kappa many copies of \mathbb{R} is the following subspace of \mathbb{R}^\kappa:

    \Sigma(\kappa)=\left\{x \in \mathbb{R}^\kappa: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \kappa \right\}

A compact space is said to be a Corson compact space if it can be embedded in \Sigma(\kappa) for some infinite cardinal \kappa.

For each x \in \Sigma(\kappa), let S(x) denote the support of the point x, i.e., S(x) is the set of all \alpha<\kappa such that x_\alpha \ne 0.

Proposition 1
Let Y be a Corson compact space. Then Y has a G_\delta point.

Proof of Proposition 1
If Y is finite, then every point is isolated and is thus a G_\delta point. Assume Y is infinite. Let \kappa be an infinite cardinal number such that Y \subset \Sigma(\kappa). For f,g \in Y, define f \le g if the following holds:

    \forall \ \alpha \in S(f), f(\alpha)=g(\alpha)

It is relatively straightforward to verify that the following three properties are satisfied:

  • f \le f for all f \in Y. (reflexivity)
  • For all f,g \in Y, if f \le g and g \le f, then f=g. (antisymmetry)
  • For all f,g,h \in Y, if f \le g and g \le h, then f \le h. (transitivity)

Thus \le as defined here is a partial order on the compact space Y. Let C \subset Y such that C is a chain with respect to \le, i.e., for all f,g \in C, f \le g or g \le f. We show that C has an upper bound (in Y) with respect to the partial order \le. We need this for an argument using Zorn’s lemma.

Let W=\bigcup_{f \in C} S(f). For each \alpha \in W, choose some f \in C such that \alpha \in S(f) and define u_\alpha=f_\alpha. For all \alpha \notin W, define u_\alpha=0. Because C is a chain, the point u is well-defined. It is also clear that f \le u for all f \in C. If u \in Y, then u is a desired upper bound of C. So assume u \notin Y. It follows that u is a limit point of C, i.e., every open set containing u contains a point of C different from u. Hence u is a limit point of Y too. Since Y is compact, u \in Y, a contradiction. Thus it must be that u \in Y. Thus every chain in the partially ordered set (Y,\le) has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order \le, i.e., there exists t \in Y such that f \le t for all f \in Y.

We now show that t is a G_\delta point in Y. Let S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}. For each p \in \mathbb{R} and for each positive integer n, let B_{p,n} be the open interval B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n}). For each positive integer n, define the open set O_n as follows:

    O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y

Note that t \in \bigcap_{n=1}^\infty O_n. Because t is a maximal element, note that if g \in Y such that g_\alpha=t_\alpha for all \alpha \in S(t), then it must be the case that g=t. Thus if g \in \bigcap_{n=1}^\infty O_n, then g_\alpha=t_\alpha for all \alpha \in S(t). We have \left\{t \right\}= \bigcap_{n=1}^\infty O_n. \blacksquare

Lemma 2
Let Y be a compact space such that for every non-empty compact subspace K of Y, there exists a G_\delta point in K. Then every non-empty open subset of Y contains a G_\delta point.

Proof of Lemma 2
Let U_1 be a non-empty open subset of the compact space Y. If there exists y \in U_1 such that \left\{y \right\} is open in Y, then y is a G_\delta point. So assume that every point of U_1 is a non-isolated point of Y. By regularity, choose an open subset U_2 of Y such that \overline{U_2} \subset U_1. Continue in the same manner and obtain a decreasing sequence U_1,U_2,U_3,\cdots of open subsets of Y such that \overline{U_{n+1}} \subset U_n for each positive integer n. Then K=\bigcap_{n=1}^\infty \overline{U_n} is a non-empty closed subset of Y and thus compact. By assumption, K has a G_\delta point, say p \in K.

Then \left\{p \right\}=\bigcap_{n=1}^\infty W_n where each W_n is open in K. For each n, let V_n be open in Y such that W_n=V_n \cap K. For each n, let V_n^*=V_n \cap U_n, which is open in Y. Then \left\{p \right\}=\bigcap_{n=1}^\infty V_n^*. This means that p is a G_\delta point in the compact space Y. Note that p \in U_1, the open set we start with. This completes the proof that every non-empty open subset of Y contains a G_\delta point. \blacksquare

Proposition 3
Let Y be a Corson compact space. Then Y has a dense set of G_\delta points.

Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of Y is also Corson compact. By Proposition 1, every compact subspace of Y has a G_\delta point. By Lemma 2, Y has a dense set of G_\delta points. \blacksquare

Corollary 4
Every Corson compact space has a dense first countable subspace.

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Blog posts on Corson compact spaces

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Reference

  1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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\copyright \ 2014 \text{ by Dan Ma}

Sigma-products of first countable spaces

A product space is never first countable if there are uncountably many factors. For example, \prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1} is not first countable. In fact any dense subspace of \mathbb{R}^{\omega_1} is not first countable. In particular, the subspace of \mathbb{R}^{\omega_1} consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the \Sigma-product of \omega_1 many copies of the real line \mathbb{R} and is denoted by \Sigma_{\alpha<\omega_1} \mathbb{R}. However, this \Sigma-product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the \Sigma-product of first countable spaces is a Frechet space.

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Fix a point a \in X. Consider the following subspace of X:

    \Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}

The above subspace of X is called the \Sigma-product of the spaces \left\{X_\alpha: \alpha \in A \right\} about the base point a. When the base point is understood, we simply say the \Sigma-product of the spaces \left\{X_\alpha: \alpha \in A \right\} and use the notation \Sigma_{\alpha \in A} X_\alpha to denote the space.

For each y \in \Sigma_{\alpha \in A} X_\alpha, define S(y) to be the set of all \alpha \in A such that y_\alpha \ne a_\alpha, i.e., the support of the point y. Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set O=\prod_{\alpha \in A} O_\alpha where each O_\alpha is an open subset of X_\alpha. The support of O, denoted by supp(O) is the finite set of all \alpha \in A such that O_\alpha \ne X_\alpha.

A space Y is said to be first countable if there exists a countable local base at each point in Y. A space Y is said to be a Frechet space if for each y \in Y and for each M \subset Y, if y \in \overline{M}, then there exists a sequence \left\{y_n: n=1,2,3,\cdots \right\} of points of M such that the sequence converges to y. Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

    Suppose each factor X_\alpha is a first countable space. Then the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is a Frechet space.

Proof of Theorem 1
Let \Sigma=\Sigma_{\alpha \in A} X_\alpha. Let M \subset \Sigma and let x \in \overline{M}. We proceed to define a sequence of points t_n \in M such that the sequence t_n converges to x. For each \alpha \in A, choose a countable local base \left\{B_{\alpha,j}: j=1,2,3,\cdots \right\} at the point x_\alpha \in X_\alpha. Assume that B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots. Then enumerate the countable set S(x) by S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}. Let C_1=\left\{\beta_{1,1} \right\}. The following set O_1 is an open subset of \Sigma.

    O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma

Note that O_1 is an open set containing x. Choose t_2 \in O_1 \cap M. Enumerate the support S(t_2) by S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}. Form the finite set C_2 by picking the first two points of S(x) and the first two points of S(t_2), i.e., C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}. Then form the following open subset of \Sigma.

    O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma

Choose t_3 \in O_2 \cap M. Enumerate the support S(t_3) by S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}. Then let C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}, i.e., picking the first three points of S(x), the first three points of S(t_2) and the first three points of S(t_3). Now, form the following open subset of \Sigma.

    O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma

Choose t_4 \in O_2 \cap M. Let this inductive process continue and we would obtain a sequence t_2,t_3,t_4,\cdots of points of M. We claim that the sequence converges to x. Before we prove the claim, let’s make a few observations about the inductive process of defining t_2,t_3,t_4,\cdots. Let C=\bigcup_{j=1}^\infty C_j.

  • Each C_j is the support of the open set O_j.
  • The sequence of open sets O_j is decreasing, i.e., O_1 \supset O_2 \supset O_3 \supset \cdots. Thus for each integer j, we have t_k \in O_j for all k \ge j.
  • The support of the point x is contained in C, i.e., S(x) \subset C.
  • The support of the each t_j is contained in C, i.e., S(t_j) \subset C.
  • In fact, C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots.
  • The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets O_j.
  • Consequently, for each j, x_\alpha=(t_j)_\alpha=a_\alpha for each \alpha \in A-C. In other words, x and each t_j agree (and agree with the base point a) on the coordinates outside of the countable set C.

Let U=\prod_{\alpha \in A} U_\alpha be a standard open set in the product space X=\prod_{\alpha \in A} X_\alpha such that x \in U. Let U^*=U \cap \Sigma. We show that for some n, t_j \in U^* for all j \ge n.

Let F=supp(U) be the support of U. Let F_1=F \cap C and F_2=F \cap (A-C). Consider the following open set:

    U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma

Note that supp(U^{**})=F_1. For each \alpha \in F_1, choose B_{\alpha,k(\alpha)} \subset U_\alpha. Let m be the maximum of all k(\alpha) where \alpha \in F_1. Then B_{\alpha,m} \subset U_\alpha for each \alpha \in F_1. Choose a positive integer p such that:

    F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}

Let n=\text{max}(m,p). It follows that there exists some n such that O_n \subset U^{**}. Then t_j \in U^{**} for all j \ge n. It is also the case that t_j \in U^{*} for all j \ge n. This is because x=t_j on the coordinates not in C. \blacksquare

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\copyright \ 2014 \text{ by Dan Ma}

Pixley-Roy hyperspaces

In this post, we introduce a class of hyperspaces called Pixley-Roy spaces. This is a well-known and well studied set of topological spaces. Our goal here is not to be comprehensive but rather to present some selected basic results to give a sense of what Pixley-Roy spaces are like.

A hyperspace refers to a space in which the points are subsets of a given “ground” space. There are more than one way to define a hyperspace. Pixley-Roy spaces were first described by Carl Pixley and Prabir Roy in 1969 (see [5]). In such a space, the points are the non-empty finite subsets of a given ground space. More precisely, let X be a T_1 space (i.e. finite sets are closed). Let \mathcal{F}[X] be the set of all non-empty finite subsets of X. For each F \in \mathcal{F}[X] and for each open subset U of X with F \subset U, we define:

    [F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}

The sets [F,U] over all possible F and U form a base for a topology on \mathcal{F}[X]. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set \mathcal{F}[X] with this topology is called a Pixley-Roy space.

The hyperspace as defined above was first defined by Pixley and Roy on the real line (see [5]) and was later generalized by van Douwen (see [7]). These spaces are easy to define and is useful for constructing various kinds of counterexamples. Pixley-Roy played an important part in answering the normal Moore space conjecture. Pixley-Roy spaces have also been studied in their own right. Over the years, many authors have investigated when the Pixley-Roy spaces are metrizable, normal, collectionwise Hausdorff, CCC and homogeneous. For a small sample of such investigations, see the references listed at the end of the post. Our goal here is not to discuss the results in these references. Instead, we discuss some basic properties of Pixley-Roy to solidify the definition as well as to give a sense of what these spaces are like. Good survey articles of Pixley-Roy are [3] and [7].

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Basic Discussion

In this section, we focus on properties that are always possessed by a Pixley-Roy space given that the ground space is at least T_1. Let X be a T_1 space. We discuss the following points:

  1. The topology defined above is a legitimate one, i.e., the sets [F,U] indeed form a base for a topology on \mathcal{F}[X].
  2. \mathcal{F}[X] is a Hausdorff space.
  3. \mathcal{F}[X] is a zero-dimensional space.
  4. \mathcal{F}[X] is a completely regular space.
  5. \mathcal{F}[X] is a hereditarily metacompact space.

Let \mathcal{B}=\left\{[F,U]: F \in \mathcal{F}[X] \text{ and } U \text{ is open in } X \right\}. Note that every finite set F belongs to at least one set in \mathcal{B}, namely [F,X]. So \mathcal{B} is a cover of \mathcal{F}[X]. For A \in [F_1,U_1] \cap [F_2,U_2], we have A \in [A,U_1 \cap U_2] \subset   [F_1,U_1] \cap [F_2,U_2]. So \mathcal{B} is indeed a base for a topology on \mathcal{F}[X].

To show \mathcal{F}[X] is Hausdorff, let A and B be finite subsets of X where A \ne B. Then one of the two sets has a point that is not in the other one. Assume we have x \in A-B. Since X is T_1, we can find open sets U, V \subset X such that x \in U, x \notin V and A \cup B-\left\{ x \right\} \subset V. Then [A,U \cup V] and [B,V] are disjoint open sets containing A and B respectively.

To see that \mathcal{F}[X] is a zero-dimensional space, we show that \mathcal{B} is a base consisting of closed and open sets. To see that [F,U] is closed, let C \notin [F,U]. Either F \not \subset C or C \not \subset U. In either case, we can choose open V \subset X with C \subset V such that [C,V] \cap [F,U]=\varnothing.

The fact that \mathcal{F}[X] is completely regular follows from the fact that it is zero-dimensional.

To show that \mathcal{F}[X] is metacompact, let \mathcal{G} be an open cover of \mathcal{F}[X]. For each F \in \mathcal{F}[X], choose G_F \in \mathcal{G} such that F \in G_F and let V_F=[F,X] \cap G_F. Then \mathcal{V}=\left\{V_F: F \in \mathcal{F}[X] \right\} is a point-finite open refinement of \mathcal{G}. For each A \in \mathcal{F}[X], A can only possibly belong to V_F for the finitely many F \subset A.

A similar argument show that \mathcal{F}[X] is hereditarily metacompact. Let Y \subset \mathcal{F}[X]. Let \mathcal{H} be an open cover of Y. For each F \in Y, choose H_F \in \mathcal{H} such that F \in H_F and let W_F=([F,X] \cap Y) \cap H_F. Then \mathcal{W}=\left\{W_F: F \in Y \right\} is a point-finite open refinement of \mathcal{H}. For each A \in Y, A can only possibly belong to W_F for the finitely many F \subset A such that F \in Y.

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More Basic Results

We now discuss various basic topological properties of \mathcal{F}[X]. We first note that \mathcal{F}[X] is a discrete space if and only if the ground space X is discrete. Though we do not need to make this explicit, it makes sense to focus on non-discrete spaces X when we look at topological properties of \mathcal{F}[X]. We discuss the following points:

  1. If X is uncountable, then \mathcal{F}[X] is not separable.
  2. If X is uncountable, then every uncountable subspace of \mathcal{F}[X] is not separable.
  3. If \mathcal{F}[X] is Lindelof, then X is countable.
  4. If \mathcal{F}[X] is Baire space, then X is discrete.
  5. If \mathcal{F}[X] has the CCC, then X has the CCC.
  6. If \mathcal{F}[X] has the CCC, then X has no uncountable discrete subspaces,i.e., X has countable spread, which of course implies CCC.
  7. If \mathcal{F}[X] has the CCC, then X is hereditarily Lindelof.
  8. If \mathcal{F}[X] has the CCC, then X is hereditarily separable.
  9. If X has a countable network, then \mathcal{F}[X] has the CCC.
  10. The Pixley-Roy space of the Sorgenfrey line does not have the CCC.
  11. If X is a first countable space, then \mathcal{F}[X] is a Moore space.

Bullet points 6 to 9 refer to properties that are never possessed by Pixley-Roy spaces except in trivial cases. Bullet points 6 to 8 indicate that \mathcal{F}[X] can never be separable and Lindelof as long as the ground space X is uncountable. Note that \mathcal{F}[X] is discrete if and only if X is discrete. Bullet point 9 indicates that any non-discrete \mathcal{F}[X] can never be a Baire space. Bullet points 10 to 13 give some necessary conditions for \mathcal{F}[X] to be CCC. Bullet 14 gives a sufficient condition for \mathcal{F}[X] to have the CCC. Bullet 15 indicates that the hereditary separability and the hereditary Lindelof property are not sufficient conditions for the CCC of Pixley-Roy space (though they are necessary conditions). Bullet 16 indicates that the first countability of the ground space is a strong condition, making \mathcal{F}[X] a Moore space.

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To see bullet point 6, let X be an uncountable space. Let \left\{F_1,F_2,F_3,\cdots \right\} be any countable subset of \mathcal{F}[X]. Choose a point x \in X that is not in any F_n. Then none of the sets F_i belongs to the basic open set [\left\{x \right\} ,X]. Thus \mathcal{F}[X] can never be separable if X is uncountable.

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To see bullet point 7, let Y \subset \mathcal{F}[X] be uncountable. Let W=\cup \left\{F: F \in Y \right\}. Let \left\{F_1,F_2,F_3,\cdots \right\} be any countable subset of Y. We can choose a point x \in W that is not in any F_n. Choose some A \in Y such that x \in A. Then none of the sets F_n belongs to the open set [A ,X] \cap Y. So not only \mathcal{F}[X] is not separable, no uncountable subset of \mathcal{F}[X] is separable if X is uncountable.

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To see bullet point 8, note that \mathcal{F}[X] has no countable open cover consisting of basic open sets, assuming that X is uncountable. Consider the open collection \left\{[F_1,U_1],[F_2,U_2],[F_3,U_3],\cdots \right\}. Choose x \in X that is not in any of the sets F_n. Then \left\{ x \right\} cannot belong to [F_n,U_n] for any n. Thus \mathcal{F}[X] can never be Lindelof if X is uncountable.

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For an elementary discussion on Baire spaces, see this previous post.

To see bullet point 9, let X be a non-discrete space. To show \mathcal{F}[X] is not Baire, we produce an open subset that is of first category (i.e. the union of countably many closed nowhere dense sets). Let x \in X a limit point (i.e. an non-isolated point). We claim that the basic open set V=[\left\{ x \right\},X] is a desired open set. Note that V=\bigcup \limits_{n=1}^\infty H_n where

    H_n=\left\{F \in \mathcal{F}[X]: x \in F \text{ and } \lvert F \lvert \le n \right\}

We show that each H_n is closed and nowhere dense in the open subspace V. To see that it is closed, let A \notin H_n with x \in A. We have \lvert A \lvert>n. Then [A,X] is open and every point of [A,X] has more than n points of the space X. To see that H_n is nowhere dense in V, let [B,U] be open with [B,U] \subset V. It is clear that x \in B \subset U where U is open in the ground space X. Since the point x is not an isolated point in the space X, U contains infinitely many points of X. So choose an finite set C with at least 2 \times n points such that B \subset C \subset U. For the the open set [C,U], we have [C,U] \subset [B,U] and [C,U] contains no point of H_n. With the open set V being a union of countably many closed and nowhere dense sets in V, the open set V is not of second category. We complete the proof that \mathcal{F}[X] is not a Baire space.

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To see bullet point 10, let \mathcal{O} be an uncountable and pairwise disjoint collection of open subsets of X. For each O \in \mathcal{O}, choose a point x_O \in O. Then \left\{[\left\{ x_O \right\},O]: O \in \mathcal{O} \right\} is an uncountable and pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] is CCC then X must have the CCC.

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To see bullet point 11, let Y \subset X be uncountable such that Y as a space is discrete. This means that for each y \in Y, there exists an open O_y \subset X such that y \in O_y and O_y contains no point of Y other than y. Then \left\{[\left\{y \right\},O_y]: y \in Y \right\} is an uncountable and pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] has the CCC, then the ground space X has no uncountable discrete subspace (such a space is said to have countable spread).

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To see bullet point 12, let Y \subset X be uncountable such that Y is not Lindelof. Then there exists an open cover \mathcal{U} of Y such that no countable subcollection of \mathcal{U} can cover Y. We can assume that sets in \mathcal{U} are open subsets of X. Also by considering a subcollection of \mathcal{U} if necessary, we can assume that cardinality of \mathcal{U} is \aleph_1 or \omega_1. Now by doing a transfinite induction we can choose the following sequence of points and the following sequence of open sets:

    \left\{x_\alpha \in Y: \alpha < \omega_1 \right\}

    \left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}

such that x_\beta \ne x_\gamma if \beta \ne \gamma, x_\alpha \in U_\alpha and x_\alpha \notin \bigcup \limits_{\beta < \alpha} U_\beta for each \alpha < \omega_1. At each step \alpha, all the previously chosen open sets cannot cover Y. So we can always choose another point x_\alpha of Y and then choose an open set in \mathcal{U} that contains x_\alpha.

Then \left\{[\left\{x_\alpha \right\},U_\alpha]: \alpha < \omega_1 \right\} is a pairwise disjoint collection of open subsets of \mathcal{F}[X]. Thus if \mathcal{F}[X] has the CCC, then X must be hereditarily Lindelof.

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To see bullet point 13, let Y \subset X. Consider open sets [A,U] where A ranges over all finite subsets of Y and U ranges over all open subsets of X with A \subset U. Let \mathcal{G} be a collection of such [A,U] such that \mathcal{G} is pairwise disjoint and \mathcal{G} is maximal (i.e. by adding one more open set, the collection will no longer be pairwise disjoint). We can apply a Zorn lemma argument to obtain such a maximal collection. Let D be the following subset of Y.

    D=\bigcup \left\{A: [A,U] \in \mathcal{G} \text{ for some open } U  \right\}

We claim that the set D is dense in Y. Suppose that there is some open set W \subset X such that W \cap Y \ne \varnothing and W \cap D=\varnothing. Let y \in W \cap Y. Then [\left\{y \right\},W] \cap [A,U]=\varnothing for all [A,U] \in \mathcal{G}. So adding [\left\{y \right\},W] to \mathcal{G}, we still get a pairwise disjoint collection of open sets, contradicting that \mathcal{G} is maximal. So D is dense in Y.

If \mathcal{F}[X] has the CCC, then \mathcal{G} is countable and D is a countable dense subset of Y. Thus if \mathcal{F}[X] has the CCC, the ground space X is hereditarily separable.

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A collection \mathcal{N} of subsets of a space Y is said to be a network for the space Y if any non-empty open subset of Y is the union of elements of \mathcal{N}, equivalently, for each y \in Y and for each open U \subset Y with y \in U, there is some A \in \mathcal{N} with x \in A \subset U. Note that a network works like a base but the elements of a network do not have to be open. The concept of network and spaces with countable network are discussed in these previous posts Network Weight of Topological Spaces – I and Network Weight of Topological Spaces – II.

To see bullet point 14, let \mathcal{N} be a network for the ground space X such that \mathcal{N} is also countable. Assume that \mathcal{N} is closed under finite unions (for example, adding all the finite unions if necessary). Let \left\{[A_\alpha,U_\alpha]: \alpha < \omega_1 \right\} be a collection of basic open sets in \mathcal{F}[X]. Then for each \alpha, find B_\alpha \in \mathcal{N} such that A_\alpha \subset B_\alpha \subset U_\alpha. Since \mathcal{N} is countable, there is some B \in \mathcal{N} such that M=\left\{\alpha< \omega_1: B=B_\alpha \right\} is uncountable. It follows that for any finite E \subset M, \bigcap \limits_{\alpha \in E} [A_\alpha,U_\alpha] \ne \varnothing.

Thus if the ground space X has a countable network, then \mathcal{F}[X] has the CCC.

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The implications in bullet points 12 and 13 cannot be reversed. Hereditarily Lindelof property and hereditarily separability are not sufficient conditions for \mathcal{F}[X] to have the CCC. See [4] for a study of the CCC property of the Pixley-Roy spaces.

To see bullet point 15, let S be the Sorgenfrey line, i.e. the real line \mathbb{R} with the topology generated by the half closed intervals of the form [a,b). For each x \in S, let U_x=[x,x+1). Then \left\{[ \left\{ x \right\},U_x]: x \in S \right\} is a collection of pairwise disjoint open sets in \mathcal{F}[S].

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A Moore space is a space with a development. For the definition, see this previous post.

To see bullet point 16, for each x \in X, let \left\{B_n(x): n=1,2,3,\cdots \right\} be a decreasing local base at x. We define a development for the space \mathcal{F}[X].

For each finite F \subset X and for each n, let B_n(F)=\bigcup \limits_{x \in F} B_n(x). Clearly, the sets B_n(F) form a decreasing local base at the finite set F. For each n, let \mathcal{H}_n be the following collection:

    \mathcal{H}_n=\left\{[F,B_n(F)]: F \in \mathcal{F}[X] \right\}

We claim that \left\{\mathcal{H}_n: n=1,2,3,\cdots \right\} is a development for \mathcal{F}[X]. To this end, let V be open in \mathcal{F}[X] with F \in V. If we make n large enough, we have [F,B_n(F)] \subset V.

For each non-empty proper G \subset F, choose an integer f(G) such that [F,B_{f(G)}(F)] \subset V and F \not \subset B_{f(G)}(G). Let m be defined by:

    m=\text{max} \left\{f(G): G \ne \varnothing \text{ and } G \subset F \text{ and } G \text{ is proper} \right\}

We have F \not \subset B_{m}(G) for all non-empty proper G \subset F. Thus F \notin [G,B_m(G)] for all non-empty proper G \subset F. But in \mathcal{H}_m, the only sets that contain F are [F,B_m(F)] and [G,B_m(G)] for all non-empty proper G \subset F. So [F,B_m(F)] is the only set in \mathcal{H}_m that contains F, and clearly [F,B_m(F)] \subset V.

We have shown that for each open V in \mathcal{F}[X] with F \in V, there exists an m such that any open set in \mathcal{H}_m that contains F must be a subset of V. This shows that the \mathcal{H}_n defined above form a development for \mathcal{F}[X].

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Examples

In the original construction of Pixley and Roy, the example was \mathcal{F}[\mathbb{R}]. Based on the above discussion, \mathcal{F}[\mathbb{R}] is a non-separable CCC Moore space. Because the density (greater than \omega for not separable) and the cellularity (=\omega for CCC) do not agree, \mathcal{F}[\mathbb{R}] is not metrizable. In fact, it does not even have a dense metrizable subspace. Note that countable subspaces of \mathcal{F}[\mathbb{R}] are metrizable but are not dense. Any uncountable dense subspace of \mathcal{F}[\mathbb{R}] is not separable but has the CCC. Not only \mathcal{F}[\mathbb{R}] is not metrizable, it is not normal. The problem of finding X \subset \mathbb{R} for which \mathcal{F}[X] is normal requires extra set-theoretic axioms beyond ZFC (see [6]). In fact, Pixley-Roy spaces played a large role in the normal Moore space conjecture. Assuming some extra set theory beyond ZFC, there is a subset M \subset \mathbb{R} such that \mathcal{F}[M] is a CCC metacompact normal Moore space that is not metrizable (see Example I in [8]).

On the other hand, Pixley-Roy space of the Sorgenfrey line and the Pixley-Roy space of \omega_1 (the first uncountable ordinal with the order topology) are metrizable (see [3]).

The Sorgenfrey line and the first uncountable ordinal are classic examples of topological spaces that demonstrate that topological spaces in general are not as well behaved like metrizable spaces. Yet their Pixley-Roy spaces are nice. The real line and other separable metric spaces are nice spaces that behave well. Yet their Pixley-Roy spaces are very much unlike the ground spaces. This inverse relation between the ground space and the Pixley-Roy space was noted by van Douwen (see [3] and [7]) and is one reason that Pixley-Roy hyperspaces are a good source of counterexamples.

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Reference

  1. Bennett, H. R., Fleissner, W. G., Lutzer, D. J., Metrizability of certain Pixley-Roy spaces, Fund. Math. 110, 51-61, 1980.
  2. Daniels, P, Pixley-Roy Spaces Over Subsets of the Reals, Topology Appl. 29, 93-106, 1988.
  3. Lutzer, D. J., Pixley-Roy topology, Topology Proc. 3, 139-158, 1978.
  4. Hajnal, A., Juahasz, I., When is a Pixley-Roy Hyperspace CCC?, Topology Appl. 13, 33-41, 1982.
  5. Pixley, C., Roy, P., Uncompletable Moore spaces, Proc. Auburn Univ. Conf. Auburn, AL, 1969.
  6. Przymusinski, T., Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, 291-297, 1981.
  7. van Douwen, E. K., The Pixley-Roy topology on spaces of subsets, Set-theoretic Topology, Academic Press, New York, 111-134, 1977.
  8. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.
  9. Tanaka, H, Normality and hereditary countable paracompactness of Pixley-Roy hyperspaces, Fund. Math. 126, 201-208, 1986.

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\copyright \ 2014 \text{ by Dan Ma}

Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}, the set of all rational numbers. Let \tau be the usual topology of the real line \mathbb{R}. The following is a base that defines a topology on \mathbb{R}.

    \mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

The real line with the topology generated by \mathcal{B} is called the Michael line and is denoted by \mathbb{M}. In essense, in \mathbb{M}, points in \mathbb{P} are made isolated and points in \mathbb{Q} retain the usual Euclidean open sets.

The Euclidean topology \tau is coarser (weaker) than the Michael line topology (i.e. \tau being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, \mathbb{M} is submetrizable (submetrized by the Euclidean topology). It is clear that \mathbb{M} is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

  1. The space \mathbb{M} is paracompact.
  2. The space \mathbb{M} is not Lindelof.
  3. The extent of the space \mathbb{M} is c where c is the cardinality of the real line.
  4. The space \mathbb{M} is not locally compact.
  5. The space \mathbb{M} is not perfectly normal, thus not metrizable.
  6. The space \mathbb{M} is not a Moore space, but has a G_\delta-diagonal.
  7. The product \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology.
  8. The product \mathbb{M} \times \mathbb{P} is metacompact.
  9. The space \mathbb{M} has a point-countable base.
  10. For each n=1,2,3,\cdots, the product \mathbb{M}^n is paracompact.
  11. The product \mathbb{M}^\omega is not normal.
  12. There exist a Lindelof space L and a separable metric space W such that L \times W is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on \mathbb{Q} coincides with the Euclidean topology on \mathbb{Q}. A set is said to be a G_\delta-set if it is the intersection of countably many open sets. By the Baire category theorem, the set \mathbb{Q} is not a G_\delta-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set \mathbb{Q} is not a G_\delta-set in the Michael line. This fact is used in Result 5.

The fact that \mathbb{Q} is not a G_\delta-set in the Euclidean real line implies that \mathbb{P} is not an F_\sigma-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let \mathcal{U} be an open cover of \mathbb{M}. We proceed to derive a locally finite open refinement \mathcal{V} of \mathcal{U}. Recall that \tau is the usual topology on \mathbb{R}. Assume that \mathcal{U} consists of open sets in the base \mathcal{B}. Let \mathcal{U}_\tau=\mathcal{U} \cap \tau. Let Y=\cup \mathcal{U}_\tau. Note that Y is a Euclidean open subspace of the real line (hence it is paracompact). Then there is \mathcal{V}_\tau \subset \tau such that \mathcal{V}_\tau is a locally finite open refinement \mathcal{V}_\tau of \mathcal{U}_\tau and such that \mathcal{V}_\tau covers Y (locally finite in the Euclidean sense). Then add to \mathcal{V}_\tau all singleton sets \left\{ x \right\} where x \in \mathbb{M}-Y and let \mathcal{V} denote the resulting open collection.

The resulting \mathcal{V} is a locally finite open collection in the Michael line \mathbb{M}. Furthermore, \mathcal{V} is also a refinement of the original open cover \mathcal{U}. \blacksquare

A similar argument shows that \mathbb{M} is hereditarily paracompact.

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Result 2

To see that \mathbb{M} is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in \mathbb{P}). For example, it is possible to construct a Cantor set entirely within \mathbb{P}.

Let C be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set C is an uncountable closed and discrete set in \mathbb{M}. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line \mathbb{M} cannot be Lindelof. \blacksquare

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space X is the smallest infinite cardinal number \mathcal{K} such that every closed and discrete set in X has cardinality \le \mathcal{K}. The extent of the space X is denoted by e(X). When the cardinal number e(X) is e(X)=\aleph_0 (the first infinite cardinal number), the space X is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When e(X)>\aleph_0, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space X is Lindelof, the extent is e(X)=\aleph_0. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of \mathbb{M} of cardinality c. Thus we have e(\mathbb{M})=c. \blacksquare

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Result 4

The Michael line \mathbb{M} is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in \mathbb{M}. \blacksquare

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Result 5

A set is said to be a G_\delta-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a G_\delta-set. In the Michael line \mathbb{M}, the set \mathbb{Q} of rational numbers is a closed set. Yet, \mathbb{Q} is not a G_\delta-set in the Michael line (see the discussion above on the Baire category theorem). Thus \mathbb{M} is not perfectly normal and hence not a metrizable space. \blacksquare

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Result 6

The diagonal of a space X is the subset of its square X \times X that is defined by \Delta=\left\{(x,x): x \in X \right\}. If the space is Hausdorff, the diagonal is always a closed set in the square. If \Delta is a G_\delta-set in X \times X, the space X is said to have a G_\delta-diagonal. It is well known that any metric space has G_\delta-diagonal. Since \mathbb{M} is submetrizable (submetrized by the usual topology of the real line), it has a G_\delta-diagonal too.

Any Moore space has a G_\delta-diagonal. However, the Michael line is an example of a space with G_\delta-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus \mathbb{M} is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. \blacksquare

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Result 7

We now show that \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology. In this proof, the following two facts are crucial:

  • The set \mathbb{P} is not an F_\sigma-set in the real line.
  • The set \mathbb{P} is dense in the real line.

Let H and K be defined by the following:

    H=\left\{(x,x): x \in \mathbb{P} \right\}
    K=\mathbb{Q} \times \mathbb{P}.

The sets H and K are disjoint closed sets in \mathbb{M} \times \mathbb{P}. We show that they cannot be separated by disjoint open sets. To this end, let H \subset U and K \subset V where U and V are open sets in \mathbb{M} \times \mathbb{P}.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval (a,b), we mean the set of all real numbers t with a<t<b. By (a,b)^*, we mean (a,b) \cap \mathbb{P}. For each x \in \mathbb{P}, choose an open interval U_x=(a,b)^* such that \left\{x \right\} \times U_x \subset U. We also assume that x is the midpoint of the open interval U_x. For each positive integer k, let P_k be defined by:

    P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}

Note that \mathbb{P}=\bigcup \limits_{k=1}^\infty P_k. For each k, let T_k=\overline{P_k} (Euclidean closure in the real line). It is clear that \bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k. On the other hand, \bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P} (otherwise \mathbb{P} would be an F_\sigma-set in the real line). So there exists T_n=\overline{P_n} such that \overline{P_n} \not\subset \mathbb{P}. So choose a rational number r such that r \in \overline{P_n}.

Choose a positive integer j such that \frac{2}{j}<\frac{1}{n}. Since \mathbb{P} is dense in the real line, choose y \in \mathbb{P} such that r-\frac{1}{j}<y<r+\frac{1}{j}. Now we have (r,y) \in K \subset V. Choose another integer m such that \frac{1}{m}<\frac{1}{j} and (r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V.

Since r \in \overline{P_n}, choose x \in \mathbb{P} such that r-\frac{1}{m}<x<r+\frac{1}{m}. Now it is clear that (x,y) \in V. The following inequalities show that (x,y) \in U.

    \lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}

The open interval U_x is chosen to have length > \frac{1}{n}. Since \lvert x-y \lvert < \frac{1}{n}, y \in U_x. Thus (x,y) \in \left\{ x \right\} \times U_x \subset U. We have shown that U \cap V \ne \varnothing. Thus \mathbb{M} \times \mathbb{P} is not normal. \blacksquare

Remark
As indicated above, the proof of Result 7 hinges on two facts about \mathbb{P}, namely that it is not an F_\sigma-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set D that is isolated is not an F_\sigma-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space D (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product \mathbb{M} \times \mathbb{P} is not paracompact since it is not normal. However, \mathbb{M} \times \mathbb{P} is metacompact.

A collection of subsets of a space X is said to be point-finite if every point of X belongs to only finitely many sets in the collection. A space X is said to be metacompact if each open cover of X has an open refinement that is a point-finite collection.

Note that \mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P}). The first \mathbb{P} in \mathbb{P} \times \mathbb{P} is discrete (a subspace of the Michael line) and the second \mathbb{P} has the Euclidean topology.

Let \mathcal{U} be an open cover of \mathbb{M} \times \mathbb{P}. For each a=(x,y) \in \mathbb{Q} \times \mathbb{P}, choose U_a \in \mathcal{U} such that a \in U_a. We can assume that U_a=A \times B where A is a usual open interval in \mathbb{R} and B is a usual open interval in \mathbb{P}. Let \mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace.

Fix x \in \mathbb{P}. For each b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}, choose some U_b \in \mathcal{U} such that b \in U_b. We can assume that U_b=\lbrace{x}\rbrace \times B where B is a usual open interval in \mathbb{P}. Let \mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace.

As a subspace of the Euclidean plane, \bigcup \mathcal{G} is metacompact. So there is a point-finite open refinement \mathcal{W} of \mathcal{G}. For each x \in \mathbb{P}, \mathcal{H}_x has a point-finite open refinement \mathcal{I}_x. Let \mathcal{V} be the union of \mathcal{W} and all the \mathcal{I}_x where x \in \mathbb{P}. Then \mathcal{V} is a point-finite open refinement of \mathcal{U}.

Note that the point-finite open refinement \mathcal{V} may not be locally finite. The vertical open intervals in \lbrace{x}\rbrace \times \mathbb{P}, x \in \mathbb{P} can “converge” to a point in \mathbb{Q} \times \mathbb{P}. Thus, metacompactness is the best we can hope for. \blacksquare

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base \mathcal{G} for a space X is said to be a point-countable base if \mathcal{G}, in addition to being a base for the space X, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for \mathbb{M}:

    \mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

where \mathcal{H} is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Alexandroff Double Circle

We discuss the Alexandroff double circle, which is a compact and non-metrizable space. A theorem about the hereditarily normality of a product space Y_1 \times Y_2 is also discussed.

Let C_1 and C_2 be the two concentric circles centered at the origin with radii 1 and 2, respectively. Specifically C_i=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 =i \right\} where i=1,2. Let X=C_1 \cup C_2. Furthermore let f:C_1 \rightarrow C_2 be the natural homeomorphism. Figure 1 below shows the underlying set.

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Figure 1 – Underlying Set

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We define a topology on X as follows:

  • Points in C_2 are isolated.
  • For each x \in C_1 and for each positive integer j, let O(x,j) be the open arc in C_1 whose center contains x and has length \frac{1}{j} (in the Euclidean topology on C_1). For each x \in C_1, an open neighborhood is of the form B(x,j) where
      \text{ }

      B(x,j)=O(x,j) \cup (f(O(x,j))-\left\{f(x) \right\}).

    The following figure shows an open neighborhood at point in C_1.

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Figure 2 – Open Neighborhood

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A List of Results

It can be verified that the open neighborhoods defined above form a base for a topology on X. We discuss the following points about the Alexandroff double circle.

  1. X is a Hausdorff space.
  2. X is not separable.
  3. X is not hereditarily Lindelof.
  4. X is compact.
  5. X is sequentially compact.
  6. X is not metrizable.
  7. X is not perfectly normal.
  8. X is completely normal (and thus hereditarily normal).
  9. X \times X is not hereditarily normal.

The proof that X \times X is not hereditarily normal can be generalized. We discuss this theorem after presenting the proof of Result 9.
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Results 1, 2, 3

It is clear that the Alexandroff double circle is a Hausdorff space. It is not separable since the outer circle C_2 consists of uncountably many singleton open subsets. For the same reason, C_2 is a non-Lindelof subspace, making the Alexandroff double circle not hereditarily Lindelof. \blacksquare

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Result 4

The property that X is compact is closely tied to the compactness of the inner circle C_1 in the Euclidean topology. Note that the subspace topology of the Alexandroff double circle on C_1 is simply the Euclidean topology. Let \mathcal{U} be an open cover of X consisting of open sets as defined above. Then there are finitely many basic open sets B(x_1,j_1), B(x_2,j_2), \cdots, B(x_n,j_n) from \mathcal{U} covering C_1. These open sets cover the entire space except for the points f(x_1), f(x_2), \cdots,f(x_n), which can be covered by finitely many open sets in \mathcal{U}. \blacksquare

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Result 5

A space W is sequentially compact if every sequence of points of W has a subsequence that converges to a point in W. The notion of sequentially compactness and compactness coincide for the class of metric spaces. However, in general these two notions are distinct.

The sequentially compactness of the Alexandroff double circle X hinges on the sequentially compactness of C_1 and C_2 in the Euclidean topology. Let \left\{x_n \right\} be a sequence of points in X. If the set \left\{x_n: n=1,2,3,\cdots \right\} is a finite set, then \left\{x_n: n>m \right\} is a constant sequence for some large enough integer m. So assume that A=\left\{x_n: n=1,2,3,\cdots \right\} is an infinite set. Either A \cap C_1 is infinite or A \cap C_2 is infinite. If A \cap C_1 is infinite, then some subsequence of \left\{x_n \right\} converges in C_1 in the Euclidean topology (hence in the Alexandroff double circle topology). If A \cap C_2 is infinite, then some subsequence of \left\{x_n \right\} converges to x \in C_2 in the Euclidean topology. Then this same subsequence converges to f^{-1}(x) in the Alexandroff double circle topology. \blacksquare

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Result 6

Note that any compact metrizable space satisfies a long list of properties, which include separable, Lindelof, hereditarily Lindelof. \blacksquare

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Result 7

A space is perfectly normal if it is normal with the additional property that every closed set is a G_\delta-set. For the Alexandroff double circle, the inner circle C_1 is not a G_\delta-set, or equivalently the outer circle C_2 is not an F_\sigma-set. To see this, suppose that C_2 is the union of countably many sets, we show that the closure of at least one of the sets goes across to the inner circle C_1. Let C_2=\bigcup \limits_{i=1}^\infty T_n. At least one of the sets is uncountable. Let T_j be one such. Consider f^{-1}(T_j), which is also uncountable and has a limit point in C_1 (in the Euclidean topology). Let t be one such point (i.e. every Euclidean open set containing t contains points of f^{-1}(T_j)). Then the point t is a member of the closure of T_j (Alexandroff double circle topology). \blacksquare

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Result 8

We first discuss the notion of separated sets. Let T be a Hausdorff space. Let E \subset T and F \subset T. The sets E and F are said to be separated (are separated sets) if E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. In other words, two sets are separated if each one does not meet the closure of the other set. In particular, any two disjoint closed sets are separated. The space T is said to be completely normal if T satisfies the property that for any two sets E and F that are separated, there are disjoint open sets U and V with E \subset U and F \subset V. Thus completely normality implies normality.

It is a well know fact that if a space is completely normal, it is hereditarily normal (actually the two notions are equivalent). Note that any metric space is completely normal. In particular, any Euclidean space is completely normal.

To show that the Alexandroff double circle X is completely normal, let E \subset X and F \subset X be separated sets. Thus we have E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. Note that E \cap C_1 and F \cap C_1 are separated sets in the Euclidean space C_1. Let G_1 and G_2 be disjoint Euclidean open subsets of C_1 with E \cap C_1 \subset G_1 and F \cap C_1 \subset G_2.

For each x \in E \cap C_1, choose open U_x (Alexandroff double circle open) with x \in U_x, U_x \cap C_1 \subset G_1 and U_x \cap \overline{F}=\varnothing. Likewise, for each y \in F \cap C_1, choose open V_y (Alexandroff double circle open) with y \in V_y, V_y \cap C_1 \subset G_2 and V_y \cap \overline{E}=\varnothing. Then let U and V be defined by the following:

    U=\biggl(\bigcup \limits_{x \in E \cap C_1} U_x \biggr) \cup \biggl(E \cap C_2 \biggr)

    \text{ }

    V= \biggl(\bigcup \limits_{y \in F \cap C_1} V_y \biggr) \cup \biggl(F \cap C_2 \biggr)

Because G_1 \cap G_2 =\varnothing, the open sets U_x and V_y are disjoint. As a result, U and V are disjoint open sets in the Alexandroff double circle with E \subset U and F \subset V.

For the sake of completeness, we show that any completely normal space is hereditarily normal. Let T be completely normal. Let Y \subset T. Let H \subset Y and K \subset Y be disjoint closed subsets of Y. Then in the space T, H and K are separated. Note that H \cap cl_T(K)=\varnothing and K \cap cl_T(H)=\varnothing (where cl_T gives the closure in T). Then there are disjoint open subsets O_1 and O_2 of T such that H \subset O_1 and K \subset O_2. Now, O_1 \cap Y and O_2 \cap Y are disjoint open sets in Y such that H \subset O_1 \cap Y and K \subset O_2 \cap Y.

Thus we have established that the Alexandroff double circle is hereditarily normal. \blacksquare

For the proof that a space is completely normal if and only if it is hereditarily normal, see Theorem 2.1.7 in page 69 of [1],
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Result 9

We produce a subspace Y \subset X \times X that is not normal. To this end, let D=\left\{d_n:n=1,2,3,\cdots \right\} be a countable subset of X such that \overline{D}-D\ne \varnothing. Let y \in \overline{D}-D. Let Y=X \times X-C_1 \times \left\{y \right\}. We show that Y is not normal.

Let H=C_1 \times (X-\left\{y \right\}) and K=C_2 \times \left\{y \right\}. These are two disjoint closed sets in Y. Let U and V be open in Y such that H \subset U and K \subset V. We show that U \cap V \ne \varnothing.

For each integer j, let U_j=\left\{x \in X: (x,d_j) \in U \right\}. We claim that each U_j is open in X. To see this, pick x \in U_j. We know (x,d_j) \in U. There exist open A and B (open in X) such that (x,d_j) \in A \times B \subset U. It is clear that x \in A \subset U_j. Thus each U_j is open.

Furthermore, we have C_1 \subset U_j for each j. Based in Result 7, C_1 is not a G_\delta-set. So we have C_1 \subset \bigcap \limits_{j=1}^\infty U_j but C_1 \ne \bigcap \limits_{j=1}^\infty U_j. There exists t \in \bigcap \limits_{j=1}^\infty U_j but t \notin C_1. Thus t \in C_2 and \left\{t \right\} is open.

Since (t,y) \in K, we have (t,y) \in V. Choose an open neighborhood B(y,k) of y such that \left\{t \right\} \times B(y,k) \subset V. since y \in \overline{D}, there exists some d_j such that (t,d_j) \in \left\{t \right\} \times B(y,k). Hence (t,d_j) \in V. Since t \in U_j, (t,d_j) \in U. Thus U \cap V \ne \varnothing. \blacksquare

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Generalizing the Proof of Result 9

The proof of Result 9 requires that one of the factors has a countable set that is not discrete and the other factor has a closed set that is not a G_\delta-set. Once these two requirements are in place, we can walk through the same proof and show that the cross product is not hereditarily normal. Thus, the statement that is proved in Result 9 is the following.

Theorem
If Y_1 has a countable subset that is not closed and discrete and if Y_2 has a closed set that is not a G_\delta-set then Y_1 \times Y_2 has a subspace that is not normal.

The theorem can be restated as:

Theorem
If Y_1 \times Y_2 is hereditarily normal, then either every countable subset of Y_1 is closed and discrete or Y_2 is perfectly normal.

The above theorem is due to Katetov and can be found in [2]. It shows that the hereditarily normality of a cross product imposes quite strong restrictions on the factors. As a quick example, if both Y_1 and Y_2 are compact, for Y_1 \times Y_2 to be hereditarily normal, both Y_1 and Y_2 must be perfectly normal.

Another example. Let W=\omega_1+1, the succesor of the first uncountable ordinal with the order topology. Note that W is not perfectly normal since the point \omega_1 is not a G_\delta point. Then for any compact space Y, W \times Y is not hereditarily normal. Let C=\omega+1, the successor of the first infinite ordinal with the order topology (essentially a convergent sequence with the limit point). The product W \times C is the Tychonoff plank and based on the discussion here is not hereditarily normal. Usually the Tychonoff plank is shown to be not hereditarily normal by removing the cornor point (\omega_1,\omega). The resulting space is the deleted Tychonoff plank and is not normal (see The Tychonoff Plank).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Przymusinski, T. C., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Stone-Cech Compactification of the Integers – Basic Facts

This is another post Stone-Cech compactification. The links for other posts on Stone-Cech compactification can be found below. In this post, we prove a few basic facts about \beta \omega, the Stone-Cech compactification of the discrete space of the non-negative integers, \omega=\left\{0,1,2,3,\cdots \right\}. We use several characterizations of Stone-Cech compactification to find out what \beta \omega is like. These characterizations are proved in the blog posts listed below. Let c denote the cardinality of the real line \mathbb{R}. We prove the following facts.

  1. The cardinality of \beta \omega is 2^c.
  2. The weight of \beta \omega is c.
  3. The space \beta \omega is zero-dimensional.
  4. Every infinite closed subset of \beta \omega contains a topological copy of \beta \omega.
  5. The space \beta \omega contains no non-trivial convergent sequence.
  6. No point of \beta \omega-\omega is an isolated point.
  7. The space \beta \omega fails to have many properties involving the existence of non-trivial convergent sequence. For example:
    \text{ }

    1. The space \beta \omega is not first countable at each point of the remainder \beta \omega-\omega.
    2. The space \beta \omega is not a Frechet space.
    3. The space \beta \omega is not a sequential space.
    4. The space \beta \omega is not sequentially compact.

    \text{ }

  8. No point of the remainder \beta \omega-\omega is a G_\delta-point.
  9. The remainder \beta \omega-\omega does not have the countable chain condition. In fact, it has a disjoint open collection of cardinality c.

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Characterization Theorems

For any completely regular space X, let C(X,I) be the set of all continuous functions from X into I=[0,1]. The Stone-Cech compactification \beta X is the subspace of the product space [0,1]^{C(X,I)} which is the closure of the image of X under the evaluation map \beta:X \rightarrow [0,1]^{C(X,I)} (for the details, see Embedding Completely Regular Spaces into a Cube).

The brief sketch of \beta \omega we present here is not based on the definition using the evaluation map. Instead we reply on some characterization theorems that are stated here (especially Theorem U3.1). These theorems uniquely describe the Stone-Cech compactification \beta X of a given completely regular space X. For example, \beta X satisfies the function extension property in Theorem C3 below. Furthermore any compactification \alpha X of X that satisfies the same property must be \beta X (Theorem U3.1). So a “C” theorem tells us a property possessed by \beta X. The corresponding “U” theorem tells us that there is only one compactification (up to equivalence) that has this property.

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    Theorem C1
    Let X be a completely regular space. Let f:X \rightarrow Y be a continuous function from X into a compact Hausdorff space Y. Then there is a continuous F: \beta X \rightarrow Y such that F \circ \beta=f.

    \text{ }

    Theorem C2
    Let X be a completely regular space. Among all compactifications of the space X, the Stone-Cech compactification \beta X of the space X is maximal with respect to the partial order \le.

    \text{ }

    Theorem U2
    The property in Theorem C2 is unique to \beta X. That is, if, among all compactifications of the space X, \alpha X is maximal with respect to the partial order \le, then \alpha X \approx \beta X.

    See Two Characterizations of Stone-Cech Compactification.
    _______________________________________________________________________________________
    \text{ }

    Theorem C3
    Let X be a completely regular space. The space X is C^*-embedded in its Stone-Cech compactification \beta X.

    \text{ }

    Theorem U3.1
    Let X be a completely regular space. Let I=[0,1]. Let \alpha X be a compactification of X such that each continuous f:X \rightarrow I can be extended to a continuous \hat{f}:\alpha X \rightarrow I. Then \alpha X must be \beta X.

    \text{ }

    Theorem U3.2
    If \alpha X is any compactification of X that satisfies the property in Theorem C3 (i.e., X is C^*-embedded in \alpha X), then \alpha X must be \beta X.

    See C*-Embedding Property and Stone-Cech Compactification.
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    \text{ }

The following discussion illustrates how we can use some of these characterizations theorem to obtain information about \beta X and \beta \omega in particular.

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Result 1 and Result 2

According to the previous post (Stone-Cech Compactification is Maximal), we have for any completely regular space X, \lvert \beta X \lvert \le 2^{2^{d(X)}} where d(X) is the density (the smallest cardinality of a dense set in X). With \omega being a countable space, \lvert \beta \omega \lvert \le 2^{2^{\omega}}=2^c.

Result 1 is established if we have 2^c \le \lvert \beta \omega \lvert. Consider the cube I^I where I is the unit interval I=[0,1]. Since the product space of c many separable space is separable (see Product of Separable Spaces), I^I is separable. Let S \subset I^I be a countable dense set. Let f:\omega \rightarrow S be a bijection. Clearly f is a continuous function from the discrete space \omega into I^I. By Theorem C1, f is extended by a continuous F:\beta \omega \rightarrow I^I. Note that the image F(\beta \omega) is dense in I^I since F(\beta \omega) contains the dense set S. On the other hand, F(\beta \omega) is compact. So F(\beta \omega)=I^I. Thus F is a surjection. The cardinality of I^I is 2^c. Thus we have 2^c \le \lvert \beta \omega \lvert.

From the same previous post (Stone-Cech Compactification is Maximal), it is shown that w(\beta X) \le 2^{d(X)}. Thus w(\beta \omega) \le 2^{\omega}=c. The same function F:\beta \omega \rightarrow I^I in the above paragraph shows that c \le w(\beta \omega) (see Lemma 2 in Stone-Cech Compactification is Maximal). Thus we have w(\beta \omega)=c \blacksquare

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Result 3

A space is said to be zero-dimensional whenever it has a base consisting of open and closed sets. The proof that \beta X is zero-dimensional comes after the following lemmas and theorems.

    Theorem 1
    Let X be a normal space. If H and K are disjoint closed subsets of X, then H and K have disjoint closures in \beta X.

Proof of Theorem 1
Let H and K be disjoint closed subsets of X. By the normality of X and by the Urysohn’s lemma, there is a continuous function g:X \rightarrow [0,1] such that g(H) \subset \left\{0 \right\} and g(K) \subset \left\{1 \right\}. By Theorem C3.1, g can be extended by G:\beta X \rightarrow [0,1]. Note that \overline{H} \subset G^{-1}(0) and \overline{K} \subset G^{-1}(1). Thus \overline{H} \cap \overline{K} = \varnothing. \blacksquare

    Theorem 2
    Let X be a completely regular space. Let H be a closed and open subset of X. Then \overline{H} (the closure of H in \beta X) is also a closed and open set in \beta X.

Proof of Theorem 2
Let H be a closed and open subset of X. Let K=X-H. Define \gamma:X \rightarrow [0,1] by letting \gamma(x)=0 for all x \in H and \gamma(x)=1 for all x \in K. Since both H and K are closed and open, the map \gamma is continuous. By Theorem C3, \gamma is extended by some continuous \Gamma:\beta X \rightarrow [0,1]. Note that \overline{H} \subset \Gamma^{-1}(0) and \overline{K} \subset \Gamma^{-1}(1). Thus H and K have disjoint closures in \beta X, i.e. \overline{H} \cap \overline{K} = \varnothing. Both H and K are closed and open in \beta X since \beta X=\overline{H} \cup \overline{K}. \blacksquare

    Lemma 3
    For every A \subset \omega, \overline{A} (the closure of A in \beta \omega) is both closed and open in \beta \omega.

Note that Lemma 3 is a corollary of Theorem 2.

    Lemma 4
    Let O \subset \beta \omega be a set that is both closed and open in \beta \omega. Then O=\overline{A} where A= O \cap \omega.

Proof of Lemma 4
Let A=O \cap \omega. Either O \subset \omega or O \cap (\beta \omega-\omega) \ne \varnothing. Thus A \ne \varnothing. We claim that O=\overline{A}. Since A \subset O, it follows that \overline{A} \subset \overline{O}=O. To show O \subset \overline{A}, pick x \in O. If x \in \omega, then x \in A. So focus on the case that x \notin \omega. It is clear that x \notin \overline{B} where B=\omega -A. But every open set containing x must contain some points of \omega. These points of \omega must be points of A. Thus we have x \in \overline{A}. \blacksquare

Proof of Result 3
Let \mathcal{A} be the set of all closed and open sets in \beta \omega. Let \mathcal{B}=\left\{\overline{A}: A \subset \omega \right\}. Lemma 3 shows that \mathcal{B} \subset \mathcal{A}. Lemma 4 shows that \mathcal{A} \subset \mathcal{B}. Thus \mathcal{A}= \mathcal{B}. We claim that \mathcal{B} is a base for \beta \omega. To this end, we show that for each open O \subset \beta \omega and for each x \in O, we can find \overline{A} \in \mathcal{B} with x \in \overline{A} \subset O. Let O be open and let x \in O. Since \beta \omega is a regular space, we can find open set V \subset \beta \omega with x \in V \subset \overline{V} \subset O. Let A=V \cap \omega.

We claim that x \in \overline{A}. Suppose x \notin \overline{A}. There exists open U \subset V such that x \in U and U misses \overline{A}. But U must meets some points of \omega, say, y \in U \cap \omega. Then y \in V \cap \omega=A, which is a contradiction. So we have x \in \overline{A}.

It is now clear that x \in \overline{A} \subset \overline{V} \subset O. Thus \beta \omega is zero-dimensional since \mathcal{B} is a base consisting of closed and open sets. \blacksquare

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Result 4 and Result 5

Result 5 is a corollary of Result 4. We first prove two lemmas before proving Result 4.

    Lemma 5
    For each infinite A \subset \omega, \overline{A} (the closure of A in \beta \omega) is a homeomorphic copy of \beta \omega and thus has cardinality 2^c.

Proof of Lemma 5
Let A \subset \omega. Let g:A \rightarrow [0,1] be any function (necessarily continuous). Let f:\omega \rightarrow [0,1] be defined by f(x)=g(x) for all x \in A and f(x)=0 for all x \in \omega-A. By Theorem C3, f can be extended by F:\beta \omega \rightarrow [0,1]. Let G=F \upharpoonright \overline{A}.

Note that the function G: \overline{A} \rightarrow [0,1] extends g:A \rightarrow [0,1]. Thus by Theorem U3.1, \overline{A} must be \beta A. Since A is a countably infinite discrete space, \beta A must be equivalent to \beta \omega. \blacksquare

    Lemma 6
    For each countably infinite A \subset \beta \omega-\omega such that A is relatively discrete, \overline{A} (the closure of A in \beta \omega) is a homeomorphic copy of \beta \omega and thus has cardinality 2^c.

Proof of Lemma 6
Let A=\left\{t_1,t_2,t_3,\cdots \right\} \subset \beta \omega -\omega such that A is discrete in the relative topology inherited from \beta \omega. There exist disjoint open sets G_1,G_2,G_3,\cdots (open in \beta \omega) such that for each j, t_j \in G_j. Since \beta \omega is zero-dimensional (Result 3), G_1,G_2,G_3,\cdots can be made closed and open.

Let f:A \rightarrow [0,1] be a continuous function. We show that f can be extended by F:\overline{A} \rightarrow [0,1]. Once this is shown, by Theorem U3.1, \overline{A} must be \beta A. Since A is a countable discrete space, \beta A must be equivalent to \beta \omega.

We first define w:\omega \rightarrow [0,1] by:

    \displaystyle w(n)=\left\{\begin{matrix}f(t_j)& \exists \ j \text{ such that } n \in \omega \cap G_j\\{0}&\text{otherwise} \end{matrix}\right.

The function w is well defined since each n \in \omega is in at most one G_j. By Theorem C3, the function w is extended by some continuous W:\beta \omega \rightarrow [0,1]. By Lemma 4, for each j, G_j=\overline{\omega \cap G_j}. Thus, for each j, t_j \in \overline{\omega \cap G_j}. Note that W is a constant function on the set \omega \cap G_j (mapping to the constant value of f(t_j)). Thus W(t_j)=f(t_j) for each j. So let F=W \upharpoonright \overline{A}. Thus F is the desired function that extends f. \blacksquare

Proof of Result 4
Let C \subset \beta \omega be an infinite closed set. Either C \cap \omega is infinite or C \cap (\beta \omega-\omega) is infinite. If C \cap \omega is infinite, then by Lemma 5, \overline{C \cap \omega} is a homeomorphic copy of \beta \omega. Now focus on the case that C_0=C \cap (\beta \omega-\omega) is infinite. We can choose inductively a countably infinite set A \subset C_0 such that A is relatively discrete. Then by Lemma 6 \overline{A} is a copy of \beta \omega that is a subset of C. \blacksquare

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Result 6

We prove that no point in the remainder \beta \omega-\omega is an isolated point. To see this, pick x \in \beta \omega-\omega and pick an arbitrary closed and open set O \subset \beta \omega with x \in O. Let V=O \cap (\beta \omega-\omega) (thus an arbitrary open set in the remainder containing x). By Lemma 4, O=\overline{A} where A=O \cap \omega. According to Lemma 5, O=\overline{A} is a copy of \beta \omega and thus has cardinality 2^c. The set V is O minus a subset of \omega. Thus V must contains 2^c many points. This means that \left\{ x \right\} can never be open in the remainder \beta \omega-\omega. In fact, we just prove that any open and closed subset of \beta \omega-\omega (thus any open subset) must have cardinality at least 2^c. \blacksquare

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Result 7

The results under Result 7 are corollary of Result 5 (there is no non-trivial convergent sequence in \beta \omega). To see Result 7.1, note that every point x in the remainder is not an isolated point and hence cannot have a countable local base (otherwise there would be a non-trivial convergent sequence converging to x).

A space Y is said to be a Frechet space if A \subset Y and for each x \in \overline{A}, there is a sequence \left\{ x_n \right\} of points of A such that x_n \rightarrow x. A set A \subset Y is said to be sequentially closed in Y if for any sequence \left\{ x_n \right\} of points of A, x_n \rightarrow x implies x \in A. A space Y is said to be a sequential space if A \subset Y is a closed set if and only if A is a sequentially closed set. If a space is Frechet, then it is sequential. It is clear that \beta \omega is not a sequential space.

A space is said to be sequentially compact if every sequence of points in this space has a convergent subsequence. Even though \beta \omega is compact, it cannot be sequentially compact.

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Result 8

Result 7.1 indicates that no point of remainder \beta \omega-\omega can have a countable local base. In fact, no point of the remainder can be a G_\delta-point (a point that is the intersection of countably many open sets). The remainder \beta \omega-\omega is a compact space (being a closed subset of \beta \omega). In a compact space, if a point is a G_\delta-point, then there is a countable local base at that point (see 3.1.F (a) on page 135 of [1] or 17F.7 on page 125 of [2]). \blacksquare

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Result 9

The space \beta \omega is a separable space since \omega is a dense set. Thus \beta \omega has the countable chain condition. However, the remainder \beta \omega-\omega does not have the countable chain condition. We show that there is a disjoint collection of c many open sets in \beta \omega-\omega.

There is a family \mathcal{A} of infinite subsets of \omega such that for every A,B \in \mathcal{A} with A \ne B, A \cap B is finite. Such a collection of sets is said to be an almost disjoint family. There is even an almost disjoint family of cardinality c (see A Space with G-delta Diagonal that is not Submetrizable). Let \mathcal{A} be such a almost disjoint family.

For each A \in \mathcal{A}, let U_A=\overline{A} and V_A=\overline{A} \cap (\beta \omega -\omega). By Lemma 3, each U_A is a closed and open set in \beta \omega. Thus each V_A is a closed and open set in the remainder \beta \omega-\omega. Note that \left\{V_A: A \in \mathcal{A} \right\} is a disjoint collection of open sets in \beta \omega-\omega. \blacksquare

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Blog Posts on Stone-Cech Compactification

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Sequentially compact spaces, II

All spaces under consideration are Hausdorff. Countably compactness and sequentially compactness are notions related to compactness. A countably compact space is one in which every countable open cover has a finite subcover, or equivalently, every countably infinite subset has a limit point. For a space X, the point p \in X is a limit point of A \subset X if every open subset of X containing p contains a point of A distinct from p. On the other hand, a space X is sequentially compact if every sequence \left\{x_n:n=1,2,3,\cdots\right\} of points of X has a subsequence that converges. Any sequentially compact space is countably compact. The converse is not true. The product space 2^I where I=[0,1] is not sequentially compact (see Sequentially compact spaces, I) . However, for sequential spaces (first countable spaces in particular), the notion of sequentially compactness and countably compactness are equivalent. For previous discussion in this blog about sequential spaces, see the links below.

Lemma
Any countably compact space that is countable in size is metrizable and thus first countable.

Proof. Let X be countably compact such that \lvert X \lvert=\aleph_0. Then X is compact (any Lindelof countably compact space is compact). In any countable space, the set of all singleton sets is a countable network. Any compact Hausdorff space with a countable network is metrizable and thus first countable. See Spaces With Countable Network. \blacksquare

Theorem
Let X be a sequential space. Then X is countably compact if and only if X is sequentially compact.

Proof. The direction \Leftarrow always holds without the space being sequential.

\Rightarrow Suppose X is countably compact. Suppose that X is not sequentially compact. Then there is a sequence \left\{x_n\right\} of points of X with no convergent subsequence. Let A be the set of all terms in this sequence, i.e. A=\left\{x_n:n=1,2,3,\cdots\right\}. Note that A is sequentially closed. Since X is sequential, A is closed in X. As a closed subset of a countably compact space, A is countably compact. By the lemma, A is first countable. Since A is an infinite compact space, A has a non-isolated point x. This means some sequence of points of A converges to x, contradicting the assumption that \left\{x_n\right\} has no convergent subsequence. Therefore X must be sequentially compact. \blacksquare

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space
An observation about sequential spaces

Reference

  1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
  3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.