Countably paracompact spaces

This post is a basic discussion on countably paracompact space. A space is a paracompact space if every open cover has a locally finite open refinement. The definition can be tweaked by saying that only open covers of size not more than a certain cardinal number \tau can have a locally finite open refinement (any space with this property is called a \tau-paracompact space). The focus here is that the open covers of interest are countable in size. Specifically, a space is a countably paracompact space if every countable open cover has a locally finite open refinement. Even though the property appears to be weaker than paracompact spaces, the notion of countably paracompactness is important in general topology. This post discusses basic properties of such spaces. All spaces under consideration are Hausdorff.

Basic discussion of paracompact spaces and their Cartesian products are discussed in these two posts (here and here).

A related notion is that of metacompactness. A space is a metacompact space if every open cover has a point-finite open refinement. For a given open cover, any locally finite refinement is a point-finite refinement. Thus paracompactness implies metacompactness. The countable version of metacompactness is also interesting. A space is countably metacompact if every countable open cover has a point-finite open refinement. In fact, for any normal space, the space is countably paracompact if and only of it is countably metacompact (see Corollary 2 below).

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Normal Countably Paracompact Spaces

A good place to begin is to look at countably paracompactness along with normality. In 1951, C. H. Dowker characterized countably paracompactness in the class of normal spaces.

Theorem 1 (Dowker’s Theorem)
Let X be a normal space. The following conditions are equivalent.

  1. The space X is countably paracompact.
  2. Every countable open cover of X has a point-finite open refinement.
  3. If \left\{U_n: n=1,2,3,\cdots \right\} is an open cover of X, there exists an open refinement \left\{V_n: n=1,2,3,\cdots \right\} such that \overline{V_n} \subset U_n for each n.
  4. The product space X \times Y is normal for any compact metric space Y.
  5. The product space X \times [0,1] is normal where [0,1] is the closed unit interval with the usual Euclidean topology.
  6. For each sequence \left\{A_n \subset X: n=1,2,3,\cdots \right\} of closed subsets of X such that A_1 \supset A_2 \supset A_3 \supset \cdots and \cap_n A_n=\varnothing, there exist open sets B_1,B_2,B_3,\cdots such that A_n \subset B_n for each n such that \cap_n B_n=\varnothing.

Dowker’s Theorem is proved in this previous post. Condition 2 in the above formulation of the Dowker’s theorem is not in the Dowker’s theorem in the previous post. In the proof for 1 \rightarrow 2 in the previous post is essentially 1 \rightarrow 2 \rightarrow 3 for Theorem 1 above. As a result, we have the following.

Corollary 2
Let X be a normal space. Then X is countably paracompact if and only of X is countably metacompact.

Theorem 1 indicates that normal countably paracompact spaces are important for the discussion of normality in product spaces. As a result of this theorem, we know that normal countably paracompact spaces are productively normal with compact metric spaces. The Cartesian product of normal spaces with compact spaces can be non-normal (an example is found here). When the normal factor is countably paracompact and the compact factor is upgraded to a metric space, the product is always normal. The connection with normality in products is further demonstrated by the following corollary of Theorem 1.

Corollary 3
Let X be a normal space. Let Y be a non-discrete metric space. If X \times Y is normal, then X is countably paracompact.

Since Y is non-discrete, there is a non-trivial convergent sequence (i.e. the sequence represents infinitely many points). Then the sequence along with the limit point is a compact metric subspace of Y. Let’s call this subspace S. Then X \times S is a closed subspace of the normal X \times Y. As a result, X \times S is normal. By Theorem 1, X is countably paracompact.

C. H. Dowker in 1951 raised the question: is every normal space countably paracompact? Put it in another way, is the product of a normal space and the unit interval always a normal space? As a result of Theorem 1, any normal space that is not countably paracompact is called a Dowker space. The search for a Dowker space took about 20 years. In 1955, M. E. Rudin showed that a Dowker space can be constructed from assuming a Souslin line. In the mid 1960s, the existence of a Souslin line was shown to be independent of the usual axioms of set theorey (ZFC). Thus the existence of a Dowker space was known to be consistent with ZFC. In 1971, Rudin constructed a Dowker space in ZFC. Rudin’s Dowker space has large cardinality and is pathological in many ways. Zoltan Balogh constructed a small Dowker space (cardinality continuum) in 1996. Various Dowker space with nicer properties have also been constructed using extra set theory axioms. The first ZFC Dowker space constructed by Rudin is found in [2]. An in-depth discussion of Dowker spaces is found in [3]. Other references on Dowker spaces is found in [4].

Since Dowker spaces are rare and are difficult to come by, we can employ a “probabilistic” argument. For example, any “concrete” normal space (i.e. normality can be shown without using extra set theory axioms) is likely to be countably paracompact. Thus any space that is normal and not paracompact is likely countably paracompact (if the fact of being normal and not paracompact is established in ZFC). Indeed, any well known ZFC example of normal and not paracompact must be countably paracompact. In the long search for Dowker spaces, researchers must have checked all the well known examples! This probability thinking is not meant to be a proof that a given normal space is countably paracompact. It is just a way to suggest a possible answer. In fact, a good exercise is to pick a normal and non-paracompact space and show that it is countably paracompact.

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Some Examples

The following lists out a few classes of spaces that are always countably paracompact.

  • Metric spaces are countably paracompact.
  • Paracompact spaces are countably paracompact.
  • Compact spaces are countably paracompact.
  • Countably compact spaces are countably paracompact.
  • Perfectly normal spaces are countably paracompact.
  • Normal Moore spaces are countably paracompact.
  • Linearly ordered spaces are countably paracompact.
  • Shrinking spaces are countably paracompact.

The first four bullet points are clear. Metric spaces are paracompact. It is clear from definition that paracompact spaces, compact and countably compact spaces are countably paracompact. One way to show perfect normal spaces are countably paracompact is to show that they satisfy condition 6 in Theorem 1 (shown here). Any Moore space is perfect (closed sets are G_\delta). Thus normal Moore space are perfectly normal and hence countably paracompact. The proof of the countably paracompactness of linearly ordered spaces can be found in [1]. See Theorem 5 and Corollary 6 below for the proof of the last bullet point.

As suggested by the probability thinking in the last section, we now look at examples of countably paracompact spaces among spaces that are “normal and not paracompact”. The first uncountable ordinal \omega_1 is normal and not paracompact. But it is countably compact and is thus countably paracompact.

Example 1
Any \Sigma-product of uncountably many metric spaces is normal and countably paracompact.

For each \alpha<\omega_1, let X_\alpha be a metric space that has at least two points. Assume that each X_\alpha has a point that is labeled 0. Consider the following subspace of the product space \prod_{\alpha<\omega_1} X_\alpha.

    \displaystyle \Sigma_{\alpha<\omega_1} X_\alpha =\left\{f \in \prod_{\alpha<\omega_1} X_\alpha: \ f(\alpha) \ne 0 \text{ for at most countably many } \alpha \right\}

The space \Sigma_{\alpha<\omega_1} X_\alpha is said to be the \Sigma-product of the spaces X_\alpha. It is well known that the \Sigma-product of metric spaces is normal, in fact collectionwise normal (this previous post has a proof that \Sigma-product of separable metric spaces is collectionwise normal). On the other hand, any \Sigma-product always contains \omega_1 as a closed subset as long as there are uncountably many factors and each factor has at least two points (see the lemma in this previous post). Thus any such \Sigma-product, including the one being discussed, cannot be paracompact.

Next we show that T=(\Sigma_{\alpha<\omega_1} X_\alpha) \times [0,1] is normal. The space T can be reformulated as a \Sigma-product of metric spaces and is thus normal. Note that T=\Sigma_{\alpha<\omega_1} Y_\alpha where Y_0=[0,1], for any n with 1 \le n<\omega, Y_n=X_{n-1} and for any \alpha with \alpha>\omega, Y_\alpha=X_\alpha. Thus T is normal since it is the \Sigma-product of metric spaces. By Theorem 1, the space \Sigma_{\alpha<\omega_1} X_\alpha is countably paracompact. \square

Example 2
Let \tau be any uncountable cardinal number. Let D_\tau be the discrete space of cardinality \tau. Let L_\tau be the one-point Lindelofication of D_\tau. This means that L_\tau=D_\tau \cup \left\{\infty \right\} where \infty is a point not in D_\tau. In the topology for L_\tau, points in D_\tau are isolated as before and open neighborhoods at \infty are of the form L_\tau - C where C is any countable subset of D_\tau. Now consider C_p(L_\tau), the space of real-valued continuous functions defined on L_\tau endowed with the pointwise convergence topology. The space C_p(L_\tau) is normal and not Lindelof, hence not paracompact (discussed here). The space C_p(L_\tau) is also homeomorphic to a \Sigma-product of \tau many copies of the real lines. By the same discussion in Example 1, C_p(L_\tau) is countably paracompact. For the purpose at hand, Example 2 is similar to Example 1. \square

Example 3
Consider R. H. Bing’s example G, which is a classic example of a normal and not collectionwise normal space. It is also countably paracompact. This previous post shows that Bing’s Example G is countably metacompact. By Corollary 2, it is countably paracompact. \square

Based on the “probabilistic” reasoning discussed at the end of the last section (based on the idea that Dowker spaces are rare), “normal countably paracompact and not paracompact” should be in plentiful supply. The above three examples are a small demonstration of this phenomenon.

Existence of Dowker spaces shows that normality by itself does not imply countably paracompactness. On the other hand, paracompact implies countably paracompact. Is there some intermediate property that always implies countably paracompactness? We point that even though collectionwise normality is intermediate between paracompactness and normality, it is not sufficiently strong to imply countably paracompactness. In fact, the Dowker space constructed by Rudin in 1971 is collectionwise normal.

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More on Countably Paracompactness

Without assuming normality, the following is a characterization of countably paracompact spaces.

Theorem 4
Let X be a topological space. Then the space X is countably paracompact if and only of the following condition holds.

  • For any decreasing sequence \left\{A_n: n=1,2,3,\cdots \right\} of closed subsets of X such that \cap_n A_n=\varnothing, there exists a decreasing sequence \left\{B_n: n=1,2,3,\cdots \right\} of open subsets of X such that A_n \subset B_n for each n and \cap_n \overline{B_n}=\varnothing.

Proof of Theorem 4
Suppose that X is countably paracompact. Suppose that \left\{A_n: n=1,2,3,\cdots \right\} is a decreasing sequence of closed subsets of X as in the condition in the theorem. Then \mathcal{U}=\left\{X-A_n: n=1,2,3,\cdots \right\} is an open cover of X. Let \mathcal{V} be a locally finite open refinement of \mathcal{U}. For each n=1,2,3,\cdots, define the following:

    B_n=\cup \left\{V \in \mathcal{V}: V \cap A_n \ne \varnothing  \right\}

It is clear that A_n \subset B_n for each n. The open sets B_n are decreasing, i.e. B_1 \supset B_2 \supset \cdots since the closed sets A_n are decreasing. To show that \cap_n \overline{B_n}=\varnothing, let x \in X. The goal is to find B_j such that x \notin \overline{B_j}. Once B_j is found, we will obtain an open set V such that x \in V and V contains no points of B_j.

Since \mathcal{V} is locally finite, there exists an open set V such that x \in V and V meets only finitely many sets in \mathcal{V}. Suppose that these finitely many open sets in \mathcal{V} are V_1,V_2,\cdots,V_m. Observe that for each i=1,2,\cdots,m, there is some j(i) such that V_i \cap A_{j(i)}=\varnothing (i.e. V_i \subset X-A_{j(i)}). This follows from the fact that \mathcal{V} is a refinement \mathcal{U}. Let j be the maximum of all j(i) where i=1,2,\cdots,m. Then V_i \cap A_{j}=\varnothing for all i=1,2,\cdots,m. It follows that the open set V contains no points of B_j. Thus x \notin \overline{B_j}.

For the other direction, suppose that the space X satisfies the condition given in the theorem. Let \mathcal{U}=\left\{U_n: n=1,2,3,\cdots \right\} be an open cover of X. For each n, define A_n as follows:

    A_n=X-U_1 \cup U_2 \cup \cdots \cup U_n

Then the closed sets A_n form a decreasing sequence of closed sets with empty intersection. Let B_n be decreasing open sets such that \bigcap_{i=1}^\infty \overline{B_i}=\varnothing and A_n \subset B_n for each n. Let C_n=X-B_n for each n. Then C_n \subset \cup_{j=1}^n U_j. Define V_1=U_1. For each n \ge 2, define V_n=U_n-\bigcup_{j=1}^{n-1}C_{j}. Clearly each V_n is open and V_n \subset U_n. It is straightforward to verify that \mathcal{V}=\left\{V_n: n=1,2,3,\cdots \right\} is a cover of X.

We claim that \mathcal{V} is locally finite in X. Let x \in X. Choose the least n such that x \notin \overline{B_n}. Choose an open set O such that x \in O and O \cap \overline{B_n}=\varnothing. Then O \cap B_n=\varnothing and O \subset C_n. This means that O \cap V_k=\varnothing for all k \ge n+1. Thus the open cover \mathcal{V} is a locally finite refinement of \mathcal{U}. \square

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We present another characterization of countably paracompact spaces that involves the notion of shrinkable open covers. An open cover \mathcal{U} of a space X is said to be shrinkable if there exists an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} of the space X such that for each U \in \mathcal{U}, \overline{V(U)} \subset U. If \mathcal{U} is shrinkable by \mathcal{V}, then we also say that \mathcal{V} is a shrinking of \mathcal{U}. Note that Theorem 1 involves a shrinking. Condition 3 in Theorem 1 (Dowker’s Theorem) can rephrased as: every countable open cover of X has a shrinking. This for any normal countably paracompact space, every countable open cover has a shrinking (or is shrinkable).

A space X is a shrinking space if every open cover of X is shrinkable. Every shrinking space is a normal space. This follows from this lemma: A space X is normal if and only if every point-finite open cover of X is shrinkable (see here for a proof). With this lemma, it follows that every shrinking space is normal. The converse is not true. To see this we first show that any shrinking space is countably paracompact. Since any Dowker space is a normal space that is not countably paracompact, any Dowker space is an example of a normal space that is not a shrinking space. To show that any shrinking space is countably paracompact, we first prove the following characterization of countably paracompactness.

Theorem 5
Let X be a space. Then X is countably paracompact if and only of every countable increasing open cover of X is shrinkable.

Proof of Theorem 5
Suppose that X is countably paracompact. Let \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an increasing open cover of X. Then there exists a locally open refinement \mathcal{V}_0 of \mathcal{U}. For each n, define V_n=\cup \left\{O \in \mathcal{V}_0: O \subset U_n \right\}. Then \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is also a locally finite refinement of \mathcal{U}. For each n, define

    G_n=\cup \left\{O \subset X: O \text{ is open and } \forall \ m > n, O \cap V_m=\varnothing \right\}

Let \mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}. It follows that G_n \subset G_m if n<m. Then \mathcal{G} is an increasing open cover of X. Observe that for each n, \overline{G_n} \cap V_m=\varnothing for all m > n. Then we have the following:

    \displaystyle \begin{aligned} \overline{G_n}&\subset X-\cup \left\{V_m: m > n \right\} \\&\subset \cup \left\{V_k: k=1,2,\cdots,n \right\} \\&\subset \cup \left\{U_k: k=1,2,\cdots,n \right\}=U_n  \end{aligned}

We have just established that \mathcal{G} is a shrinking of \mathcal{U}, or that \mathcal{U} is shrinkable.

For the other direction, to show that X is countably paracompact, we show that the condition in Theorem 4 is satisfied. Let \left\{A_1,A_2,A_3,\cdots \right\} be a decreasing sequence of closed subsets of X with empty intersection. Then \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an open cover of X where U_n=X-A_n for each n. By assumption, \mathcal{U} is shrinkable. Let \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} be a shrinking. We can assume that \mathcal{V} is an increasing sequence of open sets.

For each n, let B_n=X-\overline{V_n}. We claim that \left\{B_1,B_2,B_3,\cdots \right\} is a decreasing sequence of open sets that expand the closed sets A_n and that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. The expansion part follows from the following:

    A_n=X-U_n \subset X-\overline{V_n}=B_n

The part about decreasing follows from:

    B_{n+1}=X-\overline{V_{n+1}} \subset X-\overline{V_n}=B_n

We show that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. To this end, let x \in X. Then x \in V_n for some n. We claim that x \notin \overline{B_n}. Suppose x \in \overline{B_n}. Since V_n is an open set containing x, V_n must contain a point of B_n, say y. Since y \in B_n, y \notin \overline{V_n}. This in turns means that y \notin V_n, a contradiction. Thus we have x \notin \overline{B_n} as claimed. We have established that every point of X is not in \overline{B_n} for some n. Thus the intersection of all the \overline{B_n} must be empty. We have established the condition in Theorem 4 is satisfied. Thus X is countably paracompact. \square

Corollary 6
If X is a shrinking space, then X is countably paracompact.

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Reference

  1. Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
  2. Rudin, M. E., A Normal Space X for which X \times I is not Normal, Fund. Math., 73, 179-486, 1971. (link)
  3. Rudin, M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
  4. Wikipedia Entry on Dowker Spaces (link)

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\copyright \ 2016 \text{ by Dan Ma}

A stroll in Bing’s Example G

In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, F \times [0,1], is a normal space.

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The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let P be any uncountable set. Let Q be the set of all subsets of P. In other words, Q is the power set of P. Consider the product of \lvert Q \lvert many copies of the two element set \left\{0,1 \right\}. The usual notation of this product space is 2^Q. The elements of 2^Q are simply the functions from Q into \left\{0,1 \right\}. An arbitrary element of 2^Q is a function f that maps every subset of P to either 0 or 1.

Though the base set P can be any uncountable set, it is a good idea to visualize clearly what P is. In the remainder of this section, think of P as the real line \mathbb{R}. Then Q is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are 2^c many subsets of the real line where c is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of 2^Q. For each real number p, define the function f_p such that f_p always maps any set of real numbers that contains p to 1 and maps any set of real numbers that does not contain p to 0. For example, the following are several values of the function f_0.

    f_0([0,1])=1

    f_0([1,2])=0

    f_0(\left\{0 \right\})=1

    f_0(\mathbb{R}-\left\{0 \right\})=0

    f_0(\mathbb{R})=1

    f_0(\varnothing)=0

    f_0(\mathbb{P})=0

where \mathbb{P} is the set of all irrational numbers. Consider the subspace F_P=\left\{f_p: p \in P \right\}. Members of F_P are easy to describe. Each function in F_P maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that F_P is important is that Bing’s Example is defined by declaring all points not in F_P isolated points and by allowing all points in F_P retaining the open sets in the product topology.

Any point f in F_P determines f(q)=0 \text{ or } 1 based on membership (whether the reference point belongs to the set q). Points not in F_P have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any f in F_P maps equally to 0 or 1. So the constant functions f(q)=0 and f(q)=1 are not in F_P. Furthermore, any f such that f(q)=1 for at most countably many q would not be in F_P.

Let’s continue focusing on the product space for the time being. When F_P is considered as a subspace of the product space 2^Q, F_P is a discrete space. For each p \in P, there is an open set W_p containing f_p such that W_p contains no other points of F_P. So F_P is relatively discrete in the product space 2^Q. Of course F_P cannot be closed in 2^Q since 2^Q is a compact space. The open set W_p is defined as follows:

    W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}

It is clear that f_p \in W_p and that f_t \notin W_p for any real number t \ne p.

Two properties of the product space 2^Q would be very relevant for the discussion. By the well known Tychonoff theorem, the product space 2^Q is compact. Since P is uncountable, 2^Q always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets W_p cannot be all pairwise disjoint. So there exist at least a pair of W_p, say W_{a} and W_{b}, with nonempty intersection.

The last observation can be generalized. For each p \in P, let V_p be any open set containing f_p (open in the product topology). We observe that there are at least two a and b from P such that V_a \cap V_b \ne \varnothing. If there are only countably many distinct sets V_p, then there are uncountably many V_p that are identical and the observation is valid. So assume that there are uncountably many distinct V_p. By the CCC in the product space, there are at least two a and b with V_a \cap V_b \ne \varnothing. This observation shows that the discrete points in F_P cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint A_1, A_2 \subset F_P can be separated by disjoint open sets. To see this, define the following two open sets E_1 and E_2 in the product topology.

    q_1=\left\{p \in P: f_p \in A_1 \right\}

    q_2=\left\{p \in P: f_p \in A_2 \right\}

    E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}

    E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}

It is clear that A_1 \subset E_1 and A_2 \subset E_2. Furthermore, E_1 \cap E_2=\varnothing. This observation will be the basis for showing that Bing’s Example G is normal.

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The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on 2^Q. Let P be any uncountable set. Let Q be the set of all subsets of P. The set F_P is defined as above. Bing’s Example G is F=2^Q with points in F_P retaining the open sets in the product topology and with points not in F_P declared isolated. For some reason, in Bing’s original paper, the notation F is used even though the example is identified by G. We will follow Bing’s notation.

The subspace F_P is discrete but not closed in the product topology. However, F_P is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space F is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set F_P cannot be separated by disjoint open sets. By declaring points not in F_P isolated, the countable chain condition in the original product space 2^Q is destroyed. However, there is still a strong trace of CCC around the points in the set F_P, which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that F is normal, let H and K be disjoint closed subsets of F. To make it easy to follow, let H=A_1 \cup B_1 and K=A_2 \cup B_2 where

    A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)

    A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)

In other words, A is the non-isolated part and B is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets E_1 and E_2 where A_1 \subset E_1 and A_2 \subset E_2. Set the following open sets.

    O_1=(E_1 \cup B_1) - K

    O_2=(E_2 \cup B_2) - H

It follows that O_1 and O_2 are disjoint open sets and that A_1 \subset O_1 and A_2 \subset O_2. Thus Bing’s Example G is a normal space.

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Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space X is countably paracompact if every countable countable open cover of X has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every countable open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent (see Corollary 2 here). Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let \mathcal{U} be a countable open cover of F. Let \mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\} be the set of all open sets in \mathcal{U} that contain points in F_P. For each i, let A_i=U_i \cap F_P. From the perspective of Bing’s Example G, the sets A_i are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let O_1,O_2,O_3,\cdots be disjoint open sets such that A_i \subset O_i for each i.

We now build a point-finite open refinement of \mathcal{U}. For each i, let V_i=U_i \cap O_i. Let V=\cup_{i=1}^\infty V_i. Consider the following.

    \mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}

It follows that \mathcal{V} is an open cover of F. All points of F_P belong to the open sets V_i. Any point that is not in one of the V_i belongs to a singleton open set. It is also clear that \mathcal{V} is a refinement of \mathcal{U}. For each i, V_i \subset U_i and each singleton set is contained in some member of \mathcal{U}. It follows that each point in F belongs to at most finitely many sets in \mathcal{V}. In fact, each point belongs to exactly one set in \mathcal{V}. Each point in F_P belongs to exactly one V_i since the open sets O_i are disjoint. Any point in V belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, F \times [0,1], is a normal space.

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Previous Posts

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\copyright \ 2016 \text{ by Dan Ma}

A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

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The countable chain conditon

A topological space X is said to have the countable chain condition (to have the CCC for short) if \mathcal{U} is a disjoint collection of non-empty open subsets of X (meaning that for any A,B \in \mathcal{U} with A \ne B, we have A \cap B=\varnothing), then \mathcal{U} is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of discussion, if X has the CCC, we also say that X is a CCC space or X is CCC. A space X is separable if there exists a countable subset A of X such that A is dense in X (meaning that if U is a nonempty open subset of X, then U \cap A \ne \varnothing).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

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The strategy

Here’s the strategy for finding CCC and not separable.

    The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

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Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if c is the cardinality of continuum, then consider \left\{0,1 \right\}^{2^c}, the product of 2^c many copies of \left\{0,1 \right\}, or consider X^{2^c} where X is your favorite separable space.

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Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the \Sigma-products. To make this clear, let’s focus on a specific example. Consider X=\left\{0,1 \right\}^{c} where c is the cardinality of continuum. Consider the following subspace.

    \Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}

The subspace \Sigma(\left\{0,1 \right\}^{c}) is dense in X, thus has CCC. It is straightforward to verify that \Sigma(\left\{0,1 \right\}^{c}) is not separable.

To implement this example, find any space X which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the \Sigma-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The \Sigma-product has the countable chain condition since it is a dense subspace of the CCC space X. The \Sigma-product is not separable since there are uncountably many factors in the product space X and that each factor has at least two points. This idea had been implemented in this previous post.

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Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space X, let \mathcal{F}[X] be the set of all non-empty finite subsets of X. For F \in \mathcal{F}[X] and for any open subset U of X, let [F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}. The sets [F,U] over all 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 Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space X is uncountable, \mathcal{F}[X] is not a separable space. We need to identify the \mathcal{F}[X] that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space X with a countable network, \mathcal{F}[X] is CCC. Thus for any uncountable space X with a countable network, the Pixley-Roy space \mathcal{F}[X] is a CCC space that is not separable. The following gives a few such examples.

    \mathcal{F}[\mathbb{R}]

    \mathcal{F}[X] where X is any uncountable, separable and metrizable space.

    \mathcal{F}[X] where X is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space X that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

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Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line \mathbb{R}. Let X be any uncountable completely regular space. The product space \mathbb{R}^X always has the CCC since it is a product of separable space. Now we single out a dense subspace of \mathbb{R}^X for which there is a characterization for separability, namely the subspace C(X), which is the set of all continuous functions from X into \mathbb{R}. The subspace C(X) as a topological space is usually denoted by C_p(X). For a basic discussion of C_p(X), see this previous post.

We know precisely when C_p(X) is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space C_p(X) is separable if and only if the domain space X has a weaker (or coarser) separable metric topology (in other words, X is submetrizable with a separable metric topology).

Based on the theorem, C_p(X) is separable for any separable metric space X. Other examples of separable C_p(X) include spaces X that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus C_p(X) would be separable if X is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces X that have no weaker separable metric topology.

  • Any compact space X that is not metrizable.
  • The space X=\omega_1, the first uncountable ordinal with the order topology.
  • Any space X=C_p(Y) where Y is not separable.

The function space C_p(X) for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact X are: the first uncountable successor ordinal \omega_1+1, the double arrow space, and the product space \left\{0,1 \right\}^{\omega_1}.

It can be shown that C_p(\omega_1) is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space C_p(Y) has a weaker (or coarser) separable metric topology if and only if the domain space Y is separable.

Thus picking a non-separable space Y would guarantee that C_p(Y) has a weaker separable metric topology. As a result, C_p(C_p(Y)) is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

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Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space X that is CCC and not separable from any one of the avenues discussed here, the function space C_p(X) has no weaker separable metric topology (by Theorem 2) and the function space C_p(C_p(X)) is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

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Exercises

  1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if X is metrizable and CCC, then X is separable.
  2. Show that any dense subspace of a CCC space is also CCC.
  3. Verify that the space \Sigma(\left\{0,1 \right\}^{c}) defined in Example 2 is dense in X and is not separable.
  4. Verify that the Pixley-Roy space \mathcal{F}[\mathbb{R}] defined in Example 3 is CCC and not separable.
  5. Verify that function space C_p(\omega_1) mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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\copyright \ 2016 \text{ by Dan Ma}

Product Space – Exercise Set 1

This post presents several exercises concerning product spaces. All the concepts involved in the exercises have been discussed in the blog.

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Exercise 1

Exercise 1a
Prove or disprove:
If X and Y are both hereditarily separable, then X \times Y is hereditarily separable.

Exercise 1b
Show that if each X_\alpha is separable, then the product space \prod_{\alpha < \omega} \ X_\alpha is separable.

Exercise 1c
Prove or disprove:
If each X_\alpha is separable, then the product space \prod_{\alpha < \omega_1} \ X_\alpha is not separable.

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Exercise 2

Exercise 2a
Show that if the space X is normal, then every closed subspace of X is a normal space.

Exercise 2b
Prove or disprove:
If the space X is normal, then every dense open subspace of X is a normal space.

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Exercise 3

Consider the product space \prod_{\alpha \in W} \ X_\alpha.

Exercise 3a
Suppose that X_\alpha is compact for all but one \alpha \in W such that the non-compact factor is a Lindelof space. Show that the product space \prod_{\alpha \in W} \ X_\alpha is a normal space.

Exercise 3b
Prove or disprove:
Suppose that X_\alpha is compact for all but one \alpha \in W such that the non-compact factor is a normal space. Then the product space \prod_{\alpha \in W} \ X_\alpha is a normal space.

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Exercise 4

Exercise 4a
Let X be a compact space.
Show that if X^n is hereditarily Lindelof for all positive integer n, then X is metrizable.

Exercise 4b
Prove or disprove:
If X^n is hereditarily Lindelof for all positive integer n, then X is metrizable.

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Exercise 5

Let Y the product of uncountably many copies of the real line \mathbb{R}. If a specific example is desired, try Y=\mathbb{R}^{\omega_1} (\omega_1 many copies of \mathbb{R}) or Y=\mathbb{R}^{\mathbb{R}} (continuum many copies of \mathbb{R}). It is also OK to use a larger number of copies of the real line.

Note that the space Y is not normal (see here).

Exercise 5a
Since the product space Y is not normal, it is not Lindelof. As an exercise, find an open cover of Y that proves that Y is not Lindelof, i.e. an open cover \mathcal{U} of Y such that no countable subcollection of \mathcal{U} can cover Y.

Exercise 5b
Show that for every open cover \mathcal{U} of the space Y, there is a countable \mathcal{V} \subset \mathcal{U} of Y such that \overline{\mathcal{V}}=Y, i.e. \cup \mathcal{V} is dense in Y. Note that with this property, the space Y is said to be weakly Lindelof.

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Exercise 6

This exercise is about the product Y=\mathbb{R}^{\mathbb{R}} (continuum many copies of \mathbb{R}). Show the following.

  1. Show that Y is separable by exhibiting a countable dense set.
  2. Show that Y is not hereditarily separable by exhibiting a non-separable subspace.
  3. Show that the space Y has a closed and discrete subspace of cardinality continuum.
  4. Show that Y is not first countable.
  5. Show that Y is not a Frechet space.
  6. Show that Y is not a countably tight space.

See here for the definition of Frechet space.

See here for the definition of countably tight space.

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Exercise 7

Consider the product space Y=\mathbb{\omega}^{\omega_1}. It is not normal (see here).

Exercise 7a
Construct a dense normal subspace of Y.

Exercise 7b
Construct a dense Lindelof subspace of Y.

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\copyright \ 2016 \text{ by Dan Ma}

Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space X=\omega_1 \times I^I, which is the product of \omega_1 with the interval topology and the product space I^I=\prod_{t \in I} I where I is of course the unit interval [0,1]. The notation of \omega_1, the first uncountable ordinal, in Steen and Seebach is [0,\Omega).

Another way to notate the example X is the product space \prod_{t \in I} X_t where X_0 is \omega_1 and X_t is the unit interval I for all t>0. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of \omega_1 makes this product space an interesting example.

The following lists out the basic topological properties of the space that X=\omega_1 \times I^I are covered in [2].

  • The space X is Hausdorff and completely regular.
  • The space X is countably compact.
  • The space X is neither compact nor sequentially compact.
  • The space X is neither separable, Lindelof nor \sigma-compact.
  • The space X is not first countable.
  • The space X is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

  • The space X is not normal.
  • The space X has a dense subspace that is normal.

It follows from these bullet points that the space X is an example of a completely regular space that is not normal. Not being a normal space, X is then not metrizable. Of course there are other ways to show that X is not metrizable. One is that neither of the two factors \omega_1 or I^I is metrizable. Another is that X is not first countable.

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The space X is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example X=\omega_1 \times I^I is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example X=\omega_1 \times I^I is to show whether the second factor I^I is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space Y, the product \omega_1 \times Y is normal if and only if Y is countably tight. Thus the normality of the space X (or the lack of) hinges on whether the compact factor I^I=\prod_{t \in I} I is countably tight.

A space Y is countably tight (or has countable tightness) if for each S \subset Y and for each x \in \overline{S}, there exists some countable B \subset S such that x \in \overline{B}. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space I^I=\prod_{t \in I} I is not countably tight, we let S be the subspace of I^I consisting of points, each of which is non-zero on at most countably many coordinates. Specifically S is defined as follows:

    S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}

The set S just defined is also called the \Sigma-product of copies of unit interval I. Let g \in I^I be defined by g(t)=1 for all t \in I. It follows that g \in \overline{S}. It can also be verified that g \notin \overline{B} for any countable B \subset S. This shows that the product space I^I=\prod_{t \in I} I is not countably tight.

By Theorem 1 found in this link, the space X=\omega_1 \times I^I is not normal.

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The space X has a dense subspace that is normal

Now that we know X=\omega_1 \times I^I is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace \omega_1 \times S where S is the \Sigma-product S=\Sigma_{t \in I} I defined in the preceding section. The subspace S is dense in the product space I^I. Thus \omega_1 \times S is dense in X=\omega_1 \times I^I. The space S is normal since the \Sigma-product of separable metric spaces is normal. Furthermore, \omega_1 can be embedded as a closed subspace of S=\Sigma_{t \in I} I. Then \omega_1 \times S is homeomorphic to a closed subspace of S \times S. Since S \times S \cong S, the space \omega_1 \times S is normal.

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Reference

  1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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\copyright \ 2015 \text{ by Dan Ma}

An exercise gleaned from the proof of a theorem on pseudocompact space

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space X is a pseudocompact space if every continuous real-valued function defined on X is bounded, i.e., if f:X \rightarrow \mathbb{R} is a continuous function, then f(X) is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

    \text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space \Psi(\mathcal{A}) where \mathcal{A} is a maximal almost disjoint family of subsets of \omega (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let Y be a compact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto a space Z. Then g is a homeomorphism.

Theorem 2
Let Y be a pseudocompact space. Let g: Y \rightarrow Z be a one-to-one continuous function from Y onto Z where Z is a separable and metrizable space. Then g is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map g as described in Theorem 2, the domain Y must be compact. Then Theorem 1 will finish the job.

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Reference

  1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

.

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\copyright \ 2015 \text{ by Dan Ma}

A note on products of sequential fans

Two posts (the previous post and this post) are devoted to discussing the behavior of countable tightness in taking Cartesian products. The previous post shows that countable tightness behaves well in the product operation if the spaces are compact. In this post, we step away from the orderly setting of compact spaces. We examine the behavior of countable tightness in product of sequential fans. In this post, we show that countable tightness can easily be destroyed when taking products of sequential fans. Due to the combinatorial nature of sequential fans, the problem of determining the tightness of products of fans is often times a set-theoretic problem. In many instances, it is hard to determine the tightness of a product of two sequential fans without using extra set theory axioms beyond ZFC. The sequential fans is a class of spaces that have been studied extensively and are involved in the solutions of many problems that were seemingly unrelated. For one example, see [3].

For a basic discussion of countable tightness, see these previous post on the notion of tightness and its relation with free sequences. Also see chapter a-4 on page 15 of [4].

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Sequential fans

Let S be a non-trivial convergent sequence along with its limit point. For convenience, let \displaystyle S=\left\{0 \right\} \cup \left\{1, 2^{-1}, 3^{-1}, 4^{-1}, \cdots \right\}, considered as a subspace of the Euclidean real line. Let \kappa be a cardinal number. The set \kappa is usually taken as the set of all the ordinals that precede \kappa. The set \omega is the first infinite ordinal, or equivalently the set of all non-negative integers. Let \omega^\kappa be the set of all functions from \kappa into \omega.

There are several ways to describe a sequential fan. One way is to describe it as a quotient space. The sequential fan S(\kappa) is the topological sum of \kappa many copies of the convergent sequence S with all non-isolated points identified as one point that is called \infty. To make the discussion easier to follow, we also use the following formulation of S(\kappa):

    \displaystyle S(\kappa)=\left\{\infty \right\} \cup (\kappa \times \omega)

In this formulation, every point is \kappa \times \omega is isolated and an open neighborhood of the point \infty is of the form:

    \displaystyle B_f=\left\{\infty \right\} \cup \left\{(\alpha,n) \in \kappa \times \omega: n \ge f(\alpha) \right\} where f \in \omega^\kappa.

According to the definition of the open neighborhood B_f, the sequence (\alpha,0), (\alpha,1), (\alpha,2),\cdots converges to the point \infty for each \alpha \in \kappa. Thus the set (\left\{\alpha \right\} \times \omega) \cup \left\{\infty \right\} is a homeomorphic copy of the convergent sequence S. The set \left\{\alpha \right\} \times \omega is sometimes called a spine. Thus the space S(\kappa) is said to be the sequential fan with \kappa many spines.

The point \infty is the only non-isolated point in the fan S(\kappa). The set \mathcal{B}=\left\{B_f: f \in \omega^\kappa \right\} is a local base at the point \infty. The base \mathcal{B} is never countable except when \kappa is finite. Thus if \kappa is infinite, the fan S(\kappa) can never be first countable. In particular, for the fan S(\omega), the character at the point \infty is the cardinal number \mathfrak{d}. See page 13 in chapter a-3 of [4]. This cardinal number is called the dominating number and is introduced below in the section “The bounding number”. This is one indication that the sequential fan is highly dependent on set theory. It is hard to pinpoint the character of S(\omega) at the point \infty. For example, it is consistent with ZFC that \mathfrak{d}=\omega_1. It is also consistent that \mathfrak{d}>\omega_1.

Even though the sequential fan is not first countable, it has a relatively strong convergent property. If \infty \in \overline{A} and \infty \notin A where A \subset S(\kappa), then infinitely many points of A are present in at least one of the spine \left\{\alpha \right\} \times \omega (if this is not true, then \infty \notin \overline{A}). This means that the sequential fan is always a Frechet space. Recall that the space Y is a Frechet space if for each A \subset Y and for each x \in \overline{A}, there exists a sequence \left\{x_n \right\} of points of A converging to x.

Some of the convergent properties weaker than being a first countable space are Frechet space, sequential space and countably tight space. Let's recall the definitions. A space X is a sequential space if A \subset X is a sequentially closed set in X, then A is a closed set in X. The set A is sequentially closed in X if this condition is satisfied: if the sequence \left\{x_n \in A: n \in \omega \right\} converges to x \in X, then x \in A. A space X is countably tight (have countable tightness) if for each A \subset X and for each x \in \overline{A}, there exists a countable B \subset A such that x \in \overline{B}. See here for more information about these convergent properties. The following shows the relative strength of these properties. None of the implications can be reversed.

    First countable space \Longrightarrow Frechet space \Longrightarrow Sequential space \Longrightarrow Countably tight space

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Examples

The relatively strong convergent property of being a Frechet space is not preserved in products or squares of sequential fans. We now look at some examples.

Example 1
Consider the product space S(\omega) \times S where S is the convergent sequence defined above. The first factor is Frechet and the second factor is a compact metric space. We show that S(\omega) \times S is not sequential. To see this, consider the following subset A of S(\omega) \times S:

    \displaystyle A_f=\left\{(x,n^{-1}) \in S(\omega) \times S: n \in \omega \text{ and } x=(n,f(n)) \right\} \ \forall \ f \in \omega^\omega

    \displaystyle A=\bigcup_{f \in \omega^\omega} A_f

It follows that (\infty,0) \in \overline{A}. Furthermore, no sequence of points of A can converge to the point (\infty,0). To see this, let a_n \in A for each n. Consider two cases. One is that some spine \left\{m \right\} \times \omega contains the first coordinate of a_n for infinitely many n \in \omega. The second is the opposite of the first – each spine \left\{m \right\} \times \omega contains the first coordinate of a_n for at most finitely many n. Either case means that there is an open set containing (\infty,0) that misses infinitely many a_n. Thus the sequence a_n cannot converge to (\infty,0).

Let A_1 be the set of all sequential limits of convergent sequences of points of A. With A \subset A_1, we know that (\infty,0) \in \overline{A_1} but (\infty,0) \notin A_1. Thus A_1 is a sequentially closed subset of S(\omega) \times S that is not closed. This shows that S(\omega) \times S is not a sequential space.

The space S(\omega) \times S is an example of a space that is countably tight but not sequential. The example shows that the product of two Frechet spaces does not even have to be sequential even when one of the factors is a compact metric space. The next example shows that the product of two sequential fans does not even have to be countably tight.

Example 2
Consider the product space S(\omega) \times S(\omega^\omega). We show that it is not countably tight. To this end, consider the following subset A of S(\omega) \times S(\omega^\omega).

    \displaystyle S(\omega)=\left\{\infty \right\} \cup (\omega \times \omega)

    \displaystyle S(\omega^\omega)=\left\{\infty \right\} \cup (\omega^\omega \times \omega)

    \displaystyle A_f=\left\{(x,y) \in S(\omega) \times S(\omega^\omega): x=(n,f(n)) \text{ and } y=(f,j)  \right\} \ \forall \ f \in \omega^\omega

    \displaystyle A=\bigcup_{f \in \omega^\omega} A_f

It follows that (\infty,\infty) \in \overline{A}. We show that for any countable C \subset A, the point (\infty,\infty) \notin \overline{C}. Fix a countable C \subset A. We can assume that C=\bigcup_{i=1}^\infty A_{f_i}. Now define a function g \in \omega^\omega by a diagonal argument as follows.

Define g(0) such that g(0)>f_0(0). For each integer n>0, define g(n) such that g(n)>\text{max} \{ \ f_n(0),f_n(1),\cdots,f_n(n) \ \} and g(n)>g(n-1). Let O=B_g \times S(\omega^\omega). The diagonal definition of g ensures that O is an open set containing (\infty,\infty) such that O \cap C=\varnothing. This shows that the space S(\omega) \times S(\omega^\omega) is not countably tight.

Example 3
The space S(\omega_1) \times S(\omega_1) is not countably tight. In fact its tightness character is \omega_1. This fact follows from Theorem 1.1 in [2].

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The set-theoretic angle

Example 2 shows that S(\omega) \times S(\omega^\omega) is not countably tight even though each factor has the strong property of a Frechet space with the first factor being a countable space. The example shows that Frechetness behaves very badly with respect to the product operation. Is there an example of \kappa>\omega such that S(\omega) \times S(\kappa) is countably tight? In particular, is S(\omega) \times S(\omega_1) countably tight?

First off, if Continuum Hypothesis (CH) holds, then Example 2 shows that S(\omega) \times S(\omega_1) is not countably tight since the cardinality of \omega^{\omega} is \omega_1 under CH. So for S(\omega) \times S(\omega_1) to be countably tight, extra set theory assumptions beyond ZFC will have to be used (in fact the extra axioms will have to be compatible with the negation of CH). In fact, it is consistent with ZFC for S(\omega) \times S(\omega_1) to be countably tight. It is also consistent with ZFC for t(S(\omega) \times S(\omega_1))=\omega_1. We point out some facts from the literature to support these observations.

Consider S(\omega) \times S(\kappa) where \kappa>\omega_1. For any regular cardinal \kappa>\omega_1, it is possible that S(\omega) \times S(\kappa) is countably tight. It is also possible for the tightness character of S(\omega) \times S(\kappa) to be \kappa (of course in a different model of set theory). Thus it is hard to pin down the tightness character of the product S(\omega) \times S(\kappa). It all depends on your set theory. In the next section, we point out some facts from the literature to support these observations.

Example 3 points out that the tightness character of S(\omega_1) \times S(\omega_1) is \omega_1, i.e. t(S(\omega_1) \times S(\omega_1))=\omega_1 (this is a fact on the basis of ZFC only). What is t(S(\omega_2) \times S(\omega_2)) or t(S(\kappa) \times S(\kappa)) for any \kappa>\omega_1? The tightness character of S(\kappa) \times S(\kappa) for \kappa>\omega_1 also depends on set theory. We also give a brief explanation by pointing out some basic information from the literature.

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The bounding number

The tightness of the product S(\omega) \times S(\kappa) is related to the cardinal number called the bounding number denoted by \mathfrak{b}.

Recall that \omega^{\omega} is the set of all functions from \omega into \omega. For f,g \in \omega^{\omega}, define f \le^* g by the condition: f(n) \le g(n) for all but finitely many n \in \omega. A set F \subset \omega^{\omega} is said to be a bounded set if F has an upper bound according to \le^*, i.e. there exists some f \in \omega^{\omega} such that g \le^* f for all g \in F. Then F \subset \omega^{\omega} is an unbounded set if it is not bounded. To spell it out, F \subset \omega^{\omega} is an unbounded set if for each f \in \omega^{\omega}, there exists some g \in F such that g \not \le^* f.

Furthermore, F \subset \omega^{\omega} is a dominating set if for each f \in \omega^{\omega}, there exists some g \in F such that f \le^* g. Define the cardinal numbers \mathfrak{b} and \mathfrak{d} as follows:

    \displaystyle \mathfrak{b}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is an unbounded set} \right\}

    \displaystyle \mathfrak{d}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is a dominating set} \right\}

The cardinal number \mathfrak{b} is called the bounding number. The cardinal number \mathfrak{d} is called the dominating number. Note that continuum \mathfrak{c}, the cardinality of \omega^{\omega}, is an upper bound of both \mathfrak{b} and \mathfrak{d}, i.e. \mathfrak{b} \le \mathfrak{c} and \mathfrak{d} \le \mathfrak{c}. How do \mathfrak{b} and \mathfrak{d} relate? We have \mathfrak{b} \le \mathfrak{d} since any dominating set is also an unbounded set.

A diagonal argument (similar to the one in Example 2) shows that no countable F \subset \omega^{\omega} can be unbounded. Thus we have \omega < \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}. If CH holds, then we have \omega_1 = \mathfrak{b} = \mathfrak{d} = \mathfrak{c}. On the other hand, it is also consistent that \omega < \mathfrak{b} < \mathfrak{d} \le \mathfrak{c}.

We now relate the bounding number to the tightness of S(\omega) \times S(\kappa). The following theorem is from Theorem 1.3 in [3].

Theorem 1 – Theorem 1.3 in [3]
The following conditions hold:

  • For \omega \le \kappa <\mathfrak{b}, the space S(\omega) \times S(\kappa) is countably tight.
  • The tightness character of S(\omega) \times S(\mathfrak{b}) is \mathfrak{b}, i.e. t(S(\omega) \times S(\mathfrak{b}))=\mathfrak{b}.

Thus S(\omega) \times S(\kappa) is countably tight for any uncountable \kappa <\mathfrak{b}. In particular if \omega_1 <\mathfrak{b}, then S(\omega) \times S(\omega_1) is countably tight. According to Theorem 5.1 in [6], this is possible.

Theorem 2 – Theorem 5.1 in [6]
Let \tau and \lambda be regular cardinal numbers such that \omega_1 \le \tau \le \lambda. It is consistent with ZFC that \mathfrak{b}=\mathfrak{d}=\tau and \mathfrak{c}=\lambda.

Theorem 2 indicates that it is consistent with ZFC that the bounding number \mathfrak{b} can be made to equal any regular cardinal number. In the model of set theory in which \omega_1 <\mathfrak{b}, S(\omega) \times S(\omega_1) is countably tight. Likewise, in the model of set theory in which \omega_1 < \kappa <\mathfrak{b}, S(\omega) \times S(\kappa) is countably tight.

On the other hand, if the bounding number is made to equal an uncountable regular cardinal \kappa, then t(S(\omega) \times S(\kappa))=\kappa. In particular, t(S(\omega) \times S(\omega_1))=\omega_1 if \mathfrak{b}=\omega_1.

The above discussion shows that the tightness of S(\omega) \times S(\kappa) is set-theoretic sensitive. Theorem 2 indicates that it is hard to pin down the location of the bounding number \mathfrak{b}. Choose your favorite uncountable regular cardinal, there is always a model of set theory in which \mathfrak{b} is your favorite uncountable cardinal. Then Theorem 1 ties the bounding number to the tightness of S(\omega) \times S(\kappa). Thus the exact value of the tightness character of S(\omega) \times S(\kappa) depends on your set theory. If your favorite uncountable regular cardinal is \omega_1, then in one model of set theory consistent with ZFC, t(S(\omega) \times S(\omega_1))=\omega (when \omega_1 <\mathfrak{b}). In another model of set theory, t(S(\omega) \times S(\omega_1))=\omega_1 (when \omega_1 =\mathfrak{b}).

One comment about the character of the fan S(\omega) at the point \infty. As indicated earlier, the character at \infty is the dominating number \mathfrak{d}. Theorem 2 tells us that it is consistent that \mathfrak{d} can be any uncountable regular cardinal. So for the fan S(\omega), it is quite difficult to pinpoint the status of a basic topological property such as character of a space. This is another indication that the sequential fan is highly dependent on additional axioms beyond ZFC.

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The collectionwise Hausdorff property

Now we briefly discuss the tightness of t(S(\kappa) \times S(\kappa)) for any \kappa>\omega_1. The following is Theorem 1.1 in [2].

Theorem 3 – Theorem 1.1 in [2]
Let \kappa be any infinite regular cardinal. The following conditions are equivalent.

  • There exists a first countable < \kappa-collectionwise Hausdorff space which fails to be a \kappa-collectionwise Hausdorff space.
  • t(S(\kappa) \times S(\kappa))=\kappa.

The existence of the space in the first condition, on the surface, does not seem to relate to the tightness character of the square of a sequential fan. Yet the two conditions were proved to be equivalent [2]. The existence of the space in the first condition is highly set-theory sensitive. Thus so is the tightness of the square of a sequential fan. It is consistent that a space in the first condition exists for \kappa=\omega_2. Thus in that model of set theory t(S(\omega_2) \times S(\omega_2))=\omega_2. It is also consistent that there does not exist a space in the first condition for \kappa=\omega_2. Thus in that model, t(S(\omega_2) \times S(\omega_2))<\omega_2. For more information, see [3].

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Remarks

Sequential fans and their products are highly set-theoretic in nature and are objects that had been studied extensively. This is only meant to be a short introduction. Any interested readers can refer to the small list of articles listed in the reference section and other articles in the literature.

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Exercise

Use Theorem 3 to show that t(S(\omega_1) \times S(\omega_1))=\omega_1 by finding a space X that is a first countable < \omega_1-collectionwise Hausdorff space which fails to be a \omega_1-collectionwise Hausdorff space.

For any cardinal \kappa, a space X is \kappa-collectionwise Hausdorff (respectively < \kappa-collectionwise Hausdorff) if for any closed and discrete set A \subset X with \lvert A \lvert \le \kappa (repectively \lvert A \lvert < \kappa), the points in A can be separated by a pairwise disjoint family of open sets.

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Reference

  1. Bella A., van Mill J., Tight points and countable fan-tightness, Topology Appl., 76, (1997), 1-27.
  2. Eda K., Gruenhage G., Koszmider P., Tamano K., Todorčeviće S., Sequential fans in topology, Topology Appl., 67, (1995), 189-220.
  3. Eda K., Kada M., Yuasa Y., Tamano K., The tightness about sequential fans and combinatorial properties, J. Math. Soc. Japan, 49 (1), (1997), 181-187.
  4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  5. LaBerge T., Landver A., Tightness in products of fans and psuedo-fans, Topology Appl., 65, (1995), 237-255.
  6. Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 111-167.

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\copyright \ 2015 \text{ by Dan Ma}