The product of a perfectly normal space and a metric space is perfectly normal

The previous post gives a positive result for normality in product space. It shows that the product of a normal countably compact space and a metric space is always normal. In this post, we discuss another positive result, which is the following theorem.

Main Theorem
If X is a perfectly normal space and Y is a metric space, then X \times Y is a perfectly normal space.

As a result of this theorem, perfectly normal spaces belong to a special class of spaces called P-spaces. K. Morita defined the notion of P-space and he proved that a space Y is a Normal P-space if and only if X \times Y is normal for every metric space X (see the section below on P-spaces). Thus any perfectly normal space is a Normal P-space.

All spaces under consideration are Hausdorff. A subset A of the space X is a G_\delta-subset of the space X if A is the intersection of countably many open subsets of X. A subset B of the space X is an F_\sigma-subset of the space X if B is the union of countably many closed subsets of X. Clearly, a set A is a G_\delta-subset of the space X if and only if X-A is an F_\sigma-subset of the space X.

A space X is said to be a perfectly normal space if X is normal with the additional property that every closed subset of X is a G_\delta-subset of X (or equivalently every open subset of X is an F_\sigma-subset of X).

The perfect normality has a characterization in terms of zero-sets and cozero-sets. A subset A of the space X is said to be a zero-set if there exists a continuous function f: X \rightarrow [0,1] such that A=f^{-1}(0), where f^{-1}(0)=\left\{x \in X: f(x)=0 \right\}. A subset B of the space X is a cozero-set if X-B is a zero-set, or more explicitly if there is a continuous function f: X \rightarrow [0,1] such that B=\left\{x \in X: f(x)>0 \right\}.

It is well known that the space X is perfectly normal if and only if every closed subset of X is a zero-set, equivalently every open subset of X is a cozero-set. See here for a proof of this result. We use this result to show that X \times Y is perfectly normal.

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The Proof

Let X be a perfectly normal space and Y be a metric space. Since Y is a metric space, let \mathcal{B}=\bigcup_{j=1}^\infty \mathcal{B}_j be a base for Y such that each \mathcal{B}_j is locally finite. We show that X \times Y is perfectly normal. To that end, we show that every open subset of X \times Y is a cozero-set. Let U be an open subset of X \times Y.

For each (x,y) \in X \times Y, there exists open O_{x,y} \subset X and there exists B_{x,y} \in \mathcal{B} such that (x,y) \in O_{x,y} \times B_{x,y} \subset U. Then U is the union of all sets O_{x,y} \times B_{x,y}. Observe that B_{x,y} \in \mathcal{B}_{j} for some integer j. For each B \in \mathcal{B} such that B=B_{x,y} for some (x,y) \in X \times Y, let O(B) be the union of all corresponding open sets O_{x,y} for all applicable (x,y).

For each positive integer j, let \mathcal{W}_j be the collection of all open sets O(B) \times B such that B \in \mathcal{B}_j and B=B_{x,y} for some (x,y) \in X \times Y. Let \mathcal{V}_j=\cup \mathcal{W}_j. As a result, U=\bigcup_{j=1}^\infty \mathcal{V}_j.

Since both X and Y are perfectly normal, for each O(B) \times B \in \mathcal{W}_j, there exist continuous functions

    F_{O(B),j}: X \rightarrow [0,1]

    G_{B,j}: Y \rightarrow [0,1]

such that

    O(B)=\left\{x \in X: F_{O(B),j}(x) >0 \right\}

    B=\left\{y \in Y: G_{B,j}(y) >0 \right\}

Now define H_j: X \times Y \rightarrow [0,1] by the following:

    \displaystyle H_j(x,y)=\sum \limits_{O(B) \times B \in \mathcal{W}_j} F_{O(B),j}(x) \ G_{B,j}(y)

for all (x,y) \in X \times Y. Note that the function H_j is well defined. Since \mathcal{B}_j is locally finite in Y, \mathcal{W}_j is locally finite in X \times Y. Thus H_j(x,y) is obtained by summing a finite number of values of F_{O(B),j}(x) \ G_{B,j}(y). On the other hand, it can be shown that H_j is continuous for each j. Based on the definition of H_j, it can be readily verified that H_j(x,y)>0 for all (x,y) \in \cup \mathcal{W}_j and H_j(x,y)=0 for all (x,y) \notin \cup \mathcal{W}_j.

Define H: X \times Y \rightarrow [0,1] by the following:

    \displaystyle H(x,y)=\sum \limits_{j=1}^\infty \biggl[ \frac{1}{2^j} \ \frac{H_j(x,y)}{1+H_j(x,y)} \biggr]

It is clear that H is continuous. We claim that U=\left\{(x,y) \in X \times Y: H(x,y) >0 \right\}. Recall that the open set U is the union of all O(B) \times B \in \mathcal{W}_j for all j. Thus if (x,y) \in \cup \mathcal{W}_j for some j, then H(x,y)>0 since H_j(x,y)>0. If (x,y) \notin \cup \mathcal{W}_j for all j, H(x,y)=0 since H_j(x,y)=0 for all j. Thus the open set U is an F_\sigma-subset of X \times Y. This concludes the proof that X \times Y is perfectly normal. \square

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Remarks

The main theorem here is a classic result in general topology. An alternative proof is to show that any perfectly normal space is a P-space (definition given below). Then by Morita’s theorem, the product of any perfectly normal space and any metric space is normal (Theorem 1 below). For another proof that is elementary, see Lemma 7 in this previous post.

The notions of perfectly normal spaces and paracompact spaces are quite different. By the theorem discussed here, perfectly normal spaces are normally productive with metric spaces. It is possible for a paracompact space to have a non-normal product with a metric space. The classic example is the Michael line (discussed here).

On the other hand, there are perfectly normal spaces that are not paracompact. One example is Bing’s Example H, which is perfectly normal and not paracompact (see here).

Even though a perfectly normal space is normally productive with metric spaces, it cannot be normally productive in general. For each non-discrete perfectly normal space X, there exists a normal space Y such that X \times Y is not normal. This follows from Morita’s first conjecture (now a true statement). Morita’s first conjecture is discussed here.

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P-Space in the Sense of Morita

Morita defined the notion of P-spaces [1] and [2]. Let \kappa be a cardinal number such that \kappa \ge 1. Let \Gamma be the set of all finite ordered sequences (\alpha_1,\alpha_2,\cdots,\alpha_n) where n=1,2,\cdots and all \alpha_i < \kappa. Let X be a space. The collection \left\{F_\sigma \subset X: \sigma \in \Gamma \right\} is said to be decreasing if this condition holds: \sigma =(\alpha_1,\alpha_2,\cdots,\alpha_n) and \delta =(\alpha_1,\alpha_2,\cdots,\alpha_n, \cdots, \alpha_m) with n<m imply that F_{\delta} \subset F_{\sigma}. The space X is a P-space if for any cardinal \kappa \ge 1 and for any decreasing collection \left\{F_\sigma \subset X: \sigma \in \Gamma \right\} of closed subsets of X, there exists open set U_\sigma for each \sigma \in \Gamma such that the following conditions hold:

  • for all \sigma \in \Gamma, F_\sigma \subset U_\sigma,
  • for any infinite sequence (\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots) where each each finite subsequence \sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n) is an element of \Gamma, if \bigcap_{n=1}^\infty F_{\sigma_n}=\varnothing, then \bigcap_{n=1}^\infty U_{\sigma_n}=\varnothing.

If \kappa=1 where 1=\left\{0 \right\}. Then the index set \Gamma defined above can be viewed as the set of all positive integers. As a result, the definition of P-space with \kappa=1 implies the a condition in Dowker’s theorem (see condition 6 in Theorem 1 here). Thus any space X that is normal and a P-space is countably paracompact (or countably shrinking or that X \times Y is normal for every compact metric space or any other equivalent condition in Dowker’s theorem). The following is a theorem of Morita.

Theorem 1 (Morita)
Let X be a space. Then X is a normal P-space if and only if X \times Y is normal for every metric space Y.

In light of Theorem 1, both perfectly normal spaces and normal countably compact spaces are P-spaces (see here). According to Theorem 1 and Dowker’s theorem, it follows that any normal P-space is countably paracompact.

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Reference

  1. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
  2. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.

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\copyright \ 2017 \text{ by Dan Ma}

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}

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}

A useful lemma for proving normality

In this post we discuss a lemma (Lemma 1 below) that is useful for proving normality. In some cases, it is more natural using this lemma to prove that a space is normal than using the definition of normality. The proof of Lemma 1 is not difficult. Yet it simplifies some proofs of normality. One reason is that the derivation of two disjoint open sets that are to separate two disjoint closed sets is done in the lemma, thus simplifying the main proof at hand. The lemma is well known and is widely used in the literature. See Lemma 1.5.15 in [1]. Two advanced examples of applications are [2] and [3]. After proving the lemma, we give three elementary applications of the lemma. One of the applications is a characterization of perfectly normal spaces. This characterization is, in some cases, easier to use, e.g. making it easy to show that perfectly normal implies hereditarily normal.

In this post, we only consider spaces that are regular and T_1. A space X is regular if for each open set U \subset X and for each x \in U, there exists an open V \subset X with x \in V \subset \overline{V} \subset U. A space is T_1 if every set with only one point is a closed set.

Lemma 1
A space Y is a normal space if the following condition (Condition 1) is satisfied:

  1. For each closed subset L of Y, and for each open subset M of Y with L \subset M, there exists a sequence M_1,M_2,M_3,\cdots of open subsets of Y such that L \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for each i.

Proof of Lemma 1
Suppose the space Y satisfies condition 1. Let H and K be disjoint closed subsets of the space Y. Consider H \subset U=Y \backslash K. Using condition 1, there exists a sequence U_1,U_2,U_3,\cdots of open subsets of the space Y such that H \subset \bigcup_{i=1}^\infty U_i and \overline{U_i} \cap K=\varnothing for each i. Consider K \subset V=Y \backslash H. Similarly, there exists a sequence V_1,V_2,V_3,\cdots of open subsets of the space Y such that K \subset \bigcup_{i=1}^\infty V_i and \overline{V_i} \cap H=\varnothing for each i.

For each positive integer n, define the open sets U_n^* and V_n^* as follows:

    U_n^*=U_n \backslash \bigcup_{k=1}^n \overline{V_k}

    V_n^*=V_n \backslash \bigcup_{k=1}^n \overline{U_k}

Let P=\bigcup_{n=1}^\infty U_n^* and Q=\bigcup_{n=1}^\infty V_n^*. It is clear P and Q are open and that H \subset P and K \subset Q. We claim that P and Q are disjoint. Suppose y \in P \cap Q. Then y \in U_n^* for some n and y \in V_m^* for some m. Assume that n \le m. The fact that y \in U_n^* implies y \in U_n. The fact that y \in V_m^* implies that y \notin \overline{U_j} for all j \le m. In particular, y \notin U_n, a contradiction. Thus P \cap Q=\varnothing. This completes the proof that the space Y is normal. \blacksquare

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Spaces with nice bases

One application is that spaces with a certain type of bases satisfy condition 1 and thus are normal. For example, spaces with bases that are countable and spaces with bases that are \sigma-locally finite. Spaces with these bases are metrizable. The proof that these spaces are metrizable will be made easier if they can be shown to be normal first. The Urysohn functions (the functions described in Urysohn’s lemma) can then be used to embed the space in question into some universal space that is known to be metrizable. Using regularity and Lemma 1, it is straightforward to verify the following three propositions.

Proposition 2
Let X be a regular space with a countable base. Then X is normal.

Proposition 3
Let X be a regular space with a \sigma-locally finite base. Then X is normal.

Proposition 4
Let X be a regular space with a \sigma-discrete base. Then X is normal.

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A characterization of perfectly normal spaces

Another application of Lemma 1 is that it leads naturally to a characterization of perfect normality. Recall that a space X is perfectly normal if X is normal and perfect. A space X is perfect if every closed subset of X is a G_\delta set (i.e. the intersection of countably many open subsets of X). Equivalently a space X is perfect if and only if every open subset of X is an F_\sigma set, i.e., the union of countably many closed subsets of X. We have the following theorem.

Theorem 5
A space Y is perfectly normal if and only if the following condition holds.

  1. For each open subset M of Y, there exists a sequence M_1,M_2,M_3,\cdots of open subsets of Y such that M \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for each i.

Clearly, condition 2 is strongly than condition 1.

Proof of Theorem 5
\Longrightarrow
Suppose that the space Y is perfectly normal. Let M be a non-empty open subset of Y. Then M=\bigcup_{n=1}^\infty P_n where each P_n is a closed subset of Y. Using normality of Y, for each n, there exists open subset M_n of Y such that P_n \subset M_n \subset \overline{M_n} \subset M. Then consition 2 is satisfied.

\Longleftarrow
Suppose condition 2 holds, which implies condition 1 of Lemma 1. Then Y is normal. It is clear that condition 2 implies that every open subset of Y is an F_\sigma set. \blacksquare

The characterization of perfectly normal spaces in Theorem 5 is hereditary. This means that any subspace of a perfectly normal space is also perfectly normal. In particular, perfectly normal implies hereditarily normal. Thus we have the following theorem.

Theorem 6
Condition 2 in Theorem 5 is hereditary, i.e., if a space satisfies Condition 2, every subspace satisfies Condition 2. Therefore if the space Y is a perfectly normal space, then every subspace of Y is also perfectly normal. In particular, if Y is perfectly normal, then Y is hereditarily normal (i.e. every subspace of Y is normal).

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Normality is hereditary with respect to F_\sigma subsets

Normality is not a hereditary notion. Lemma 1 can used to show that normality is hereditary with respect to F_\sigma subspaces.

Theorem 7
Let Y be a normal space. Then every F_\sigma subspace of Y is normal.

Proof of Theorem 7
Let H be a subspace of Y such that H=\bigcup_{n=1}^\infty P_n where each P_n is closed subset of Y. Let L be a closed subset of H and let M be an open subset of H such that L \subset M. We need to find M_1,M_2,M_3,\cdots, open in H, such that L \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for all i (closure of M_i is within H).

Let U be an open subset of Y such that M=U \cap H. For each positive integer n, let H_n=P_n \cap L. Obviously H_n is closed in H. It is also the case that H_n is closed in Y. To see this, let p \in Y be a limit point of H_n. Then p is a limit point of P_n. Hence p \in P_n since P_n is closed in Y. We now have p \in H. The point p is also a limit point of L. Thus p \in L since L is closed in H. Now we have p \in H_n=P_n \cap L, proving that H_n is closed in Y.

Now we have H_n \subset U for all n. By Lemma 1, for each n, there exists a sequence U_{n,1},U_{n,2},U_{n,3},\cdots of open subsets of Y such that H_n \subset \bigcup_{j=1}^\infty U_{n,j} and \overline{U_{n,j}} \subset U for all j. Note that L=\bigcup_{n=1}^\infty H_n. Rename M_{n,j}=U_{n,j} \cap H over all n,j by the sequence M_1,M_2,M_3,\cdots. Then L \subset \bigcup_{i=1}^\infty M_i. It also follows that \overline{M_i} \subset M for all i (closure of M_i is within H). This completes the proof that the F_\sigma set H is normal. \blacksquare

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Gruenhage, G., Normality in X^2 for complete X, Trans. Amer. Math. Soc., 340 (2), 563-586, 1993.
  3. Nyikos, P., A compact nonmetrizable space P such that P^2 is completely normal, Topology Proc., 2, 359-363, 1977.

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\copyright \ 2014-2015 \text{ by Dan Ma} Revised April 14, 2015

Bing’s Example H

In a previous post we introduced Bing’s Example G, a classic example of a normal but not collectionwise normal space. Other properties of Bing’s Example G include: completely normal, not perfectly normal and not metacompact. This is an influential example introduced in an influential paper of R. H. Bing in 1951 (see [1]). In the same paper, another example called Example H was introduced. This space has some of the same properties of Example G, except that it is perfectly normal. In this post, we define and discuss Example H.

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Defining Bing’s Example H

Throughout the discussion in this post, we use \omega to denote the first infinite ordinal, i.e., \omega =\left\{0,1,2,3,\cdots \right\}. Let P be any uncountable set. Let Q be the set of all subsets of the set P, i.e., it is the power set of P. Let H be the set of all functions f:Q \rightarrow \omega. In other words, the set H is the Cartesian product \prod \limits_{q \in Q} \omega. But the topology on H is not the product topology.

For each p \in P, consider the function f_p:Q \rightarrow 2=\left\{0,1 \right\} such that for each q \in Q:

    f_p(q) = \begin{cases} 1, & \mbox{if } p \in q \\ 0, & \mbox{if } p \notin q \end{cases}

Let H_P=\left\{f_p: p \in P \right\}. Now define a topology on the set H by the following:

  • Each point of H-H_P is an isolated point.
  • Each point f_p \in H_P has basic open sets of the form U(p,W,n) defined as follows:

      U(p,W,n)=\left\{f_p \right\} \cup D(p,W,n)

      D(p,W,n)=\left\{f \in H: \forall q \in Q, f(q) \ge n \text{ and } \forall q \in W, f(q) \equiv f_p(q) \ (\text{mod} \ 2) \right\}

    where p \in P, W \subset Q is finite, and n \in \omega.

If a and b are integers, the a \equiv b \ (\text{mod} \ 2) means that a-b is divisible by 2. The congruence equation f(q) \equiv f_p(q) \ (\text{mod} \ 2) means that f(q) is an even integer if f_p(q)=0. On the other hand, f(q) \equiv f_p(q) \ (\text{mod} \ 2) means that f(q) is an odd integer if f_p(q)=1.

The set D(p,W,n) seems to mimic a basic open set of the point f_p in the product topology: for each point in D(p,W,n), the value of each coordinate is an integer \ge n and the values for finitely many coordinates are fixed to agree with the function f_p modulo 2. Adding the point f_p to D(p,W,n), we have a basic open set U(p,W,n).

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Basic Discussion

The points in H-H_P are the isolated points in the space H. The points in H_P are the non-isolated points (limit points). The space H is a Hausdorff space. Another interesting point is that the set H_P is a closed and discrete set in the space H.

To see that H is Hausdorff, let h_1, h_2 \in H with h_1 \ne h_2. Consider the case that h_1 is an isolated point and h_2=f_p for some p \in P. Let n be the minimum of all h_1(q) over all q \in Q. Let O_1=\left\{h_1 \right\} and O_2=U(p,W,n+1) where W \subset Q is any finite set. Then O_1 and O_2 are disjoint open set containing h_1 and h_2, respectively.

Now consider the case that h_1=f_p and h_2=f_{p'} where p \ne p'. Let O_1=U(p,W,0) and O_2=U(p',W,0) where W=\left\{ \left\{ p \right\},\left\{ p' \right\} \right\}. Then O_1 and O_2 are disjoint open set containing h_1 and h_2, respectively.

The set H_P is a closed and discrete set in the space H. It is closed since H-H_P consists of isolated points. To see that H_P is discrete, note that U(p,W,0), where W=\left\{ \left\{ p \right\} \right\}, is an open set with f_p \in U(p,W,0) and f_{p'} \notin U(p,W,0) for all p' \ne p.

In the sections below, we show that the space H is normal, completely normal (thus hereditarily normal), and is perfectly normal. Furthermore, we show that it is not collectionwise Hausdorff (hence not collectionwise normal) and not meta-lindelof (hence not metacompact).

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Bing’s Example H is Normal

In the next section, we show that Bing’s Example H is completely normal (i.e. any two separated sets can be separated by disjoint open sets). Note that any two disjoint closed sets are separated sets.

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Bing’s Example H is Completely Normal

Let X be a space. Let A \subset X and B \subset X. The sets A and B are separated sets if A \cap \overline{B}=\varnothing=\overline{A} \cap B. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space X is said to be completely normal if for every two separated sets A and B in X, there exist disjoint open subsets U and V of X such that A \subset U and B \subset V. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space X, X is completely normal if and only if X is hereditarily normal. For more about completely normality, see [3] and [6].

Let S and T be separated sets in the space H, i.e.,

    S \cap \overline{T}=\varnothing=\overline{S} \cap T

We consider two cases. Case 1 is that one of the sets consists entirely of isolated points. Assume that S \subset H-H_P. Let O_1=S. For each x \in T, choose an open set V_x with x \in V_x and V_x \cap \overline{S}=\varnothing. Let O_2=\bigcup \limits_{x \in T} V_x. Then O_1 and O_2 are disjoint open sets containing S and T respectively.

Now consider Case 2 where S_1=S \cap H_P \ne \varnothing and T_1=T \cap H_P \ne \varnothing. Consider the sets q_1 and q_2 defined as follows:

    q_1=\left\{p \in P: f_p \in S_1 \right\}

    q_2=\left\{p \in P: f_p \in T_1 \right\}

Let W=\left\{q_1,q_2 \right\}. Let Y_1 and Y_2 be the following open sets:

    Y_1=\bigcup \limits_{p \in q_1} U(p,W,0)

    Y_2=\bigcup \limits_{p \in q_2} U(p,W,0)

Immediately, we know that S_1 \subset Y_1, T_1 \subset Y_2 and Y_1 \cap Y_2=\varnothing. Let S_2=S \cap (H-H_P) and T_2=T \cap (H-H_P) (both of which are open). Let O_1 and O_2 be the following open sets:

    O_1=(Y_1 \cup S_2)-\overline{T}

    O_2=(Y_2 \cup T_2)-\overline{S}

Then O_1 and O_2 are disjoint open sets containing S and T respectively.

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Bing’s Example H is Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is a G_\delta-set (i.e. the intersection of countably many open subsets). All we need to show here is that every closed subset is a G_\delta-set.

Let C \subset H be a closed set. Of course, if C consists entirely of isolated points, then we are done. So assume that C \cap H_P \ne \varnothing. Let q*=\left\{p \in P: f_p \in C \right\}. Let O=C \cap (H-H_P), which is open. For each positive integer n, define the open set Y_n as follows:

    Y_n=O \cup \biggl( \bigcup \limits_{p \in q*} U(p,\left\{q* \right\},n) \biggr)

Immediately we have C \subset Y_n for each n. Let g \in \bigcap \limits_{n=1}^\infty Y_n. We claim that g \in C. Suppose g \notin C. Then g \notin O. It follows that for each n g \in U(p_n,\left\{q* \right\},n) for some p_n \in q*. Recall that U(p_n,\left\{q* \right\},n)=\left\{f_{p_n} \right\} \cup D(p_n,\left\{q* \right\},n).

The assumption that g \notin C implies that g \ne f_{p_n} for all n. Then g \in D(p_n,\left\{q* \right\},n) for all n. By the definition of D(p_n,\left\{q* \right\},n), it follows that for all q \in Q, g(q) \ge n for all positive integer n. This is a contradiction. So it must be the case that g \in C. This completes the proof that Bing’s Example H is perfectly normal.

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Collectionwise Normal Spaces

Let X be a space. Let \mathcal{A} be a collection of subsets of X. We say \mathcal{A} is pairwise disjoint if A \cap B=\varnothing whenever A,B \in \mathcal{A} with A \ne B. We say \mathcal{A} is discrete if for each x \in X, there is an open set O containing x such that O intersects at most one set in \mathcal{A}.

The space X is said to be collectionwise normal if for every discrete collection \mathcal{D} of closed subsets fo X, there is a pairwise disjoint collection \left\{U_D: D \in \mathcal{D} \right\} of open subsets of X such that D \subset U_D for each D \in \mathcal{D}. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus both Bing’s Example G and Example H are not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. As shown below Bing’s Example H is actually not collectionwise Hausdorff.

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Bing’s Example H is not Collectionwise Hausdorff

To prove that Bing’s Example H is not collectionwise Hausdorff, we need an intermediate result (Lemma 1) that is based on an infinitary combinatorial result called the Delta-system lemma.

A family \mathcal{A} of sets is called a Delta-system (or \Delta-system) if there exists a set r, called the root of the \Delta-system, such that for any A,B \in \mathcal{A} with A \ne B, we have A \cap B=r. The following is a version of the Delta-system lemma (see Theorem 1.5 in p. 49 of [2]).

    Delta-System Lemma

      Let \mathcal{A} be an uncountable family of finite sets. Then there exists an uncountable \mathcal{B} \subset \mathcal{A} such that \mathcal{B} is a \Delta-system.
    Lemma 1

      Let P_0 \subset P be any uncountable subset. For each p \in P_0, let U(p,W_p,n_p) be a basic open subset containing f_p. Then there exists an uncountable P_1 \subset P_0 such that \bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing.

Proof of Lemma 1
Let \mathcal{A}=\left\{W_p: p \in P_0 \right\}. We need to break this up into two cases – \mathcal{A} is a countable family of finite sets or an uncountable family of finite sets. The first case is relatively easy to see. The second case requires using the Delta-system lemma.

Suppose that \mathcal{A} is countable. Then there exists an uncountable R \subset P_0 such that for all p,t \in R with p \ne t, we have W_p=W_t=W and n_p=n_t=n. Suppose that W=\left\{q_1,q_2,\cdots,q_m \right\}. By inductively working on the sets q_j, we can obtain an uncountable set P_1 \subset R such that for all p,t \in P_1 with p \ne t, we have f_p(q_j)=f_t(q_j) for each j=1,2,\cdots,m. Clearly, we have:

    \bigcap \limits_{p \in P_1} U(p,W,n) \ne \varnothing

To show the above, just define a function h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\} such that h(q_j)=f_p(q_j) for all j=1,2,\cdots,m for one particular p \in P_1. Then h belongs to the intersection.

Suppose that \mathcal{A} is uncountable. By the Delta-system lemma, there is an uncountable R \subset P_0 and there exists a finite set r \subset Q such that for all p,t \in R with p \ne t, we have W_p \cap W_t=r. Suppose that r=\left\{q_1,q_2,\cdots,q_m \right\}. As in the previous case, work inductively on the sets q_j, we can obtain an uncountable S \subset R such that for all p,t \in S with p \ne t, we have f_p(q_j)=f_t(q_j) for each j=1,2,\cdots,m. Now narrow down to an uncountable P_1 \subset S such that n_p=n_t=n for all p,t \in P_1 with p \ne t. We now show that

    \bigcap \limits_{p \in P_1} U(p,W_p,n) \ne \varnothing

To define a function h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\} that belongs to the above intersection, we define h so that h matches f_t (mod 2) with one particular t \in P_1 on the set r=\left\{q_1,q_2,\cdots,q_m \right\}. Note that W_p-r are disjoint over all p \in P_1. So h can be defined on W_p-r to match f_p (mod 2). For any remaining values in the domain, define h freely to be at least the integer n. Then the function h belongs to the intersection.

With the two cases established, the proof of Lemma 1 is completed. \blacksquare

The fact that Example H is not collectionwise Hausdorff is a corollary of Lemma 1. The set H_P is a discrete collection of points in the space H. It follows that H_P cannot be separated by disjoint open sets. For each p \in P, let U(p,W_p,n_p) be a basic open set containing the point f_p. By Lemma 1, there is an uncountable P_1 \subset P such that \bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing. Thus there can be no disjoint collection of open sets in H that separate the points in H_P.

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Bing’s Example H is not Metacompact

Let X be a space. A collection \mathcal{A} of subsets of X is said to be a point-finite (point-countable) collection if every point of X belongs to only finitely (countably) many sets in \mathcal{A}. A space X is said to be a metacompact space if every open cover \mathcal{U} of X has a point-finite open refinement \mathcal{V}. A space X is said to be a meta-Lindelof space if every open cover \mathcal{U} of X has a point-countable open refinement \mathcal{V}. Clearly, every metacompact space is meta-Lindelof.

It follows from Lemma 1 that Example H is not meta-Lindelof. Thus Example H is not metacompact. To see that it is not meta-Lindelof, for each f_p \in H_P, let U_{f_p}=U(p,\left\{\left\{p \right\} \right\},0), and for each x \in H-H_P, let U_x=\left\{x \right\}. Let \mathcal{U} be the following open cover of H:

    \mathcal{U}=\left\{U_x: x \in H \right\}

Each f_p \in H_P belongs to only one set in \mathcal{U}, namely U_{f_p}. So for any open refinement \mathcal{V} of \mathcal{U} (consisting of basic open sets), we have uncountably many open sets of the form U(p,W_p,n_p). By Lemma 1, we can find uncountably many such open sets with non-empty intersection. So no open refinement of \mathcal{U} can be point-countable.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

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\copyright \ 2014 \text{ by Dan Ma}

A subspace of Bing’s example G

Bing’s Example G is the first example of a topological space that is normal but not collectionwise normal (see [1]). Example G was an influential example from an influential paper. The Example G and its subspaces had been extensively studied. In addition to being normal and not collectionwise normal, Example G is not perfectly normal and not metacompact. See the previous post “Bing’s Example G” for a basic discussion of Example G. In this post we focus on one subspace of Example G examined by Michael in [3]. This subspace is normal, not collectionwise normal and not perfectly normal just like Example G. However it is metacompact. In [3], Michael proved that any metacompact collectionwise normal space is paracompact (metacompact was called pointwise paracompact in that paper). This subspace of Example G demonstrates that collectionwise normality in Michael’s theorem cannot be replaced by normality.

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Bing’s Example G

For a more detailed discussion of Bing’s Example G in this blog, see the blog post “Bing’s Example G”. For the sake of completeness, we repeat the definition of Example G. Let P be any uncountable set. Let Q be the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Obviously 2^Q is simply the Cartesian product of \lvert Q \lvert many copies of the two-point discrete space \left\{0,1 \right\}, i.e., \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. The following is another topology on 2^Q:

    \tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}

Bing’s Example G is the set F=2^Q with the topology \tau^*. In other words, each x \in F-F_P is made an isolated point and points in F_P retain the usual product open sets.

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Michael’s Subspace of Example G

For each f \in F, let supp(f) be the support of f, i.e., supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}. Michael in [3] considered the following subspace of F.

    M=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}

Michael in [3] used the letter G to denote the space M. We choose another letter to distinguish it from Example G. The subspace M consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for only finitely many q \in Q. The space M is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points f \in F that have values of 1 for infinitely many q \in Q, we obtain a subspace that is metacompact. We discuss the following points:

  • The space M is normal.
  • The space M is not collectionwise Hausdorff and hence not collectionwise normal.
  • The space M is metacompact.
  • The space M is not perfectly normal.

The space M is normal since the space F that is Example G is hereditarily normal (see the section called Bing’s Example G is Completely Normal in the post “Bing’s Example G”).

To show that the space M is not collectionwise Hausdorff, it is helpful to first look at M as a subspace of the product space 2^Q. The product space 2^Q has the countable chain condition (CCC) since it is a product of separable spaces. Note that M is dense in the product space 2^Q. Thus M as a subspace of the product space has the CCC.

In the space M, the set F_P is still a closed and discrete set. In the space M, open sets containing points of F_P are the same as product open sets in 2^Q relative to the set M. Since M has CCC (as a subspace of the product space 2^Q), M cannot have uncountably many pairwise disjoint open sets containing points of F_P (in either the product topology or the Example G subspace topology). It follows that M is not collectionwise Hausdorff. If it were, there would be uncountably many pairwise disjoint product open sets separating points in F_P, which is not possible.

To see that M is metacompact, let \mathcal{U} be an open cover of M. For each p \in P, choose U_p \in \mathcal{U} such that f_p \in U_p. For each p \in P, let W_p=\left\{f \in M: f(\left\{p \right\})=1 \right\}. Let \mathcal{V} be the following:

    \mathcal{V}=\left\{U_p \cap W_p: p \in P \right\} \cup \left\{\left\{x \right\}: x \in M-F_P \right\}

Note that \mathcal{V} is a point-finite open refinement of \mathcal{U}. Each U_p \cap W_p contains only one point of F_P, namely f_p. On the other hand, each f \in M with finite support can belong to at most finitely many U_p \cap W_p.

The space M is not perfectly normal. This point is alluded to in [3] by Michael and elsewhere in the literature, e.g. in Bing’s paper (see [1]) and in Engelking’s general topology text (see 5.53 on page 338 of [2]). In fact Michael indicated that one can obtain a perfectly normal example with the aforementioned properties using Example H defined in [1] instead of using the subspace M defined here in this post.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
  4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Bing’s Example G

Bing’s Example G is an example of a topological space that is normal but not collectionwise normal. It was introduced in an influential paper of R. H. Bing in 1951 (see [1]). This paper has a metrization theorem that is now called Bing’s metrization theorem (any regular space is metrizable if and only if it has a \sigma-discrete base). The paper also introduced the notion of collectionwise normality and discussed the roles it plays in metrization theory (e.g. a Moore space is metrizable if and only if it is collectionwise normal). Example G was an influential example from an influential paper. It became the basis of construction for many other counterexamples (see [5] for one example). Investigations were also conducted by looking at various covering properties among subspaces of Example G (see [2] and [4] are two examples).

In this post we prove some basic results about Bing’s Example G. Some of the results we prove are found in Bing’s 1951 paper. The other results shown here are usually mentioned without proof in various places in the literature.

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Bing’s Example G – Definition

Let P be any uncountable set. Let Q be the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Another notation for 2^Q is the Cartesian product \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. We now consider another topology on 2^Q generated by the following base:

    \mathcal{B}=\tau \cup \left\{\left\{x \right\}: x \in F-F_P \right\}

Bing’s Example G is the set F=2^Q with the topology generated by the base \mathcal{B}. In other words, each x \in F-F_P is made an isolated point and points in F_P retain the usual product open sets.

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Bing’s Example G – Initial Discussion

Bing’s Example G, i.e. the space F as defined above, is obtained by altering the topology of the product space of 2^{\lvert \mathcal{K} \lvert} many copies of the two-point discrete space where \mathcal{K} is the cardinality of the power set of the uncountable index set P we start with. Out of this product space, a set F_P of points is carefully chosen such that F_P has the same cardinality as P and such that F_P is relatively discrete in the product space. Points in F_P are made to retain the product topology and all points outside of F_P are declared as isolated points.

We now show that the set F_P is a discrete set in the space F. For each p \in P, let W_p be the open set defined by

    W_p=\left\{f \in F: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}.

It is clear that f_p is the only point of F_P belonging to W_p. Therefore, in the Example G topology, the set F_P is discrete and closed . In the section “Bing’s Example G is not Collectionwise Hausdorff” below, we show below that F_P cannot be separated by any pairwise disjoint collection of open sets.

The character at a point is the minimum cardinality of a local base at that point. The character at a point in F_P in the Example G topology agrees with the product topology. Points in F_P have character \lvert Q \lvert=2^{\lvert P \lvert}. Specifically if the starting P has cardinality \omega_1, then points in F_P have character 2^{\omega_1}. Thus Example G has large character and cannot be a Moore space (any Moore space has a countable base at every point).

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Bing’s Example G is Normal

Let H_1 and H_2 be disjoint closed subsets of F. The easy case is that one of H_1 and H_2 is a subset of F-F_P, say H_1 \subset F-F_P. Then H_1 is a closed and open set in F. Then H_1 and F-H_1 are disjoint open sets containing H_1 and H_2, respectively. So we can assume that both H_1 \cap F_P \ne \varnothing and H_2 \cap F_P \ne \varnothing.

Let A_1=H_1 \cap F_P and A_2=H_2 \cap F_P. Let q_1=\left\{p \in P: f_p \in A_1 \right\} and q_2=\left\{p \in P: f_p \in A_2 \right\}. Define the following open sets:

    U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}
    U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}

Because H_1 \cap H_2=\varnothing, we have A_1 \subset U_1 and A_2 \subset U_2. Furthermore, U_1 \cap U_2=\varnothing. Let B_1=H_1 \cap (F-F_P) and B_2=H_2 \cap (F-F_P), which are open since they consist of isolated points. Then O_1=(U_1 \cup B_1)-H_2 and O_2=(U_2 \cup B_2)-H_1 are disjoint open subsets of F with H_1 \subset O_1 and H_2 \subset O_2.

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Collectionwise Normal Spaces

Let X be a space. Let \mathcal{A} be a collection of subsets of X. We say \mathcal{A} is pairwise disjoint if A \cap B=\varnothing whenever A,B \in \mathcal{A} with A \ne B. We say \mathcal{A} is discrete if for each x \in X, there is an open set O containing x such that O intersects at most one set in \mathcal{A}.

The space X is said to be collectionwise normal if for every discrete collection \mathcal{D} of closed subsets fo X, there is a pairwise disjoint collection \left\{U_D: D \in \mathcal{D} \right\} of open subsets of X such that D \subset U_D for each D \in \mathcal{D}. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus Bing’s Example G is not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. Bing’s Example is actually not collectionwise Hausdorff.

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Bing’s Example G is not Collectionwise Hausdorff

The discrete set F_P cannot be separated by disjoint open sets. For each p \in P, let O_p be an open subset of F such that p \in O_p. We show that the open sets O_p cannot be pairwise disjoint. For each p \in P, choose an open set L_p in the product topology of 2^Q such that p \in L_p \subset O_p. The product space 2^Q is a product of separable spaces, hence has the countable chain condition (CCC). Thus the open sets L_p cannot be pairwise disjoint. Thus L_t \cap L_s \ne \varnothing and O_t \cap O_s \ne \varnothing for at least two points s,t \in P.

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Bing’s Example G is Completely Normal

The proof for showing Bing’s Example G is normal can be modified to show that it is completely normal. First some definitions. Let X be a space. Let A \subset X and B \subset X. The sets A and B are separated sets if A \cap \overline{B}=\varnothing=\overline{A} \cap B. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space X is said to be completely normal if for every two separated sets A and B in X, there exist disjoint open subsets U and V of X such that A \subset U and B \subset V. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space X, X is completely normal if and only if X is hereditarily normal. For more about completely normality, see [3] and [6].

Let H_1 \subset F and H_2 \subset F such that H_1 \cap \overline{H_2}=\varnothing=\overline{H_1} \cap H_2. We consider two cases. One is that one of H_1 and H_2 is a subset of F-F_P. The other is that both H_1 \cap F_P \ne \varnothing and H_2 \cap F_P \ne \varnothing.

The first case. Suppose H_1 \subset F-F_P. Then H_1 consists of isolated points and is an open subset of F. For each x \in H_2 \cap F_P, choose an open subset V_x of F such that x \in V_x and V_x contains no points of F_P-\left\{ x \right\} and V_x \cap \overline{H_1}=\varnothing. For each x \in H_2 \cap (F-F_P), let V_x=\left\{x \right\}. Let V be the union of all V_x where x \in H_2. Let U=H_1. Then U and V are disjoint open sets with H_1 \subset U and H_2 \subset V.

The second case. Suppose A_1=H_1 \cap F_P \ne \varnothing and A_2=H_2 \cap F_P \ne \varnothing. Let q_1=\left\{p \in P: f_p \in A_1 \right\} and q_2=\left\{p \in P: f_p \in A_2 \right\}. Define the following open sets:

    U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}
    U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}

Because H_1 \cap H_2=\varnothing, we have A_1 \subset U_1 and A_2 \subset U_2. Furthermore, U_1 \cap U_2=\varnothing. Let B_1=H_1 \cap (F-F_P) and B_2=H_2 \cap (F-F_P), which are open since they consist of isolated points. Then O_1=(U_1 \cup B_1)-\overline{H_2} and O_2=(U_2 \cup B_2)-\overline{H_1} are disjoint open subsets of F with H_1 \subset O_1 and H_2 \subset O_2.

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Bing’s Example G is not Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is G_\delta (i.e. the intersection of countably many open subsets). The set F_P of non-isolated points is a closed set in F. We show that F_P cannot be a G_\delta-set. Before we do so, we need to appeal to a fact about the product space 2^Q.

According to the Tychonoff theorem, the product space 2^Q is a compact space since it is a product of compact spaces. On the other hand, 2^Q is a product of uncountably many factors and is thus not first countable. It is a well known fact that in a compact Hausdorff space, if a point is a G_\delta-point, then there is a countable local base at that point (i.e. the space is first countable at that point). Thus no point of the compact product space 2^Q can be a G_\delta-point. Since points of F_P retain the open sets of the product topology, no point of F_P can be a G_\delta-point in the Bing’s Example G topology.

For each p \in P, let W_p be open in F such that f_p \in W_p and W_p contains no points F_P-\left\{f_p \right\}. For example, we can define W_p as in the above section “Bing’s Example G – Initial Discussion”.

Suppose that F_P is a G_\delta-set. Then F_P=\bigcap \limits_{i=1}^\infty O_i where each O_i is an open subset of F. Now for each p \in P, we have \left\{f_p \right\}=\bigcap \limits_{i=1}^\infty (O_i \cap W_p), contradicting the fact that the point f_p cannot be a G_\delta-point in the space F (and in the product space 2^Q). Thus F_P is not a G_\delta-set in the space F, leading to the conclusion that Bing’s Example G is not perfectly normal.

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Bing’s Example G is not Metacompact

A space M is said to have caliber \omega_1 if for every uncountable collection \left\{U_\alpha: \alpha < \omega_1 \right\} of non-empty open subsets of M, there is an uncountable A \subset \omega_1 such that \bigcap \left\{U_\alpha: \alpha \in A \right\} \ne \varnothing. Any product of separable spaces has this property (see Topological Spaces with Caliber Omega 1). Thus the product space 2^Q has caliber \omega_1. Thus in the product space 2^Q, no collection of uncountably many non-empty open sets can be a point-finite collection (in fact cannot even be point-countable).

To see that the Example G is not metacompact, let \mathcal{W}=\left\{W_p: p \in P \right\} be a collection of open sets such that for p \in P, f_p \in W_p, W_p is open in the product topology of 2^Q and W_p contains no points F_P-\left\{f_p \right\}. For example, we can define W_p as in the above section “Bing’s Example G – Initial Discussion”.

Let W=\bigcup \mathcal{W}. Let \mathcal{V}=\mathcal{W} \cup \left\{\left\{ x \right\}: x \in F-W \right\}. Any open refinement of \mathcal{V} would contain uncountably many open sets in the product topology and thus cannot be point-finite. Thus the space F cannot be metacompact.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Burke, D. K., A note on R. H. Bing’s example G, Top. Conf. VPI, Lectures Notes in Mathematics, 375, Springer Verlag, New York, 47-52, 1974.
  3. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  4. Lewis, I. W., On covering properties of subspaces of R. H. Bing’s Example G, Gen. Topology Appl., 7, 109-122, 1977.
  5. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
  6. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012