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}

Definitions of Collectionwise Normal Spaces

The notion of collectionwise normal spaces is a property stronger than normal spaces. There are several ways of defining the notion of collectionwise normality. We show that they are equivalent.

We only consider spaces in which any set with only one point is considered a closed set (i.e. T_1 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 for all A,B \in \mathcal{A} where A \ne B. We say \mathcal{A} is a discrete collection of subsets of X if for each x \in X, there is an open neighborhood O with x \in O such that O meets at most one element of \mathcal{A}. When \mathcal{A} is such a collection, we also say that \mathcal{A} is discrete in X (or discrete if X is understood).

    Definition 1
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of closed subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is pairwise disjoint and A \subset U_A for each A \in \mathcal{A}.

In other words, every discrete collection of closed sets can be separated by pairwise disjoint open sets. Clearly any collectionwise normal space is normal. 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.

Some authors define a collectionwise normal space as one in which every discrete collection of sets can be separated by pairwise disjoint open sets. We have the following definition.

    Definition 2
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is pairwise disjoint and A \subset U_A for each A \in \mathcal{A}.

It is clear that Definition 2 implies Definition 1. For any discrete collection \mathcal{A} of subsets, \mathcal{A}^*=\left\{\overline{A}: A \in \mathcal{A} \right\} is also a discrete collection. Thus Definition 1 implies Definition 2. The following is another way of defining collectionwise normal.

    Definition 3
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of closed subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is a discrete collection and A \subset U_A for each A \in \mathcal{A}.

Since discrete collection is pairwise disjoint collection of sets, Definition 3 implies Definition 1. We show that Definition 1 implies Definition 3. Let X be a collectionwise normal space according to Definition 1. Let \mathcal{A} be a discrete collection of closed subsets of X. Let \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} be as in Definition 1. Let H=X-\cup \mathcal{U} and K=\cup \mathcal{A}. Note that H and K are disjoint closed subsets of X.

Suppose H=\varnothing. Then the sets in \mathcal{U} are both open and closed. Thus \mathcal{U} is the desired discrete collection of open sets separating sets in \mathcal{A}. So assume that H \ne \varnothing. Since X is a normal space, we have O_H and O_K, disjoint open subsets of X, such that H \subset O_H and K \subset O_K. For each A \in \mathcal{A}, let V_A=U_A \cap O_K. Let \mathcal{V}=\left\{V_A: A \in \mathcal{A} \right\}.

We claim that \mathcal{V} is a discrete collection of open sets separating sets in \mathcal{A}. Let x \in X. Suppose x \notin H. We have x \in U_A for some A \in \mathcal{A}. Then U_A meets only one set of \mathcal{A}, namely V_A. So assume x \in H. Then x \in O_H and O_H does not meet any V_A. Thus \mathcal{A} is a discrete collection of open sets. It is clear that A \subset V_A for each A \in \mathcal{A}.

Thus all three statements defining the notion of collectionwise normal spaces are equivalent. It is clear that any space satisfying any one of these collectionwise normal definitions is a normal space. The notion of collectionwise normality is stronger than normality. Bing’s Example G is a normal space that is not collectionwise normal (see “Bing’s Example G”).

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Reference

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

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\copyright \ \ 2012

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

Bing’s G

This post is a basic discussion of Bing’s Example G. The original post was published on 10/27/2009 and is now replaced by a new post. The following link will take you there. Thank you.