A stroll in Bing’s Example G

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

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


The Product Space Angle

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

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

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



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

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




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

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

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

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

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

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

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

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

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

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

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

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

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


The Topology of Bing’s Example G

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

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

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

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

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

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

    O_1=(E_1 \cup B_1) - K

    O_2=(E_2 \cup B_2) - H

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


Bing’s Example G is Countably Paracompact

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

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

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

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

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

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


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\copyright \ 2016 \text{ by Dan Ma}


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