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, , 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 be any uncountable set. Let be the set of all subsets of . In other words, is the power set of . Consider the product of many copies of the two element set . The usual notation of this product space is . The elements of are simply the functions from into . An arbitrary element of is a function that maps every subset of to either 0 or 1.
Though the base set can be any uncountable set, it is a good idea to visualize clearly what is. In the remainder of this section, think of as the real line . Then 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 many subsets of the real line where is the cardinality of the continuum.
To further visualize the product space, let’s look at a particular subspace of . For each real number , define the function such that always maps any set of real numbers that contains to 1 and maps any set of real numbers that does not contain to 0. For example, the following are several values of the function .
where is the set of all irrational numbers. Consider the subspace . Members of are easy to describe. Each function in maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that is important is that Bing’s Example is defined by declaring all points not in isolated points and by allowing all points in retaining the open sets in the product topology.
Any point in determines based on membership (whether the reference point belongs to the set ). Points not in have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any in maps equally to 0 or 1. So the constant functions and are not in . Furthermore, any such that for at most countably many would not be in .
Let’s continue focusing on the product space for the time being. When is considered as a subspace of the product space , is a discrete space. For each , there is an open set containing such that contains no other points of . So is relatively discrete in the product space . Of course cannot be closed in since is a compact space. The open set is defined as follows:
It is clear that and that for any real number .
Two properties of the product space would be very relevant for the discussion. By the well known Tychonoff theorem, the product space is compact. Since is uncountable, 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 cannot be all pairwise disjoint. So there exist at least a pair of , say and , with nonempty intersection.
The last observation can be generalized. For each , let be any open set containing (open in the product topology). We observe that there are at least two and from such that . If there are only countably many distinct sets , then there are uncountably many that are identical and the observation is valid. So assume that there are uncountably many distinct . By the CCC in the product space, there are at least two and with . This observation shows that the discrete points in 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 can be separated by disjoint open sets. To see this, define the following two open sets and in the product topology.
It is clear that and . Furthermore, . 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 . Let be any uncountable set. Let be the set of all subsets of . The set is defined as above. Bing’s Example G is with points in retaining the open sets in the product topology and with points not in declared isolated. For some reason, in Bing’s original paper, the notation is used even though the example is identified by G. We will follow Bing’s notation.
The subspace is discrete but not closed in the product topology. However, 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 is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set cannot be separated by disjoint open sets. By declaring points not in isolated, the countable chain condition in the original product space is destroyed. However, there is still a strong trace of CCC around the points in the set , which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.
To show that is normal, let and be disjoint closed subsets of . To make it easy to follow, let and where
In other words, is the non-isolated part and is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets and where and . Set the following open sets.
It follows that and are disjoint open sets and that and . 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 is countably paracompact if every countable countable open cover of 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 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 be a countable open cover of . Let be the set of all open sets in that contain points in . For each , let . From the perspective of Bing’s Example G, the sets 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 be disjoint open sets such that for each .
We now build a point-finite open refinement of . For each , let . Let . Consider the following.
It follows that is an open cover of . All points of belong to the open sets . Any point that is not in one of the belongs to a singleton open set. It is also clear that is a refinement of . For each , and each singleton set is contained in some member of . It follows that each point in belongs to at most finitely many sets in . In fact, each point belongs to exactly one set in . Each point in belongs to exactly one since the open sets are disjoint. Any point in 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, , is a normal space.
Introduction of Bing’s Example G