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 ). 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.
Defining Bing’s Example H
Throughout the discussion in this post, we use to denote the first infinite ordinal, i.e., . Let be any uncountable set. Let be the set of all subsets of the set , i.e., it is the power set of . Let be the set of all functions . In other words, the set is the Cartesian product . But the topology on is not the product topology.
For each , consider the function such that for each :
Let . Now define a topology on the set by the following:
- Each point of is an isolated point.
- Each point has basic open sets of the form defined as follows:
where , is finite, and .
If and are integers, the means that is divisible by . The congruence equation means that is an even integer if . On the other hand, means that is an odd integer if .
The set seems to mimic a basic open set of the point in the product topology: for each point in , the value of each coordinate is an integer and the values for finitely many coordinates are fixed to agree with the function modulo . Adding the point to , we have a basic open set .
The points in are the isolated points in the space . The points in are the non-isolated points (limit points). The space is a Hausdorff space. Another interesting point is that the set is a closed and discrete set in the space .
To see that is Hausdorff, let with . Consider the case that is an isolated point and for some . Let be the minimum of all over all . Let and where is any finite set. Then and are disjoint open set containing and , respectively.
Now consider the case that and where . Let and where . Then and are disjoint open set containing and , respectively.
The set is a closed and discrete set in the space . It is closed since consists of isolated points. To see that is discrete, note that , where , is an open set with and for all .
In the sections below, we show that the space 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).
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.
Bing’s Example H is Completely Normal
Let be a space. Let and . The sets and are separated sets if . 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 is said to be completely normal if for every two separated sets and in , there exist disjoint open subsets and of such that and . Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space , is completely normal if and only if is hereditarily normal. For more about completely normality, see  and .
Let and be separated sets in the space , i.e.,
We consider two cases. Case 1 is that one of the sets consists entirely of isolated points. Assume that . Let . For each , choose an open set with and . Let . Then and are disjoint open sets containing and respectively.
Now consider Case 2 where and . Consider the sets and defined as follows:
Let . Let and be the following open sets:
Immediately, we know that , and . Let and (both of which are open). Let and be the following open sets:
Then and are disjoint open sets containing and respectively.
Bing’s Example H is Perfectly Normal
A space is perfectly normal if it is normal and that every closed subset is a -set (i.e. the intersection of countably many open subsets). All we need to show here is that every closed subset is a -set.
Let be a closed set. Of course, if consists entirely of isolated points, then we are done. So assume that . Let . Let , which is open. For each positive integer , define the open set as follows:
Immediately we have for each . Let . We claim that . Suppose . Then . It follows that for each for some . Recall that .
The assumption that implies that for all . Then for all . By the definition of , it follows that for all , for all positive integer . This is a contradiction. So it must be the case that . This completes the proof that Bing’s Example H is perfectly normal.
Collectionwise Normal Spaces
Let be a space. Let be a collection of subsets of . We say is pairwise disjoint if whenever with . We say is discrete if for each , there is an open set containing such that intersects at most one set in .
The space is said to be collectionwise normal if for every discrete collection of closed subsets fo , there is a pairwise disjoint collection of open subsets of such that for each . Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of ). 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.
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 of sets is called a Delta-system (or -system) if there exists a set , called the root of the -system, such that for any with , we have . The following is a version of the Delta-system lemma (see Theorem 1.5 in p. 49 of ).
Let be an uncountable family of finite sets. Then there exists an uncountable such that is a -system.
Let be any uncountable subset. For each , let be a basic open subset containing . Then there exists an uncountable such that .
Proof of Lemma 1
Let . We need to break this up into two cases – 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 is countable. Then there exists an uncountable such that for all with , we have and . Suppose that . By inductively working on the sets , we can obtain an uncountable set such that for all with , we have for each . Clearly, we have:
To show the above, just define a function such that for all for one particular . Then belongs to the intersection.
Suppose that is uncountable. By the Delta-system lemma, there is an uncountable and there exists a finite set such that for all with , we have . Suppose that . As in the previous case, work inductively on the sets , we can obtain an uncountable such that for all with , we have for each . Now narrow down to an uncountable such that for all with . We now show that
To define a function that belongs to the above intersection, we define so that matches (mod 2) with one particular on the set . Note that are disjoint over all . So can be defined on to match (mod 2). For any remaining values in the domain, define freely to be at least the integer . Then the function belongs to the intersection.
With the two cases established, the proof of Lemma 1 is completed.
The fact that Example H is not collectionwise Hausdorff is a corollary of Lemma 1. The set is a discrete collection of points in the space . It follows that cannot be separated by disjoint open sets. For each , let be a basic open set containing the point . By Lemma 1, there is an uncountable such that . Thus there can be no disjoint collection of open sets in that separate the points in .
Bing’s Example H is not Metacompact
Let be a space. A collection of subsets of is said to be a point-finite (point-countable) collection if every point of belongs to only finitely (countably) many sets in . A space is said to be a metacompact space if every open cover of has a point-finite open refinement . A space is said to be a meta-Lindelof space if every open cover of has a point-countable open refinement . 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 , let , and for each , let . Let be the following open cover of :
Each belongs to only one set in , namely . So for any open refinement of (consisting of basic open sets), we have uncountably many open sets of the form . By Lemma 1, we can find uncountably many such open sets with non-empty intersection. So no open refinement of can be point-countable.
- Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
- Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.