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 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 be any uncountable set. Let be the set of all subsets of . Let be the set of all functions . Another notation for is the Cartesian product . For each , define the function by the following:

, if and if
Let . Let be the set of all open subsets of in the product topology. We now consider another topology on generated by the following base:
Bing’s Example G is the set with the topology generated by the base . In other words, each is made an isolated point and points in retain the usual product open sets.
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Bing’s Example G – Initial Discussion
Bing’s Example G, i.e. the space as defined above, is obtained by altering the topology of the product space of many copies of the twopoint discrete space where is the cardinality of the power set of the uncountable index set we start with. Out of this product space, a set of points is carefully chosen such that has the same cardinality as and such that is relatively discrete in the product space. Points in are made to retain the product topology and all points outside of are declared as isolated points.
We now show that the set is a discrete set in the space . For each , let be the open set defined by
.
It is clear that is the only point of belonging to . Therefore, in the Example G topology, the set is discrete and closed . In the section “Bingâ€™s Example G is not Collectionwise Hausdorff” below, we show below that 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 in the Example G topology agrees with the product topology. Points in have character . Specifically if the starting has cardinality , then points in have character . 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 and be disjoint closed subsets of . The easy case is that one of and is a subset of , say . Then is a closed and open set in . Then and are disjoint open sets containing and , respectively. So we can assume that both and .
Let and . Let and . Define the following open sets:
Because , we have and . Furthermore, . Let and , which are open since they consist of isolated points. Then and are disjoint open subsets of with and .
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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 [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 cannot be separated by disjoint open sets. For each , let be an open subset of such that . We show that the open sets cannot be pairwise disjoint. For each , choose an open set in the product topology of such that . The product space is a product of separable spaces, hence has the countable chain condition (CCC). Thus the open sets cannot be pairwise disjoint. Thus and for at least two points .
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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 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 [3] and [6].
Let and such that . We consider two cases. One is that one of and is a subset of . The other is that both and .
The first case. Suppose . Then consists of isolated points and is an open subset of . For each , choose an open subset of such that and contains no points of and . For each , let . Let be the union of all where . Let . Then and are disjoint open sets with and .
The second case. Suppose and . Let and . Define the following open sets:
Because , we have and . Furthermore, . Let and , which are open since they consist of isolated points. Then and are disjoint open subsets of with and .
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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 (i.e. the intersection of countably many open subsets). The set of nonisolated points is a closed set in . We show that cannot be a set. Before we do so, we need to appeal to a fact about the product space .
According to the Tychonoff theorem, the product space is a compact space since it is a product of compact spaces. On the other hand, 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 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 can be a point. Since points of retain the open sets of the product topology, no point of can be a point in the Bing’s Example G topology.
For each , let be open in such that and contains no points . For example, we can define as in the above section “Bing’s Example G – Initial Discussion”.
Suppose that is a set. Then where each is an open subset of . Now for each , we have , contradicting the fact that the point cannot be a point in the space (and in the product space ). Thus is not a set in the space , 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 is said to have caliber if for every uncountable collection of nonempty open subsets of , there is an uncountable such that . Any product of separable spaces has this property (see Topological Spaces with Caliber Omega 1). Thus the product space has caliber . Thus in the product space , no collection of uncountably many nonempty open sets can be a pointfinite collection (in fact cannot even be pointcountable).
To see that the Example G is not metacompact, let be a collection of open sets such that for , , is open in the product topology of and contains no points . For example, we can define as in the above section “Bing’s Example G – Initial Discussion”.
Let . Let . Any open refinement of would contain uncountably many open sets in the product topology and thus cannot be pointfinite. Thus the space cannot be metacompact.
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Reference
 Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175186, 1951.
 Burke, D. K., A note on R. H. Bing’s example G, Top. Conf. VPI, Lectures Notes in Mathematics, 375, Springer Verlag, New York, 4752, 1974.
 Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
 Lewis, I. W., On covering properties of subspaces of R. H. Bing’s Example G, Gen. Topology Appl., 7, 109122, 1977.
 Michael, E., Pointfinite and locally finite coverings, Canad. J. Math., 7, 275279, 1955.
 Willard, S., General Topology, AddisonWesley Publishing Company, 1970.
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