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:
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 two-point 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 “Bing’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 non-isolated 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 non-empty 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 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 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 point-finite. Thus the space
cannot be metacompact.
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Reference
- Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 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, 47-52, 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, 109-122, 1977.
- Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
- Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.
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