In the previous post called Compact Subspaces of Bing’s Example G, we give a characterization of the compact subspaces of Bing’s Example G (the space is denoted by the letter ). In this post, we discuss how this characterization of compact subsets of Bing’s G can shed light on certain subspaces of Bing’s G. This post should be read (or studied) alongside the previous post on the characterization of compact sets of . The only thing we repeat is the definition of Bing’s Example G.
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Defining Bing’s Example G
First we repeat the definition of Bing’s Example G. Let be any uncountable set. Let be the power set of , i.e., the set of all subsets of . Let be the set of all functions . Obviously is simply the Cartesian product of many copies of the two-point discrete space , i.e., . For each , define the function by the following:
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, if and if
Let . Let be the set of all open subsets of in the product topology. The following is another topology on :
Bing’s Example G is the set with the topology . In other words, each is made an isolated point and points in retain the usual product open sets.
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Compact Subspaces of Bing’s Example G
We do not repeat the results given in the previous post. For the purposes in this post, the following observation is crucial (see Theorem 2 in the previous post).
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Any infinite compact subspace of is the union of a finite set and finitely many other compact sets each of which is a compact set with only one limit point.
Any compact subset of that has exactly one limit point is a member of the collection of sets for some (see Theorem 1 in the previous post). For , the set is defined by:
where is the collection of all closed subsets of each of which has the point as the only limit point. For the results shown below, it suffices to work with a member of some when we work with an infinite compact subset of .
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Some Subspaces of Example G
For each , let be the support of , i.e., . For any infinite cardinal number , we consider the following subspace:
The subspace consists of all points and all other such that for less than many . So the support of these functions is small (in relation to the size of the domain ). Among these subspaces, of particular interest are the following two subspaces:
The subspace was discussed by Michael in [1] and is discussed in our blog in the post called A subspace of Bing’s example G. Michael in [1] used the letter to denote the space . We choose another letter to distinguish it from Example G. The subspace consists of all points and all other such that for only finitely many . Just like Example G, the space is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points that have values of for infinitely many , we obtain a subspace that is metacompact.
We show the following claim about the subspace :
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Proposition 1
All compact subsets of the space are finite.
Proof of Proposition 1
In light of the comment in the preceding section, we only need to show for any compact subset of such that for some , is finite.
Suppose is infinite for . Choose such that for . Note that is a closed subset of and is thus compact.
For each let . Since each , each has cardinality less than . Thus has cardinality less than too. On the other hand, let . Since has cardinality equal to , we can pick .
Right away we know that and for all . Let which is an open set that contains . But for all . Thus the following collection
is an open cover of that has no finite subcover, contradicting the fact that is compact. Thus for any , for any compact set , is finite. In other words, for any compact subset of with only one limit point, must be finite. It follows that in the space , all compact sets are finite.
Note that for all infinite cardinal numbers . Thus the compact subsets of all such subspaces are finite. In particular, for the subspace , there are no infinite compact subsets. Thus we have the following two easy propositions.
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Proposition 2
For any infinite cardinal number , all compact subsets of the space are finite.
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Proposition 3
In particular, all compact subsets of the space are finite.
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Another Proposition
The proof of Proposition 1 can be modified to show that every “small” subspace of the space has no limit point. We have the following proposition.
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Proposition 4
For any with , the subspace has no limit point (i.e. cluster point) in .
Once Proposition 4 is established, we have the following two propositions.
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Proposition 5
Let be any infinite cardinal number . Then for any set with , the subspace has no limit point (i.e. cluster point) in .
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Proposition 6
In particular, for any with , the subspace has no limit point (i.e. cluster point) in .
The limit points in Bing’s Example G have large character. In the subspaces of Bing’s Example G discussed here, the characters at the points of are still large. In these subspaces, the closure of any “small” subset cannot reach the limit points in the set . So even by narrowing the focus on just the subspaces of points with “small” support, we still obtain subspaces that have large characters. For example, (the subspace with finite support on the isolated points) is not only not first countable; it cannot even have any convergent sequence. In fact, any long as a subset is small (cardinality less than the cardinality of ), the closure cannot reach any limit points at all.
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Reference
- Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
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