In a previous post, we discuss basic properties of Bing’s Example G, a classic and influential example of a normal but not collectionwise normal space. In another post, we discuss a subspace of Bing’s Example G which is also normal but not collectionwise normal but is metacompact. In this post, we further discuss Bing’s Example G by characterizing its compact subspaces.
The ideas discussed here can be found in [1] with a lot of details omitted. In this post, we provide all the necessary details in understanding these ideas. In the next post called Some subspaces of Bing’s Example G, we discuss some subspaces of Bing’s Example G based on the characterization of compact sets given in this post.
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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 twopoint discrete space , i.e., . 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. 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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Basic Open Sets in Bing’s Example G
To facilitate the discussion below, we now fix now some notation for basic open sets of the points . For any finite , the following describes the basic open sets containing the point .
If , the open set can be denoted by
Recall that for any space and for any and for any point , the point is a limit point of if every open subset of containing contains a point of that is different from .
Because points of retain the product open sets, points of are the only limit points in the space . The set is a closed and discrete set in the space . To see this, consider the open set where and . It contains and does not contain for any .
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One Example of Compact Subsets of Bing’s Example G
Since the set is a closed and discrete set in the space , any compact subset of can contain at most finitely many points of . We first present an example of an infinite compact subset of whose intersection with has only one point (i.e. the compact set has only one limit point).
For each and for each , define a function by the following:
Essentially, the function agrees with the function except on the point . For each , consider the following subset of the space .
We show that for a fixed , the functions are distinct for distinct . So for , we show that . By definition, it is clear that

and .
Thus the set consists of distinct elements. Since is uncountable, is uncountable. If is the cardinal number , then .
We make the following claims about the set .
 For each , the set contains no point of .
 Every open set containing contains all but finitely many points of .
 The point is the only limit point of .
 For each , is compact.
 Thus, is like the onepoint compactification of .
To see the first bullet point, clearly for all . The function and the function differ at the point . It is also the case that for , for all . Suppose that for some . There are two cases to consider: or . Suppose . Then . So . But . Thus . This means that . Now let . First . On the other hand, , leading to a contradiction. It can be shown that assuming will also lead to a contradiction too. Thus for any with , .
To see the second bullet point, let be a finite set and let be an arbitrary basic open set containing . For any , for all , and , implying that . Thus every open set containing contains all but finitely many points of .
The second bullet shows that is a limit point of . We now show that is the only limit point of . Let . Consider the basic open set , which contains . For each with , , showing that . Thus the open set can contain at most one point of , namely . So cannot be a limit point of for all .
Thus the set is a compact set in the space . It has only one limit point, namely the point . Viewing as a space by itslef, the open set at the point is cofinite (second bullet point). Thus is like the onepoint compactification of .
We have demonstrated a specific example of infinite compact subsets of the space . In the characterization of compact sets in the next section, we can always refer to and know that the sets described by the characterization below exist.
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Characterizing Compact Subsets of Bing’s Example G
Let be the collection of all infinite closed subsets of the space . For each , let be the following collection:
Note that the set as discussed above is a member of . Let be the following set:
Note that since . It turns out that characterizes the compact subsets with as the only limit point. We now prove the following theorem.

Theorem 1

Let . Then is an infinite compact subset of the space with being the only limit point of if and only if .
Proof of Theorem 1
Suppose that is an infinite compact subset of the space with being the only limit point of . Suppose . Then for some , the set
is infinite. Now choose an infinite subset where for . Consider the open set . Note that and for all . Thus is an open containing that misses infinitely many points of , a contradiction. Thus .
Suppose . Let be finitely many sets from . Consider the basic open set containing . Recall that this is just the set of all that agree with on the elements . Because , for each , the following set is finite.
For each , and for each , . Thus the open set contains all but finitely many points of . So is a limit point of .
We now show that is the only limit point of . Let . We show that is not a limit point of . Let . Consider the basic open set . Note that and . Consider the set :
The set is finite. For each , and thus . So the open set can contain at most finitely many points of . Thus is not a limit point of . We have shown that is an infinite compact subset of with as the only limit point.
Theorem 1 characterizes the compact subsets of Bing’s Example G with only one limit point. We now characterize all compact subsets of the space . We have the following theorem.

Theorem 2

Let . Then is a compact subspace of if and only if is the union of finitely many sets in where
with being the collection of all finite subsets of .
Proof of Theorem 2
The direction is clear. For the direction , let be a compact subset of . If is finite, then . So assume that is infinite. Since the set is closed and discrete in , can only contain finitely many points of , say .
Choose open sets such that for each , and such that for . For each , let . Let be the set of points of not in any of the . Note that and . So is the union of finitely many sets in the collection .
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Comment
Essentially an infinite compact subset of Bing’s Example G is the union of finitely many sets from finitely many . In some cases, when working with compact subsets of Example G, it is sufficient to work with sets from for one arbitrary . See the next post for an example.
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Reference
 Boone, J. R., Some characterizations of paracompactness in kspaces, Fund. Math., 72, 145155, 1971.
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