Some subspaces of Bing’s Example G

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 F). 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 F. 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 P be any uncountable set. Let Q be the power set of P, i.e., the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Obviously 2^Q is simply the Cartesian product of \lvert Q \lvert many copies of the two-point discrete space \left\{0,1 \right\}, i.e., \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. The following is another topology on 2^Q:

    \tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}

Bing’s Example G is the set F=2^Q with the topology \tau^*. In other words, each x \in F-F_P is made an isolated point and points in F_P 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).

    Any infinite compact subspace of F 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 F that has exactly one limit point is a member of the collection of sets \mathcal{K}_p for some p \in P (see Theorem 1 in the previous post). For p \in P, the set \mathcal{K}_p is defined by:

    \mathcal{K}_p=\left\{K \in \mathcal{C}_p: \forall q \in Q, \left\{f \in K: f(q) \ne f_p(q) \right\} \text{ is finite} \right\}

where \mathcal{C}_p is the collection of all closed subsets of F each of which has the point f_p as the only limit point. For the results shown below, it suffices to work with a member of some \mathcal{K}_p when we work with an infinite compact subset of F.

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Some Subspaces of Example G

For each f \in F, let supp(f) be the support of f, i.e., supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}. For any infinite cardinal number \theta \le \lvert Q \lvert, we consider the following subspace:

    M_{\theta}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\theta \right\}

The subspace M_{\theta} consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for less than \theta many q \in Q. So the support of these functions is small (in relation to the size of the domain Q). Among these subspaces, of particular interest are the following two subspaces:

    M_{\lvert Q \lvert}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\lvert Q \lvert \right\}
    M_{\omega}=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}

The subspace M_{\omega} 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 G to denote the space M_{\omega}. We choose another letter to distinguish it from Example G. The subspace M_{\omega} consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for only finitely many q \in Q. Just like Example G, the space M is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points f \in F that have values of 1 for infinitely many q \in Q, we obtain a subspace that is metacompact.

We show the following claim about the subspace M_{\lvert Q \lvert}:

    Proposition 1
    All compact subsets of the space M_{\lvert Q \lvert} are finite.

Proof of Proposition 1
In light of the comment in the preceding section, we only need to show for any compact subset K of F such that K \in \mathcal{K}_p for some p \in P, K \cap M_{\lvert Q \lvert} is finite.

Suppose K \cap M_{\lvert Q \lvert} is infinite for K \in \mathcal{K}_p. Choose \left\{g_1,g_2,g_3,\cdots \right\} \subset K \cap M_{\lvert Q \lvert} such that g_i \ne g_j for i \ne j. Note that K_0=\left\{g_1,g_2,g_3,\cdots \right\} \cup \left\{f_p \right\} is a closed subset of K and is thus compact.

For each j let Q_j=\left\{q \in Q: g_j(q)=1 \right\}. Since each g_j \in M_{\lvert Q \lvert}, each Q_j has cardinality less than \lvert Q \lvert. Thus Q_1 \cup Q_2 \cup \cdots has cardinality less than \lvert Q \lvert too. On the other hand, let Q_\omega=\left\{q \in Q: p \in q \right\}. Since Q_\omega has cardinality equal to \lvert Q \lvert, we can pick r \in Q_\omega-(Q_1 \cup Q_2 \cup \cdots).

Right away we know that f_p(r)=1 and g_j(r)=0 for all j. Let V=\left\{f \in 2^Q: f(r)=1 \right\} which is an open set that contains f_p. But g_j \notin V for all j. Thus the following collection

    \left\{V \right\} \cup \left\{ \left\{g_j \right\}:j=1,2,3,\cdots \right\}

is an open cover of K_0 that has no finite subcover, contradicting the fact that K_0 is compact. Thus for any p \in P, for any compact set K \in \mathcal{K}_p, K \cap M_{\lvert Q \lvert} is finite. In other words, for any compact subset K of F with only one limit point, K \cap M_{\lvert Q \lvert} must be finite. It follows that in the space M_{\lvert Q \lvert}, all compact sets are finite. \blacksquare

Note that M_\theta \subset M_{\lvert Q \lvert} for all infinite cardinal numbers \theta < \lvert Q \lvert. Thus the compact subsets of all such subspaces M_\theta are finite. In particular, for the subspace M_{\omega}, there are no infinite compact subsets. Thus we have the following two easy propositions.

    Proposition 2
    For any infinite cardinal number \theta < \lvert Q \lvert, all compact subsets of the space M_{\theta} are finite.
    Proposition 3
    In particular, all compact subsets of the space M_{\omega} 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 M_{\lvert Q \lvert} has no limit point. We have the following proposition.

    Proposition 4
    For any Y \subset M_{\lvert Q \lvert} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\lvert Q \lvert}.

Once Proposition 4 is established, we have the following two propositions.

    Proposition 5
    Let \theta be any infinite cardinal number \theta < \lvert Q \lvert. Then for any set Y \subset M_{\theta} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\theta}.
    Proposition 6
    In particular, for any Y \subset M_{\omega} with \lvert Y \lvert <\lvert Q \lvert, the subspace Y has no limit point (i.e. cluster point) in M_{\omega}.

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 F_P are still large. In these subspaces, the closure of any “small” subset cannot reach the limit points in the set F_P. 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, M_{\omega} (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 Q), the closure cannot reach any limit points at all.

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Reference

  1. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.

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\copyright \ 2014 \text{ by Dan Ma}

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