# 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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$\copyright \ 2014 \text{ by Dan Ma}$

• A handy example is $X=[0,1] \cup \mathbb{Q}$ with the usual subspace topology from the real line. Here $\mathbb{Q}$ is the set of all rational numbers.