# Compact Subspaces of Bing’s Example G

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 $P$ 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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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 $f_p \in F_P$. For any finite $S \subset Q$, the following describes the basic open sets containing the point $f_p \in F_P$.

$U(f_p,S)=\left\{f \in 2^Q: \forall q \in S,f(q)=f_p(q) \right\}$

If $S=\left\{q_1,q_2,\cdots,q_n \right\}$, the open set $U(f_p,S)$ can be denoted by

$V(f_p,q_1,q_2,\cdots,q_n)=U(f_p,S)$

Recall that for any space $X$ and for any $A \subset X$ and for any point $p \in X$, the point $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ that is different from $p$.

Because points of $F_P$ retain the product open sets, points of $F_P$ are the only limit points in the space $F$. The set $F_P$ is a closed and discrete set in the space $F$. To see this, consider the open set $V(f_p,q)$ where $q=\left\{p \right\}$ and $p \in P$. It contains $f_p \in F_P$ and does not contain $f_t$ for any $t \ne p$.

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One Example of Compact Subsets of Bing’s Example G

Since the set $F_P$ is a closed and discrete set in the space $F$, any compact subset of $F$ can contain at most finitely many points of $F_P$. We first present an example of an infinite compact subset of $F$ whose intersection with $F_P$ has only one point (i.e. the compact set has only one limit point).

For each $p \in P$ and for each $q \in Q$, define a function $H(p,q):Q \rightarrow 2$ by the following:

$H(p,q)(r) = \begin{cases} f_p(r), & \mbox{if } r \in Q-\left\{q \right\} \\ \ne f_p(r), & \mbox{if } r=q \end{cases}$

Essentially, the function $H(p,q)$ agrees with the function $f_p$ except on the point $q$. For each $p \in P$, consider the following subset of the space $F$.

$D(p)=\left\{H(p,q): q \in Q \right\}$

We show that for a fixed $p$, the functions $H(p,q)$ are distinct for distinct $q$. So for $q \ne q'$, we show that $H(p,q) \ne H(p,q')$. By definition, it is clear that

$H(p,q)(q')=f_p(q')$ and $H(p,q')(q') \ne f_p(q')$.

Thus the set $D(p)$ consists of distinct elements. Since $Q$ is uncountable, $D(p)$ is uncountable. If $\lvert P \lvert$ is the cardinal number $\theta$, then $\lvert D(p) \lvert=\lvert Q \lvert=2^\theta$.

We make the following claims about the set $D(p)$.

• For each $p \in P$, the set $D(p)$ contains no point of $F_P$.
• Every open set containing $f_p$ contains all but finitely many points of $D(p)$.
• The point $f_p$ is the only limit point of $D(p)$.
• For each $p \in P$, $A(p)=D(p) \cup \left\{f_p \right\}$ is compact.
• Thus, $A(p)$ is like the one-point compactification of $D(p)$.

To see the first bullet point, clearly $f_p \ne H(p,q)$ for all $q \in Q$. The function $f_p$ and the function $H(p,q)$ differ at the point $q$. It is also the case that for $t \ne p$, $f_t \ne H(p,q)$ for all $q \in Q$. Suppose that $f_t=H(p,q)$ for some $q \in Q$. There are two cases to consider: $t \in q$ or $t \notin q$. Suppose $t \in q$. Then $f_t(q)=1$. So $H(p,q)(q)=f_t(q)=1$. But $H(p,q)(q) \ne f_p(q)$. Thus $f_p(q)=0$. This means that $p \notin q$. Now let $r=q \cup \left\{p \right\}-\left\{t \right\}$. First $H(p,q)(r)=f_p(r)=1$. On the other hand, $H(p,q)(r)=f_p(r)=0$, leading to a contradiction. It can be shown that assuming $t \notin q$ will also lead to a contradiction too. Thus for any $t \in P$ with $t \ne p$, $f_t \notin D(p)$.

To see the second bullet point, let $S \subset Q$ be a finite set and let $U(f_p,S)$ be an arbitrary basic open set containing $f_p$. For any $q \subset Q-S$, for all $r \in S$, $r \ne q$ and $H(p,q)(r)=f_p(r)$, implying that $H(p,q) \in U(f_p,S)$. Thus every open set containing $f_p$ contains all but finitely many points of $D(p)$.

The second bullet shows that $f_p$ is a limit point of $D(p)$. We now show that $f_p$ is the only limit point of $D(p)$. Let $t \in P-\left\{p \right\}=v$. Consider the basic open set $V(f_t,v)$, which contains $f_t$. For each $q \in Q$ with $q \ne v$, $H(p,q)(v)=f_p(v)=0$, showing that $H(p,q) \notin V(f_t,v)$. Thus the open set $V(f_t,v)$ can contain at most one point of $D(p)$, namely $H(p,v)$. So $f_t$ cannot be a limit point of $D(p)$ for all $t \in P-\left\{p \right\}$.

Thus the set $A(p)$ is a compact set in the space $F$. It has only one limit point, namely the point $f_p$. Viewing $A(p)$ as a space by itslef, the open set at the point $f_p$ is co-finite (second bullet point). Thus $A(p)$ is like the one-point compactification of $D(p)$.

We have demonstrated a specific example of infinite compact subsets of the space $F$. In the characterization of compact sets in the next section, we can always refer to $A(p)$ and know that the sets described by the characterization below exist.

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Characterizing Compact Subsets of Bing’s Example G

Let $\mathcal{C}$ be the collection of all infinite closed subsets of the space $F$. For each $p \in P$, let $\mathcal{C}_p$ be the following collection:

$\mathcal{C}_p=\left\{C \in \mathcal{C}: f_p \text{ is the only limit point of } C \text{ and if } t \ne p, f_t \notin C \right\}$

Note that the set $A(p)=D(p) \cup \left\{f_p \right\}$ as discussed above is a member of $\mathcal{C}_p$. Let $\mathcal{K}_p$ be the following set:

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

Note that $\mathcal{K}_p \ne \varnothing$ since $A(p) \in \mathcal{K}_p$. It turns out that $\mathcal{K}_p$ characterizes the compact subsets with $f_p$ as the only limit point. We now prove the following theorem.

Theorem 1

Let $p \in P$. Then $K$ is an infinite compact subset of the space $F$ with $f_p$ being the only limit point of $K$ if and only if $K \in \mathcal{K}_p$.

Proof of Theorem 1
$\Longrightarrow$
Suppose that $K$ is an infinite compact subset of the space $F$ with $f_p$ being the only limit point of $K$. Suppose $K \notin \mathcal{K}_p$. Then for some $q \in Q$, the set

$T_q=\left\{f \in K: f(q) \ne f_p(q) \right\}$

is infinite. Now choose an infinite subset $\left\{f_1,f_2,f_3,\cdots \right\} \subset T_q$ where $f_i \ne f_j$ for $i \ne j$. Consider the open set $V(f_p,q)=\left\{f \in 2^Q: f(q)=f_p(q) \right\}$. Note that $f_p \in V(f_p,q)$ and $f_j \notin V(f_p,q)$ for all $j$. Thus $V(f_p,q)$ is an open containing $f_p$ that misses infinitely many points of $K$, a contradiction. Thus $K \in \mathcal{K}_p$.

$\Longleftarrow$
Suppose $K \in \mathcal{K}_p$. Let $q_1,q_2,\cdots,q_n \in Q$ be finitely many sets from $Q$. Consider the basic open set $V(f_p,q_1,\cdots,q_n)$ containing $f_p$. Recall that this is just the set of all $f \in 2^Q$ that agree with $f_p$ on the elements $q_1,q_2,\cdots,q_n \in Q$. Because $K \in \mathcal{K}_p$, for each $q_j$, the following set is finite.

$A_j=\left\{f \in K: f(q_j) \ne f_p(q_j) \right\}$

For each $f \in K-(A_1 \cup A_2 \cup \cdots \cup A_n)$, and for each $j$, $f(q_j)=f_p(q_j)$. Thus the open set $V(f_p,q_1,\cdots,q_n)$ contains all but finitely many points of $K$. So $f_p$ is a limit point of $K$.

We now show that $f_p$ is the only limit point of $K$. Let $t \in P-\left\{p \right\}$. We show that $f_t$ is not a limit point of $K$. Let $r=P-\left\{p \right\}$. Consider the basic open set $V(f_t,r)$. Note that $f_t \in V(f_t,r)$ and $f_p(r)=0$. Consider the set $A$:

$A=\left\{f \in K: f(r) \ne f_p(r) \right\}$

The set $A$ is finite. For each $f \in K-A$, $f(r)=f_p(r)=0 \ne 1=f_t(r)$ and thus $f \notin V(f_t,r)$. So the open set $V(f_t,r)$ can contain at most finitely many points of $K$. Thus $f_t$ is not a limit point of $K$. We have shown that $K$ is an infinite compact subset of $F$ with $f_p$ as the only limit point. $\blacksquare$

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 $F$. We have the following theorem.

Theorem 2

Let $K \subset F$. Then $K$ is a compact subspace of $F$ if and only if $K$ is the union of finitely many sets in $\mathcal{K}$ where

$\mathcal{K}=\biggl( \bigcup \limits_{p \in P} \mathcal{K}_p \biggr) \cup \mathcal{H}$

with $\mathcal{H}$ being the collection of all finite subsets of $F$.

Proof of Theorem 2
The direction $\Longleftarrow$ is clear. For the direction $\Longrightarrow$, let $K$ be a compact subset of $F$. If $K$ is finite, then $K \in \mathcal{H}$. So assume that $K$ is infinite. Since the set $F_P$ is closed and discrete in $F$, $K$ can only contain finitely many points of $F_P$, say $f_{p_1},f_{p_2},\cdots,f_{p_n}$.

Choose open sets $U_1,U_2,\cdots,U_n$ such that for each $j$, $f_{p_j} \in U_j$ and such that $\overline{U_i} \cap \overline{U_j} = \varnothing$ for $i \ne j$. For each $j$, let $K_j=\overline{U_j} \cap K$. Let $H$ be the set of points of $K$ not in any of the $K_j$. Note that $H \in \mathcal{H}$ and $K_j \in \mathcal{K}_{p_j}$. So $K$ is the union of finitely many sets in the collection $\mathcal{K}$. $\blacksquare$

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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 $\mathcal{K}_p$. In some cases, when working with compact subsets of Example G, it is sufficient to work with sets from $\mathcal{K}_p$ for one arbitrary $p \in P$. See the next post for an example.

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Reference

1. Boone, J. R., Some characterizations of paracompactness in k-spaces, Fund. Math., 72, 145-155, 1971.

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