A stroll in Bing’s Example G

In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

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The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. In other words, $Q$ is the power set of $P$. Consider the product of $\lvert Q \lvert$ many copies of the two element set $\left\{0,1 \right\}$. The usual notation of this product space is $2^Q$. The elements of $2^Q$ are simply the functions from $Q$ into $\left\{0,1 \right\}$. An arbitrary element of $2^Q$ is a function $f$ that maps every subset of $P$ to either 0 or 1.

Though the base set $P$ can be any uncountable set, it is a good idea to visualize clearly what $P$ is. In the remainder of this section, think of $P$ as the real line $\mathbb{R}$. Then $Q$ is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are $2^c$ many subsets of the real line where $c$ is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of $2^Q$. For each real number $p$, define the function $f_p$ such that $f_p$ always maps any set of real numbers that contains $p$ to 1 and maps any set of real numbers that does not contain $p$ to 0. For example, the following are several values of the function $f_0$.

$f_0([0,1])=1$

$f_0([1,2])=0$

$f_0(\left\{0 \right\})=1$

$f_0(\mathbb{R}-\left\{0 \right\})=0$

$f_0(\mathbb{R})=1$

$f_0(\varnothing)=0$

$f_0(\mathbb{P})=0$

where $\mathbb{P}$ is the set of all irrational numbers. Consider the subspace $F_P=\left\{f_p: p \in P \right\}$. Members of $F_P$ are easy to describe. Each function in $F_P$ maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that $F_P$ is important is that Bing’s Example is defined by declaring all points not in $F_P$ isolated points and by allowing all points in $F_P$ retaining the open sets in the product topology.

Any point $f$ in $F_P$ determines $f(q)=0 \text{ or } 1$ based on membership (whether the reference point belongs to the set $q$). Points not in $F_P$ have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any $f$ in $F_P$ maps equally to 0 or 1. So the constant functions $f(q)=0$ and $f(q)=1$ are not in $F_P$. Furthermore, any $f$ such that $f(q)=1$ for at most countably many $q$ would not be in $F_P$.

Let’s continue focusing on the product space for the time being. When $F_P$ is considered as a subspace of the product space $2^Q$, $F_P$ is a discrete space. For each $p \in P$, there is an open set $W_p$ containing $f_p$ such that $W_p$ contains no other points of $F_P$. So $F_P$ is relatively discrete in the product space $2^Q$. Of course $F_P$ cannot be closed in $2^Q$ since $2^Q$ is a compact space. The open set $W_p$ is defined as follows:

$W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$

It is clear that $f_p \in W_p$ and that $f_t \notin W_p$ for any real number $t \ne p$.

Two properties of the product space $2^Q$ would be very relevant for the discussion. By the well known Tychonoff theorem, the product space $2^Q$ is compact. Since $P$ is uncountable, $2^Q$ always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets $W_p$ cannot be all pairwise disjoint. So there exist at least a pair of $W_p$, say $W_{a}$ and $W_{b}$, with nonempty intersection.

The last observation can be generalized. For each $p \in P$, let $V_p$ be any open set containing $f_p$ (open in the product topology). We observe that there are at least two $a$ and $b$ from $P$ such that $V_a \cap V_b \ne \varnothing$. If there are only countably many distinct sets $V_p$, then there are uncountably many $V_p$ that are identical and the observation is valid. So assume that there are uncountably many distinct $V_p$. By the CCC in the product space, there are at least two $a$ and $b$ with $V_a \cap V_b \ne \varnothing$. This observation shows that the discrete points in $F_P$ cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint $A_1, A_2 \subset F_P$ can be separated by disjoint open sets. To see this, define the following two open sets $E_1$ and $E_2$ in the product topology.

$q_1=\left\{p \in P: f_p \in A_1 \right\}$

$q_2=\left\{p \in P: f_p \in A_2 \right\}$

$E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

It is clear that $A_1 \subset E_1$ and $A_2 \subset E_2$. Furthermore, $E_1 \cap E_2=\varnothing$. This observation will be the basis for showing that Bing’s Example G is normal.

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The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on $2^Q$. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. The set $F_P$ is defined as above. Bing’s Example G is $F=2^Q$ with points in $F_P$ retaining the open sets in the product topology and with points not in $F_P$ declared isolated. For some reason, in Bing’s original paper, the notation $F$ is used even though the example is identified by G. We will follow Bing’s notation.

The subspace $F_P$ is discrete but not closed in the product topology. However, $F_P$ is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space $F$ is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set $F_P$ cannot be separated by disjoint open sets. By declaring points not in $F_P$ isolated, the countable chain condition in the original product space $2^Q$ is destroyed. However, there is still a strong trace of CCC around the points in the set $F_P$, which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that $F$ is normal, let $H$ and $K$ be disjoint closed subsets of $F$. To make it easy to follow, let $H=A_1 \cup B_1$ and $K=A_2 \cup B_2$ where

$A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)$

$A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)$

In other words, $A$ is the non-isolated part and $B$ is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets $E_1$ and $E_2$ where $A_1 \subset E_1$ and $A_2 \subset E_2$. Set the following open sets.

$O_1=(E_1 \cup B_1) - K$

$O_2=(E_2 \cup B_2) - H$

It follows that $O_1$ and $O_2$ are disjoint open sets and that $A_1 \subset O_1$ and $A_2 \subset O_2$. Thus Bing’s Example G is a normal space.

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Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space $X$ is countably paracompact if every countable countable open cover of $X$ has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every countable open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent (see Corollary 2 here). Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let $\mathcal{U}$ be a countable open cover of $F$. Let $\mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\}$ be the set of all open sets in $\mathcal{U}$ that contain points in $F_P$. For each $i$, let $A_i=U_i \cap F_P$. From the perspective of Bing’s Example G, the sets $A_i$ are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let $O_1,O_2,O_3,\cdots$ be disjoint open sets such that $A_i \subset O_i$ for each $i$.

We now build a point-finite open refinement of $\mathcal{U}$. For each $i$, let $V_i=U_i \cap O_i$. Let $V=\cup_{i=1}^\infty V_i$. Consider the following.

$\mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}$

It follows that $\mathcal{V}$ is an open cover of $F$. All points of $F_P$ belong to the open sets $V_i$. Any point that is not in one of the $V_i$ belongs to a singleton open set. It is also clear that $\mathcal{V}$ is a refinement of $\mathcal{U}$. For each $i$, $V_i \subset U_i$ and each singleton set is contained in some member of $\mathcal{U}$. It follows that each point in $F$ belongs to at most finitely many sets in $\mathcal{V}$. In fact, each point belongs to exactly one set in $\mathcal{V}$. Each point in $F_P$ belongs to exactly one $V_i$ since the open sets $O_i$ are disjoint. Any point in $V$ belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

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Previous Posts

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$\copyright \ 2016 \text{ by Dan Ma}$

Bing’s Example H

In a previous post we introduced Bing’s Example G, a classic example of a normal but not collectionwise normal space. Other properties of Bing’s Example G include: completely normal, not perfectly normal and not metacompact. This is an influential example introduced in an influential paper of R. H. Bing in 1951 (see [1]). In the same paper, another example called Example H was introduced. This space has some of the same properties of Example G, except that it is perfectly normal. In this post, we define and discuss Example H.

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Defining Bing’s Example H

Throughout the discussion in this post, we use $\omega$ to denote the first infinite ordinal, i.e., $\omega =\left\{0,1,2,3,\cdots \right\}$. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of the set $P$, i.e., it is the power set of $P$. Let $H$ be the set of all functions $f:Q \rightarrow \omega$. In other words, the set $H$ is the Cartesian product $\prod \limits_{q \in Q} \omega$. But the topology on $H$ is not the product topology.

For each $p \in P$, consider the function $f_p:Q \rightarrow 2=\left\{0,1 \right\}$ such that for each $q \in Q$:

$f_p(q) = \begin{cases} 1, & \mbox{if } p \in q \\ 0, & \mbox{if } p \notin q \end{cases}$

Let $H_P=\left\{f_p: p \in P \right\}$. Now define a topology on the set $H$ by the following:

• Each point of $H-H_P$ is an isolated point.
• Each point $f_p \in H_P$ has basic open sets of the form $U(p,W,n)$ defined as follows:

$U(p,W,n)=\left\{f_p \right\} \cup D(p,W,n)$

$D(p,W,n)=\left\{f \in H: \forall q \in Q, f(q) \ge n \text{ and } \forall q \in W, f(q) \equiv f_p(q) \ (\text{mod} \ 2) \right\}$

where $p \in P$, $W \subset Q$ is finite, and $n \in \omega$.

If $a$ and $b$ are integers, the $a \equiv b \ (\text{mod} \ 2)$ means that $a-b$ is divisible by $2$. The congruence equation $f(q) \equiv f_p(q) \ (\text{mod} \ 2)$ means that $f(q)$ is an even integer if $f_p(q)=0$. On the other hand, $f(q) \equiv f_p(q) \ (\text{mod} \ 2)$ means that $f(q)$ is an odd integer if $f_p(q)=1$.

The set $D(p,W,n)$ seems to mimic a basic open set of the point $f_p$ in the product topology: for each point in $D(p,W,n)$, the value of each coordinate is an integer $\ge n$ and the values for finitely many coordinates are fixed to agree with the function $f_p$ modulo $2$. Adding the point $f_p$ to $D(p,W,n)$, we have a basic open set $U(p,W,n)$.

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Basic Discussion

The points in $H-H_P$ are the isolated points in the space $H$. The points in $H_P$ are the non-isolated points (limit points). The space $H$ is a Hausdorff space. Another interesting point is that the set $H_P$ is a closed and discrete set in the space $H$.

To see that $H$ is Hausdorff, let $h_1, h_2 \in H$ with $h_1 \ne h_2$. Consider the case that $h_1$ is an isolated point and $h_2=f_p$ for some $p \in P$. Let $n$ be the minimum of all $h_1(q)$ over all $q \in Q$. Let $O_1=\left\{h_1 \right\}$ and $O_2=U(p,W,n+1)$ where $W \subset Q$ is any finite set. Then $O_1$ and $O_2$ are disjoint open set containing $h_1$ and $h_2$, respectively.

Now consider the case that $h_1=f_p$ and $h_2=f_{p'}$ where $p \ne p'$. Let $O_1=U(p,W,0)$ and $O_2=U(p',W,0)$ where $W=\left\{ \left\{ p \right\},\left\{ p' \right\} \right\}$. Then $O_1$ and $O_2$ are disjoint open set containing $h_1$ and $h_2$, respectively.

The set $H_P$ is a closed and discrete set in the space $H$. It is closed since $H-H_P$ consists of isolated points. To see that $H_P$ is discrete, note that $U(p,W,0)$, where $W=\left\{ \left\{ p \right\} \right\}$, is an open set with $f_p \in U(p,W,0)$ and $f_{p'} \notin U(p,W,0)$ for all $p' \ne p$.

In the sections below, we show that the space $H$ is normal, completely normal (thus hereditarily normal), and is perfectly normal. Furthermore, we show that it is not collectionwise Hausdorff (hence not collectionwise normal) and not meta-lindelof (hence not metacompact).

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Bing’s Example H is Normal

In the next section, we show that Bing’s Example H is completely normal (i.e. any two separated sets can be separated by disjoint open sets). Note that any two disjoint closed sets are separated sets.

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Bing’s Example H is Completely Normal

Let $X$ be a space. Let $A \subset X$ and $B \subset X$. The sets $A$ and $B$ are separated sets if $A \cap \overline{B}=\varnothing=\overline{A} \cap B$. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space $X$ is said to be completely normal if for every two separated sets $A$ and $B$ in $X$, there exist disjoint open subsets $U$ and $V$ of $X$ such that $A \subset U$ and $B \subset V$. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space $X$, $X$ is completely normal if and only if $X$ is hereditarily normal. For more about completely normality, see [3] and [6].

Let $S$ and $T$ be separated sets in the space $H$, i.e.,

$S \cap \overline{T}=\varnothing=\overline{S} \cap T$

We consider two cases. Case 1 is that one of the sets consists entirely of isolated points. Assume that $S \subset H-H_P$. Let $O_1=S$. For each $x \in T$, choose an open set $V_x$ with $x \in V_x$ and $V_x \cap \overline{S}=\varnothing$. Let $O_2=\bigcup \limits_{x \in T} V_x$. Then $O_1$ and $O_2$ are disjoint open sets containing $S$ and $T$ respectively.

Now consider Case 2 where $S_1=S \cap H_P \ne \varnothing$ and $T_1=T \cap H_P \ne \varnothing$. Consider the sets $q_1$ and $q_2$ defined as follows:

$q_1=\left\{p \in P: f_p \in S_1 \right\}$

$q_2=\left\{p \in P: f_p \in T_1 \right\}$

Let $W=\left\{q_1,q_2 \right\}$. Let $Y_1$ and $Y_2$ be the following open sets:

$Y_1=\bigcup \limits_{p \in q_1} U(p,W,0)$

$Y_2=\bigcup \limits_{p \in q_2} U(p,W,0)$

Immediately, we know that $S_1 \subset Y_1$, $T_1 \subset Y_2$ and $Y_1 \cap Y_2=\varnothing$. Let $S_2=S \cap (H-H_P)$ and $T_2=T \cap (H-H_P)$ (both of which are open). Let $O_1$ and $O_2$ be the following open sets:

$O_1=(Y_1 \cup S_2)-\overline{T}$

$O_2=(Y_2 \cup T_2)-\overline{S}$

Then $O_1$ and $O_2$ are disjoint open sets containing $S$ and $T$ respectively.

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Bing’s Example H is Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is a $G_\delta$-set (i.e. the intersection of countably many open subsets). All we need to show here is that every closed subset is a $G_\delta$-set.

Let $C \subset H$ be a closed set. Of course, if $C$ consists entirely of isolated points, then we are done. So assume that $C \cap H_P \ne \varnothing$. Let $q*=\left\{p \in P: f_p \in C \right\}$. Let $O=C \cap (H-H_P)$, which is open. For each positive integer $n$, define the open set $Y_n$ as follows:

$Y_n=O \cup \biggl( \bigcup \limits_{p \in q*} U(p,\left\{q* \right\},n) \biggr)$

Immediately we have $C \subset Y_n$ for each $n$. Let $g \in \bigcap \limits_{n=1}^\infty Y_n$. We claim that $g \in C$. Suppose $g \notin C$. Then $g \notin O$. It follows that for each $n$ $g \in U(p_n,\left\{q* \right\},n)$ for some $p_n \in q*$. Recall that $U(p_n,\left\{q* \right\},n)=\left\{f_{p_n} \right\} \cup D(p_n,\left\{q* \right\},n)$.

The assumption that $g \notin C$ implies that $g \ne f_{p_n}$ for all $n$. Then $g \in D(p_n,\left\{q* \right\},n)$ for all $n$. By the definition of $D(p_n,\left\{q* \right\},n)$, it follows that for all $q \in Q$, $g(q) \ge n$ for all positive integer $n$. This is a contradiction. So it must be the case that $g \in C$. This completes the proof that Bing’s Example H is perfectly normal.

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Collectionwise Normal Spaces

Let $X$ be a space. Let $\mathcal{A}$ be a collection of subsets of $X$. We say $\mathcal{A}$ is pairwise disjoint if $A \cap B=\varnothing$ whenever $A,B \in \mathcal{A}$ with $A \ne B$. We say $\mathcal{A}$ is discrete if for each $x \in X$, there is an open set $O$ containing $x$ such that $O$ intersects at most one set in $\mathcal{A}$.

The space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{D}$ of closed subsets fo $X$, there is a pairwise disjoint collection $\left\{U_D: D \in \mathcal{D} \right\}$ of open subsets of $X$ such that $D \subset U_D$ for each $D \in \mathcal{D}$. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus both Bing’s Example G and Example H are not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. As shown below Bing’s Example H is actually not collectionwise Hausdorff.

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Bing’s Example H is not Collectionwise Hausdorff

To prove that Bing’s Example H is not collectionwise Hausdorff, we need an intermediate result (Lemma 1) that is based on an infinitary combinatorial result called the Delta-system lemma.

A family $\mathcal{A}$ of sets is called a Delta-system (or $\Delta$-system) if there exists a set $r$, called the root of the $\Delta$-system, such that for any $A,B \in \mathcal{A}$ with $A \ne B$, we have $A \cap B=r$. The following is a version of the Delta-system lemma (see Theorem 1.5 in p. 49 of [2]).

Delta-System Lemma

Let $\mathcal{A}$ be an uncountable family of finite sets. Then there exists an uncountable $\mathcal{B} \subset \mathcal{A}$ such that $\mathcal{B}$ is a $\Delta$-system.
Lemma 1

Let $P_0 \subset P$ be any uncountable subset. For each $p \in P_0$, let $U(p,W_p,n_p)$ be a basic open subset containing $f_p$. Then there exists an uncountable $P_1 \subset P_0$ such that $\bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing$.

Proof of Lemma 1
Let $\mathcal{A}=\left\{W_p: p \in P_0 \right\}$. We need to break this up into two cases – $\mathcal{A}$ is a countable family of finite sets or an uncountable family of finite sets. The first case is relatively easy to see. The second case requires using the Delta-system lemma.

Suppose that $\mathcal{A}$ is countable. Then there exists an uncountable $R \subset P_0$ such that for all $p,t \in R$ with $p \ne t$, we have $W_p=W_t=W$ and $n_p=n_t=n$. Suppose that $W=\left\{q_1,q_2,\cdots,q_m \right\}$. By inductively working on the sets $q_j$, we can obtain an uncountable set $P_1 \subset R$ such that for all $p,t \in P_1$ with $p \ne t$, we have $f_p(q_j)=f_t(q_j)$ for each $j=1,2,\cdots,m$. Clearly, we have:

$\bigcap \limits_{p \in P_1} U(p,W,n) \ne \varnothing$

To show the above, just define a function $h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\}$ such that $h(q_j)=f_p(q_j)$ for all $j=1,2,\cdots,m$ for one particular $p \in P_1$. Then $h$ belongs to the intersection.

Suppose that $\mathcal{A}$ is uncountable. By the Delta-system lemma, there is an uncountable $R \subset P_0$ and there exists a finite set $r \subset Q$ such that for all $p,t \in R$ with $p \ne t$, we have $W_p \cap W_t=r$. Suppose that $r=\left\{q_1,q_2,\cdots,q_m \right\}$. As in the previous case, work inductively on the sets $q_j$, we can obtain an uncountable $S \subset R$ such that for all $p,t \in S$ with $p \ne t$, we have $f_p(q_j)=f_t(q_j)$ for each $j=1,2,\cdots,m$. Now narrow down to an uncountable $P_1 \subset S$ such that $n_p=n_t=n$ for all $p,t \in P_1$ with $p \ne t$. We now show that

$\bigcap \limits_{p \in P_1} U(p,W_p,n) \ne \varnothing$

To define a function $h:Q \rightarrow \left\{n,n+1,n+2,\cdots \right\}$ that belongs to the above intersection, we define $h$ so that $h$ matches $f_t$ (mod 2) with one particular $t \in P_1$ on the set $r=\left\{q_1,q_2,\cdots,q_m \right\}$. Note that $W_p-r$ are disjoint over all $p \in P_1$. So $h$ can be defined on $W_p-r$ to match $f_p$ (mod 2). For any remaining values in the domain, define $h$ freely to be at least the integer $n$. Then the function $h$ belongs to the intersection.

With the two cases established, the proof of Lemma 1 is completed. $\blacksquare$

The fact that Example H is not collectionwise Hausdorff is a corollary of Lemma 1. The set $H_P$ is a discrete collection of points in the space $H$. It follows that $H_P$ cannot be separated by disjoint open sets. For each $p \in P$, let $U(p,W_p,n_p)$ be a basic open set containing the point $f_p$. By Lemma 1, there is an uncountable $P_1 \subset P$ such that $\bigcap \limits_{p \in P_1} U(p,W_p,n_p) \ne \varnothing$. Thus there can be no disjoint collection of open sets in $H$ that separate the points in $H_P$.

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Bing’s Example H is not Metacompact

Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a point-finite (point-countable) collection if every point of $X$ belongs to only finitely (countably) many sets in $\mathcal{A}$. A space $X$ is said to be a metacompact space if every open cover $\mathcal{U}$ of $X$ has a point-finite open refinement $\mathcal{V}$. A space $X$ is said to be a meta-Lindelof space if every open cover $\mathcal{U}$ of $X$ has a point-countable open refinement $\mathcal{V}$. Clearly, every metacompact space is meta-Lindelof.

It follows from Lemma 1 that Example H is not meta-Lindelof. Thus Example H is not metacompact. To see that it is not meta-Lindelof, for each $f_p \in H_P$, let $U_{f_p}=U(p,\left\{\left\{p \right\} \right\},0)$, and for each $x \in H-H_P$, let $U_x=\left\{x \right\}$. Let $\mathcal{U}$ be the following open cover of $H$:

$\mathcal{U}=\left\{U_x: x \in H \right\}$

Each $f_p \in H_P$ belongs to only one set in $\mathcal{U}$, namely $U_{f_p}$. So for any open refinement $\mathcal{V}$ of $\mathcal{U}$ (consisting of basic open sets), we have uncountably many open sets of the form $U(p,W_p,n_p)$. By Lemma 1, we can find uncountably many such open sets with non-empty intersection. So no open refinement of $\mathcal{U}$ can be point-countable.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.

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

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

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

Bing’s Example G

Bing’s Example G is an example of a topological space that is normal but not collectionwise normal. It was introduced in an influential paper of R. H. Bing in 1951 (see [1]). This paper has a metrization theorem that is now called Bing’s metrization theorem (any regular space is metrizable if and only if it has a $\sigma$-discrete base). The paper also introduced the notion of collectionwise normality and discussed the roles it plays in metrization theory (e.g. a Moore space is metrizable if and only if it is collectionwise normal). Example G was an influential example from an influential paper. It became the basis of construction for many other counterexamples (see [5] for one example). Investigations were also conducted by looking at various covering properties among subspaces of Example G (see [2] and [4] are two examples).

In this post we prove some basic results about Bing’s Example G. Some of the results we prove are found in Bing’s 1951 paper. The other results shown here are usually mentioned without proof in various places in the literature.

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Bing’s Example G – Definition

Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Another notation for $2^Q$ is the Cartesian product $\prod \limits_{q \in Q} \left\{0,1 \right\}$. 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. We now consider another topology on $2^Q$ generated by the following base:

$\mathcal{B}=\tau \cup \left\{\left\{x \right\}: x \in F-F_P \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology generated by the base $\mathcal{B}$. 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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Bing’s Example G – Initial Discussion

Bing’s Example G, i.e. the space $F$ as defined above, is obtained by altering the topology of the product space of $2^{\lvert \mathcal{K} \lvert}$ many copies of the two-point discrete space where $\mathcal{K}$ is the cardinality of the power set of the uncountable index set $P$ we start with. Out of this product space, a set $F_P$ of points is carefully chosen such that $F_P$ has the same cardinality as $P$ and such that $F_P$ is relatively discrete in the product space. Points in $F_P$ are made to retain the product topology and all points outside of $F_P$ are declared as isolated points.

We now show that the set $F_P$ is a discrete set in the space $F$. For each $p \in P$, let $W_p$ be the open set defined by

$W_p=\left\{f \in F: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$.

It is clear that $f_p$ is the only point of $F_P$ belonging to $W_p$. Therefore, in the Example G topology, the set $F_P$ is discrete and closed . In the section “Bing’s Example G is not Collectionwise Hausdorff” below, we show below that $F_P$ cannot be separated by any pairwise disjoint collection of open sets.

The character at a point is the minimum cardinality of a local base at that point. The character at a point in $F_P$ in the Example G topology agrees with the product topology. Points in $F_P$ have character $\lvert Q \lvert=2^{\lvert P \lvert}$. Specifically if the starting $P$ has cardinality $\omega_1$, then points in $F_P$ have character $2^{\omega_1}$. Thus Example G has large character and cannot be a Moore space (any Moore space has a countable base at every point).

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Bing’s Example G is Normal

Let $H_1$ and $H_2$ be disjoint closed subsets of $F$. The easy case is that one of $H_1$ and $H_2$ is a subset of $F-F_P$, say $H_1 \subset F-F_P$. Then $H_1$ is a closed and open set in $F$. Then $H_1$ and $F-H_1$ are disjoint open sets containing $H_1$ and $H_2$, respectively. So we can assume that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$.

Let $A_1=H_1 \cap F_P$ and $A_2=H_2 \cap F_P$. Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$. Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$
$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$, we have $A_1 \subset U_1$ and $A_2 \subset U_2$. Furthermore, $U_1 \cap U_2=\varnothing$. Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$, which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-H_2$ and $O_2=(U_2 \cup B_2)-H_1$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$.

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Collectionwise Normal Spaces

Let $X$ be a space. Let $\mathcal{A}$ be a collection of subsets of $X$. We say $\mathcal{A}$ is pairwise disjoint if $A \cap B=\varnothing$ whenever $A,B \in \mathcal{A}$ with $A \ne B$. We say $\mathcal{A}$ is discrete if for each $x \in X$, there is an open set $O$ containing $x$ such that $O$ intersects at most one set in $\mathcal{A}$.

The space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{D}$ of closed subsets fo $X$, there is a pairwise disjoint collection $\left\{U_D: D \in \mathcal{D} \right\}$ of open subsets of $X$ such that $D \subset U_D$ for each $D \in \mathcal{D}$. Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [3]). Thus Bing’s Example G is not paracompact.

When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff. Bing’s Example is actually not collectionwise Hausdorff.

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Bing’s Example G is not Collectionwise Hausdorff

The discrete set $F_P$ cannot be separated by disjoint open sets. For each $p \in P$, let $O_p$ be an open subset of $F$ such that $p \in O_p$. We show that the open sets $O_p$ cannot be pairwise disjoint. For each $p \in P$, choose an open set $L_p$ in the product topology of $2^Q$ such that $p \in L_p \subset O_p$. The product space $2^Q$ is a product of separable spaces, hence has the countable chain condition (CCC). Thus the open sets $L_p$ cannot be pairwise disjoint. Thus $L_t \cap L_s \ne \varnothing$ and $O_t \cap O_s \ne \varnothing$ for at least two points $s,t \in P$.

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Bing’s Example G is Completely Normal

The proof for showing Bing’s Example G is normal can be modified to show that it is completely normal. First some definitions. Let $X$ be a space. Let $A \subset X$ and $B \subset X$. The sets $A$ and $B$ are separated sets if $A \cap \overline{B}=\varnothing=\overline{A} \cap B$. Essentially, any two disjoint sets are separated sets if and only if none of them contains limit points (i.e. accumulation points) of the other set. A space $X$ is said to be completely normal if for every two separated sets $A$ and $B$ in $X$, there exist disjoint open subsets $U$ and $V$ of $X$ such that $A \subset U$ and $B \subset V$. Any two disjoint closed sets are separated sets. Thus any completely normal space is normal. It is well known that for any regular space $X$, $X$ is completely normal if and only if $X$ is hereditarily normal. For more about completely normality, see [3] and [6].

Let $H_1 \subset F$ and $H_2 \subset F$ such that $H_1 \cap \overline{H_2}=\varnothing=\overline{H_1} \cap H_2$. We consider two cases. One is that one of $H_1$ and $H_2$ is a subset of $F-F_P$. The other is that both $H_1 \cap F_P \ne \varnothing$ and $H_2 \cap F_P \ne \varnothing$.

The first case. Suppose $H_1 \subset F-F_P$. Then $H_1$ consists of isolated points and is an open subset of $F$. For each $x \in H_2 \cap F_P$, choose an open subset $V_x$ of $F$ such that $x \in V_x$ and $V_x$ contains no points of $F_P-\left\{ x \right\}$ and $V_x \cap \overline{H_1}=\varnothing$. For each $x \in H_2 \cap (F-F_P)$, let $V_x=\left\{x \right\}$. Let $V$ be the union of all $V_x$ where $x \in H_2$. Let $U=H_1$. Then $U$ and $V$ are disjoint open sets with $H_1 \subset U$ and $H_2 \subset V$.

The second case. Suppose $A_1=H_1 \cap F_P \ne \varnothing$ and $A_2=H_2 \cap F_P \ne \varnothing$. Let $q_1=\left\{p \in P: f_p \in A_1 \right\}$ and $q_2=\left\{p \in P: f_p \in A_2 \right\}$. Define the following open sets:

$U_1=\left\{f \in F: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$
$U_2=\left\{f \in F: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

Because $H_1 \cap H_2=\varnothing$, we have $A_1 \subset U_1$ and $A_2 \subset U_2$. Furthermore, $U_1 \cap U_2=\varnothing$. Let $B_1=H_1 \cap (F-F_P)$ and $B_2=H_2 \cap (F-F_P)$, which are open since they consist of isolated points. Then $O_1=(U_1 \cup B_1)-\overline{H_2}$ and $O_2=(U_2 \cup B_2)-\overline{H_1}$ are disjoint open subsets of $F$ with $H_1 \subset O_1$ and $H_2 \subset O_2$.

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Bing’s Example G is not Perfectly Normal

A space is perfectly normal if it is normal and that every closed subset is $G_\delta$ (i.e. the intersection of countably many open subsets). The set $F_P$ of non-isolated points is a closed set in $F$. We show that $F_P$ cannot be a $G_\delta$-set. Before we do so, we need to appeal to a fact about the product space $2^Q$.

According to the Tychonoff theorem, the product space $2^Q$ is a compact space since it is a product of compact spaces. On the other hand, $2^Q$ is a product of uncountably many factors and is thus not first countable. It is a well known fact that in a compact Hausdorff space, if a point is a $G_\delta$-point, then there is a countable local base at that point (i.e. the space is first countable at that point). Thus no point of the compact product space $2^Q$ can be a $G_\delta$-point. Since points of $F_P$ retain the open sets of the product topology, no point of $F_P$ can be a $G_\delta$-point in the Bing’s Example G topology.

For each $p \in P$, let $W_p$ be open in $F$ such that $f_p \in W_p$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$. For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Suppose that $F_P$ is a $G_\delta$-set. Then $F_P=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is an open subset of $F$. Now for each $p \in P$, we have $\left\{f_p \right\}=\bigcap \limits_{i=1}^\infty (O_i \cap W_p)$, contradicting the fact that the point $f_p$ cannot be a $G_\delta$-point in the space $F$ (and in the product space $2^Q$). Thus $F_P$ is not a $G_\delta$-set in the space $F$, leading to the conclusion that Bing’s Example G is not perfectly normal.

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Bing’s Example G is not Metacompact

A space $M$ is said to have caliber $\omega_1$ if for every uncountable collection $\left\{U_\alpha: \alpha < \omega_1 \right\}$ of non-empty open subsets of $M$, there is an uncountable $A \subset \omega_1$ such that $\bigcap \left\{U_\alpha: \alpha \in A \right\} \ne \varnothing$. Any product of separable spaces has this property (see Topological Spaces with Caliber Omega 1). Thus the product space $2^Q$ has caliber $\omega_1$. Thus in the product space $2^Q$, no collection of uncountably many non-empty open sets can be a point-finite collection (in fact cannot even be point-countable).

To see that the Example G is not metacompact, let $\mathcal{W}=\left\{W_p: p \in P \right\}$ be a collection of open sets such that for $p \in P$, $f_p \in W_p$, $W_p$ is open in the product topology of $2^Q$ and $W_p$ contains no points $F_P-\left\{f_p \right\}$. For example, we can define $W_p$ as in the above section “Bing’s Example G – Initial Discussion”.

Let $W=\bigcup \mathcal{W}$. Let $\mathcal{V}=\mathcal{W} \cup \left\{\left\{ x \right\}: x \in F-W \right\}$. Any open refinement of $\mathcal{V}$ would contain uncountably many open sets in the product topology and thus cannot be point-finite. Thus the space $F$ cannot be metacompact.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Burke, D. K., A note on R. H. Bing’s example G, Top. Conf. VPI, Lectures Notes in Mathematics, 375, Springer Verlag, New York, 47-52, 1974.
3. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
4. Lewis, I. W., On covering properties of subspaces of R. H. Bing’s Example G, Gen. Topology Appl., 7, 109-122, 1977.
5. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
6. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

Bing’s G

This post is a basic discussion of Bing’s Example G. The original post was published on 10/27/2009 and is now replaced by a new post. The following link will take you there. Thank you.