# 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.

____________________________________________________________________

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 $U(p,W,n)$.

____________________________________________________________________

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).

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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 $H_P$.

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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

|

# Product of Spaces with Countable Chain Condition

A topological space is said to be separable if it has a countable dense subset. In any separable space, there cannot exist uncountably many pairwise disjoint open subsets (if that is the case, any dense set will have to be uncountable). A topological space $X$ is said to have countable chain condition (ccc) if every disjoint collection of open subsets of $X$ is countable. Thus any separable space is a space with ccc. We take a look at what happen when we take product of spaces with these two properties.

The product of two separable spaces is always a separable space. If $A$ is a countable dense set in the space $X$ and $B$ is a countable dense set in the space $Y$, then $A \times B$ is a countable dense set in the product space $X \times Y$. It follows that the product of finitely many separable spaces is separable. When the number of factors is infinite, the cardinality of the continuum is the dividing line. The product of separable spaces is sometimes separable (when the number of factors is less than or equal to continuum) and is sometimes not separable (when the number of factors is greater than continuum). For a discussion of this result, see Product of Separable Spaces in this blog, or see [1] and [3].

Is the product of two ccc spaces a space with ccc? It turns out that this question is independent of ZFC. That is, this question cannot be answered on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC (see [2], page 50). However it can be proven in ZFC that the product of any number of separable spaces has countable chain condition. In proving this result, Delta-system lemma is used.

___________________________________________________________________________

Delta-System Lemma

A collection $\mathcal{D}$ of sets is said to be a Delta-system (or $\Delta$-system) if there is a set $D$ such that for every $A,B \in \mathcal{D}$ with $A \ne B$, we have $A \cap B = D$. When such set $D$ exists, it is called the root of the Delta-system $\mathcal{D}$. The following is the statement of Delta-system lemma.

Lemma 1 – Delta-System Lemma
For every uncountable collection $\mathcal{A}$ of finite sets, there is an uncountable $\mathcal{D} \subset \mathcal{A}$ such that $\mathcal{D}$ is a $\Delta$-system.

The statement of Delta-system lemma presented here is a special case (see [2], page 49) for the general version.

___________________________________________________________________________

Products of CCC Spaces

Even though the question of whether the product of two ccc spaces has ccc cannot be answered just within ZFC, we have a theorem that can gives us quite a bit of clarity. We have an amazing result that whenever the product of two ccc spaces has ccc, the product of any number of ccc spaces has ccc. Note that when countable chain condition is preserved by taking product with two factors, ccc is preserved by taking product with any finite number of factors. The following theorem shows that whenever ccc is preserved by taking product with any finite number of factors, ccc is preserved by taking the product of any number of factors. As a corollary, we have the result that the product of any number of separable spaces has ccc.

By manipulating the number of factors, we can easily obtain a space that has ccc but is not separable. For example, let $\mathcal{K}$ be any cardinal that is larger than continuum. Then $\left\{0,1\right\}^\mathcal{K}$ is a space that has ccc but is not separable.

Theorem 2
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of spaces such that $\prod \limits_{\alpha \in F} X_\alpha$ has countable chain condition for every finite $F \subset T$. Then $\prod \limits_{\alpha \in T} X_\alpha$ has countable chain condition.

Proof
In proving the product space $\prod \limits_{\alpha \in T} X_\alpha$ having ccc, it suffices to work with basic open sets of the form $\prod \limits_{\alpha \in T} U_\alpha$ where $U_\alpha=X_\alpha$ for all but finitely many $\alpha \in T$. Let $\mathcal{U}$ be an uncountable collection of such basic open sets such that open sets in $\mathcal{U}$ are pairwise disjoint.

For each $G = \prod \limits_{\alpha \in T} U_\alpha \in \mathcal{U}$, let $A(G)$ be the finite set such that $A(G) \subset T$, and such that $\alpha \in A(G)$ if and only if $U_\alpha \ne X_\alpha$. Let $\mathcal{A}$ be the set of all such $A(G)$. By Delta-system lemma, there is an uncountable $\mathcal{D} \subset \mathcal{A}$ such that $\mathcal{D}$ is a Delta-system. Let $D$ be the root of this Delta-system.

The root of the Delta-system cannot be non-empty. If $D = \varnothing$, then for any $A(G_1) \in \mathcal{D}$ and $A(G_2) \in \mathcal{D}$ where $A(G_1) \ne A(G_2)$ and $G_1,G_2 \in \mathcal{U}$, we have $A(G_1) \cap A(G_2) = \varnothing$, which leads to $G_1 \cap G_2 \ne \varnothing$. Thus $D \ne \varnothing$.

Let $\mathcal{U}^*$ be the collection of all $G \in \mathcal{U}$ such that $A(G) \in \mathcal{D}$. For each $G = \prod \limits_{\alpha \in T} U_\alpha \in \mathcal{U}^*$, let $p(G)=\prod \limits_{\alpha \in D} U_\alpha$ (i.e. $p$ is the projection map). Let $\mathcal{U}^{**}$ be the collection of all $p(G)$ where $G \in \mathcal{U}^*$.

We have the following observation:

• Observation. For any $A(G_1),A(G_2) \in \mathcal{D}$ with $A(G_1) \ne A(G_2)$, $\alpha \in A(G_1)-D$ implies $\alpha \notin A(G_2)-D$ and $\alpha \in A(G_2)-D$ implies $\alpha \notin A(G_1)-D$.

It follows from the above observation that the map $p$ is a one-to-one map from $\mathcal{U}^*$ into $\mathcal{U}^{**}$. Then $\mathcal{U}^{**}$ is an uncountable collection of open subsets of $\prod \limits_{\alpha \in D} X_\alpha$, which has ccc by hypothesis. So there exists $p(G_1),p(G_2) \in \mathcal{U}^{**}$ with $p(G_1) \cap p(G_2) \ne \varnothing$. The above observation allows us to define a point $x \in G_1 \cap G_2$, contradicting the assumption that $\mathcal{U}$ is a pairwise disjoint collection. Thus the entire product space $\prod \limits_{\alpha \in T} X_\alpha$ must have countable chain condition. $\blacksquare$

Corollary 3
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of separable spaces. Then $\prod \limits_{\alpha \in T} X_\alpha$ has countable chain condition.

Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has ccc). $\blacksquare$

___________________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Kunen, K., Set Theory, An Introduction to Independence Proofs, North-Holland, Amsterdam, 1980.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

___________________________________________________________________________

$\copyright \ 2012$