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

# The Sorgenfrey plane is subnormal

The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form $[a,b)$. The Sorgenfrey line is one of the most important counterexamples in general topology. One of the often recited facts about this counterexample is that the Sorgenfrey plane (the square of the Sorgengfrey line) is not normal. We show that, though far from normal, the Sorgenfrey plane is subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal. The Sorgenfrey plane is an example of a subnormal space that is not normal.

In the proof of the non-normality of the Sorgenfrey plane in this previous post, one of the two disjoint closed subsets of the Sorgenfrey plane that cannot be separated by disjoint open sets is countable. Thus the Sorgenfrey plane is not only not normal; it is not pseudonormal (also discussed in this previous post). A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The examples of the Sorgenfrey plane and $\omega_1 \times (\omega_1+1)$ show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal but not subnormal (see this previous post for the non-subnormality).

A space $Y$ is said to be a perfect space if every closed subset of $Y$ is a $G_\delta$ subset of $Y$ (equivalently, every open subset of $Y$ is an $F_\sigma$-subset of $Y$). It is clear that any perfect space is subnormal. We show that the Sorgenfrey plane is perfect. There are subnormal spaces that are not perfect (see the example below).

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The Sorgenfrey plane is perfect

Let $S$ denote the Sorgenfrey line, i.e., the real line $\mathbb{R}$ topologized using the base of half-open intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x . The Sorgenfrey plane is the product space $S \times S$. We show the following:

Proposition 1
The Sorgenfrey line $S$ is perfect.

Proof of Proposition 1
Let $U$ be a non-empty subset of $S$. We show that $U$ is a $F_\sigma$-set. Let $U_0$ be the interior of $U$ in the usual topology. In other words, $U_0$ is the following set:

$U_0=\left\{x \in U: \exists \ (a,b) \text{ such that } x \in (a,b) \text{ and } (a,b) \subset U \right\}$

The real line with the usual topology is perfect. Thus $U_0=\bigcup_{n=1}^\infty F_n$ where each $F_n$ is a closed subset of the real line $\mathbb{R}$. Since the Sorgenfrey topology is finer than the usual topology, each $F_n$ is also closed in the Sorgenfrey line.

Consider $Y=U-U_0$. We claim that $Y$ is countable. Suppose $Y$ is uncountable. Since the Sorgenfrey line is hereditarily Lindelof, there exists $y \in Y$ such that $y$ is a limit point of $Y$ (see Corollary 2 in this previous post). Since $y \in Y \subset U$, $[y,t) \subset U$ for some $t$. Note that $(y,t) \subset U_0$, which means that no point of the open interval $(y,t)$ can belong to $Y$. On the other hand, since $y$ is a limit point of $Y$, $y for some $w \in Y$, a contradiction. Thus $Y$ must be countable. It follows that $U$ is the union of countably many closed subsets of $S$. $\blacksquare$

Proposition 2
If $X$ is perfect and $Y$ is metrizable, then $X \times Y$ is perfect.

Proof of Proposition 2
Let $X$ be perfect. Let $Y$ be a space with a base $\mathcal{B}=\bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$, in addition to being a collection of basic open sets, is a discrete collection. The existence of such a base is equivalent to metrizability, a well known result called Bing’s metrization theorem (see Theorem 4.4.8 in [1]). Let $U$ be a non-empty open subset of $X \times Y$. We show that it is an $F_\sigma$-set in $X \times Y$. For each $x \in U$, there is some open subset $V$ of $X$ and there is some $W \in \mathcal{B}$ such that $x \in V \times W$ and $V \times \overline{W} \subset U$. Thus $U$ is the union of a collection of sets of the form $V \times \overline{W}$. Thus we have:

$U=\bigcup \mathcal{O} \text{ where } \mathcal{O}=\left\{ V_\alpha \times \overline{W_\alpha}: \alpha \in A \right\}$

for some index set $A$. For each positive integer $m$, let $\mathcal{O}_m$ be defined by

$\mathcal{O}_m=\left\{V_\alpha \times \overline{W_\alpha} \in \mathcal{O}: W_\alpha \in \mathcal{B}_m \right\}$

For each $\alpha \in A$, let $V_\alpha=\bigcup_{n=1}^\infty V_{\alpha,n}$ where each $V_{\alpha,n}$ is a closed subset of $X$. For each pair of positive integers $n$ and $m$, define $\mathcal{O}_{n,m}$ by

$\mathcal{O}_{n,m}=\left\{V_{\alpha,n} \times \overline{W_\alpha}: V_\alpha \times \overline{W_\alpha} \in \mathcal{O}_m \right\}$

We claim that each $\mathcal{O}_{n,m}$ is a discrete collection of sets in the space $X \times Y$. Let $(a,b) \in X \times Y$. Since $\mathcal{B}_m$ is discrete, there exists some open subset $H_b$ of $Y$ with $b \in H_b$ such that $H_b$ can intersect at most one $\overline{W}$ where $W \in \mathcal{B}_m$. Then $X \times H_b$ is an open subset of $X \times Y$ with $(a,b) \in X \times H_b$ such that $X \times H_b$ can intersect at most one set of the form $V_{\alpha,n} \times \overline{W_\alpha}$. Then $C_{n,m}=\bigcup \mathcal{O}_{n,m}$ is a closed subset of $X \times Y$. It is clear that $U$ is the union of $C_{n,m}$ over all countably many possible pairs $n,m$. Thus $U$ is an $F_\sigma$-set in $X \times Y$. $\blacksquare$

Proposition 3
The Sorgenfrey plane $S \times S$ is perfect.

Proof of Proposition 3
To get ready for the proof, consider the product spaces $X_1=\mathbb{R} \times S$ and $X_2=S \times \mathbb{R}$ where $\mathbb{R}$ has the usual topology. By both Proposition 1 and Proposition 2, both $X_1$ and $X_2$ are perfect. Also note that the Sorgenfrey plane topology is finer than the topologies for both $X_1$ and $X_2$. Thus a closed set in $X_1$ (in $X_2$) is also a closed set in $S \times S$. It follows that any $F_\sigma$-set in $X_1$ (in $X_2$) is also an $F_\sigma$-set in $S \times S$.

Let $U$ be a non-empty subset of $S \times S$. We show that $U$ is a $F_\sigma$-set. We assume that $U$ is the union of basic open sets of the form $[a,b) \times [a,b)$. Consider the sets $U_1$ and $U_2$ defined by:

$U_1=\left\{x \in U: \exists \ (a,b) \times [a,b) \text{ such that } x \in (a,b) \times [a,b) \text{ and } (a,b) \times [a,b) \subset U \right\}$

$U_2=\left\{x \in U: \exists \ [a,b) \times (a,b) \text{ such that } x \in [a,b) \times (a,b) \text{ and } [a,b) \times (a,b) \subset U \right\}$

Note that $U_1$ is the interior of $U$ when $U$ is considered as a subspace of $X_1$. Likewise, $U_2$ is the interior of $U$ when $U$ is considered as a subspace of $X_2$. Since both $X_1$ and $X_2$ are perfect, $U_1$ and $U_2$ are $F_\sigma$ in $X_1$ and $X_2$, respectively. Hence both $U_1$ and $U_2$ are $F_\sigma$-sets in $S \times S$.

Let $Y=U-(U_1 \cup U_2)$. We claim that $Y$ is an $F_\sigma$-set in $S \times S$. Proposition 3 is established when this claim is proved. To get ready to prove this claim, for each $x=(x_1,x_2) \in S \times S$, and for each positive integer $k$, let $B_k(x)$ be the half-open square $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$. Then $\mathcal{B}(x)=\left\{B_k(x): k=1,2,3,\cdots \right\}$ is a local base at the point $x$. For each positive integer $k$, define $Y_k$ by

$Y_k=\left\{y=(y_1,y_2) \in Y: B_k(y) \subset U \right\}$

Clearly $Y=\bigcup_{k=1}^\infty Y_k$. We claim that each $Y_k$ is closed in $S \times S$. Suppose $x=(x_1,x_2) \in S \times S-Y_k$. In relation to the point $x$, $Y_k$ can be broken into several subsets as follows:

$Y_{k,1}=\left\{y=(y_1,y_2) \in Y_k: y_1=x_1 \text{ and } y_2 \ne x_2 \right\}$

$Y_{k,2}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 = x_2 \right\}$

$Y_{k,\varnothing}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 \ne x_2 \right\}$

Since $x \notin Y_k$, it follows that $Y_k=Y_{k,1} \cup Y_{k,2} \cup Y_{k,\varnothing}$. We show that for each of these three sets, there is an open set containing the point $x$ that is disjoint from the set.

Consider $Y_{k,1}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,1}$, then we are done. So assume $B_k(x) \cap Y_{k,1} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,1}$. Note that $t_1=x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_2

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,1}=\varnothing$. Suppose $g \in G \cap Y_{k,1}$. Then $g_1=x_1$ and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

Note that $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. Thus $H$ is a subset of the interior of $U$ (as a subspace of $X_2$). We have $H \subset U_2$. It follows that $t \in H$ since

$x_1=g_1=t_1$

$x_2

On the other hand, $t \in Y_{k,1} \subset Y_k \subset Y$. Hence $t \notin U_2$, a contradiction. Thus the claim that $G \cap Y_{k,1}=\varnothing$ must be true.

The case $Y_{k,2}$ is symmetrical to the case $Y_{k,1}$. Thus by applying a similar argument, there is an open set containing the point $x$ that is disjoint from the set $Y_{k,2}$.

Now consider the case $Y_{k,\varnothing}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,\varnothing}$, then we are done. So assume $B_k(x) \cap Y_{k,\varnothing} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,\varnothing}$. Note that $t_1>x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,\varnothing}=\varnothing$. Suppose $g \in G \cap Y_{k,\varnothing}$. Then $x_1 and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

As in the previous case, $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. As before, $H \subset U_2$. We also have a contradiction in that $t \in H$ (based on the following)

$x_1

$x_2

and on the one hand and $t \in Y_{k,\varnothing} \subset Y=U-(U_1 \cup U_2)$. Thus the claim that $G \cap Y_{k,\varnothing}=\varnothing$ is true. Take the intersection of the three open sets from the three cases, we have an open set containing $x$ that is disjoint from $Y_k$. Thus $Y_k$ is closed in $S \times S$ and $Y=\bigcup_{k=1}^\infty Y_k$ is $F_\sigma$ in $S \times S$ . $\blacksquare$

Remarks
The authors of [2] showed that any finite power of the Sorgenfrey line is perfect. The proof in [2] is an inductive proof: if $S^n$ is perfect, then $S^{n+1}$ is perfect. We take the inductive proof in [2] and adapt it for the Sorgenfrey plane. The authors in [2] also proved that for a sequence of spaces $X_1,X_2,X_3,\cdots$ such that the product of any finite number of these spaces is perfect, the product $\prod_{n=1}^\infty X_n$ is perfect. Then $S^\omega$ is perfect.

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A non-perfect example

Any perfect space is subnormal. Subnormal spaces do not have to be perfect. In fact subnormal non-normal spaces do not have to be perfect. From a perfect space that is not normal (e.g. the Sorgenfrey plane), one can generate a subnormal and non-normal space that is not perfect. Let $X$ be a subnormal and non-normal space. Let $Y$ be a normal space that is not perfectly normal. There are many possible choices for $Y$. If a specific example is needed, one can take $Y=\omega_1$ with the order topology. Let $X \bigoplus Y$ be the disjoint sum (union) of $X$ and $Y$. The presence of $Y$ destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint $G_\delta$-sets.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Heath, R. W., Michael, E., A property of the Sorgenfrey line, Compositio Math., 23, 185-188, 1971.

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

# Normal x compact needs not be subnormal

In this post, we revisit a counterexample that was discussed previously in this blog. A previous post called “Normal x compact needs not be normal” shows that the Tychonoff product of two normal spaces needs not be normal even when one of the factors is compact. The example is $\omega_1 \times (\omega_1+1)$. In this post, we show that $\omega_1 \times (\omega_1+1)$ fails even to be subnormal. Both $\omega_1$ and $\omega_1+1$ are spaces of ordinals. Thus they are completely normal (equivalent to hereditarily normal). The second factor is also a compact space. Yet their product is not only not normal; it is not even subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal.

A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal (see this previous post). The Sorgenfrey plane is an example of a subnormal space that is not pseudonormal (see here). Thus the two weak forms of normality (pseudonormal and subnormal) are not equivalent.

The same two disjoint closed sets that prove the non-normality of $\omega_1 \times (\omega_1+1)$ are also used for proving non-subnormality. The two closed sets are:

$H=\left\{(\alpha,\alpha): \alpha<\omega_1 \right\}$

$K=\left\{(\alpha,\omega_1): \alpha<\omega_1 \right\}$

The key tool, as in the proof for non-normality, is the Pressing Down Lemma ([1]). The lemma has been used in a few places in this blog, especially for proving facts about $\omega_1$ (e.g. this previous post on the first uncountable ordinal). Lemma 1 below is a lemma that is derived from the Pressing Down Lemma.

Pressing Down Lemma
Let $S$ be a stationary subset of $\omega_1$. Let $f:S \rightarrow \omega_1$ be a pressing down function, i.e., $f$ satisfies: $\forall \ \alpha \in S, f(\alpha)<\alpha$. Then there exists $\alpha<\omega_1$ such that $f^{-1}(\alpha)$ is a stationary set.

Lemma 1
Let $L=\left\{(\alpha,\alpha) \in \omega_1 \times \omega_1: \alpha \text{ is a limit ordinal} \right\}$. Suppose that $L \subset \bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open subset of $\omega_1 \times \omega_1$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty O_n$ for some $\gamma<\omega_1$.

Proof of Lemma 1
For each $n$ and for each $\alpha<\omega_1$ where $\alpha$ is a limit, choose $g_n(\alpha)<\alpha$ such that $[g_n(\alpha),\alpha] \times [g_n(\alpha),\alpha] \subset O_n$. The function $g_n$ can be chosen since $O_n$ is open in the product $\omega_1 \times \omega_1$. By the Pressing Down Lemma, for each $n$, there exists $\gamma_n < \omega_1$ and there exists a stationary set $S_n \subset \omega_1$ such that $g_n(\alpha)=\gamma_n$ for all $\alpha \in S_n$. It follows that $[\gamma_n,\omega_1) \times [\gamma_n,\omega_1) \subset O_n$ for each $n$. Choose $\gamma<\omega_1$ such that $\gamma_n<\gamma$ for all $n$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset O_n$ for each $n$. $\blacksquare$

Theorem 2
The product space $\omega_1 \times (\omega_1+1)$ is not subnormal.

Proof of Theorem 2
Let $H$ and $K$ be defined as above. Suppose $H \subset \bigcap_{n=1}^\infty U_n$ and $K \subset \bigcap_{n=1}^\infty V_n$ where each $U_n$ and each $V_n$ are open in $\omega_1 \times (\omega_1+1)$. Without loss of generality, we can assume that $U_n \cap (\omega_1 \times \left\{\omega_1 \right\})=\varnothing$, i.e., $U_n$ is open in $\omega_1 \times \omega_1$ for each $n$. By Lemma 1, $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$ for some $\gamma<\omega_1$.

Choose $\beta>\gamma$ such that $\beta$ is a successor ordinal. Note that $(\beta,\omega_1) \in \bigcap_{n=1}^\infty V_n$. For each $n$, there exists some $\delta_n<\omega_1$ such that $\left\{\beta \right\} \times [\delta_n,\omega_1] \subset V_n$. Choose $\delta<\omega_1$ such that $\delta >\delta_n$ for all $n$ and that $\delta >\gamma$. Note that $\left\{\beta \right\} \times [\delta,\omega_1) \subset \bigcap_{n=1}^\infty V_n$. It follows that $\left\{\beta \right\} \times [\delta,\omega_1) \subset [\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$. Thus there are no disjoint $G_\delta$ sets separating $H$ and $K$. $\blacksquare$

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Reference

1. Kunen, K., Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York, 1980.

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

# An exercise involving non-normal spaces

A space is normal if any two disjoint closed subsets of the space can be separated by disjoint open sets. A space is pseudonormal if any two disjoint closed subsets of the space, one of which is countable, can be separated by disjoint open sets. In this post, we present an interesting exercise that deals with non-normal spaces:

Take a space that is not normal. Then determine whether it is pseudonormal. You can supply your own examples or you can start with several non-normal spaces listed below. Once you have a list, determine which ones are psuedonormal and which ones are not.

To make the exercise more interesting, we propose that the focus is on spaces that are $T_1$ (i.e. singleton sets are closed) and regular. Since regular Lindelof spaces are normal, we will be certain that any non-normal (and regular) space is not Lindelof.

In the previous post called Pseudonormal spaces, we identify four spaces that are known to be non-normal. Three of these spaces are not normal because one countable closed set and another closed set cannot be separated, hence not pseudonormal (one is the Sorgenfrey plane and one is the Niemmytzkis’ plane). The fourth non-normal space is pseudonormal.

Here’s a list of several other non-normal spaces previously discussed in this blog.

• The Tychonoff Plank.
• The sigma-product of $\omega_1$ many copies of $\omega_1+1$.
• The product space $\omega^{\omega_1}$.
• The product of the Michael line and the space of irrationals.
• The product of countably many copies of the Michael line.
• The product of a Lindelof space and a Bernstein set.
• The Pixley-Roy space $\mathcal{F}[\mathbb{R}]$.
• Mrowka space, defined on a maximal almost disjoint family of subsets of $\omega$.

Readers are welcome to submit other examples of non-normal spaces. Submit examples by entering a comment below. Submitted examples that are different from the ones listed above will be appended to this post.

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

# Pseudonormal spaces

When two disjoint closed sets in a topological space cannot be separated by disjoint open sets, the space fails to be a normal space. When one of the two closed sets is countable, the space fails to satisfy a weaker property than normality. A space $X$ is said to be a pseudonormal space if $H$ and $K$ can always be separated by two disjoint open sets whenever $H$ and $K$ are disjoint closed subsets of $X$ and one of them is countable. In this post, we discuss several non-normal spaces that actually fail to be pseudonormal. We also give an example of a pseudonormal space that is not normal.

We work with spaces that are at minimum $T_1$ spaces, i.e., spaces in which singleton sets are closed. Then any pseudonormal space is regular. To see this, let $X$ be $T_1$ and pseudonormal. For any closed subset $C$ of $X$ and for any point $x \in X-C$, we can always separate the disjoint closed sets $\left\{ x \right\}$ and $C$ by disjoint open sets. This is one reason why we insist on having $T_1$ separation axiom as a starting point. We now show some examples of spaces that fail to be pseudonormal.

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Some Non-Pseudonormal Examples

All three examples in this section are spaces where the failure of normality is exhibited by the inability of separating a countable closed set and another disjoint closed set.

Example 1
This example of a non-normal space that fails to be pseudonormal is defined in the previous post called An Example of a Completely Regular Space that is not Normal. This is an example of a Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular space that is not normal. We state the definition of the space and present a proof that it is not pseudonormal.

Let $E$ be the set of all points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $y \ge 0$. For each real number $x$, define the following sets:

$V_x=\left\{(x,y) \in E: 0 \le y \le 2 \right\}$

$D_x=\left\{(s,s-x) \in E: x \le s \le x+2 \right\}$

$O_x=V_x \cup D_x$

The set $V_x$ is the vertical line of height 2 at the point $(x,0)$. The set $D_x$ is the line originating at $(x,0)$ and going in the Northeast direction reaching the same vertical height as $V_x$ as shown in the following figure.

The topology on $E$ is defined by the following:

• Each point $(x,y) \in E$ where $y>0$ is isolated.
• For each point $(x,0) \in E$, a basic open set is of the form $O_x - F$ where $(x,0) \notin F$ and $F$ is a finite subset of $O_x$.

The x-axis in this example is a closed and discrete set of cardinality continuum. Amy two disjoint subsets of the x-axis are disjoint closed sets. The two closed sets that cannot be separated are:

$H=\left\{(x,0) \in E: x \text{ is rational} \right\}$

$K=\left\{(x,0) \in E: x \text{ is irrational} \right\}$

For each $(x,0)$, let $W_x=O_x-F_x$ where $F_x \subset O_x$ is finite and $(x,0) \notin F_x$. Furthermore, break up $F_x$ by letting $F_{x,d}=F_x \cap D_x$ and $F_{x,v}=F_x \cap V_x$. Let $U$ and $V$ be defined by:

$U_H=\bigcup \limits_{(x,0) \in H} W_x$

$U_K=\bigcup \limits_{(x,0) \in K} W_x$

The open sets $U_H$ and $U_K$ are essentially arbitrary open sets containing $H$ and $K$ respectively. We claims that $U_H \cap U_K \ne \varnothing$.

Define the projection map $\tau_1:\mathbb{R}^2 \rightarrow \mathbb{R}$ by $\tau_1(x,y)=x$. Let $A$ and $B$ be defined by:

$A=\bigcup \left\{\tau_1(F_{x,d}): (x,0) \in H \right\}$

$B=\left\{(x,0) \in K: (x,0) \notin A \right\}$

The set $A$ is countable. So the set $B$ is uncountable. Choose $(x,0) \in B$. Choose $(a,0) \in H$ on the left of $(x,0)$ and close enough to $(x,0)$ such that $V_x \cap D_a=\left\{t \right\}$ and $t \notin F_{x,v}$. This means that

$t \in V_x \cup D_x -F_x=O_x-F_x=W_x$

$t \in V_a \cup D_a -F_a=O_a-F_a=W_a$.

Thus $U_H \cap U_K \ne \varnothing$. We have shown that the space $E$ is not pseudonormal and thus not normal.

Example 2
The Sorgenfrey line is the real line $\mathbb{R}$ topologized by the base consisting of half open and half closed intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x < b \right\}$. In this post, we use $S$ to denote the real line $\mathbb{R}$ with this topology.

The Sorgenfrey line $S$ is a classic example of a normal space whose square $S \times S$ is not normal. In the Sorgenfrey plane $S \times S$, the set $\left\{(x,-x) \in S \times S: x \in \mathbb{R} \right\}$ is a closed and discrete set and is called the anti-diagonal. The proof presented in this previous post shows that the following two disjoint closed subsets of $S \times S$

$H=\left\{(x,-x) \in S \times S: x \text{ is rational} \right\}$

$K=\left\{(x,-x) \in S \times S: x \text{ is irrational} \right\}$

cannot be separated by disjoint open sets. The argument is based on the fact that the real line with the usual topology is of second category. The key point in the argument is that the set of the irrationals cannot be the union of countably many closed and nowhere dense sets (in the usual topology of the real line).

Thus $S \times S$ fails to be pseudonormal. This example shows that normality can fail to be preserved by taking Cartesian product in such a way that even pseudonormality cannot be achieved in the Cartesian product!

Example 3
Another example of a non-normal space that fails to be pseudonormal is the Niemmytzkis’ plane (Example 2 in in this previous post). The underlying set is $N=\left\{(x,y) \in \mathbb{R} \times \mathbb{R}: y \ge 0 \right\}$. The points lying above the x-axis have the usual Euclidean open neighborhoods. A point $(x,0)$ in the x-axis has as neighborhoods $\left\{(x,0) \right\}$ together with the interior of a disc in the upper half plane that is tangent at the point $(x,0)$. Consider the following the two disjoint closed sets on the x-axis:

$H=\left\{(x,0): x \text{ is rational} \right\}$

$K=\left\{(x,0): x \text{ is irrational} \right\}$

The disjoint closed sets $H$ and $K$ cannot be separated by disjoint open sets (see Niemytzki’s Tangent Disc Topology in [2], Example 82). Like Example 2 above, the argument that $H$ and $K$ cannot be separated is also a Baire category argument.

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An Example of Pseudonormal but not Normal

Example 4
One way to find such a space is to look for spaces that are non-normal and see which one is pseudonormal. On the other hand, in a pseudonormal space, countable closed sets are easily separated from other disjoint closed sets. One space in which “countable” is nice is the first uncountable ordinal $\omega_1$ with the order topology. But $\omega_1$ is normal. So we look at the Cartesian product $\omega_1 \times (\omega_1 +1)$. The second factor is the successor ordinal to $\omega_1$ or as a space that is obtained by tagging one more point to $\omega_1$ that is considered greater than all the points in $\omega_1$. Let’s use $X \times Y=\omega_1 \times (\omega_1 +1)$ to denote this space.

The space $X \times Y$ is not normal (shown in this previous post). In the previous post, $X \times Y$ is presented as an example showing that the product of a normal space with a compact space needs not be normal. However, in this case at least, the product is pseudonormal.

Let $\alpha < \omega_1$. Then the square $\alpha \times \alpha$ as a subspace of $X \times Y$ is a countable space and a first countable space. So it has a countable base (second countable) and thus metrizable, and in particular normal. Any countable subset of $X \times Y$ is contained in one of these countable squares, making it easy to separate a countable closed set from another closed set.

Let $H$ and $K$ be disjoint closed sets in $X \times Y$ such that $H$ is countable. Then there is some successor ordinal $\mu < \omega_1$ ($\mu=\alpha+1$ for some ordinal $\alpha<\omega_1$) such that $H \subset \mu \times \mu$. Based on the discussion in the preceding paragraph, there are disjoint open sets $O_H$ and $O_K$ in $\mu \times \mu$ such that $H \subset O_H$ and $(K \cap (\mu \times \mu)) \subset O_K$. With $\mu$ being a successor ordinal, the square $\mu \times \mu$ is both closed and open in $X \times Y$. Then the following sets

$V_H=O_H$

$V_K=O_K \cup (X \times Y-\mu \times \mu)$

are disjoint open sets in $X \times Y$ separating $H$ and $K$.

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In each of Examples 1, 2 and 3 discussed above, there is a closed and discrete set of cardinality continuum (the x-axis in Examples 1 and 3 and the anti-diagonal in Example 2). So the extent of each of these three spaces is continuum. Note that the extent of a space is the maximum cardinality of a closed and discrete subset.

In each of these examples, it just so happens that it is not possible to separate the rationals from the irrationals in the x-axis or the anti-diagonal by disjoint open sets, making each example not only not normal but also not pseudonormal.

What if we consider a smaller subset of the x-axis or anti-diagonal? For example, consider an uncountable set of cardinality less than continuum. Then what can we say about the pseudonormality or normality of the resulting subspaces? For Example 1, the picture is clear cut.

In Example 1, the argument that $H$ and $K$ cannot be separated is a “countable vs. uncountable” argument. The argument will work as long as $H$ is a countable dense set in the x-axis (dense in the usual topology) and $K$ is any uncountable set.

For Example 2 and Example 3, the argument that $H$ and $K$ cannot be separated is not a “countable vs. uncountable” argument and instead is a Baire category argument. The fact that one of the closed sets is the irrationals is a crucial point. On the other hand, both Example 2 and Example 3 (especially Example 3) are set-theoretic sensitive examples. For Example 2 and Example 3, the normality of the resulting smaller subspaces is dependent on some extra axioms beyond ZFC. For pseudonormality, it could be set-theoretic sensitive too. We give some indication here why this is so.

Let $S$ be the Sorgenfrey line as in Example 2 above. Assuming Martin’s Axiom and the negation of the continuum hypothesis (abbreviated by MA + not CH), for any uncountable $X \subset S$ with $\lvert X \lvert < c$, $X \times X$ is normal but not paracompact (see Example 6.3 in [1] and see [3]). Even though $X \times X$ is not exactly a comparable example, this example shows that restricting to a smaller subset on the anti-diagonal seems to make the space normal.

Example 3 has an illustrious history with respect to the normal Moore space conjecture. There is not surprise that extra set-theory axioms are used. For any subset $B$ of the x-axis, let $N(B)$ be the space defined as in Example 3 above except that only points of $B$ are used on the x-axis. Assuming MA + not CH, for any uncountable $B$ that is of cardinality less than continuum, it can be shown that $N(B)$ is normal non-metrizable Moore space (see Example F in [4]). So by assuming extra axiom of MA + not CH, we cannot get a non-pseudonormal example out of Example 3 by restricting to a smaller uncountable subset of the x-axis. Under other set-theoretic axioms, there exists no normal non-metrizable Moore space. Just because this is a set-theoretic sensitive example, it is conceivable that $N(B)$ could be a space that is not pseudonormal under some other axioms.

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Reference

1. Burke, D. K., Covering Properties, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 347-422, 1984.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc, Amsterdam, New York, 1995.
3. Przymusinski, T. C., A Lindelof space $X$ such that $X \times X$ is normal but not paracompact, Fund. Math., 91, 161-165, 1973.
4. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.

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

# Pixley-Roy hyperspaces

In this post, we introduce a class of hyperspaces called Pixley-Roy spaces. This is a well-known and well studied set of topological spaces. Our goal here is not to be comprehensive but rather to present some selected basic results to give a sense of what Pixley-Roy spaces are like.

A hyperspace refers to a space in which the points are subsets of a given “ground” space. There are more than one way to define a hyperspace. Pixley-Roy spaces were first described by Carl Pixley and Prabir Roy in 1969 (see [5]). In such a space, the points are the non-empty finite subsets of a given ground space. More precisely, let $X$ be a $T_1$ space (i.e. finite sets are closed). Let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For each $F \in \mathcal{F}[X]$ and for each open subset $U$ of $X$ with $F \subset U$, we define:

$[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$

The sets $[F,U]$ over all possible $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The hyperspace as defined above was first defined by Pixley and Roy on the real line (see [5]) and was later generalized by van Douwen (see [7]). These spaces are easy to define and is useful for constructing various kinds of counterexamples. Pixley-Roy played an important part in answering the normal Moore space conjecture. Pixley-Roy spaces have also been studied in their own right. Over the years, many authors have investigated when the Pixley-Roy spaces are metrizable, normal, collectionwise Hausdorff, CCC and homogeneous. For a small sample of such investigations, see the references listed at the end of the post. Our goal here is not to discuss the results in these references. Instead, we discuss some basic properties of Pixley-Roy to solidify the definition as well as to give a sense of what these spaces are like. Good survey articles of Pixley-Roy are [3] and [7].

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Basic Discussion

In this section, we focus on properties that are always possessed by a Pixley-Roy space given that the ground space is at least $T_1$. Let $X$ be a $T_1$ space. We discuss the following points:

1. The topology defined above is a legitimate one, i.e., the sets $[F,U]$ indeed form a base for a topology on $\mathcal{F}[X]$.
2. $\mathcal{F}[X]$ is a Hausdorff space.
3. $\mathcal{F}[X]$ is a zero-dimensional space.
4. $\mathcal{F}[X]$ is a completely regular space.
5. $\mathcal{F}[X]$ is a hereditarily metacompact space.

Let $\mathcal{B}=\left\{[F,U]: F \in \mathcal{F}[X] \text{ and } U \text{ is open in } X \right\}$. Note that every finite set $F$ belongs to at least one set in $\mathcal{B}$, namely $[F,X]$. So $\mathcal{B}$ is a cover of $\mathcal{F}[X]$. For $A \in [F_1,U_1] \cap [F_2,U_2]$, we have $A \in [A,U_1 \cap U_2] \subset [F_1,U_1] \cap [F_2,U_2]$. So $\mathcal{B}$ is indeed a base for a topology on $\mathcal{F}[X]$.

To show $\mathcal{F}[X]$ is Hausdorff, let $A$ and $B$ be finite subsets of $X$ where $A \ne B$. Then one of the two sets has a point that is not in the other one. Assume we have $x \in A-B$. Since $X$ is $T_1$, we can find open sets $U, V \subset X$ such that $x \in U$, $x \notin V$ and $A \cup B-\left\{ x \right\} \subset V$. Then $[A,U \cup V]$ and $[B,V]$ are disjoint open sets containing $A$ and $B$ respectively.

To see that $\mathcal{F}[X]$ is a zero-dimensional space, we show that $\mathcal{B}$ is a base consisting of closed and open sets. To see that $[F,U]$ is closed, let $C \notin [F,U]$. Either $F \not \subset C$ or $C \not \subset U$. In either case, we can choose open $V \subset X$ with $C \subset V$ such that $[C,V] \cap [F,U]=\varnothing$.

The fact that $\mathcal{F}[X]$ is completely regular follows from the fact that it is zero-dimensional.

To show that $\mathcal{F}[X]$ is metacompact, let $\mathcal{G}$ be an open cover of $\mathcal{F}[X]$. For each $F \in \mathcal{F}[X]$, choose $G_F \in \mathcal{G}$ such that $F \in G_F$ and let $V_F=[F,X] \cap G_F$. Then $\mathcal{V}=\left\{V_F: F \in \mathcal{F}[X] \right\}$ is a point-finite open refinement of $\mathcal{G}$. For each $A \in \mathcal{F}[X]$, $A$ can only possibly belong to $V_F$ for the finitely many $F \subset A$.

A similar argument show that $\mathcal{F}[X]$ is hereditarily metacompact. Let $Y \subset \mathcal{F}[X]$. Let $\mathcal{H}$ be an open cover of $Y$. For each $F \in Y$, choose $H_F \in \mathcal{H}$ such that $F \in H_F$ and let $W_F=([F,X] \cap Y) \cap H_F$. Then $\mathcal{W}=\left\{W_F: F \in Y \right\}$ is a point-finite open refinement of $\mathcal{H}$. For each $A \in Y$, $A$ can only possibly belong to $W_F$ for the finitely many $F \subset A$ such that $F \in Y$.

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More Basic Results

We now discuss various basic topological properties of $\mathcal{F}[X]$. We first note that $\mathcal{F}[X]$ is a discrete space if and only if the ground space $X$ is discrete. Though we do not need to make this explicit, it makes sense to focus on non-discrete spaces $X$ when we look at topological properties of $\mathcal{F}[X]$. We discuss the following points:

1. If $X$ is uncountable, then $\mathcal{F}[X]$ is not separable.
2. If $X$ is uncountable, then every uncountable subspace of $\mathcal{F}[X]$ is not separable.
3. If $\mathcal{F}[X]$ is Lindelof, then $X$ is countable.
4. If $\mathcal{F}[X]$ is Baire space, then $X$ is discrete.
5. If $\mathcal{F}[X]$ has the CCC, then $X$ has the CCC.
6. If $\mathcal{F}[X]$ has the CCC, then $X$ has no uncountable discrete subspaces,i.e., $X$ has countable spread, which of course implies CCC.
7. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily Lindelof.
8. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily separable.
9. If $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.
10. The Pixley-Roy space of the Sorgenfrey line does not have the CCC.
11. If $X$ is a first countable space, then $\mathcal{F}[X]$ is a Moore space.

Bullet points 6 to 9 refer to properties that are never possessed by Pixley-Roy spaces except in trivial cases. Bullet points 6 to 8 indicate that $\mathcal{F}[X]$ can never be separable and Lindelof as long as the ground space $X$ is uncountable. Note that $\mathcal{F}[X]$ is discrete if and only if $X$ is discrete. Bullet point 9 indicates that any non-discrete $\mathcal{F}[X]$ can never be a Baire space. Bullet points 10 to 13 give some necessary conditions for $\mathcal{F}[X]$ to be CCC. Bullet 14 gives a sufficient condition for $\mathcal{F}[X]$ to have the CCC. Bullet 15 indicates that the hereditary separability and the hereditary Lindelof property are not sufficient conditions for the CCC of Pixley-Roy space (though they are necessary conditions). Bullet 16 indicates that the first countability of the ground space is a strong condition, making $\mathcal{F}[X]$ a Moore space.

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To see bullet point 6, let $X$ be an uncountable space. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $\mathcal{F}[X]$. Choose a point $x \in X$ that is not in any $F_n$. Then none of the sets $F_i$ belongs to the basic open set $[\left\{x \right\} ,X]$. Thus $\mathcal{F}[X]$ can never be separable if $X$ is uncountable.

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To see bullet point 7, let $Y \subset \mathcal{F}[X]$ be uncountable. Let $W=\cup \left\{F: F \in Y \right\}$. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $Y$. We can choose a point $x \in W$ that is not in any $F_n$. Choose some $A \in Y$ such that $x \in A$. Then none of the sets $F_n$ belongs to the open set $[A ,X] \cap Y$. So not only $\mathcal{F}[X]$ is not separable, no uncountable subset of $\mathcal{F}[X]$ is separable if $X$ is uncountable.

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To see bullet point 8, note that $\mathcal{F}[X]$ has no countable open cover consisting of basic open sets, assuming that $X$ is uncountable. Consider the open collection $\left\{[F_1,U_1],[F_2,U_2],[F_3,U_3],\cdots \right\}$. Choose $x \in X$ that is not in any of the sets $F_n$. Then $\left\{ x \right\}$ cannot belong to $[F_n,U_n]$ for any $n$. Thus $\mathcal{F}[X]$ can never be Lindelof if $X$ is uncountable.

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For an elementary discussion on Baire spaces, see this previous post.

To see bullet point 9, let $X$ be a non-discrete space. To show $\mathcal{F}[X]$ is not Baire, we produce an open subset that is of first category (i.e. the union of countably many closed nowhere dense sets). Let $x \in X$ a limit point (i.e. an non-isolated point). We claim that the basic open set $V=[\left\{ x \right\},X]$ is a desired open set. Note that $V=\bigcup \limits_{n=1}^\infty H_n$ where

$H_n=\left\{F \in \mathcal{F}[X]: x \in F \text{ and } \lvert F \lvert \le n \right\}$

We show that each $H_n$ is closed and nowhere dense in the open subspace $V$. To see that it is closed, let $A \notin H_n$ with $x \in A$. We have $\lvert A \lvert>n$. Then $[A,X]$ is open and every point of $[A,X]$ has more than $n$ points of the space $X$. To see that $H_n$ is nowhere dense in $V$, let $[B,U]$ be open with $[B,U] \subset V$. It is clear that $x \in B \subset U$ where $U$ is open in the ground space $X$. Since the point $x$ is not an isolated point in the space $X$, $U$ contains infinitely many points of $X$. So choose an finite set $C$ with at least $2 \times n$ points such that $B \subset C \subset U$. For the the open set $[C,U]$, we have $[C,U] \subset [B,U]$ and $[C,U]$ contains no point of $H_n$. With the open set $V$ being a union of countably many closed and nowhere dense sets in $V$, the open set $V$ is not of second category. We complete the proof that $\mathcal{F}[X]$ is not a Baire space.

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To see bullet point 10, let $\mathcal{O}$ be an uncountable and pairwise disjoint collection of open subsets of $X$. For each $O \in \mathcal{O}$, choose a point $x_O \in O$. Then $\left\{[\left\{ x_O \right\},O]: O \in \mathcal{O} \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ is CCC then $X$ must have the CCC.

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To see bullet point 11, let $Y \subset X$ be uncountable such that $Y$ as a space is discrete. This means that for each $y \in Y$, there exists an open $O_y \subset X$ such that $y \in O_y$ and $O_y$ contains no point of $Y$ other than $y$. Then $\left\{[\left\{y \right\},O_y]: y \in Y \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then the ground space $X$ has no uncountable discrete subspace (such a space is said to have countable spread).

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To see bullet point 12, let $Y \subset X$ be uncountable such that $Y$ is not Lindelof. Then there exists an open cover $\mathcal{U}$ of $Y$ such that no countable subcollection of $\mathcal{U}$ can cover $Y$. We can assume that sets in $\mathcal{U}$ are open subsets of $X$. Also by considering a subcollection of $\mathcal{U}$ if necessary, we can assume that cardinality of $\mathcal{U}$ is $\aleph_1$ or $\omega_1$. Now by doing a transfinite induction we can choose the following sequence of points and the following sequence of open sets:

$\left\{x_\alpha \in Y: \alpha < \omega_1 \right\}$

$\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$

such that $x_\beta \ne x_\gamma$ if $\beta \ne \gamma$, $x_\alpha \in U_\alpha$ and $x_\alpha \notin \bigcup \limits_{\beta < \alpha} U_\beta$ for each $\alpha < \omega_1$. At each step $\alpha$, all the previously chosen open sets cannot cover $Y$. So we can always choose another point $x_\alpha$ of $Y$ and then choose an open set in $\mathcal{U}$ that contains $x_\alpha$.

Then $\left\{[\left\{x_\alpha \right\},U_\alpha]: \alpha < \omega_1 \right\}$ is a pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then $X$ must be hereditarily Lindelof.

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To see bullet point 13, let $Y \subset X$. Consider open sets $[A,U]$ where $A$ ranges over all finite subsets of $Y$ and $U$ ranges over all open subsets of $X$ with $A \subset U$. Let $\mathcal{G}$ be a collection of such $[A,U]$ such that $\mathcal{G}$ is pairwise disjoint and $\mathcal{G}$ is maximal (i.e. by adding one more open set, the collection will no longer be pairwise disjoint). We can apply a Zorn lemma argument to obtain such a maximal collection. Let $D$ be the following subset of $Y$.

$D=\bigcup \left\{A: [A,U] \in \mathcal{G} \text{ for some open } U \right\}$

We claim that the set $D$ is dense in $Y$. Suppose that there is some open set $W \subset X$ such that $W \cap Y \ne \varnothing$ and $W \cap D=\varnothing$. Let $y \in W \cap Y$. Then $[\left\{y \right\},W] \cap [A,U]=\varnothing$ for all $[A,U] \in \mathcal{G}$. So adding $[\left\{y \right\},W]$ to $\mathcal{G}$, we still get a pairwise disjoint collection of open sets, contradicting that $\mathcal{G}$ is maximal. So $D$ is dense in $Y$.

If $\mathcal{F}[X]$ has the CCC, then $\mathcal{G}$ is countable and $D$ is a countable dense subset of $Y$. Thus if $\mathcal{F}[X]$ has the CCC, the ground space $X$ is hereditarily separable.

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A collection $\mathcal{N}$ of subsets of a space $Y$ is said to be a network for the space $Y$ if any non-empty open subset of $Y$ is the union of elements of $\mathcal{N}$, equivalently, for each $y \in Y$ and for each open $U \subset Y$ with $y \in U$, there is some $A \in \mathcal{N}$ with $x \in A \subset U$. Note that a network works like a base but the elements of a network do not have to be open. The concept of network and spaces with countable network are discussed in these previous posts Network Weight of Topological Spaces – I and Network Weight of Topological Spaces – II.

To see bullet point 14, let $\mathcal{N}$ be a network for the ground space $X$ such that $\mathcal{N}$ is also countable. Assume that $\mathcal{N}$ is closed under finite unions (for example, adding all the finite unions if necessary). Let $\left\{[A_\alpha,U_\alpha]: \alpha < \omega_1 \right\}$ be a collection of basic open sets in $\mathcal{F}[X]$. Then for each $\alpha$, find $B_\alpha \in \mathcal{N}$ such that $A_\alpha \subset B_\alpha \subset U_\alpha$. Since $\mathcal{N}$ is countable, there is some $B \in \mathcal{N}$ such that $M=\left\{\alpha< \omega_1: B=B_\alpha \right\}$ is uncountable. It follows that for any finite $E \subset M$, $\bigcap \limits_{\alpha \in E} [A_\alpha,U_\alpha] \ne \varnothing$.

Thus if the ground space $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.

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The implications in bullet points 12 and 13 cannot be reversed. Hereditarily Lindelof property and hereditarily separability are not sufficient conditions for $\mathcal{F}[X]$ to have the CCC. See [4] for a study of the CCC property of the Pixley-Roy spaces.

To see bullet point 15, let $S$ be the Sorgenfrey line, i.e. the real line $\mathbb{R}$ with the topology generated by the half closed intervals of the form $[a,b)$. For each $x \in S$, let $U_x=[x,x+1)$. Then $\left\{[ \left\{ x \right\},U_x]: x \in S \right\}$ is a collection of pairwise disjoint open sets in $\mathcal{F}[S]$.

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A Moore space is a space with a development. For the definition, see this previous post.

To see bullet point 16, for each $x \in X$, let $\left\{B_n(x): n=1,2,3,\cdots \right\}$ be a decreasing local base at $x$. We define a development for the space $\mathcal{F}[X]$.

For each finite $F \subset X$ and for each $n$, let $B_n(F)=\bigcup \limits_{x \in F} B_n(x)$. Clearly, the sets $B_n(F)$ form a decreasing local base at the finite set $F$. For each $n$, let $\mathcal{H}_n$ be the following collection:

$\mathcal{H}_n=\left\{[F,B_n(F)]: F \in \mathcal{F}[X] \right\}$

We claim that $\left\{\mathcal{H}_n: n=1,2,3,\cdots \right\}$ is a development for $\mathcal{F}[X]$. To this end, let $V$ be open in $\mathcal{F}[X]$ with $F \in V$. If we make $n$ large enough, we have $[F,B_n(F)] \subset V$.

For each non-empty proper $G \subset F$, choose an integer $f(G)$ such that $[F,B_{f(G)}(F)] \subset V$ and $F \not \subset B_{f(G)}(G)$. Let $m$ be defined by:

$m=\text{max} \left\{f(G): G \ne \varnothing \text{ and } G \subset F \text{ and } G \text{ is proper} \right\}$

We have $F \not \subset B_{m}(G)$ for all non-empty proper $G \subset F$. Thus $F \notin [G,B_m(G)]$ for all non-empty proper $G \subset F$. But in $\mathcal{H}_m$, the only sets that contain $F$ are $[F,B_m(F)]$ and $[G,B_m(G)]$ for all non-empty proper $G \subset F$. So $[F,B_m(F)]$ is the only set in $\mathcal{H}_m$ that contains $F$, and clearly $[F,B_m(F)] \subset V$.

We have shown that for each open $V$ in $\mathcal{F}[X]$ with $F \in V$, there exists an $m$ such that any open set in $\mathcal{H}_m$ that contains $F$ must be a subset of $V$. This shows that the $\mathcal{H}_n$ defined above form a development for $\mathcal{F}[X]$.

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Examples

In the original construction of Pixley and Roy, the example was $\mathcal{F}[\mathbb{R}]$. Based on the above discussion, $\mathcal{F}[\mathbb{R}]$ is a non-separable CCC Moore space. Because the density (greater than $\omega$ for not separable) and the cellularity ($=\omega$ for CCC) do not agree, $\mathcal{F}[\mathbb{R}]$ is not metrizable. In fact, it does not even have a dense metrizable subspace. Note that countable subspaces of $\mathcal{F}[\mathbb{R}]$ are metrizable but are not dense. Any uncountable dense subspace of $\mathcal{F}[\mathbb{R}]$ is not separable but has the CCC. Not only $\mathcal{F}[\mathbb{R}]$ is not metrizable, it is not normal. The problem of finding $X \subset \mathbb{R}$ for which $\mathcal{F}[X]$ is normal requires extra set-theoretic axioms beyond ZFC (see [6]). In fact, Pixley-Roy spaces played a large role in the normal Moore space conjecture. Assuming some extra set theory beyond ZFC, there is a subset $M \subset \mathbb{R}$ such that $\mathcal{F}[M]$ is a CCC metacompact normal Moore space that is not metrizable (see Example I in [8]).

On the other hand, Pixley-Roy space of the Sorgenfrey line and the Pixley-Roy space of $\omega_1$ (the first uncountable ordinal with the order topology) are metrizable (see [3]).

The Sorgenfrey line and the first uncountable ordinal are classic examples of topological spaces that demonstrate that topological spaces in general are not as well behaved like metrizable spaces. Yet their Pixley-Roy spaces are nice. The real line and other separable metric spaces are nice spaces that behave well. Yet their Pixley-Roy spaces are very much unlike the ground spaces. This inverse relation between the ground space and the Pixley-Roy space was noted by van Douwen (see [3] and [7]) and is one reason that Pixley-Roy hyperspaces are a good source of counterexamples.

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Reference

1. Bennett, H. R., Fleissner, W. G., Lutzer, D. J., Metrizability of certain Pixley-Roy spaces, Fund. Math. 110, 51-61, 1980.
2. Daniels, P, Pixley-Roy Spaces Over Subsets of the Reals, Topology Appl. 29, 93-106, 1988.
3. Lutzer, D. J., Pixley-Roy topology, Topology Proc. 3, 139-158, 1978.
4. Hajnal, A., Juahasz, I., When is a Pixley-Roy Hyperspace CCC?, Topology Appl. 13, 33-41, 1982.
5. Pixley, C., Roy, P., Uncompletable Moore spaces, Proc. Auburn Univ. Conf. Auburn, AL, 1969.
6. Przymusinski, T., Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, 291-297, 1981.
7. van Douwen, E. K., The Pixley-Roy topology on spaces of subsets, Set-theoretic Topology, Academic Press, New York, 111-134, 1977.
8. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.
9. Tanaka, H, Normality and hereditary countable paracompactness of Pixley-Roy hyperspaces, Fund. Math. 126, 201-208, 1986.

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

# 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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