# When all powers of a space are normal

It follows from the Tychonoff Theorem that if a topological space $X$ is a compact Hausdorff space, then the product space $X^\tau$ is compact for any cardinal number $\tau$. Hence for any compact Hausdorff space $X$, the product space $X^\tau$ is normal for any cardinal number $\tau$. The converse is also true. If $X^\tau$ is normal for any cardinal number $\tau$, then $X$ must be compact. This is a theorem that is due to Noble (see [5]). The proof in [5] is a corollary resulted from a long chain of previous results of Noble and others. Many authors had produced simpler and more direct proofs of Noble’s theorem (e.g. [2], [3] and [4]). All these more direct proofs make use of the fact that the product space $\omega^{\omega_1}$ is not normal (due to A. H. Stone). All of them except [2] make use of other strong topological results in order to derive Noble’s theorem. In [2], Engelking established Noble’s theorem by an elementary proof. In this post, we present the proof in [2] in full details. Noble’s theorem is also given in Engelking’s textbook as an exercise (see 3.12.15 on p. 233 in [1]).

Before proceeding to the main theorem, let’s set up some notation for working with the product space $\prod_{\alpha \in S} X_\alpha$. For $x \in \prod_{\alpha \in S} X_\alpha$, the $\alpha^{th}$ coordinate of $x$ is denoted by $x_\alpha$ or $(x)_\alpha$. For $M \subset S$, the map $P_M: \prod_{\alpha \in S} X_\alpha \rightarrow \prod_{\alpha \in M} X_\alpha$ is the natural projection map. In the product space $\prod_{\alpha \in S} X_\alpha$, standard basic open sets are of the form $\prod_{\alpha \in S} O_\alpha$ where $O_\alpha= X_\alpha$ for all but finitely many $\alpha$. We use $supp(\prod_{\alpha \in S} O_\alpha)$ to denote the set of the finite number of $\alpha \in S$ where $O_\alpha \ne X_\alpha$.

Noble’s Theorem

If each power of a space $X$ is normal, then $X$ is compact.

Proof
Suppose that $X^\tau$ is normal for all cardinal numbers $\tau$. Suppose that $X$ is not compact. Then there exists a collection $\mathcal{H}=\left\{H_\alpha: \alpha \in S \right\}$ of closed subsets of $X$ such that $\mathcal{H}$ has the finite intersection property but has empty intersection. Let $H=\prod_{\alpha \in S} H_\alpha$, which is a subspace of the product space $\prod_{\alpha \in S} X_\alpha$ where each $X_\alpha=X$. We can also denote the product space $\prod_{\alpha \in S} X_\alpha$ by $X^\tau$ where $\tau=\lvert S \lvert$.

Let $K=\left\{k \in X^{\lvert S \lvert}: \forall \ \beta, \gamma \in S, \ k_\beta=k_\gamma \right\}$. Note that $K$ is commonly referred to as the diagonal of the product space in question. Both $H$ and $K$ are closed sets in the product space $X^\tau$. Because the collection $\mathcal{H}$ has empty intersection, $H$ and $K$ are disjoint closed sets. Since $X^\tau$ is normal, there exist disjoint open subsets $U$ and $V$ of $X^\tau$ such that $H \subset U$ and $K \subset V$.

Let $x_1 \in H$. Let $O_1$ be a basic standard open set with $x \in O_1 \subset U$. Let $S_1=supp(O_1)$. Then we have $P^{-1}_{S_1}(P_{S_1}(x_1)) \subset U$. Since $\mathcal{H}$ has the finite intersection property, choose $a_1 \in \bigcap_{\alpha \in S_1} H_\alpha$. Then define $x_2 \in H$ such that $(x_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(x_2)_\alpha=(x_1)_\alpha$ for all $\alpha \in S-S_1$.

Let $O_2$ be a basic standard open set with $x_2 \in O_2 \subset U$. Let $S_2=supp(O_2)$. By making $O_2$ a smaller open set if necessary, we can have $S_1 \subset S_2$. Then we have $P^{-1}_{S_2}(P_{S_2}(x_2)) \subset U$. Choose $a_2 \in \bigcap_{\alpha \in S_2} H_\alpha$. Then define $x_3 \in H$ such that $(x_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(x_3)_\alpha=(x_2)_\alpha$ for all $\alpha \in S-S_2$.

After this inductive process is completed, we can obtain:

• a sequence $x_1,x_2,x_3,\cdots$ of points of $H=\prod_{\alpha \in S} H_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3, \subset \cdots$ of finite subsets of the index set $S$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for each $n \ge 2$, $(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$.

By A. H. Stone’s theorem (Theorem 5 in [6]), $X$ cannot contain a closed copy of $\mathbb{N}$ (the space of the positive integers with the discrete topology). A proof that $\mathbb{N}^{\omega_1}$ is also found in this post. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is infinite or finite.

Case 1
Assume that $A$ is an infinite set. Then $A$ has a limit point $a$, meaning that every open subset of $X$ containing $a$ contains some $a_n$ different from $a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a$ for all $\alpha \in S-S_n$

It is the case that $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, since $y_n$ agrees with $x_n$ on the finite set $S_n$. Let $t \in K$ such that $t_\alpha=a$ for all $\alpha \in S$. It follows that $t$ is a limit point of $\left\{y_2,y_3,y_4,\cdots \right\}$. Thus $t \in \overline{U}$. Since $t \in K \subset V$, the open set $V$ would have to contain points of $U$. But $U$ and $V$ are supposed to be disjoint open subsets of the product space $\prod_{\alpha \in S} X_\alpha$. Thus we have a contradiction.

Case 2
Assume that $A$ is a finite set. Then for some $m$, $a_j=a_m$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha \in S} X_\alpha$ such that

• $(y_n)_\alpha=(x_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$
• $(y_n)_\alpha=a_m$ for all $\alpha \in S-S_n$

Let $t \in K$ such that $t_\alpha=a_m$ for all $\alpha \in S$. Then $y_n=t$ for all $n \ge m+1$. As in Case 1, $y_n \in P^{-1}_{S_n}(P_{S_n}(x_n)) \subset U$ for all $n$, implying that $t \in K \cap U$. This is a contradiction, since $K \subset V$ and $U$ and $V$ are supposed to be disjoint.

Both cases lead to a contradiction. Thus if all powers of $X$ is normal, $X$ must be compact. This completes the proof of the theorem.

____________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., An elementary proof of Noble’s theorem on normality of powers, Comment. Math. Univ. Carolinae, 29.4, 677-678, 1988.
3. Franklin, S. P., Walker, R. C., Normalit of powers implies compactness, Proc. Amer. Math. Soc., 36, 295-296, 1972.
4. Keesling, J., Normality and infinite product spaces, Adv. in. Math., 9, 90-92, 1972.
5. Noble, N., Products with closed projections, II, Trans. Amer. Math. Soc., 160, 169-183, 1971.
6. Ross, K. A., Stone, A. H. Products of separable spaces, Amer. Math. Monthly, 71, 398-403, 1964.

____________________________________________________________________

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

# Looking for a closed and discrete subspace of a product space

I had long suspected that there probably is an uncountable closed and discrete subset of the product space of uncountably many copies of the real line. Then I found the statement in the Encyclopedia of General Topology (page 76 in [3]) that “for every infinite cardinal $\mathcal{K}$, the product $D(\mathcal{K})^{2^{\mathcal{K}}}$ includes $D(2^{\mathcal{K}})$ as a closed subspace” where $D(\tau)$ is the discrete space of cardinality of $\tau$. When $\mathcal{K}=\aleph_0$ (the first infinite cardinal), there is a closed and discrete subset of cardinality $c=2^{\aleph_0}$ in the product space $\mathbb{N}^{c}$ (the product space of continuum many copies of a countable discrete space). Despite the fact that this product space is a separable space, a closed and discrete set of cardinality continuum is hiding in the product space $\mathbb{N}^{c}$. What is more amazing is that this result gives us a glimpse into the working of the product topology with uncountably many factors. There are easily defined discrete subspaces of $\mathbb{N}^{c}$. But these discrete subspaces are not closed in the product space, making the result indicated here a remarkable one.

The Encyclopedia of General Topology points to two references [2] and [4]. I could not find these papers online. It turns out that Engelking, the author of [2], included this fact as an exercise in his general topology textbook (see Exercise 3.1.H (a) in [1]). This post presents a proof of this fact based on the hints that are given in [1]. To make the argument easier to follow, the proof uses some of the hints in a slightly different form.

____________________________________________________________________

The Exercise

Let $I=[0,1]$ be the closed unit interval. Let $X=I$ be the unit interval with the discrete topology. Let $\omega$ be the set of all nonnegative integers with the discrete topology. Let $Y=\prod_{t \in I} Y_t$ where each $Y_t=\omega$. We can also denote $Y$ by $\omega^I$. The problem is to show that the discrete space $X$ can be embedded as a closed and discrete subspace of $Y$.

____________________________________________________________________

A Solution

For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that

• $t \in O_{t,j}$ for each $j$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$),
• $\left\{t \right\}=\bigcap \limits_{j=1}^\infty O_{t,j}$.

For any $t \in I-\left\{0,1 \right\}$, we can make $O_{t,j}$ open intervals of the form $(a,b)$. For $t=0$, $O_{0,j}$ have the form $[0,b)$. For $t=1$, $O_{1,j}$ have the form $(a,1]$.

The above sequences of open intervals help define a homeomorphic embedding of the discrete space $X$ into the product $Y$. For each $t \in I$, define the function $f_t:I \rightarrow \omega$ by letting:

$f_t(x) = \begin{cases} 0 & \mbox{if } x=t \\ 1 & \mbox{if } x \in I-O_{t,1} \\ 2 & \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \cdots \\ j & \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \text{etc} \end{cases}$

for each $x \in X$. We now define the embedding $E:X \rightarrow Y=\omega^I$ by letting:

$E(x)=< f_t(x) >_{t \in I}$

for each $x \in X$. For each point $x \in X$, $E(x)$ is the point in the product space such that the $t$ coordinate of $E(x)$ is the value of the function $f_t$ evaluated at $x$. Let $\mathcal{F}=\left\{f_t: t \in I \right\}$.

The mapping $E$ is an evaluation map (called diagonal map in [1]). It is a homeomorphism if the following three conditions are met:

• If each $f_t \in \mathcal{F}$ is continuous, then $E$ is continuous.
• If the family of functions $\mathcal{F}$ separates points in $X$, then $E$ is injective (i.e. a one-to-one function).
• If $\mathcal{F}$ separates points from closed sets in $X$, then the inverse $E^{-1}$ is also continuous.

The first point is easily seen. Note that both the domain and the range of $f_t:X \rightarrow \omega$ have the discrete topology. Thus these functions are continuous. To see the second point, let $p,q \in X$ with $p \ne q$. Note that $f_p(p)=0$ while $f_p(q) \ne 0$.

Since $X$ is discrete, any subset of $X$ is closed. To see the third point, let $C \subset X$ and $p \notin C$. Once again, $f_p(p)=0$ while $f_p(x) \ne 0$ for all $x \in C$. Clearly $f_p(p) \notin \overline{f(C)}$ (closure in the discrete space $\omega$). Thus $E^{-1}$ is also continuous. For more details about why $E$ is an embedding, see the previous post called The Evaluation Map.

Let $W=E(X)$. Since $E$ is a homeomorphism, $W$ is a discrete subspace of the product space $Y$. We only need to show $W$ is closed in $Y$. Let $k \in Y$ such that $k=< k_t >_{t \in I} \ \notin W$. We show that there is an open neighborhood of $k$ that misses the set $W$. There are two cases to consider. One is that $k_r \ne 0$ for all $r \in I$. The other is that $k_r=0$ for some $r \in I$. The first case is more involved.

Case 1
Suppose that $k_r \ne 0$ for all $r \in I$. First define a local base $\mathcal{B}_{k}$ of the point $k$. Let $H \subset I$ be finite and let $G_{H}$ be defined by:

$G_{H}=\left\{b \in Y: \forall \ h \in H, b_h=k_h \right\}$

Let $\mathcal{B}_{k}$ be the set of all possible $G_{H}$. For an arbitrary $G_{H}$, consider $E^{-1}(G_H)$. By definition, $E^{-1}(G_H)=\left\{c \in I: E(c) \in G_H \right\}$. To make the argument below easier to see, let’s further describe $E^{-1}(G_H)$.

\displaystyle \begin{aligned} E^{-1}(G_H)&=\left\{c \in I: E(c) \in G_H \right\} \\&=\left\{c \in I: \forall \ h \in H, f_h(c)=k_h \ne 0 \right\} \\&=\left\{c \in I: \forall \ h \in H, c \in I-O_{h,k_h} \right\} \\&=\bigcap \limits_{h \in H} I-O_{h,k_h} \\&=I-\bigcup \limits_{h \in H} O_{h,k_h} \end{aligned}

The last description above indicates that if $x \in X$ and if $x \in E^{-1}(G_H)$, then $x \notin \bigcup \limits_{h \in H} O_{h,k_h}$. To wrap up Case 1, we would like to produce one particular $H$. Consider the open cover $\left\{O_{t,k_t}: t \in I \right\}$ of $I$. Since $I$ is compact in the usual topology, there is a finite $H \subset I$ such that $I=\bigcup \limits_{h \in H} O_{h,k_h}$. This means that for this particular finite set $H$, $E^{-1}(G_H)=\varnothing$. Putting it in another way, the open neighborhood $B_H$ of $k$ contains no point of $E(x)$. The proof for Case 1 is completed.

Case 2
Suppose that $k_r=0$ for some $r \in I$. Since $k \notin W=E(X)$, in particular $k \ne E(r)=< f_t(r) >_{t \in I}$. So for some $q \in I$, $k_q \ne f_q(r)$. Now define the following open set containing $k$.

$G=\left\{b \in Y: b_r=0 \text{ and } b_q=k_q \right\}$

Note that $E(r) \notin G$ since $k_q \ne f_q(r)$. Furthermore, for each $p \in I-\left\{r \right\}$, $E(p) \notin G$ since $f_r(p) \ne 0$. Thus $G$ is an open neighborhood of $k$ containing no point of $E(X)$. The proof for Case 2 is completed.

We have shown that the image of the discrete space $X$ under the homeomorphism $E$ is closed in the product space $Y=\prod_{t \in I} Y_t=\omega^I$.

____________________________________________________________________

The preceding proof shows that the product space of continuum many copies of $\omega$ contains a closed and discrete subspace of cardinality continuum. This is a remarkable result. At a glance, it is not entirely clear that a closed and discrete set of this large size can be found in the product space in question.

One immediate consequence is that the product space $Y=\prod_{t \in I} Y_t=\omega^I$ is not normal since it is a separable space. By Jones’ lemma, any separable normal space cannot have a closed and discrete subset of cardinality continuum. However, if the goal is only to show non-normality, we only need to show that $\omega^{\omega_1}$ is not normal (a proof is found in this post). Thus the value of the preceding proof is to demonstrate how to produce a closed and discrete subspace of cardinality continuum in the product space in question.

____________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Engelking, R., On the double circumference of Alexandroff, Bull. Acad. Polon. Sci., 16, 629-634, 1968.
3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
4. Juhasz, I., On closed discrete subspace of product spaces, Bull. Acad. Polon. Sci., 17, 219-223, 1969.

____________________________________________________________________

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

# 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.

____________________________________________________________________

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

# 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.

____________________________________________________________________

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.

____________________________________________________________________

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 $K$.

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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

# 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].

____________________________________________________________________

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 $F \in Y$.

____________________________________________________________________

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 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.

__________________________________

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.

__________________________________

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.

__________________________________

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.

__________________________________

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.

__________________________________

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.

__________________________________

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).

__________________________________

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.

__________________________________

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.

__________________________________

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.

__________________________________

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 $\mathcal{F}[S]$.

__________________________________

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 $\mathcal{F}[X]$.

____________________________________________________________________

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.

____________________________________________________________________

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.

____________________________________________________________________

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

# 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.

____________________________________________________________________

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.

____________________________________________________________________

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

|

# Compact Subspaces of Bing’s Example G

In a previous post, we discuss basic properties of Bing’s Example G, a classic and influential example of a normal but not collectionwise normal space. In another post, we discuss a subspace of Bing’s Example G which is also normal but not collectionwise normal but is metacompact. In this post, we further discuss Bing’s Example G by characterizing its compact subspaces.

The ideas discussed here can be found in [1] with a lot of details omitted. In this post, we provide all the necessary details in understanding these ideas. In the next post called Some subspaces of Bing’s Example G, we discuss some subspaces of Bing’s Example G based on the characterization of compact sets given in this post.

____________________________________________________________________

Defining Bing’s Example G

First we repeat the definition of Bing’s Example G. Let $P$ be any uncountable set. Let $Q$ be the power set of $P$, i.e., the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Obviously $2^Q$ is simply the Cartesian product of $\lvert Q \lvert$ many copies of the two-point discrete space $\left\{0,1 \right\}$, i.e., $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$, $f_p(q)=1$ if $p \in q$ and $f_p(q)=0$ if $p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:

$\tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology $\tau^*$. In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

____________________________________________________________________

Basic Open Sets in Bing’s Example G

To facilitate the discussion below, we now fix now some notation for basic open sets of the points $f_p \in F_P$. For any finite $S \subset Q$, the following describes the basic open sets containing the point $f_p \in F_P$.

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

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

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

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

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

____________________________________________________________________

One Example of Compact Subsets of Bing’s Example G

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

____________________________________________________________________

Characterizing Compact Subsets of Bing’s Example G

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

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

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

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

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

Theorem 1

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

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

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

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

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

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

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

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

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

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

Theorem 1 characterizes the compact subsets of Bing’s Example G with only one limit point. We now characterize all compact subsets of the space $F$. We have the following theorem.

Theorem 2

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

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

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

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

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

___________________________________________________________________________________

Comment

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

___________________________________________________________________________________

Reference

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

____________________________________________________________________

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

# Tietze-Urysohn-like theorems for completely regular spaces

Completely regular spaces (also called Tychonoff spaces) are topological spaces that come with a guarantee of having continuous real-valued functions in sufficient quantity. Thus the class of completely regular spaces is an ideal setting for many purposes that require the use of continuous real-valued functions (one example is working with function spaces). In a completely regular space $X$, for any closed set $B$ and for any point $x$ not in the closed set $B$, there always exists a continuous function $f:X \rightarrow [0,1]$ such that $f(x)=0$ and $f$ maps $B$ to $1$. It turns out that in such a space, we can replace the point $x$ with any compact set $A$ that is disjoint from the closed set $B$. This is a useful tool for proving theorems as well as for constructing objects. In this post, we discuss and prove this result (which resembles Urysohn’s lemma) and another useful fact about completely regular spaces that works very much like Tietze’s extension theorem. Specifically we prove the following results.

Theorem 1

Let $X$ be a completely regular space. For any compact set $A \subset X$ and for any closed set $B \subset X$ that is disjoint from $A$, there exists a continuous function $f:X \rightarrow [0,1]$ such that

• $f(x)=0$ for all $x \in A$,
• $f(x)=1$ for all $x \in B$.
Theorem 2

Let $X$ be a completely regular space. For any compact set $A \subset X$, any continuous function $f:A \rightarrow \mathbb{R}$ can be continuously extended over $X$, i.e., there exists a continuous function $\hat{f}:X \rightarrow \mathbb{R}$ such that $\hat{f}(x)=f(x)$ for all $x \in A$ (in symbol we write $\hat{f} \upharpoonright A=f$).

A space $X$ is normal if any two disjoint closed sets $A \subset X$ and $B \subset X$ can be separated by disjoint open sets, i.e., $A \subset U$ and $B \subset V$ for some disjoint open subsets $U$ and $V$ of $X$. Normal spaces are usually have the additional requirement that singleton sets are closed (i.e. $T_1$ spaces).

These two theorems remind us of two important tools for normal spaces, namely Urysohn’s lemma and Tietze’s extension theorem.

Urysohn’s lemma indicates that for any two disjoint closed sets in a normal space, the space can be mapped continuously to the closed unit interval $[0,1]$ such that one closed set is mapped to $0$ and the other closed set is mapped to $1$. Theorem 1 is like a weakened version of Urysohn’s lemma in that one of the two disjoint closed sets must be compact.

Tietze’s extension theorem indicates that in a normal space, any continuous real-valued function defined on a closed subspace can be extended to the entire space. Theorem 2 is like a weakened version of Tiezte’s extension theorem in that the continuous extension only works for continuous functions defined on a compact subspace.

So if one only works in a completely regular space, one can still apply these two theorems about normal spaces (the weakened versions of course). For the sake of completeness, we state these two theorems about normal spaces.

Urysohn’s Lemma

Let $X$ be a normal space. For any two disjoint closed sets $A \subset X$ and $B \subset X$, there exists a continuous function $f:X \rightarrow [0,1]$ such that

• $f(x)=0$ for all $x \in A$,
• $f(x)=1$ for all $x \in B$.
Tietze’s Extension Theorem

Let $X$ be a normal space. For any closed set $A \subset X$, any continuous function $f:A \rightarrow \mathbb{R}$ can be continuously extended over $X$, i.e., there exists a continuous function $\hat{f}:X \rightarrow \mathbb{R}$ such that $\hat{f}(x)=f(x)$ for all $x \in A$ (in symbol we write $\hat{f} \upharpoonright A=f$).

____________________________________________________________________

Proof of Theorem 1

We now prove Theorem 1. Let $X$ be a completely regular space. Let $A \subset X$ and $B \subset X$ be two disjoint closed sets where $A$ is compact. For each $x \in A$, there exists a continuous function $f_x:X \rightarrow [0,1]$ such that $f_x(x)=0$ and $f_x(B) \subset \left\{1 \right\}$. The following collection is an open cover of the compact set $A$.

$\left\{f_x^{-1}([0,\frac{1}{10})): x \in A \right\}$

Finitely many sets in this collection would cover $A$ since $A$ is compact. Choose $x_1,x_2,\cdots,x_n \in A$ such that $A \subset \bigcup \limits_{j=1}^n f_{x_n}^{-1}([0,\frac{1}{10}))$. Define $h:X \rightarrow [0,1]$ by, for each $x \in X$, letting $h(x)$ be the minimum of $f_{x_1}(x),\cdots,f_{x_n}(x)$. It can be shown that the function $h$, being the minimum of finitely many continuous real-valued functions, is continuous. Furthermore, we have:

• $A \subset h^{-1}([0,\frac{1}{10}))$, and
• $h(B) \subset \left\{1 \right\}$

Now define $w:X \rightarrow [0,1]$ by, for each $x \in X$, letting $w(x)$ be as follows:

$\displaystyle w(x)=\frac{10}{9} \cdot \biggl[ \text{max} \left\{h(x)-\frac{1}{10},0\right\} \biggr]$

It can be shown that the function $w$ is continuous. It is clear that $w(x)=0$ for all $x \in A$ and $w(x)=1$ for all $x \in B$. $\blacksquare$

____________________________________________________________________

Proof of Theorem 2

Interestingly, the proof of Theorem 2 given here uses Tietze’s extension theorem even Theorem 2 is described earlier as a weakened version of Tietze’s extension theorem. Beside using Tietze’s extension theorem, we also use the fact that any completely regular space can be embedded in a cube (see the previous post called Embedding Completely Regular Spaces into a Cube).

The proof is quite short once all the deep results that are used are understood. Let $X$ be a completely regular space. Then $X$ can be embedded in a cube, which is a product of the closed unit interval $[0,1]$. Thus $X$ is homeomorphic to a subspace of the following product space

$Y=\prod \limits_{a \in S} I_a$

for some index set $S$ where $I_a=[0,1]$ for all $a \in S$. We can now regard $X$ as a subspace of the compact space $Y$. Let $A \subset X$ be a compact subset of $X$. Let $f:A \rightarrow \mathbb{R}$ be a continuous function.

The set $A$ is a subset of $X$ and can also be regarded as a subspace of the compact space $Y$, which is normal. Hence Tietze’s extension theorem is applicable in $Y$. Let $\bar{f}:Y \rightarrow \mathbb{R}$ be a continuous extension of $f$. Let $\hat{f}=\bar{f} \upharpoonright X$. Then $\hat{f}$ is the required continuous extension. $\blacksquare$

____________________________________________________________________

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

# One way to find collectionwise normal spaces

Collectionwise normality is a property that is weaker than paracompactness and stronger than normality (see the implications below). Normal spaces need not be collectionwise normal. Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). We discuss one instance when normal spaces are collectionwise normal, giving a way to obtain collectionwise normal spaces that are not paracompact.

$\text{ }$
$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

___________________________________________________________________________________

Collectionwise Normal Spaces

A normal space is one in which any two disjoint closed sets can be separated by disjoint open sets. By induction, in a normal space any finite number of disjoint closed sets can be separated by disjoint open sets. Of course, the inductive reasoning cannot be carried over to the case of infinitely many disjoint closed sets. In the real line with the usual topology, the singleton sets $\left\{x \right\}$, where $x$ is rational, are disjoint closed sets that cannot be simultaneously separated by disjoint open sets. In order to separate an infinite collection of disjoint closed sets, it makes sense to restrict on the type of collections of closed sets. A space $X$ is collectionwise normal if every discrete collection of closed subsets of $X$ can be separated by pairwise disjoint open subsets of $X$. The following is a more specific definition.

Definition
A space $X$ is collectionwise normal if for every discrete collection $\mathcal{A}$ of closed subsets of $X$, there exists a pairwise disjoint collection $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ of open subsets of $X$ such that $A \subset U_A$ for each $A \in \mathcal{A}$.

For more details about the definitions of collectionwise normality, see “Definitions of Collectionwise Normal Spaces”. The implications displayed above are repeated below. None of the arrows is reversible.

$\text{ }$
$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

As indicated above, Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). The propositions in the next section can be used to obtain collectionwise normal spaces that are not paracompact.

___________________________________________________________________________________

When Normal implies Collectionwise Normal

Being able to simultaneously separate any discrete collection of closed sets is stronger than the property of merely being able to separate finite collection of disjoint closed sets. It turns out that the stronger property of collectionwise normality is required only for separating uncountable discrete collections of closed sets. As the following lemma shows, normality is sufficient to separate any countable discrete collection of closed sets.

Lemma 1
Let $X$ be a normal space. Then for every discrete collection $\left\{C_1,C_2,C_3,\cdots \right\}$ of closed subsets of $X$, there exists a pairwise disjoint collection $\left\{O_1,O_2,O_3,\cdots \right\}$ of open subsets of $X$ such that $C_i \subset O_i$ for each $i$.

Proof of Lemma 1
Let $\left\{C_1,C_2,C_3,\cdots \right\}$ be a discrete collection of closed subsets of $X$. For each $i$, choose disjoint open sets $U_i$ and $V_i$ such that $C_i \subset U_i$ and $\cup \left\{C_j: j \ne i \right\} \subset V_i$. Let $O_1=U_1$. For each $i>1$, let $O_i=U_i \cap V_1 \cap \cdots \cap V_{i-1}$. It follows that $O_i \cap O_j = \varnothing$ for all $i \ne j$. It is also clear that for each $i$, $C_i \subset O_i$. $\blacksquare$

We have the following propositions.

Proposition 2
Let $X$ be a normal space. If all discrete collections of closed subsets of $X$ are at most countable, then $X$ is collectionwise normal.

Proposition 3
Let $X$ be a normal space. If all closed and discrete subsets of $X$ are at most countable (such a space is said to have countable extent), then $X$ is collectionwise normal.

Proposition 4
Any normal and countably compact space is collectionwise normal.

Proposition 2 follows from Lemma 1. As noted in Proposition 3, any space in which all closed and discrete subsets are countable is said to have countable extent. It is easy to verify that $X$ has countable extent if and only if all discrete collections of closed subsets of $X$ are at most countable. If $X$ is a countably compact space, then every infinite subset of $X$ has a limit point. Thus Proposition 4 follows from the fact that any countably compact space has countable extent.

___________________________________________________________________________________

Paracompact Spaces

One way to find a collectionwise normal space that is not paracompact is to find a non-paracompact space that satisfies Propositions 3, 4 or 5. For example, $\omega_1$, the space of all countable ordinals with the order topology, is not paracompact. Since $\omega_1$ is normal and countably compact, it is collectionwise normal by Proposition 4. For a basic discussion of $\omega_1$ as a topological space, see “The First Uncountable Ordinal”.

As the following theorem shows, paracompact spaces are collectionwise normal. Thus the class of collectionwise normal spaces includes all metric spaces and paracompact spaces.

Theorem 5
If a space $X$ is paracompact, then $X$ is collectionwise normal.

Proof of Theorem 5
Let $X$ be a paracompact space. Let $\mathcal{A}$ be a discrete collection of closed subsets of $X$. For each $x \in A$, let $O_x$ be open such that $x \in O_x$ and $O_x$ meets at most one element of $\mathcal{A}$. Let $\mathcal{O}=\left\{O_x: x \in X \right\}$. By the paracompactness of $X$, $\mathcal{O}$ has a locally finite open refinement $\mathcal{V}=\left\{V_x: x \in X \right\}$ such that $V_x \subset O_x$ for each $x \in X$.

For each $A \in \mathcal{A}$, let $W_A=\cup \left\{\overline{V}: V \in \mathcal{V} \text{ and } \overline{V} \cap A=\varnothing \right\}$ and let $U_A=X-W_A$. Each $W_A$ is a closed set since $\mathcal{V}$ is locally finite. Thus each $U_A$ is open. Furthermore, for each $A \in \mathcal{A}$, $A \subset U_A$. It is easily checked that $\left\{U_A: A \in \mathcal{A} \right\}$ is pairwise disjoint. $\blacksquare$

___________________________________________________________________________________

Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

___________________________________________________________________________________

$\copyright \ \ 2012$

# Definitions of Collectionwise Normal Spaces

The notion of collectionwise normal spaces is a property stronger than normal spaces. There are several ways of defining the notion of collectionwise normality. We show that they are equivalent.

We only consider spaces in which any set with only one point is considered a closed set (i.e. $T_1$ spaces). Let $X$ be a space. Let $\mathcal{A}$ be a collection of subsets of $X$. We say $\mathcal{A}$ is pairwise disjoint if $A \cap B = \varnothing$ for all $A,B \in \mathcal{A}$ where $A \ne B$. We say $\mathcal{A}$ is a discrete collection of subsets of $X$ if for each $x \in X$, there is an open neighborhood $O$ with $x \in O$ such that $O$ meets at most one element of $\mathcal{A}$. When $\mathcal{A}$ is such a collection, we also say that $\mathcal{A}$ is discrete in $X$ (or discrete if $X$ is understood).

Definition 1
A space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{A}$ of closed subsets of $X$, there is a collection $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ of open subsets of $X$ such that $\mathcal{U}$ is pairwise disjoint and $A \subset U_A$ for each $A \in \mathcal{A}$.

In other words, every discrete collection of closed sets can be separated by pairwise disjoint open sets. Clearly any collectionwise normal space is normal. 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.

Some authors define a collectionwise normal space as one in which every discrete collection of sets can be separated by pairwise disjoint open sets. We have the following definition.

Definition 2
A space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{A}$ of subsets of $X$, there is a collection $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ of open subsets of $X$ such that $\mathcal{U}$ is pairwise disjoint and $A \subset U_A$ for each $A \in \mathcal{A}$.

It is clear that Definition 2 implies Definition 1. For any discrete collection $\mathcal{A}$ of subsets, $\mathcal{A}^*=\left\{\overline{A}: A \in \mathcal{A} \right\}$ is also a discrete collection. Thus Definition 1 implies Definition 2. The following is another way of defining collectionwise normal.

Definition 3
A space $X$ is said to be collectionwise normal if for every discrete collection $\mathcal{A}$ of closed subsets of $X$, there is a collection $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ of open subsets of $X$ such that $\mathcal{U}$ is a discrete collection and $A \subset U_A$ for each $A \in \mathcal{A}$.

Since discrete collection is pairwise disjoint collection of sets, Definition 3 implies Definition 1. We show that Definition 1 implies Definition 3. Let $X$ be a collectionwise normal space according to Definition 1. Let $\mathcal{A}$ be a discrete collection of closed subsets of $X$. Let $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ be as in Definition 1. Let $H=X-\cup \mathcal{U}$ and $K=\cup \mathcal{A}$. Note that $H$ and $K$ are disjoint closed subsets of $X$.

Suppose $H=\varnothing$. Then the sets in $\mathcal{U}$ are both open and closed. Thus $\mathcal{U}$ is the desired discrete collection of open sets separating sets in $\mathcal{A}$. So assume that $H \ne \varnothing$. Since $X$ is a normal space, we have $O_H$ and $O_K$, disjoint open subsets of $X$, such that $H \subset O_H$ and $K \subset O_K$. For each $A \in \mathcal{A}$, let $V_A=U_A \cap O_K$. Let $\mathcal{V}=\left\{V_A: A \in \mathcal{A} \right\}$.

We claim that $\mathcal{V}$ is a discrete collection of open sets separating sets in $\mathcal{A}$. Let $x \in X$. Suppose $x \notin H$. We have $x \in U_A$ for some $A \in \mathcal{A}$. Then $U_A$ meets only one set of $\mathcal{A}$, namely $V_A$. So assume $x \in H$. Then $x \in O_H$ and $O_H$ does not meet any $V_A$. Thus $\mathcal{A}$ is a discrete collection of open sets. It is clear that $A \subset V_A$ for each $A \in \mathcal{A}$.

Thus all three statements defining the notion of collectionwise normal spaces are equivalent. It is clear that any space satisfying any one of these collectionwise normal definitions is a normal space. The notion of collectionwise normality is stronger than normality. Bing’s Example G is a normal space that is not collectionwise normal (see “Bing’s Example G”).

___________________________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

___________________________________________________________________________________

$\copyright \ \ 2012$