The Sorgenfrey plane is subnormal

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

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

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

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

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

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The Sorgenfrey plane is perfect

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

Proposition 1
The Sorgenfrey line S is perfect.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_2<t_2 \right\}

The set G is an open set containing the point x. We claim that G \cap Y_{k,1}=\varnothing. Suppose g \in G \cap Y_{k,1}. Then g_1=x_1 and x_2<g_2<t_2. Consider the following set:

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

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

    x_1=g_1=t_1

    x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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

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

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

    G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1<t_1 \text{ and }y_2<t_2 \right\}

The set G is an open set containing the point x. We claim that G \cap Y_{k,\varnothing}=\varnothing. Suppose g \in G \cap Y_{k,\varnothing}. Then x_1<g_1<t_1 and x_2<g_2<t_2. Consider the following set:

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

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

    x_1<g_1<t_1<x_1+\frac{1}{k}<g_1+\frac{1}{k}

    x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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

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

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A non-perfect example

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

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Reference

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

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\copyright \ 2014 \text{ by Dan Ma}

Pseudonormal spaces

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

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

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Some Non-Pseudonormal Examples

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

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

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

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

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

    O_x=V_x \cup D_x

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

The topology on E is defined by the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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An Example of Pseudonormal but not Normal

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

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

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

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

    V_H=O_H

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

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

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Some Comments about Examples 1 – 3

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

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

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

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

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

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

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

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Reference

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

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\copyright \ 2014 \text{ by Dan Ma}

Pixley-Roy hyperspaces

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

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

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

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

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

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Basic Discussion

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

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

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

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

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

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

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

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

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More Basic Results

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Examples

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

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

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

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Reference

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

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\copyright \ 2014 \text{ by Dan Ma}

Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line S=\mathbb{R} where the topology is generated by a base consisting the half open intervals of the form [a,b). The Sorgenfrey plane is the square S \times S.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let X be a regular space. A development for X is a sequence \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots of open covers of X such that for each x \in X, and for each open subset U of X with x \in U, there exists one cover \mathcal{G}_n satisfying the condition that for any open set V \in \mathcal{G}_n, x \in V \Rightarrow V \subset U. When X has a development, X is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots, the open sets in \mathcal{G}_n are considered “smaller” as the index n increases. In fact, this is how a development is defined for a metric space, where \mathcal{G}_n consists of all open balls with diameters less than \frac{1}{n}. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a G_\delta-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a G_\delta-diagonal (the diagonal \Delta=\left\{(x,x): x \in X \right\} is a G_\delta-set in X \times X). It is a well known theorem that every compact space with a G_\delta-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that X is compact and has a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots. Then each \mathcal{G}_n has a finite subcover \mathcal{H}_n. Then \bigcup_{n=1}^\infty \mathcal{H}_n is a countable base for X. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that X is paracompact and has a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots. Then each \mathcal{G}_n has a locally finite open refinement \mathcal{H}_n. Then \bigcup_{n=1}^\infty \mathcal{H}_n is a \sigma-locally finite base for X. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a \sigma-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a \sigma-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

    Bing’s Metrization Theorem
    A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space X is said to be collectionwise normal if X is a T_1-space and for every discrete collection \left\{W_\alpha: \alpha \in A \right\} of closed sets in X, there exists a discrete collection \left\{V_\alpha: \alpha \in A \right\} of open subsets of X such that W_\alpha \subset V_\alpha. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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Reference

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

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\copyright \ \ 2012

When is a Pseudocompact Space Metrizable?

Compactness, countably compactness and pseudocompactness are three successively weaker properties. It follows easily from definitions that

(A) \ \ \ \ \ \text{compact} \Rightarrow \text{countably compact} \Rightarrow \text{pseudocompact}

None of these arrows can be reversed. It is well known that either compactness or countably complactness plus having a G_\delta-diagonal implies metrizability. We have:

(B) \ \ \ \ \ \text{compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}

(C) \ \ \ \ \ \text{countably compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}

A question can be asked whether these results can be extended to pseudocompactness.

Question (D) \ \ \ \ \ \text{pseudocompact compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable?}

The answer to this question is no. The space defined using a maximal almost disjoint family of subsets of \omega is an example of a non-metrizable pseudocompact space with a G_\delta-diagonal (discussed in this post). In this post we show that if we strengthen “having a G_\delta-diagonal” to being submetrizable, we have a theorem. Specifically, we show:

(E) \ \ \ \ \ \text{pseudocompact} + \text{submetrizable} \Rightarrow \text{metrizable}

For the result of (B), see this post. For the result of (C), see this post. In this post, we discuss the basic properties of pseudocompactness that build up to the result of (E). All spaces considered here are at least Tychonoff (i.e. completely regular). For any basic notions not defined here, see [1] or [2].

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Pseudocompact Spaces

A space X is said to be pseudocompact if every real-valued continuous function defined on X is a bounded function. Any real-valued continuous function defined on a compact space must be bounded (and is thus pseudocomppact). If there were an unbounded real-valued continuous function defined on a space X, then X would have a countably infinite discrete set (thus not countably compact). Thus countably compact implies pseudocompact, as indicated by (A).

A space X is submetrizable if there is a coarser (i.e. weaker) topology that is a metrizable topology. Specifically the topological space (X,\tau) is submetrizable if there is another topology \tau^* that can be defined on X such that \tau^* \subset \tau and (X,\tau^*) is metrizable. The Sorgenfrey line is non-metrizable and yet the Sorgenfrey topology has a weaker topology that is metrizable, namely the Euclidean topology of the real line.

The following two theorems characterizes pseudocompact spaces in terms of locally finite open family of open sets (Theorem 1) and the finite intersection property (Theorem 2). Both theorems are found in Engelking (Theorem 3.10.22 and Theorem 3.10.23 in page 207 of [1]). Theorem 3 states that in a pseudocompact space, closed domains are pseudocompact (the definition of closed domain is stated before the theorem). Theorem 4 is the main theorem (result E stated above).

Theorem 1
Let X be a space. The following conditions are equivalent:

  1. The space X is pseudocompact.
  2. If \mathcal{V} is a locally finite family of non-empty open subsets of X, then \mathcal{V} is finite.
  3. If \mathcal{V} is a locally finite open cover of X, then \mathcal{V} is finite.
  4. If \mathcal{V} is a locally finite open cover of X, then \mathcal{V} has a finite subcover.

Proof
1 \Rightarrow 2
Suppose that condition 2 does not hold. Then there is an infinite locally finite family of non-empty open sets \mathcal{V} such that \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}. We wish to define an unbounded continuous function using \mathcal{V}.

This is where we need to invoke the assumption of complete regularity. For each n choose a point x_n \in V_n. Then for each n, there is a continuous function f_n:X \rightarrow [0,n] such that f_n(x_n)=n and f_n(X-V_n) \subset \left\{ 0 \right\}. Define f:X \rightarrow [0,\infty) by f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots.

Because \mathcal{V} is locally finite, the function f is essentially pointwise the sum of finitely many f_n. In other words, for each x \in X, for some positive integer N, f_j(x)=0 for all j \ge N. Thus the function f is well defined and is continuous at each x \in X. Note that for each x_n, f(x_n) \ge n, showing that it is unbounded.

The directions 2 \Rightarrow 3 and 3 \Rightarrow 4 are clear.

4 \Rightarrow 1
Let g:X \rightarrow \mathbb{R} be a continuous function. We want to show that g is a bounded function. Consider the open family \mathcal{O}=\left\{\cdots,O_{-3},O_{-2},O_{-1},O_0,O_1,O_2,O_3,\cdots \right\} where each O_n=g^{-1}((n,n+2)). Note that \mathcal{O} is a locally finite family in X since its members O_n=g^{-1}((n,n+2)) are inverse images of members of a locally finite family in the range space \mathbb{R}. By condition 4, \mathcal{O} has a finite subcover, leading to the conclusion that g is a bounded function. \blacksquare

Theorem 2
Let X be a space. The following conditions are equivalent:

  1. The space X is pseudocompact.
  2. If \mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\} is a family of non-empty open subsets of X such that O_n \supset O_{n+1} for each n, then \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing.
  3. If \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is a family of non-empty open subsets of X such that \mathcal{V} has the finite intersection property, then \bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing.

Proof
1 \Rightarrow 2
Suppose that X is pseudocompact. Suppose \mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\} satisfies the hypothesis of condition 2. If there is some positive integer m such that O_n=O_m for all n \ge m, then we are done. So assume that O_n are distinct for infinitely many n. According to condition 2 in Theorem 1, \mathcal{O} must not be a locally finite family. Then there exists a point x \in X such that every open set containing x must meet infinitely many O_n. This implies that x \in \overline{O_n} for infinitely many n. Thus x \in \bigcap \limits_{n=1}^\infty \overline{O_n}.

2 \Rightarrow 3
Suppose \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is a family of non-empty open sets with the finite intersection property as in the hypothesis of 3. Then let O_1=V_1, O_2=V_1 \cap V_2, O_3=V_1 \cap V_2 \cap V_3, and so on. By condition 2, we have \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing, which implies \bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing.

3 \Rightarrow 1
Let g:X \rightarrow \mathbb{R} be a continuous function such that g is unbounded. For each positive integer n, let V_n=\left\{x \in X: \lvert g(x) \lvert > n \right\}. Clearly the open sets V_n have the finite intersection property. Because g is unbounded, it follows that \bigcap \limits_{n=1}^\infty \overline{V_n} = \varnothing. \blacksquare

Let X be a space. Let A \subset X. The interior of A, denoted by \text{int}(A), is the set of all points x \in X such that there exists an open set O with x \in O \subset A. Points of \text{int}(A) are called the interior points of A. A subset C \subset X is said to be a closed domain if C=\overline{\text{int}(C)}. It is clear that C is a closed domain if and only if C is the closure of an open set.

Theorem 3
The property of being a pseudocompact space is hereditary with respect to subsets that are closed domains.

Proof
Let X be a pseudocompact space. We show that \overline{U} is pseudocompact for any nonempty open set U \subset X. Let Y=\overline{U} where U is a non-empty open subset of X. Let S_1 \supset S_2 \supset S_3 \supset \cdots be a decreasing sequence of open subsets of Y. Note that each S_i contains points of the open set U. Let O_i=S_i \cap U for each i. Note that the open sets O_i form a decreasing sequence of open sets in the pseudocompact space X. By Theorem 2, we have \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing (closure here is with respect to X). Note that points in \bigcap \limits_{n=1}^\infty \overline{O_n} are also points in \bigcap \limits_{n=1}^\infty \overline{S_n} (closure with respect to Y). By Theorem 2, Y=\overline{U} is pseudocompact. \blacksquare

Theorem 4 (Statement E above)
Let X be a pseudocompact submetrizable space. Then X is metrizable.

Proof
Let (X,\tau) be a pseudocompact submetrizable space. Then there exists topology \tau^* on X such (X,\tau^*) is metrizable and \tau^* \subset \tau. We show that \tau \subset \tau^*, leading to the conclusion that (X,\tau) is also metrizable. If A \subset X, we denote the closure of A in (X,\tau) by cl_{\tau}(A) and the closure of A in (X,\tau^*) by cl_{\tau^*}(A).

To show that \tau \subset \tau^*, we show any closed set with respect to the topology \tau is also a closed set with respect to the topology \tau^*. Let C be a closed set in (X,\tau). Consider the family \mathcal{W}=\left\{cl_{\tau}(U): U \in \tau \text{ and } C \subset U \right\}. We make the following claims.

Claim 1. C=\bigcap \left\{W: W \in \mathcal{W} \right\}.

Claim 2. Each W \in \mathcal{W} is pseudocompact in (X,\tau).

Claim 3. Each W \in \mathcal{W} is pseudocompact in (X,\tau^*).

Claim 4. Each W \in \mathcal{W} is compact in (X,\tau^*).

We now discuss each of these four claims. For Claim 1, it is clear that C \subset \bigcap \left\{W: W \in \mathcal{W} \right\}. The reverse set inclusion follows from the fact that X is a regular space. Claim 2 follows from Theorem 3. Note that sets in \mathcal{W} are closed domains in the pseudocompact space (X,\tau).

If sets in \mathcal{W} are pseudocompact in the larger topology \tau, they would be pseudocompact in the weaker topology \tau^* too. Thus Claim 3 is established. In a metrizable space, compactness and weaker notions such as countably compactness and pseudocompactness coincide. Because they are pseudocompact subsets, sets in \mathcal{W} are compact in the metrizable space (X,\tau^*). Thus Claim 4 is established.

It follows that C is closed in (X,\tau^*) since it is the intersection of compact sets in (X,\tau^*). Thus (X,\tau) is identical to (X,\tau^*), implying that (X,\tau) is metrizable. \blacksquare

Reference

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

Jones’ Lemma

Jones’ lemma is a great tool in working with normal spaces. It is useful, under cerntain conditions, to show the non-normality of a space. The lemma establishes an upper bound for the cardinality of closed and discrete sets in any separable and normal space. Thus, whenever you have a separable space with a closed and discrete set whose cardinality exceeds the upper bound, you have a non-normal space (see examples discussed below). One way to prove the Jones’ lemma is to explore the set-theoretic relationship between the density (the minimum cardinality of a dense set) and the cardinality of closed and discrete sets in normal spaces. We sketch a proof of this lemma and give some examples. We also state an extension of Jones’ lemma. All spaces under consideration are at least Hausdorff.

Let X be a space. A subset D of X is said to be a closed and discrete set in X if D is a closed set in X and D, in the relative topology, is a discrete space. Good basic references are [1] and [3]. For more detailed information about cardinal functions, see [2].
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Jones’ Lemma
Let X be a separable and normal space. Then for any set D that is a closed and discrete set in X, we have \displaystyle 2^{\lvert D \lvert} \le 2^{\omega}.

Corollary 1
Let X be a separable and normal space. Then for any set D that is a closed and discrete set in X, we have \displaystyle \lvert D \lvert < 2^{\omega}.

Jones' lemma, as stated above, is essentially saying that the cardinality of continuum is an upper bound of the cardinalities of the power sets of closed and discrete sets in any separable and normal space. The corollary says that the cardinality of continuum is an upper bound of the cardinalities of closed and discrete sets in any separable and normal space. As indicated at the beginning, the corollary is a great way for checking the non-normality of a separable space.

We now give a sketch of the proof Jones' lemma. Let C(X) be the set of all continuous real-valued functions defined on the space X. Suppose X is a separable space. A key point is that the cardinality of C(X) is sandwiched between the two cardinalities in the lemma:

\displaystyle (1) \ \ \ \ \ 2^{\lvert D \lvert} \le \lvert C(X) \lvert \le 2^{\omega}

The first inequality in (1) says that there are at least as many continuous real-valued functions as there are subsets of the closed and discrete set D. To see this, we appeal for some help from Urysohn's lemma. For each E \subset D, E and D-E are disjoint closed sets in X. By Urysohn's lemma, there is a continuous function f_E:X \rightarrow [0,1] such that f_E maps E to 1 and f_E maps D-E to 0. Note that the mapping \Psi: \mathcal{P}(D) \rightarrow C(X) defined by \Psi(E)=f_E is a one-to-one map, where \mathcal{P}(D) is the collection of all subsets of D.

The second inequality in (1) says that there are at most continuum many continuous real-valued functions defined on the space X. In other words, the number of continuous functions is capped by the cardinality continuum (actually equals continuum in this case, but one inequality is all we need here). Let G be a countable dense subset of X. Consider the map \rho:C(X) \rightarrow \mathbb{R}^{G} defined by \rho(f)=f \upharpoonright G, which is the function f restricted to the set G. The notation \mathbb{R}^{G} refers to the set of all functions from the set G into \mathbb{R}.

The key point here is that any continuous functions f:X \rightarrow \mathbb{R} and g:X \rightarrow \mathbb{R}, if they agree on the countable dense set G (f \upharpoonright G=g \upharpoonright G), then f=g on the whole space X.So \rho is a one-to-one map. Some elementary cardinal arithmetic shows that \lvert \mathbb{R}^G \lvert=2^\omega. Thus the second inequality in (1) is established.

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Examples

Let S be the Sorgenfrey line. This is the real line with the topology generated by half open intervals of the form [a,b). The Sorgenfrey line is a classic example of a normal space whose square is not normal. The Sorgenfrey plane S \times S is separable with a closed and discrete set of cardinality continuum. By the corollary of Jones’ lemma, the Sorgenfrey plane is not normal. The Sorgenfrey line is discussed in greater details in this post.

The tangent disc space is another example of a separable space with a closed and discrete set of size continuum, hence ensuring that it is a non-normal space.

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Generalization

The generalization involves the notion of density and the notion of extent. The density of an infinite Hausdorff space X is the smallest cardinal number of the form \lvert G \lvert where G is a dense subset of X. This cardinal number is denoted by d(X). For any separable space X, we have d(X)=\omega. If the space X is finite, then the convention adopted by many authors is that d(X)=\omega.

The extent of a space X is the smallest infinite cardinal \mathcal{K} such that every closed and discrete set in X has cardinality \le \mathcal{K}. The extent of the space X is denoted by e(X). For more detailed information about cardinal functions, see [2].

The following are the generalization of Jones’ lemma.

Jones’ Lemma
Let X be a normal space. Then for any set D that is a closed and discrete set in X, we have \displaystyle 2^{\lvert D \lvert} \le 2^{d(X)}.

Corollary 2
Let X be a normal space. Then for any set D that is a closed and discrete set in X, we have \displaystyle \lvert D \lvert < 2^{d(X)}, which implies \displaystyle e(X) \le 2^{d(X)}.

Corollary 2 suggests that the cardinal number 2^{d(X)} is an upper bound of the cardinalities of closed and discrete sets in any normal space X. As a result of this, the cardinal number 2^{d(X)} dominates the extent e(X) in a normal space.
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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Hodel, R., Cardinal Functions I, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 1-61.
  3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Examples of Lindelof Spaces that are not Hereditarily Lindelof

We observe from the following statement two examples of Lindelof spaces that are not hereditarily Lindelof.

  • Any product space contains a discrete subspace having the same cardinality as the number of factor spaces.

Using the above observation, by choosing the factor spaces judiciously, the product of uncountably many spaces is a handy way of obtaining Lindelof spaces (in some cases \sigma-compact spaces) that are not hereditarily Lindelof. For definition and basic information about product spaces, see this previous post.

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All spaces under consideration are at least Hausdorff. For each \alpha \in A, let X_\alpha be a space with at least two points. For each \alpha \in A, fix two points p_\alpha, q_\alpha \in X_\alpha. Then the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha contains a discrete subspace Y that has the same cardinality as the cardinality of the index set A.

For each \alpha \in A, define y_\alpha \in \prod \limits_{\alpha \in A} X_\alpha by the following:

\displaystyle (1) \ \ \ \ \ \ y_\alpha(\gamma)=\left\{\begin{matrix}p_\alpha&\ \gamma=\alpha\\{q_\alpha}&\ \gamma \ne \alpha \end{matrix}\right.

Let Y=\left\{ y_\alpha: \alpha \in A\right\}. It follows that \lvert Y \lvert = \lvert A \lvert and that Y is a discrete space.

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Whenever the index set A is uncountable, the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha contains an uncountable discrete subspace. Thus even if the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha is Lindelof, one of its subspace Y cannot be Lindelof. Taking the product of uncountably many factor spaces is a handy way to obtain Lindelof space that is not hereditarily Lindelof. Some examples are shown below.

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Examples

Let the index set A be uncountable. To make the product space Lindelof, we can make every one of its factor X_\alpha compact. Thus the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha is compact and not hereditarily Lindelof.

Thus the product space [0,1]^{\omega_1}, the product of \omega_1 many copies of the unit interval, is compact and not hereditarily Lindelof. Another example is \left\{ 0,1 \right\}^{\omega_1}, the product of \omega_1 many copies of \left\{ 0,1 \right\}

Another way to make the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha Lindelof is to make some of the factors compact such that the product of the remaining non-compact factors is Lindelof. Then the product space is essentially the product of a compact space and a Lindelof space, which is always Lindelof.

For example, let X_0=\mathbb{R} and let X_\alpha=[0,1] for all \alpha with 0<\alpha<\omega_1. Then the product space \displaystyle \prod \limits_{\alpha \in A} X_\alpha is Lindelof since it is essentially the product of a compact space and a Lindelof space. However, the product \displaystyle \prod \limits_{\alpha \in A} X_\alpha is not hereditarily Lindelof.

In fact, the product space in the previous paragraph is \sigma-compact (i.e. the union of countably many compact sets). To make the example not \sigma-compact, simply make the first factor space a non-locally compact Lindelof space. For example, use the Sorgenfrey line or the space of the irrational numbers.