Revisiting example 106 from Steen and Seebach

Posted on December 25, 2021 by Dan Ma

The example 106 from Counterexamples in Topology by Steen and Seebach [3] is the space $\omega_1 \times I^I$ where the first factor $\omega_1$ is the space of countable ordinals with the usual order topology and the second factor $I^I$ is the product of continuum many copies of the unit interval $I=[0,1]$ .

This space was previously discussed in this site. One of the key results from that discussion is that $\omega_1 \times I^I$ is not normal, a result not shown in Steen and Seebach. The proof that was given in this site (see here) is based on an article published in 1976 [1], long before the publication date of the first edition of Steen and Seebach in 1970. It turns out that the non-normality of $\omega_1 \times I^I$ was given as an exercise in Steen and Seebach in the problem section at the end of the book (problem 127 in page 211, Dover edition). Problem 127: Show that $[0, \Omega) \times I^I$ is not normal. This indicates that the result not shown in Steen and Seebach was because it was given as a problem and not because the tool for solving it was not yet available. The fact that it is given as an exercise also means that there is a more basic proof of the non-normality of $\omega_1 \times I^I$ . So, once this is realized, I set out to find a simpler proof or at least one that does not rely on the result from [1]. Interestingly, this proof brings out a broader discussion that is worthwhile and goes beyond the example at hand. The goal here is to examine the more basic proof and the broader discussion.

A Classic Example

Before talking about the promised proof, we consider the product of $\omega_1$ and its immediate successor.

As noted at the beginning, the space $\omega_1$ is the set of all countable ordinals with the order topology. The ordinal $\omega_1+1$ is the immediate successor of $\omega_1$ . It can be regarded as the result of adding one more point to $\omega_1$ . The extra point is $\omega_1$ , i.e., $\omega_1 +1=\omega_1 \cup \{ \omega_1 \}$ with $\omega_1$ greater than all points $\beta < \omega_1$ . The ordinal $\omega_1+1$ with the order topology is a compact space. Using interval notation, $\omega_1=[0, \omega_1)$ and $\omega_1+1=[0, \omega_1]$ . As ordinals, $\omega_1$ is the first uncountable ordinal and $\omega_1+1$ is the first uncountable successor ordinal. For more information, see here.

The product $[0, \omega_1) \times [0, \omega_1]$ is a classic example of a product of a normal space (the first factor) and a compact space (the second factor) that is not normal. This example and others like it show that normality is easily broken upon taking product even if one of the factors is as nice as a compact space. The non-normality of $[0, \omega_1) \times [0, \omega_1]$ is discussed here. In that proof, two disjoint closed sets $H$ and $K$ are given such that they cannot be separated by disjoint open sets. The $H$ and $K$ are:

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

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

The Basic Proof

To show that $\omega_1 \times I^I$ is not normal, we show that one of its closed subspaces is not normal. That closed subspace is $[0, \omega_1) \times [0, \omega_1]$ . To this end, we show that $[0, \omega_1]$ can be embedded in the product space $I^I$ . With a non-normal closed subspace, it follows that $\omega_1 \times I^I$ is not normal. The remainder of the proof is to give the embedding.

We show that $[0, \omega_1]$ can be embedded as a closed subspace of $I^{\omega_1}$ , the product of $\omega_1$ many copies of $I$ . This means that $[0, \omega_1]$ is also a closed subspace of $I^I$ .

For each $\beta < \omega_1$ , define $T_\beta: \omega_1 \rightarrow I$ as follows:

$T_\beta(\gamma) = \begin{cases} 1 & \ \ \ \mbox{if } \gamma < \beta \\ 0 & \ \ \ \mbox{if } \gamma \ge \beta \end{cases}$

Furthermore, define $T: \omega_1 \rightarrow I$ by letting $T(\beta)=1$ for all $\beta < \omega_1$ . Consider the correspondence $\beta \rightarrow T_\beta$ with $\beta < \omega_1$ and $\omega_1 \rightarrow T$ . The mapping is clearly one-to-one from $[0, \omega_1]$ onto $\{ T_\beta: \beta < \omega_1 \} \cup \{ T \}$ . Upon closer inspection, the mapping in each direction is continuous (this is a good exercise to walk through). Thus, the mapping is a homeomorphism. It follows that $[0, \omega_1]$ can be considered a subspace of $I^{\omega_1}$ . Since $[0, \omega_1]$ is compact, it must be a closed subspace. With the cardinality of $\omega_1$ being less than or equal to continuum, it follows that $[0, \omega_1]$ can be embedded as a closed subspace of $I^I$ .

Stone-Cech Compactification

The first broader discussion is that of Stone-Cech compactification. More specifically, $\beta \omega_1=\omega_1+1$ , i.e., the Stone-Cech compactification of the first uncountable ordinal is its immediate successor.

To see that $\beta \omega_1=\omega_1+1$ , note that every continuous function defined on $[0,\omega_1)$ is bounded and is eventually constant (see result B here). As a result, every continuous function defined on $[0,\omega_1)$ can be extended to a continuous function defined on $[0, \omega_1]$ . For any continuous function $f: \omega_1 \rightarrow \mathbb{R}$ , we can simply define $f(\omega_1)$ to be the eventual constant value. A subspace $W$ of a space $Y$ is $C^*$ -embedded in $Y$ if every bounded continuous real-valued function on $W$ can be extended to $Y$ . According to theorem 19.12 in [4], if $Y$ is a compactification of $X$ and if $X$ is $C^*$ -embedded in $Y$ , then $Y$ is the Stone-Cech compactification of $X$ . Thus $[0,\omega_1)$ is $C^*$ -embedded in $[0,\omega_1]$ and $[0,\omega_1]$ is the Stone-Cech compactification of $[0,\omega_1)$ . In this instance, the Stone-Cech compactification agrees with the one-point compactification. Consider the following class theorem about normality in product space. The theorem is Corollary 3.4 in the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

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

The space $X$ is paracompact.
The product space $X \times \beta X$ is normal.

Based on the discussion presented above, the non-normality of $\omega_1 \times I^I$ is due to the non-normality of $[0, \omega_1) \times [0, \omega_1]$ . Based on this theorem, the non-normality of $[0, \omega_1) \times [0, \omega_1]$ is due to the non-paracompactness of $[0, \omega_1)$ . See result G here for a proof that $[0, \omega_1)$ is not paracompact.

The discussion up to this point points to two ways to prove that $\omega_1 \times I^I$ is not normal. One way is the basic proof indicated above. The other way is to use Theorem 1, along with the homeomorphic embedding from $[0, \omega_1]$ into $I^I$ , the fact that $\beta \omega_1=\omega_1+1$ and the fact that $[0, \omega_1)$ is not paracompact. Both are valuable. The first way is basic and is a constructive proof. Because it is more hands-on, it is a better proof to learn from. The second way provides a broader perspective that is informative but requires quoting a couple of fairly deep results. Perhaps it is best used as a second proof for perspective.

Countable Tightness

The essence of the basic proof above goes like this: if the space $Y$ contains a copy of $\omega_1+1=[0, \omega_1]$ , then the product space $[0, \omega_1) \times Y$ is not normal. The contrapositive statement would be the following:

Corollary
Let $Y$ be a space. If the product space $\omega_1 \times Y$ is normal, then $Y$ cannot contain a copy of $\omega_1+1$ .

In the space of $\omega_1+1=[0, \omega_1]$ , note the following about the last point: $\omega_1 \in \overline{[0, \omega_1)}$ but $\omega_1 \notin \overline{C}$ for any countable $C \subset [0, \omega_1)$ , i.e., the last point is the limit point of the set of all the points preceding it but is not in the closure of any countable set. This means that the space $\omega_1+1=[0, \omega_1]$ does not have countable tightness (or is not countably tight). See here for definition. The property of countable tightness is hereditary. If $Y$ contains a copy of $\omega_1+1$ , then $Y$ is not countably tight (or is uncountably tight). This brings us to the following theorem.

Theorem 2
Let $Y$ be an infinite compact space. Then $\omega_1 \times Y$ is normal if and only if $Y$ has countable tightness.

Whenever we consider the normality of a product with the first factor being $\omega_1$ and the second factor being a compact space, the real story is the tightness of that compact space. If the tightness is countable, the product is normal. Otherwise, the product is not normal. The theorem is another reason that $\omega_1 \times I^I$ is not normal. Instead of embedding $[0, \omega_1]$ into $I^I$ , we can actually show that $I^I$ does not have countable tightness. This is the approach that was taken in this previous post.

Theorem 2 is the result from 1976 alluded to earlier [1]. A proof of Theorem 2 is found in this previous post. For results concerning normality in a product space with a compact factor (the other factor does not have to be $\omega_1$ ), see the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Reference

Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.
Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

$\text{ }$

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2021 – Dan Ma

Counterexample 106 from Steen and Seebach

Posted on August 18, 2015 by Dan Ma

As the title suggests, this post discusses counterexample 106 in the well known book Counterexamples in Topology by Steen and Seebach [2]. We extend the discussion by adding two facts not found in the book.

The counterexample 106 is the space $X=\omega_1 \times I^I$ , which is the product of $\omega_1$ with the interval topology (ordered topology) and the product space $I^I=\prod_{t \in I} I$ where $I$ is of course the unit interval $[0,1]$ . The notation of $\omega_1$ , the first uncountable ordinal, in Steen and Seebach is $[0,\Omega)$ .

Another way to notate the example $X$ is the product space $\prod_{t \in I} X_t$ where $X_0$ is $\omega_1$ and $X_t$ is the unit interval $I$ for all $t>0$ . Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of $\omega_1$ makes this product space an interesting example.

The basic topological properties of the space $X=\omega_1 \times I^I$ that are covered in [2] are:

The space $X$ is Hausdorff and completely regular.
The space $X$ is countably compact.
The space $X$ is neither compact nor sequentially compact.
The space $X$ is neither separable, Lindelof nor $\sigma$ -compact.
The space $X$ is not first countable.
The space $X$ is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

The space $X$ is not normal.
The space $X$ has a dense subspace that is normal.

It follows from these bullet points that the space $X$ is an example of a completely regular space that is not normal. Not being a normal space, $X$ is then not metrizable. Of course there are other ways to show that $X$ is not metrizable. One is that neither of the two factors $\omega_1$ or $I^I$ is metrizable. Another is that $X$ is not first countable.

The space $X$ is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example $X=\omega_1 \times I^I$ is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example $X=\omega_1 \times I^I$ is to show whether the second factor $I^I$ is a countably tight space.

The tool in [1] is this theorem: for any compact space $Y$ , the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. For a proof of this theorem, see here. Thus the normality of the space $X$ (or the lack of) hinges on whether the compact factor $I^I=\prod_{t \in I} I$ is countably tight.

A space $Y$ is countably tight (or has countable tightness) if for each $S \subset Y$ and for each $x \in \overline{S}$ , there exists some countable $B \subset S$ such that $x \in \overline{B}$ . The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space $I^I=\prod_{t \in I} I$ is not countably tight, we define $S$ as follows. For each finite $A \subset I=[0,1]$ , define $f_A: I \rightarrow I$ such that $f_A$ maps $A$ to 1 and maps $I \backslash A$ to 0. Let $S$ be the set of $f_A$ for all possible finite $A \subset I$ . Let $f:I \rightarrow I$ be defined by $f(x)=1$ for all $x \in I$ .

It follows that $f \in \overline{S}$ . We claim that for any countable $B \subset S$ , $f \notin \overline{B}$ . Let $B=\{f_{A_1},f_{A_2}, \cdots \} \subset S$ be countable where each $A_j$ is finite. Then choose $a \in I \backslash \bigcup_{j} A_j$ . Consider the open set $U=\prod_{t \in I} W_t$ where $W_t=I$ for $t \ne a$ and $W_a=(0.5,1]$ . Then $f \in U$ and $U \cap B=\varnothing$ . Thus $f \notin \overline{B}$ . This shows that the product space $I^I=\prod_{t \in I} I$ is not countably tight.

By Theorem 1 found in this link, the space $X=\omega_1 \times I^I$ is not normal.

The space $X$ has a dense subspace that is normal

Now that we know $X=\omega_1 \times I^I$ is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace $\omega_1 \times S$ where $S$ is the $\Sigma$ -product $S=\Sigma_{t \in I} I$ where $S$ is the space of all $f \in I^I$ such that for each $f$ , $f(x) \ne 0$ for at most countably many $x \in I$ .

The subspace $S$ is dense in the product space $I^I$ . Thus $\omega_1 \times S$ is dense in $X=\omega_1 \times I^I$ . The space $S$ is normal since the $\Sigma$ -product of separable metric spaces is normal (see here). Furthermore, $\omega_1$ can be embedded as a closed subspace of $S=\Sigma_{t \in I} I$ . Then $\omega_1 \times S$ is homeomorphic to a closed subspace of $S \times S$ . Note that $S \times S \cong S$ . Since $\omega_1 \times S$ can be embedded as a closed subspace of the normal space $S$ , the space $\omega_1 \times S$ is normal.

Reference

Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

$\copyright$ 2015 Dan Ma

Revised January 28, 2021

Normality in the powers of countably compact spaces

Posted on May 17, 2015 by Dan Ma

Let $\omega_1$ be the first uncountable ordinal. The topology on $\omega_1$ we are interested in is the ordered topology, the topology induced by the well ordering. The space $\omega_1$ is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of $\omega_1$ . We also make comments on normality in the powers of a countably compact non-compact space.

Let $\omega$ be the first infinite ordinal. It is well known that $\omega^{\omega_1}$ , the product space of $\omega_1$ many copies of $\omega$ , is not normal (a proof can be found in this earlier post). This means that any product space $\prod_{\alpha<\kappa} X_\alpha$ , with uncountably many factors, is not normal as long as each factor $X_\alpha$ contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space $\prod_{\alpha<\kappa} X_\alpha$ , the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of $\omega$ ). In other words, the interesting case is that each factor $X_\alpha$ is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in $X^{\kappa}$ where $X$ is a countably non-compact space. In this post we start with the space $X=\omega_1$ of the countable ordinals. We examine $\omega_1$ power $\omega_1^{\omega_1}$ as well as the countable power $\omega_1^{\omega}$ . The former is not normal while the latter is normal. The proof that $\omega_1^{\omega}$ is normal is an application of the normality of $\Sigma$ -product of the real line.

____________________________________________________________________

The uncountable product

Theorem 1
The product space $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal.

Theorem 1 follows from Theorem 2 below. For any space $X$ , a collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for any finite $\mathcal{F} \subset \mathcal{C}$ , the intersection $\cap \mathcal{F} \ne \varnothing$ . Such a collection $\mathcal{C}$ is called an f.i.p collection for short. It is well known that a space $X$ is compact if and only collection $\mathcal{C}$ of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space $X=\omega_1$ , there is an f.i.p. collection of cardinality $\omega_1$ using its linear order. For each $\alpha<\omega_1$ , let $C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}$ . Let $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ . It is a collection of closed subsets of $X=\omega_1$ . It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space $X$ with an f.i.p. collection of cardinality $\omega_1$ that has empty intersection, the product space $X^{\omega_1}$ is not normal.

Theorem 2
Let $X$ be a countably compact space. Suppose that there exists a collection $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ of closed subsets of $X$ such that $\mathcal{C}$ has the finite intersection property and that $\mathcal{C}$ has empty intersection. Then the product space $X^{\omega_1}$ is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space $X^{\omega_1}=\prod_{\alpha<\omega_1} X$ , we mean a set of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that each $O_\alpha$ is an open subset of $X$ and that $O_\alpha=X$ for all but finitely many $\alpha<\omega_1$ . Given a standard basic open set $O=\prod_{\alpha<\omega_1} O_\alpha$ , the notation $\text{Supp}(O)$ refers to the finite set of $\alpha$ for which $O_\alpha \ne X$ . For any set $M \subset \omega_1$ , the notation $\pi_M$ refers to the projection map from $\prod_{\alpha<\omega_1} X$ to the subproduct $\prod_{\alpha \in M} X$ . Each element $d \in X^{\omega_1}$ can be considered a function $d: \omega_1 \rightarrow X$ . By $(d)_\alpha$ , we mean $(d)_\alpha=d(\alpha)$ .

For each $t \in X$ , let $f_t: \omega_1 \rightarrow X$ be the constant function whose constant value is $t$ . Consider the following subspaces of $X^{\omega_1}$ .

$H=\prod_{\alpha<\omega_1} C_\alpha$

$\displaystyle K=\left\{f_t: t \in X \right\}$

Both $H$ and $K$ are closed subsets of the product space $X^{\omega_1}$ . Because the collection $\mathcal{C}$ has empty intersection, $H \cap K=\varnothing$ . We show that $H$ and $K$ cannot be separated by disjoint open sets. To this end, let $U$ and $V$ be open subsets of $X^{\omega_1}$ such that $H \subset U$ and $K \subset V$ .

Let $d_1 \in H$ . Choose a standard basic open set $O_1$ such that $d_1 \in O_1 \subset U$ . Let $S_1=\text{Supp}(O_1)$ . Since $S_1$ is the support of $O_1$ , it follows that $\pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U$ . Since $\mathcal{C}$ has the finite intersection property, there exists $a_1 \in \bigcap_{\alpha \in S_1} C_\alpha$ .

Define $d_2 \in H$ such that $(d_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(d_2)_\alpha=(d_1)_\alpha$ for all $\alpha \in \omega_1-S_1$ . Choose a standard basic open set $O_2$ such that $d_2 \in O_2 \subset U$ . Let $S_2=\text{Supp}(O_2)$ . It is possible to ensure that $S_1 \subset S_2$ by making more factors of $O_2$ different from $X$ . We have $\pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U$ . Since $\mathcal{C}$ has the finite intersection property, there exists $a_2 \in \bigcap_{\alpha \in S_2} C_\alpha$ .

Now choose a point $d_3 \in H$ such that $(d_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(d_3)_\alpha=(d_2)_\alpha$ for all $\alpha \in \omega_1-S_2$ . Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

a sequence $d_1,d_2,d_3,\cdots$ of point of $H=\prod_{\alpha<\omega_1} C_\alpha$ ,
a sequence $S_1 \subset S_2 \subset S_3 \subset \cdots$ of finite subsets of $\omega_1$ ,
a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for all $n \ge 2$ , $(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $\pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ . Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$ . Either $A$ is finite or $A$ is infinite. Let’s examine the two cases.

Case 1
Suppose that $A$ is infinite. Since $X$ is countably compact, $A$ has a limit point $a$ . That means that every open set containing $a$ contains some $a_n \ne a$ . For each $n \ge 2$ , define $y_n \in \prod_{\alpha< \omega_1} X$ such that

$(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$ ,
$(y_n)_\alpha=a$ for all $\alpha \in \omega_1-S_n$

From the induction step, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$ . Let $t=f_a \in K$ , the constant function whose constant value is $a$ . It follows that $t$ is a limit of $\left\{y_1,y_2,y_3,\cdots \right\}$ . This means that $t \in \overline{U}$ . Since $t \in K \subset V$ , $U \cap V \ne \varnothing$ .

Case 2
Suppose that $A$ is finite. Then there is some $m$ such that $a_m=a_j$ for all $j \ge m$ . For each $n \ge 2$ , define $y_n \in \prod_{\alpha< \omega_1} X$ such that

$(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$ ,
$(y_n)_\alpha=a_m$ for all $\alpha \in \omega_1-S_n$

As in Case 1, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$ . Let $t=f_{a_m} \in K$ , the constant function whose constant value is $a_m$ . It follows that $t=y_n$ for all $n \ge m+1$ . Thus $U \cap V \ne \varnothing$ .

Both cases show that $U \cap V \ne \varnothing$ . This completes the proof the product space $X^{\omega_1}$ is not normal. $\blacksquare$

____________________________________________________________________

The countable product

Theorem 3
The product space $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is collectionwise normal, which is stronger than normality. The proof makes use of the $\Sigma$ -product of $\kappa$ many copies of $\mathbb{R}$ , which is the following subspace of the product space $\mathbb{R}^{\kappa}$ .

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^{\kappa}: x(\alpha) \ne 0 \text{ for at most countably many } \alpha<\kappa \right\}$

It is well known that $\Sigma(\kappa)$ is collectionwise normal (see this earlier post). We show that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is a closed subspace of $\Sigma(\kappa)$ where $\kappa=\omega_1$ . Thus $\omega_1^{\omega}$ is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space $\omega_1$ is embedded as a closed subspace of $\Sigma(\omega_1)$ .

For each $\beta<\omega_1$ , define $f_\beta:\omega_1 \rightarrow \mathbb{R}$ such that $f_\beta(\gamma)=1$ for all $\gamma<\beta$ and $f_\beta(\gamma)=0$ for all $\beta \le \gamma <\omega_1$ . Let $W=\left\{f_\beta: \beta<\omega_1 \right\}$ . We show that $W$ is a closed subset of $\Sigma(\omega_1)$ and $W$ is homeomorphic to $\omega_1$ according to the mapping $f_\beta \rightarrow W$ .

First, we show $W$ is closed by showing that $\Sigma(\omega_1)-W$ is open. Let $y \in \Sigma(\omega_1)-W$ . We show that there is an open set containing $y$ that contains no points of $W$ .

Suppose that for some $\gamma<\omega_1$ , $y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}$ . Consider the open set $Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1)$ where $Q_\alpha=\mathbb{R}$ except that $Q_\gamma=O$ . Then $y \in Q$ and $Q \cap W=\varnothing$ .

So we can assume that for all $\gamma<\omega_1$ , $y_\gamma \in \left\{0, 1 \right\}$ . There must be some $\theta$ such that $y_\theta=1$ . Otherwise, $y=f_0 \in W$ . Since $y \ne f_\theta$ , there must be some $\delta<\gamma$ such that $y_\delta=0$ . Now choose the open interval $T_\theta=(0.9,1.1)$ and the open interval $T_\delta=(-0.1,0.1)$ . Consider the open set $M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1)$ such that $M_\alpha=\mathbb{R}$ except for $M_\theta=T_\theta$ and $M_\delta=T_\delta$ . Then $y \in M$ and $M \cap W=\varnothing$ . We have just established that $W$ is closed in $\Sigma(\omega_1)$ .

Consider the mapping $f_\beta \rightarrow W$ . Based on how it is defined, it is straightforward to show that it is a homeomorphism between $\omega_1$ and $W$ .

Claim 2
The $\Sigma$ -product $\Sigma(\omega_1)$ has the interesting property it is homeomorphic to its countable power, i.e.

$\Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}$

Because each element of $\Sigma(\omega_1)$ is nonzero only at countably many coordinates, concatenating countably many elements of $\Sigma(\omega_1)$ produces an element of $\Sigma(\omega_1)$ . Thus Claim 2 can be easily verified. With above claims, we can see that

$\displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)$

Thus $\omega_1^{\omega}$ is a closed subspace of $\Sigma(\omega_1)$ . Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that $\omega_1^{\omega}$ is normal. $\blacksquare$

____________________________________________________________________

The normality in the powers of $X$

We have established that $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal. Hence any higher uncountable power of $\omega_1$ is not normal. We have also established that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ , the countable power of $\omega_1$ is normal (in fact collectionwise normal). Hence any finite power of $\omega_1$ is normal. However $\omega_1^{\omega}$ is not hereditarily normal. One of the exercises below is to show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let $X$ be a countably compact space has an f.i.p. collection $\mathcal{C}$ of closed sets such that $\bigcap \mathcal{C}=\varnothing$ . Then $X^{\kappa}$ is not normal where $\kappa=\lvert \mathcal{C} \lvert$ .

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let $X$ be a countably compact space that is not compact. Then there exists a cardinal number $\kappa$ such that $X^{\kappa}$ is not normal and $X^{\tau}$ is normal for all cardinal number $\tau<\kappa$ .

By the non-compactness of $X$ , there exists an f.i.p. collection $\mathcal{C}$ of closed subsets of $X$ such that $\bigcap \mathcal{C}=\varnothing$ . Let $\kappa$ be the least cardinality of such an f.i.p. collection. By Theorem 4, that $X^{\kappa}$ is not normal. Because $\kappa$ is least, any smaller power of $X$ must be normal.

Theorem 6
Let $X$ be a space that is not countably compact. Then $X^{\kappa}$ is not normal for any cardinal number $\kappa \ge \omega_1$ .

Since the space $X$ in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, $\omega^{\omega_1}$ is not normal. Then $\omega^{\omega_1}$ is a closed subspace of $X^{\omega_1}$ .

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space $X$ , $X^{\kappa}$ is not normal for some cardinal number $\kappa$ . The $\kappa$ from either Theorem 5 or Theorem 6 is at least $\omega_1$ . Interestingly for some spaces, the $\kappa$ can be much smaller. For example, for the Sorgenfrey line, $\kappa=2$ . For some spaces (e.g. the Michael line), $\kappa=\omega$ .

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space $X$ is normal, then $X$ is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of $X$ is just another way of discussing Theorem 7. According to Theorem 7, if $X$ is not compact, some power of $X$ is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in $X^{\kappa}$ for more examples of countably compact non-compact $X$ . One particular interesting example would be a countably compact non-compact $X$ such that the least power $\kappa$ for non-normality in $X^{\kappa}$ is more than $\omega_1$ . A possible candidate could be the second uncountable ordinal $\omega_2$ . By Theorem 2, $\omega_2^{\omega_2}$ is not normal. The issue is whether the $\omega_1$ power $\omega_2^{\omega_1}$ and countable power $\omega_2^{\omega}$ are normal.

____________________________________________________________________

Exercises

Exercise 1
Show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Exercise 2
Show that the mapping $f_\beta \rightarrow W$ in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space $\omega_1$ is a closed subspace of the $\Sigma$ -product of the real line. Show that $\omega_1$ can be embedded in the $\Sigma$ -product of arbitrary spaces.

For each $\alpha<\omega_1$ , let $X_\alpha$ be a space with at least two points. Let $p \in \prod_{\alpha<\omega_1} X_\alpha$ . The $\Sigma$ -product of the spaces $X_\alpha$ is the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$ .

$\Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}$

The point $p$ is the center of the $\Sigma$ -product. Show that the space $\Sigma(X_\alpha)$ contains $\omega_1$ as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that $\omega_1^{\omega}$ is normal.

____________________________________________________________________
$\copyright \ 2015 \text{ by Dan Ma}$

Comparing two function spaces

Posted on August 5, 2014 by Dan Ma

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$ . Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function $f$ defined on either $\omega_1$ or $\omega_1+1$ is eventually constant, i.e., there exists some $\alpha<\omega_1$ such that the function $f$ is constant on the ordinals beyond $\alpha$ . Now consider the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ . Thus individually, elements of these two function spaces appear identical. Any $f \in C_p(\omega_1)$ matches a function $f^* \in C_p(\omega_1+1)$ where $f^*$ is the result of adding the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$ . This fact may give the impression that the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

____________________________________________________________________

Tightness

One topological property that is different between $C_p(\omega_1)$ and $C_p(\omega_1+1)$ is that of tightness. The function space $C_p(\omega_1+1)$ is countably tight, while $C_p(\omega_1)$ is not countably tight.

Let $X$ be a space. The tightness of $X$ , denoted by $t(X)$ , is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$ , there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$ . When $t(X)=\omega$ , we say that $X$ has countable tightness or is countably tight. When $t(X)>\omega$ , we say that $X$ has uncountable tightness or is uncountably tight.

First, we show that the tightness of $C_p(\omega_1)$ is greater than $\omega$ . For each $\alpha<\omega_1$ , define $f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\}$ such that $f_\alpha(\beta)=0$ for all $\beta \le \alpha$ and $f_\alpha(\beta)=1$ for all $\beta>\alpha$ . Let $g \in C_p(\omega_1)$ be the function that is identically zero. Then $g \in \overline{F}$ where $F$ is defined by $F=\left\{f_\alpha: \alpha<\omega_1 \right\}$ . It is clear that for any countable $B \subset F$ , $g \notin \overline{B}$ . Thus $C_p(\omega_1)$ cannot be countably tight.

The space $\omega_1+1$ is a compact space. The fact that $C_p(\omega_1+1)$ is countably tight follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$ .

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for $C_p(\omega_1+1)$ .

____________________________________________________________________

The Frechet-Urysohn property

In fact, $C_p(\omega_1+1)$ has a property that is stronger than countable tightness. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space (see this previous post). Of course, $C_p(\omega_1)$ not being countably tight means that it is not a Frechet-Urysohn space.

____________________________________________________________________

Normality

The function space $C_p(\omega_1+1)$ is not normal. If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ would have countable extent. However, there exists an uncountable closed and discrete subset of $C_p(\omega_1+1)$ (see this previous post). On the other hand, $C_p(\omega_1)$ is Lindelof. The fact that $C_p(\omega_1)$ is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if $X$ is a space consisting of ordinals such that $X$ is first countable and countably compact, then $C_p(X)$ is Lindelof.

____________________________________________________________________

Embedding one function space into the other

The two function space $C_p(\omega_1+1)$ and $C_p(\omega_1)$ are very different topologically. However, one of them can be embedded into the other one. The space $\omega_1+1$ is the continuous image of $\omega_1$ . Let $g: \omega_1 \longrightarrow \omega_1+1$ be a continuous surjection. Define a map $\psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1)$ by letting $\psi(f)=f \circ g$ . It is shown in this previous post that $\psi$ is a homeomorphism. Thus $C_p(\omega_1+1)$ is homeomorphic to the image $\psi(C_p(\omega_1+1))$ in $C_p(\omega_1)$ . The map $g$ is also defined in this previous post.

The homeomposhism $\psi$ tells us that the function space $C_p(\omega_1)$ , though Lindelof, is not hereditarily normal.

On the other hand, the function space $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$ . Note that $C_p(\omega_1+1)$ is countably tight, which is a hereditary property.

____________________________________________________________________

Remark

There is a mapping that is alluded to at the beginning of the post. Each $f \in C_p(\omega_1)$ is associated with $f^* \in C_p(\omega_1+1)$ which is obtained by appending the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$ . It may be tempting to think of the mapping $f \rightarrow f^*$ as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

____________________________________________________________________

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Buzyakova, R. Z., In search of Lindelof $C_p$ ‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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

Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

Posted on August 4, 2014 by Dan Ma

This is another post that discusses what $C_p(X)$ is like when $X$ is a compact space. In this post, we discuss the example $C_p(\omega_1+1)$ where $\omega_1+1$ is the first compact uncountable ordinal. Note that $\omega_1+1$ is the successor to $\omega_1$ , which is the first (or least) uncountable ordinal. The function space $C_p(\omega_1+1)$ is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by $C_p(X)$ for all compact spaces $X$ . The second property is possessed by all compact scattered spaces. After we discuss $C_p(\omega_1+1)$ , we discuss briefly the general results for $C_p(X)$ .

____________________________________________________________________

Initial discussion

The function space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$ . In fact, $C_p(\omega_1+1)$ is homeomorphic to a subspace of the following subspace of $\mathbb{R}^{\omega_1}$ :

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

The subspace $\Sigma(\omega_1)$ is the $\Sigma$ -product of $\omega_1$ many copies of the real line $\mathbb{R}$ . The $\Sigma$ -product of separable metric spaces is monolithic (see here). The $\Sigma$ -product of first countable spaces is Frechet-Urysohn (see here). Thus $\Sigma(\omega_1)$ has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space $C_p(\omega_1+1)$ has both of these properties. The key to the discussion is then to show that $C_p(\omega_1+1)$ is homeopmophic to a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

____________________________________________________________________

Connection to $\Sigma$ -product

We show that the function space $C_p(\omega_1+1)$ is homeomorphic to a subspace of the $\Sigma$ -product of $\omega_1$ many copies of the real lines. Let $Y_0$ be the following subspace of $C_p(\omega_1+1)$ :

$Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}$

Every function in $Y_0$ has non-zero values at only countably points of $\omega_1+1$ . Thus $Y_0$ can be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

By Theorem 1 in this previous post, $C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}$ , i.e, the function space $C_p(\omega_1+1)$ is homeomorphic to the product space $Y_0 \times \mathbb{R}$ . On the other hand, the product $Y_0 \times \mathbb{R}$ can also be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ . Basically adding one additional factor of the real line to $Y_0$ still results in a subspace of the $\Sigma$ -product. Thus we have:

$C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Thus $C_p(\omega_1+1)$ possesses all the hereditary properties of $\Sigma(\omega_1)$ . Another observation we can make is that $\Sigma(\omega_1)$ is not hereditarily normal. The function space $C_p(\omega_1+1)$ is not normal (see here). The $\Sigma$ -product $\Sigma(\omega_1)$ is normal (see here). Thus $\Sigma(\omega_1)$ is not hereditarily normal.

____________________________________________________________________

A closer look at $C_p(\omega_1+1)$

In fact $C_p(\omega_1+1)$ has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let $h$ be a homeomorphism from $C_p(\omega_1+1)$ onto $Y_0 \times \mathbb{R}$ . For each $\alpha<\omega_1$ , let $H_\alpha$ be defined as follows:

$H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}$

Clearly $H_\alpha \subset Y_0$ . Furthermore $H_\alpha$ can be considered as a subspace of $\mathbb{R}^\omega$ and is thus metrizable. Let $A$ be a countable subset of $C_p(\omega_1+1)$ . Then $h(A) \subset H_\alpha \times \mathbb{R}$ for some $\alpha<\omega_1$ . The set $H_\alpha \times \mathbb{R}$ is metrizable. The set $H_\alpha \times \mathbb{R}$ is also a closed subset of $Y_0 \times \mathbb{R}$ . Then $\overline{A}$ is contained in $H_\alpha \times \mathbb{R}$ and is therefore metrizable. We have shown that the closure of every countable subspace of $C_p(\omega_1+1)$ is metrizable. In other words, every separable subspace of $C_p(\omega_1+1)$ is metrizable. This property follows from the fact that $C_p(\omega_1+1)$ is strongly monolithic.

____________________________________________________________________

Monolithicity and Frechet-Urysohn property

As indicated at the beginning, the $\Sigma$ -product $\Sigma(\omega_1)$ is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space $C_p(\omega_1+1)$ is both strongly monolithic and Frechet-Urysohn.

Let $\tau$ be an infinite cardinal. A space $X$ is $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $nw(\overline{A}) \le \tau$ . A space $X$ is monolithic if it is $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is monolithic if and only of for every subspace $Y$ of $X$ , the density of $Y$ equals to the network weight of $Y$ , i.e., $d(Y)=nw(Y)$ . A longer discussion of the definition of monolithicity is found here.

A space $X$ is strongly $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $w(\overline{A}) \le \tau$ . A space $X$ is strongly monolithic if it is strongly $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is strongly monolithic if and only if for every subspace $Y$ of $X$ , the density of $Y$ equals to the weight of $Y$ , i.e., $d(Y)=w(Y)$ .

In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.

In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space $C_p([0,1])$ . It is clear that all metrizable spaces are strongly monolithic.

The function space $C_p(\omega_1+1)$ is not separable. Since it is strongly monolithic, every separable subspace of $C_p(\omega_1+1)$ is metrizable. We can see this by knowing that $C_p(\omega_1+1)$ is a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ , or by using the homeomorphism $h$ as in the previous section.

For any compact space $X$ , $C_p(X)$ is countably tight (see this previous post). In the case of the compact uncountable ordinal $\omega_1+1$ , $C_p(\omega_1+1)$ has the stronger property of being Frechet-Urysohn. A space $Y$ is said to be a Frechet-Urysohn space (also called a Frechet space) if for each $y \in Y$ and for each $M \subset Y$ , if $y \in \overline{M}$ , then there exists a sequence $\left\{y_n \in M: n=1,2,3,\cdots \right\}$ such that the sequence converges to $y$ . As we shall see below, $C_p(X)$ is rarely Frechet-Urysohn.

____________________________________________________________________

General discussion

For any compact space $X$ , $C_p(X)$ is monolithic but does not have to be strongly monolithic. The monolithicity of $C_p(X)$ follows from the following theorem, which is Theorem II.6.8 in [1].

Theorem 1
Then the function space $C_p(X)$ is monolithic if and only if $X$ is a stable space.

See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space $X$ is stable if for any continuous image $Y$ of $X$ , the weak weight of $Y$ , denoted by $ww(Y)$ , coincides with the network weight of $Y$ , denoted by $nw(Y)$ . In [1], $ww(Y)$ is notated by $iw(Y)$ . The cardinal function $ww(Y)$ is the minimum cardinality of all $w(T)$ , the weight of $T$ , for which there exists a continuous bijection from $Y$ onto $T$ .

All compact spaces are stable. Let $X$ be compact. For any continuous image $Y$ of $X$ , $Y$ is also compact and $ww(Y)=w(Y)$ , since any continuous bijection from $Y$ onto any space $T$ is a homeomorphism. Note that $ww(Y) \le nw(Y) \le w(Y)$ always holds. Thus $ww(Y)=w(Y)$ implies that $ww(Y)=nw(Y)$ . Thus we have:

Corollary 2
Let $X$ be a compact space. Then the function space $C_p(X)$ is monolithic.

However, the strong monolithicity of $C_p(\omega_1+1)$ does not hold in general for $C_p(X)$ for compact $X$ . As indicated above, $C_p([0,1])$ is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of $C_p(X)$ .

Theorem 3
Let $X$ be a space. Then $C_p(X)$ is strongly monolithic if and only if $X$ is simple.

A space $X$ is $\tau$ -simple if whenever $Y$ is a continuous image of $X$ , if the weight of $Y$ $\le \tau$ , then the cardinality of $Y$ $\le \tau$ . A space $X$ is simple if it is $\tau$ -simple for all infinite cardinal numbers $\tau$ . Interestingly, any separable metric space that is uncountable is not $\omega$ -simple. Thus $[0,1]$ is not $\omega$ -simple and $C_p([0,1])$ is not strongly monolithic, according to Theorem 3.

For compact spaces $X$ , $C_p(X)$ is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].

Theorem 4
Let $X$ be a compact space. Then the following conditions are equivalent.

$C_p(X)$ is a Frechet-Urysohn space.
$C_p(X)$ is a k-space.
The compact space $X$ is a scattered space.

A space $X$ is a scattered space if for every non-empty subspace $Y$ of $X$ , there exists an isolated point of $Y$ (relative to the topology of $Y$ ). Any space of ordinals is scattered since every non-empty subset has a least element. Thus $\omega_1+1$ is a scattered space. On the other hand, the unit interval $[0,1]$ with the Euclidean topology is not scattered. According to this theorem, $C_p([0,1])$ cannot be a Frechet-Urysohn space.

____________________________________________________________________

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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

Cp(omega 1 + 1) is not normal

Posted on August 2, 2014 by Dan Ma

In this and subsequent posts, we consider $C_p(X)$ where $X$ is a compact space. Recall that $C_p(X)$ is the space of all continuous real-valued functions defined on $X$ and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is $\omega_1+1$ , the first compact uncountable ordinal. There are many interesting results about the function space $C_p(\omega_1+1)$ . In this post we show that $C_p(\omega_1+1)$ is not normal. An even more interesting fact about $C_p(\omega_1+1)$ is that $C_p(\omega_1+1)$ does not have any dense normal subspace [1].

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$ . The set $\omega_1$ is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space $\omega_1+1$ is the first compact uncountable ordinal. In proving that $C_p(\omega_1+1)$ is not normal, a theorem that is due to D. P. Baturov is utilized [2]. This theorem is also proved in this previous post.

For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).

The fact that $C_p(\omega_1+1)$ is not normal is established by the following two points.

If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ has countable extent, i.e. every closed and discrete subspace of $C_p(\omega_1+1)$ is countable.
There exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$ .

We discuss each of the bullet points separately.

The function space $C_p(\omega_1+1)$ is a dense subspace of $\mathbb{R}^{\omega_1}$ , the product of $\omega_1$ many copies of $\mathbb{R}$ . According to a result of D. P. Baturov [2], any dense normal subspace of the product of $\omega_1$ many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus $C_p(\omega_1+1)$ cannot be normal if the second bullet point above is established.

Now we show that there exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$ . For each $\alpha$ with $0<\alpha<\omega_1$ , define $h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by:

$h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1 \end{cases}$

Clearly, $h_\alpha \in C_p(\omega_1 +1)$ for each $\alpha$ . Let $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ . We show that $H$ is a closed and discrete subspace of $C_p(\omega_1 +1)$ . The fact that $H$ is closed in $C_p(\omega_1 +1)$ is establish by the following claim.

Let $h \in C_p(\omega_1 +1) \backslash H$ . We wish to establish the following claim. Once the claim is established, it follows that $H$ is a closed subset of $C_p(\omega_1 +1)$ .

Claim 1
There exists an open subset $U$ of $C_p(\omega_1 +1)$ such that $h \in U$ and $U \cap H=\varnothing$ .

Consider the two mutually exclusive cases. Case 1. There exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$ . Case 2. $h(\omega_1+1) \subset \left\{0,1 \right\}$ .

For Case 1, let $U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}$ . Clearly $h \in U$ and $U \cap H=\varnothing$ .

Now assume Case 2. Within this case, there are three sub cases. Case 2.1. $h$ is a constant function with value 0. Case 2.2. $h$ is a constant function with value 1. Case 2.3. $h$ is not a constant function.

Case 2.1. If $h(\alpha)=0$ for all $\alpha \le \omega_1$ , then consider the open set $U$ where $U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}$ . Clearly $h \in U$ and $U \cap H=\varnothing$ .

Case 2.2. Suppose $h$ is a constant function with value 1. Then let $U$ be the open set: $U=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in \mathbb{R} \backslash \left\{0 \right\} \right\}$ . It is clear that no function in $H$ can be in $U$ .

Case 2.3. Suppose $h$ is not a constant function. This case be broken down into two cases. Case 2.3.1. $h(\omega_1)=1$ . Case 2.3.2. $h(\omega_1)=0$ .

Case 2.3.1. Just like in Case 2.2, let $U=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in \mathbb{R} \backslash \left\{0 \right\} \right\}$ . Then $h \in U$ and $U \cap H=\varnothing$ .

Case 2.3.2. Assume that $h(\omega_1)=0$ . Since $h$ is not a constant function, it must takes on a value of 1 at some point. Let $\alpha<\omega_1$ be the largest such that $h(\alpha)=1$ . This $\alpha$ exists because $h$ is continuous and $h(\omega_1)=0$ . This case can be further broken into 2 cases. Case 2.3.2.1. There exists $\beta<\alpha$ such that $h(\beta)=0$ . Case 2.3.2.2. $h(\beta)=1$ for all $\beta<\alpha$ .

Case 2.3.2.1. Define $U=\left\{f \in C_p(\omega_1 +1): f(\beta) \in (-0.1,0.1) \text{ and } f(\alpha) \in (0.9,1.1) \right\}$ . Note that $h \in U$ and $U \cap H=\varnothing$ .

Case 2.3.2.2. In this case, $h(\beta)=1$ for all $\beta \le \alpha$ and $h(\gamma)=0$ for all $\alpha<\gamma \le \omega_1$ . This means that $h=h_\alpha$ . This is a contradiction since $h \notin H$ .

In all the cases except the last one, Claim 1 is true. The last case is not possible. Thus Claim 1 is established. The set $H$ is a closed subset of $C_p(\omega_1 +1)$ .

Next we show that $H$ is discrete in $C_p(\omega_1 +1)$ . Fix $h_\alpha$ where $0<\alpha<\omega_1$ . Let $W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}$ . It is clear that $h_\alpha \in W$ . Furthermore, $h_\gamma \notin W$ for all $\alpha < \gamma$ and $h_\gamma \notin W$ for all $\gamma <\alpha$ . Thus $W$ is open such that $\left\{h_\alpha \right\}=W \cap H$ . This completes the proof that $H$ is discrete.

We have established that $H$ is an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$ . This implies that $C_p(\omega_1 +1)$ is not normal.

Remarks

The set $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ as defined above is closed and discrete in $C_p(\omega_1 +1)$ . However, the set $H$ is not discrete in a larger subspace of the product space. The set $H$ is also a subset of the following $\Sigma$ -product:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

Because $\Sigma(\omega_1)$ is the $\Sigma$ -product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, $\Sigma(\omega_1)$ would have countable extent. Thus the set $H$ cannot be closed and discrete in $\Sigma(\omega_1)$ . We can actually see this directly. Let $\alpha<\omega_1$ be a limit ordinal. Define $t:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by $t(\beta)=1$ for all $\beta<\alpha$ and $t(\beta)=0$ for all $\beta \ge \alpha$ . Clearly $t \notin C_p(\omega_1 +1)$ and $t \in \Sigma(\omega_1)$ . Furthermore, $t \in \overline{H}$ (the closure is taken in $\Sigma(\omega_1)$ ).

The function space $C_p(\omega_1)$ , in contrast, is a Lindelof space and hence a normal space. If we restrict the above defined functions $h_\alpha$ to just $\omega_1$ , would the resulting functions form a closed and discrete set in $C_p(\omega_1)$ ? For each $\alpha$ with $0<\alpha<\omega_1$ , let $g_\alpha=h_\alpha \upharpoonright \omega_1$ . Let $G=\left\{g_\alpha: 0<\alpha<\omega_1 \right\}$ .

Is $G$ a closed and discrete subset of $C_p(\omega_1)$ ? It turns out that $G$ is a discrete subspace of $C_p(\omega_1)$ (relatively discrete). However it is not closed in $C_p(\omega_1)$ . Let $g:\omega_1 \rightarrow \{0, 1\}$ that takes on the constant value of 1. It follows that $g \in \overline{G}$ (the closure is in $C_p(\omega_1)$ ).

It seems that the argument above for showing $H$ is closed and discrete in $C_p(\omega_1+1)$ can be repeated for $G$ . Note that the argument for $H$ relies on the fact that the functions $h_\alpha$ takes on a value at the point $\omega_1$ . So the same argument cannot show that $G$ is a closed and discrete set. Thus $G$ is not discrete in $C_p(\omega_1)$ . Because $C_p(\omega_1)$ is Lindelof (hence normal), it has countable extent. It follows that any uncountable discrete subspace of $C_p(\omega_1)$ cannot be closed in $C_p(\omega_1)$ (the set $G$ is a demonstration). Any uncountable closed subset of $C_p(\omega_1)$ cannot be closed.

Reference

Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

$\copyright$ 2014-2018 – Dan Ma

Revised 9/17/2018

Normal x compact needs not be subnormal

Posted on May 7, 2014 by Dan Ma

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

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

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

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

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

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

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

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

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

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

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

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

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

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

____________________________________________________________________

Reference

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

____________________________________________________________________

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

The normality of the product of the first uncountable ordinal with a compact factor

Posted on March 18, 2014 by Dan Ma

The product of a normal space with a compact space needs not be normal. For example, the product space $\omega_1 \times (\omega_1+1)$ is not normal where $\omega_1$ is the first uncountable ordinal with the order topology and $\omega_1+1$ is the immediate successor of $\omega_1$ (see this post). However, $\omega_1 \times I$ is normal where $I=[0,1]$ is the unit interval with the usual topology. The topological story here is that $I$ has countable tightness while the compact space $\omega_1+1$ does not. In this post, we prove the following theorem:

Theorem 1

$Y$

The product space $\omega_1 \times Y$ is normal.
$Y$ has countable tightness, i.e., $t(Y)=\omega$ .

Theorem 1 is a special case of the theorem found in [4]. The proof for the direction of countable tightness of $Y$ implies $\omega_1 \times Y$ is normal given in [4] relies on a theorem in another source. In this post we attempt to fill in some of the gaps. For the direction $2 \Longrightarrow 1$ , we give a complete proof. For the direction $1 \Longrightarrow 2$ , we essentially give the same proof as in [4], proving it by using a series of lemmas (stated below).

The authors in [2] studied the normality of $X \times \omega_1$ where $X$ is not necessarily compact. The necessary definitions are given below. All spaces are at least Hausdorff.

____________________________________________________________________

Definitions and Lemmas

Let $X$ be a topological space. The tightness of $X$ , denoted by $t(X)$ , is the least infinite cardinal number $\kappa$ such that for any $A \subset X$ and for any $x \in \overline{A}$ , there exists a $B \subset A$ such that $x \in \overline{B}$ and $\lvert B \lvert \le \kappa$ . When $t(X)=\omega$ , we say $X$ has countable tightness or is countably tight. When $t(X)>\omega$ , we say $X$ has uncountably tightness or is uncountably tight. An handy example of a space with uncountably tightness is $\omega_1+1=\omega_1 \cup \left\{\omega_1 \right\}$ . This space has uncountable tightness at the point $\omega_1$ . All first countable spaces and all Frechet spaces have countable tightness. The concept of countable tightness and tightness in general are discussed in more details here.

A sequence $\left\{x_\alpha: \alpha<\tau \right\}$ of points of a space $X$ is said to be a free sequence if for each $\alpha<\tau$ , $\overline{\left\{x_\beta: \beta<\alpha \right\}} \cap \overline{\left\{x_\beta: \beta \ge \alpha \right\}}=\varnothing$ . When a free sequence is indexed by the cardinal number $\tau$ , the free sequence is said to have length $\tau$ . The cardinal function $F(X)$ is the least infinite cardinal $\kappa$ such that if $\left\{x_\alpha \in X: \alpha<\tau \right\}$ is a free sequence of length $\tau$ , then $\tau \le \kappa$ . The concept of tightness was introduced by Arkhangelskii and he proved that $t(X)=F(X)$ (see p. 15 of [3]). This fact implies the following lemma.

Lemma 2

$X$

$t(X) \ge \tau$

$\left\{x_\alpha \in X: \alpha<\tau \right\}$

$\tau$

A proof of Lemma 2 can be found here.

The proof of the direction $1 \Longrightarrow 2$ also uses the following lemmas.

Lemma 3

$Y$

$\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$

Lemma 4

$X$

$H$

$K$

$X$

$H$

$K$

$\beta X$

For Lemma 3, see 3.12.20(c) on p. 237 of [1]. For Lemma 4, see 3.6.4 on p. 173 of [1].

____________________________________________________________________

Proof of Theorem 1

$1 \Longrightarrow 2$
Let $X=\omega_1 \times Y$ . Suppose that $X$ is normal. Suppose that $Y$ has uncountable tightness, i.e., $t(Y) \ge \omega_1$ . By Lemma 2, there exists a free sequence $\left\{y_\alpha \in Y: \alpha<\omega_1 \right\}$ . For each $\beta<\omega_1$ , let $C_\beta=\left\{y_\alpha: \alpha>\beta \right\}$ . Then the collection $\left\{\overline{C_\beta}: \beta<\omega_1 \right\}$ has the finite intersection property. Since $Y$ is compact, $\bigcap_{\beta<\omega_1} \overline{C_\beta} \ne \varnothing$ . Let $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$ . Consider the following closed subsets of $X=\omega_1 \times Y$ .

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

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

We claim that $H \cap K=\varnothing$ . Suppose that $(\alpha,p) \in H \cap K$ . Either $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ or $p \in \overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$ . The latter case is not possible. Note that $[0,\alpha] \times Y$ is an open set containing $(\alpha,p)$ . This open set cannot contain points of the form $(\delta,p)$ where $\delta \ge \alpha+1$ . So the first case $p \in \overline{\left\{y_\delta: \delta< \alpha+1 \right\}}$ must hold. Since $p \in \bigcap_{\beta<\omega_1} \overline{C_\beta}$ , $p \in \overline{C_\alpha}=\overline{\left\{y_\delta: \delta \ge \alpha+1 \right\}}$ , a contradiction. So $H$ and $K$ are disjoint closed subsets of $X=\omega_1 \times Y$ .

Now consider $\beta X$ , the Stone-Cech compactification of $X=\omega_1 \times Y$ . By Lemma 3, $\beta X=\beta (\omega_1 \times Y)=(\omega_1+1) \times Y$ . Let $H^*=\overline{H}$ and $K^*=\overline{K}$ (closures in $\beta X$ ). We claim that $(\omega_1,p) \in H^* \cap K^*$ . Let $O=(\theta,\omega_1] \times V$ be an open set in $\beta X$ with $(\omega_1,p) \in O$ . Note that $p \in \overline{C_\theta}=\left\{y_\delta: \delta>\theta \right\}$ . Thus $V \cap \overline{C_\theta} \ne \varnothing$ . Choose $\delta>\theta$ such that $y_\delta \in V$ . We have $(\delta,y_\delta) \in (\theta,\omega_1] \times V$ and $(\delta,y_\delta) \in H^*$ . On the other hand, $(\delta,p) \in K^*$ . Thus $(\omega_1,p) \in H^* \cap K^*$ , a contradiction. Since $X=\omega_1 \times Y$ is normal, Lemma 4 indicates that $H$ and $K$ should have disjoint closures in $\beta X=(\omega_1+1) \times Y$ . Thus $Y$ has countable tightness.

$2 \Longrightarrow 1$
Suppose $t(Y)=\omega$ . Let $H$ and $K$ be disjoint closed subsets of $\omega_1 \times Y$ . The following series of claims will complete the proof:

Claim 1
For each $y \in Y$ , there exists an $\alpha<\omega_1$ such that either $W_{H,y} \subset \alpha+1$ or $W_{K,y} \subset \alpha+1$ where

$W_{H,y}=\left\{\delta<\omega_1: (\delta,y) \in H \right\}$

$W_{K,y}=\left\{\delta<\omega_1: (\delta,y) \in K \right\}$

Proof of Claim 1
Let $y \in Y$ . The set $V=\omega_1 \times \left\{y \right\}$ is a copy of $\omega_1$ . It is a known fact that in $\omega_1$ , there cannot be two disjoint closed and unbounded sets. Let $V_H=V \cap H$ and $V_K=V \cap K$ . If $V_H \ne \varnothing$ and $V_K \ne \varnothing$ , they cannot be both unbounded in $V$ . Thus the claim follows if both $V_H \ne \varnothing$ and $V_K \ne \varnothing$ . Now suppose only one of $V_H$ and $V_K$ is non-empty. If the one that is non-empty is bounded, then the claim follows. Suppose the one that is non-empty is unbounded, say $V_K$ . Then $W_{H,y}=\varnothing$ and the claim follows.

Claim 2
For each $y \in Y$ , there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that one and only one of the following holds:

$H \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (1)$

$K \cap (\omega_1 \times \overline{O_y}) \subset (\alpha+1) \times \overline{O_y} \ \ \ \ \ \ \ \ (2)$

Proof of Claim 2
Let $y \in Y$ . Let $\alpha<\omega_1$ be as in Claim 1. Assume that $W_{H,y} \subset \alpha+1$ . We want to show that there exists an open set $O_y \subset Y$ with $y \in O_y$ such that (1) holds. Suppose that for each open $O \subset Y$ with $y \in O$ , there is a $q \in \overline{O}$ and there exists $\delta_q>\alpha$ such that $(\delta_q,q) \in H$ . Let $S$ be the set of all such points $q$ . Then $y \in \overline{S}$ . Since $Y$ has countable tightness, there exists countable $T \subset S$ such that $y \in \overline{T}$ . Since $T$ is countable, choose $\gamma >\omega_1$ such that $\alpha<\delta_q<\gamma$ for all $q \in T$ . Note that $[\alpha,\gamma] \times \left\{y \right\}$ does not contain points of $H$ since $W_{H,y} \subset \alpha+1$ . For each $\theta \in [\alpha,\gamma]$ , the point $(\theta,y)$ has an open neighborhood that contains no point of $H$ . Since $[\alpha,\gamma] \times \left\{y \right\}$ is compact, finitely many of these neighborhoods cover $[\alpha,\gamma] \times \left\{y \right\}$ . Let these finitely many open neighborhoods be $M_i \times N_i$ where $i=1,\cdots,m$ . Let $N=\bigcap_{i=1}^m N_i$ . Then $y \in N$ and $N$ would contain a point of $T$ , say $q$ . Then $(\delta_q,q) \in M_i \times N_i$ for some $i$ , a contradiction. Note that $(\delta_q,q)$ is a point of $H$ . Thus there exists an open $O_y \subset Y$ with $y \in O_y$ such that (1) holds. This completes the proof of Claim 2.

Claim 3
For each $y \in Y$ , there exists an $\alpha<\omega_1$ and there exists an open set $O_y \subset Y$ with $y \in O_y$ such that there are disjoint open subsets $Q_H$ and $Q_K$ of $\omega_1 \times \overline{O_y}$ with $H \cap (\omega_1 \times \overline{O_y}) \subset Q_H$ and $K \cap (\omega_1 \times \overline{O_y}) \subset Q_K$ .

Proof of Claim 3
Let $y \in Y$ . Let $\alpha$ and $O_y$ be as in Claim 2. Assume (1) in the statement of Claim 2 holds. Note that $(\alpha+1) \times \overline{O_y}$ is a product of two compact spaces and is thus compact (and normal). Let $R_{H,y}$ and $R_{K,y}$ be disjoint open sets in $(\alpha+1) \times \overline{O_y}$ such that $H \cap (\alpha+1) \times \overline{O_y} \subset R_{H,y}$ and $K \cap (\alpha+1) \times \overline{O_y} \subset R_{K,y}$ . Note that $[\alpha+1,\omega_1) \times \overline{O_y}$ contains no points of $H$ . Then $Q_{H,y}=R_{H,y}$ and $Q_{K,y}=R_{K,y} \cup [\alpha+1,\omega_1) \times \overline{O_y}$ are the desired open sets. This completes the proof of Claim 3.

To make the rest of the proof easier to see, we prove the following claim , which is a general fact that is cleaner to work with. Claim 4 describes precisely (in a topological way) what is happening at this point in the proof.

Claim 4
Let $Z$ be a space. Let $C$ and $D$ be disjoint closed subsets of $Z$ . Suppose that $\left\{U_1,U_2,\cdots,U_m \right\}$ is a collection of open subsets of $Z$ covering $C \cup D$ such that for each $i=1,2,\cdots,m$ , only one of the following holds:

$C \cap \overline{U_i} \ne \varnothing \text{ and } D \cap \overline{U_i}=\varnothing$

$C \cap \overline{U_i} = \varnothing \text{ and } D \cap \overline{U_i} \ne \varnothing$

Then there exist disjoint open subsets of $Z$ separating $C$ and $D$ .

Proof of Claim 4
Let $U_C=\cup \left\{U_i: \overline{U_i} \cap C \ne \varnothing \right\}$ and $U_D=\cup \left\{U_i: \overline{U_i} \cap D \ne \varnothing \right\}$ . Note that $\overline{U_C}=\cup \left\{\overline{U_i}: \overline{U_i} \cap C \ne \varnothing \right\}$ . Likewise, $\overline{U_D}=\cup \left\{\overline{U_i}: \overline{U_i} \cap D \ne \varnothing \right\}$ . Let $V_C=U_C-\overline{U_D}$ and $V_D=U_D-\overline{U_C}$ . Then $V_C$ and $V_D$ are disjoint open sets. Furthermore, $C \subset V_C$ and $D \subset V_D$ . This completes the proof of Claim 4.

Now back to the proof of Theorem 1. For each $y \in Y$ , let $O_y$ , $Q_{H,y}$ and $Q_{K,y}$ be as in Claim 3. Since $Y$ is compact, there exists $\left\{y_1,y_2,\cdots,y_n \right\} \subset Y$ such that $\left\{O_{y_1},O_{y_2},\cdots,O_{y_n} \right\}$ is a cover of $Y$ . For each $i=1,\cdots,n$ , let $L_i=Q_{H,y_i} \cap (\omega_1 \times O_y)$ and $M_i=Q_{K,y_i} \cap (\omega_1 \times O_y)$ . Note that both $L_i$ and $M_i$ are open in $\omega_1 \times Y$ . To apply Claim 4, rearrange the open sets $L_i$ and $M_i$ and re-label them as $U_1,U_2,\cdots,U_m$ . By letting $Z=\omega_1 \times Y$ , $C=H$ and $D=K$ , the open sets $U_i$ satisfy Claim 4. Tracing the $U_i$ to $L_j$ or $M_j$ and then to $Q_{H,y_j}$ and $Q_{K,y_j}$ , it is clear that the two conditions in Claim 4 are satisfied:

$H \cap \overline{U_i} \ne \varnothing \text{ and } K \cap \overline{U_i}=\varnothing$

$H \cap \overline{U_i} = \varnothing \text{ and } K \cap \overline{U_i} \ne \varnothing$

Then by Claim 4, the disjoint closed sets $H$ and $K$ can be separated by two disjoint open subsets of $\omega_1 \times Y$ . $\blacksquare$

The theorem proved in [4] is essentially the statement that for any compact space $Y$ , the product $\kappa^+ \times Y$ is normal if and only $t(Y) \le \kappa$ . Here $\kappa^+$ is the first ordinal of the next cardinal that is greater than $\kappa$ .

____________________________________________________________________

Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Gruenhage, G., Nogura, T., Purisch, S., Normality of $X \times \omega_1$ , Topology and its Appl., 39, 263-275, 1991.
Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

____________________________________________________________________

Pseudonormal spaces

Posted on February 17, 2014 by Dan Ma

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 $K$ .

____________________________________________________________________

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.

____________________________________________________________________

Reference

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.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc, Amsterdam, New York, 1995.
Przymusinski, T. C., A Lindelof space $X$ such that $X \times X$ is normal but not paracompact, Fund. Math., 91, 161-165, 1973.
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.

____________________________________________________________________

The Michael Line and the Continuum Hypothesis

Posted on November 4, 2012 by Dan Ma

There exist a Lindelof space and a separable metric space such that their Cartesian product is not normal (discussed in the post “Bernstein Sets and the Michael Line”). The separable metric space is a Bernstein set, a subspace of the real line that is far from being a complete metric space. However, this example is constructed without using any additional set theory axiom beyond the Zermelo-Fraenkel axioms plus the axiom of choice (abbreviated ZFC). A natural question is whether there exists a Lindelof space and a complete metric space such that their product is not normal. In particular, does there exist a Lindelof space $L$ such that the product of $L$ with the space of all irrational numbers is not normal? As of the writing of this post, it is still unknown that such a Lindelof space can exist in just ZFC alone without applying additional set theory axiom. However, such a Lindelof space can be constructed from various additional axioms (e.g. continuum hypothesis or Martin’s axiom). In this post, we present an example of such construction using the continuum hypothesis (the statement that the cardinality of the real line is the same as the first uncountable cardinal $\aleph_1$ ).

Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the set of irrational numbers with the usual topology inherited from the real line. It is a classical result that the product $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). The Lindelof example we wish to discuss is an uncountable Lindelof subspace $L$ of $\mathbb{M}$ such that $L$ contains the set $\mathbb{Q}$ of rational numbers. The same proof that $\mathbb{M} \times \mathbb{P}$ is not normal will show that $L \times \mathbb{P}$ is not normal.

See the following posts for a basic discussion of the Michael line:

“Michael Line Basics”

“Finite and Countable Products of the Michael Line”

“Bernstein Sets and the Michael Line”

___________________________________________________________________________________

Luzin Sets

The Lindelof space $X$ we want to find is a subset of the real line that is called a Luzin set. Before defining Luzin sets, recall some definitions. Let $Y$ be a space. Let $A \subset Y$ . The set $A$ is said to be nowhere dense in $Y$ if for every non-empty open subset $U$ of $Y$ , there is a non-empty open subset $V$ of $Y$ such that $V \subset U$ and $V$ misses $A$ (equivalently, the closure of $A$ has no interior). The set $A$ is of first category in $Y$ if it is the union of countably many nowhere dense sets.

To define Luzin sets, we focus on the Euclidean space $\mathbb{R}$ . Let $A \subset \mathbb{R}$ . The set $A$ is said to be a Luzin set if for every set $W \subset \mathbb{R}$ that is of first category in the real line, $A \cap W$ is at most countable. The Russian mathematician Luzin in 1914 constructed such an uncountable Luzin set using continuum hypothesis (CH). A good reference for Luzin sets is [4]. We have the following theorem.

Theorem 1
Assume CH. There exists an uncountable Luzin set.

Proof of Theorem 1
There are continuum many closed nowhere dense subsets of the real line. Since we assume the continuum hypothesis, we can enumerate these sets in a sequence of length $\omega_1$ . Let $\left\{F_\alpha: \alpha < \omega_1 \right\}$ be the set of all closed nowhere dense sets in the real line. Choose a real number $x_0 \notin F_0$ to start. For each $\alpha$ with $0 < \alpha <\omega_1$ , choose a real number $x_\alpha$ not in the following set:

$\left\{x_\beta: \beta<\alpha \right\} \cup \bigcup \limits_{\beta<\alpha} F_\beta$

The above set is a countable union of closed nowhere dense sets of the real line. As a complete metric space, the real line cannot be of first category. In fact, according to the Baire category theorem, the complement of a set of first category (such as the one described above) is dense in the real line. So such an $x_\alpha$ can always be selected at each $\alpha<\omega_1$ . Then $X=\left\{x_\alpha: \alpha<\omega_1 \right\}$ is a Luzin set. $\blacksquare$

Now that we have a way of constructing an uncountable Luzin sets, the following observations provide some useful facts for our problem at hand.

Nowhere dense sets and sets of first category are "thin" sets. Any "thin" set can intersect with a Luzin set with only countably many points. Thus any "co-thin" set contains all but countably many points of a Luzin set. For example, let $A$ be an uncountable Luzin set. Then if $F$ is a closed nowhere dense set in the real line, then $\mathbb{R}-F$ contains all but countably many points of $A$ . Furthermore, if $F_1,F_2,F_3,\cdots,$ are closed nowhere dense subsets of the real line, then $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ contains all but countably many points of the Luzin set $A$ .

Note that the set $\mathbb{R}-F$ in the preceding paragraph is a dense open set. Thus the complement of a closed nowhere dense set is a dense open set. Note that the set $\mathbb{R}- \bigcup \limits_{i=1}^\infty F_i$ in the preceding paragraph is a dense $G_\delta$ -set. Thus the complement of the union of countably many closed nowhere dense sets is a dense $G_\delta$ -set. Thus the observation in the preceding paragraph gives the following proposition:

Proposition 2
Given an uncountable Luzin set $A$ and given a dense $G_\delta$ subset $H$ of the real line, $H$ contains all but countably many points of $A$ .

In fact, Proposition 2 not only hold in the real line, it also holds in any uncountable dense subset of the real line.

Proposition 3
Let $A$ be an uncountable Luzin set. Let $Y \subset \mathbb{R}$ be uncountable and dense in the real line such that $A \cap Y$ is uncountable. Given a dense $G_\delta$ subset $H$ of $Y$ , $H$ contains all but countably many points of $A \cap Y$ .

Proof of Proposition 3
We want to show that $Y-H$ can only contain countably many points of $A$ . Let $H=\bigcap \limits_{i=1}^\infty O_i$ where each $O_i$ is open and dense in $Y$ . Then for each $i$ , let $U_i$ be open in the real line such that $U_i \cap Y=O_i$ . Each $U_i$ is open and dense in the real line. Thus $H^*=\bigcap \limits_{i=1}^\infty U_i$ contains all but countably many points of the Luzin set $A$ . Note the following set inclusion:

$H=\bigcap \limits_{i=1}^\infty U_i \cap Y=\bigcap \limits_{i=1}^\infty O_i \subset \bigcap \limits_{i=1}^\infty U_i=H^*$

Suppose that $Y-H$ contains uncountably many points of $A$ . Then these points, except for countably many points, must belong to $H^*=\bigcap \limits_{i=1}^\infty U_i$ . The above set inclusion shows that these points must belong to $H$ too, a contradiction. Thus $Y-H$ can only contain countably many points of $A$ , equivalently the $G_\delta$ -set $H$ contains all but countably many points of $A \cap Y$ . $\blacksquare$

The following proposition follows from Proposition 3 and is a useful fact that will help us see that the product of an uncountable Luzin set and $\mathbb{P}$ is not normal.

Proposition 4
Let $Y$ be an uncountable Luzin set such that $\mathbb{Q} \subset Y$ . Then $Y-\mathbb{Q}$ cannot be an $F_\sigma$ -set in the Euclidean space $Y$ , equivalently $\mathbb{Q}$ cannot be a $G_\delta$ -set in the space $Y$ .

Proof of Proposition 4
By Proposition 3, any dense $G_\delta$ -subset of $Y$ must be co-countable. $\blacksquare$

The following proposition is another useful observation about Luzin sets. Let $A \subset \mathbb{R}$ . Let $D \subset \mathbb{R}$ be a countable dense subset of the real line. The set $A$ is said to be concentrated about $D$ if for every open subset $O$ of the real line such that $D \subset O$ , $O$ contains all but countably many points of $A$ . The following proposition can be readily checked based on the definition of Luzin sets.

Proposition 5
For any $A \subset \mathbb{R}$ , $A$ is a Luzin set if and only if $A$ is concentrated about every countable dense subset of the real line.

___________________________________________________________________________________

Lindelof Subspace of The Michael Line

Let $A$ be an uncountable Luzin set. We can assume that $A$ is dense in the real line. If not, just add a countble subset of $\mathbb{P}$ that is dense in the real line. Let $L=A \cup \mathbb{Q}$ . It is clear that adding countably many points to a Luzin set still results in a Luzin set. Thus $L$ is also a Luzin set. Now consider $L$ as a subspace of the Michael line $\mathbb{M}$ . Then points of $L-\mathbb{Q}$ are discrete and points in $\mathbb{Q}$ have Euclidean open neighborhoods. By Proposition 5, the set $L$ is concentrated about every countable dense subset of the real line. In particular, it is concentrated about $\mathbb{Q}$ . Thus as a subspace of the Michael line, $L$ is a Lindelof space, since every open set containing $\mathbb{Q}$ contains all but countably many points of $L$ .

___________________________________________________________________________________

The Non-Normal Product $L \times \mathbb{P}$

We highlight the following two facts about the Luzin set $L=A \cup \mathbb{Q}$ as discussed in the preceding section.

$L-\mathbb{Q}$ is not an $F_\sigma$ -set in $L$ (as Euclidean space).
$A=L-\mathbb{Q}$ is dense in the real line.

The first bullet point follows from Proposition 4. The second bullet point is clear since we assume the Luzin set $A$ we start with is dense. Recall that when thinking of $L$ as a subspace of the Michael line, $L-\mathbb{Q}$ are isolated and $\mathbb{Q}$ retains the usual real line open sets. Because of the above two bullet points, $L \times \mathbb{P}$ is not normal. The proof that $L \times \mathbb{P}$ is not normal is the corollary of the proof that $\mathbb{M} \times \mathbb{P}$ is not normal. Note that in the proof for showing $\mathbb{M} \times \mathbb{P}$ is not normal, the two crucial points about the proof are that the isolated points of the Michael line cannot be an $F_\sigma$ -set and are dense in the real line (found in “Michael Line Basics”).

___________________________________________________________________________________

Michael Space

The example $L \times \mathbb{P}$ that we construct here was hinted in footnote 4 in [6]. In a later publication, E. Michael constructed an uncountable Lindelof subspace of the Michael line (see Lemma 3.1 in [5]). That construction should produce a similar set as the Luzin sets since the approach in [5] is a mirror image of the Luzin set construction. The approach in the Luzin set construction in Theorem 1 is to pick points not in the union of countably many closed nowhere dense sets, while the approach in [5] was to pick points in dense $G_\delta$ -sets in a transfinite induction process.

A Michael space is a Lindelof space whose product with $\mathbb{P}$ is not normal. The example shown here shows that under CH, there exists a Michael space. However, the question of whether there exists a Michael space in ZFC is still unsolved. This is called the Michael problem. A recent mention of this unsolved problem is [3] (page 160). A Michael space can also be constructed using Martin’s axiom (see [1]).

A space is said to be a productively Lindelof space if its product with every Lindelof space is Lindelof. Is $\mathbb{P}$ a productively Lindelof space? As we see here, under CH the answer is no. Another way of looking at the Michael problem: is it possible to show that $\mathbb{P}$ is not productively Lindelof in ZFC alone?

___________________________________________________________________________________

Reference

Alster, K., The product of a Lindelof space with the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., 110 (1990) 543-547.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
Michael, E., The product of a normal space and a metric space need not be normal, Bull. Amer. Math. Soc., 69 (1963) 375-376.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

___________________________________________________________________________________

Dan Ma's Topology Blog

Expository Articles In General Topology

Category Archives: First uncountable ordinal

Revisiting example 106 from Steen and Seebach

Counterexample 106 from Steen and Seebach

Normality in the powers of countably compact spaces

Comparing two function spaces

Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

Cp(omega 1 + 1) is not normal

Normal x compact needs not be subnormal

The normality of the product of the first uncountable ordinal with a compact factor

Pseudonormal spaces

The Michael Line and the Continuum Hypothesis