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.

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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

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.

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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 $C_p(\omega_1+1)$ .

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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.

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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.

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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.

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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.

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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.

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

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 $C_p(X)$ .

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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 $\Sigma(\omega_1)$ .

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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.

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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.

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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.

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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.

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Reference

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

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

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 $K$ . $\blacksquare$

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Reference

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

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

Normal dense subspaces of products of “omega 1” many separable metric factors

Posted on April 19, 2014 by Dan Ma

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii (see Problem I.5.25 in [2]). One partial positive answer is a theorem attributed to Corson: if $Y$ is a normal dense subspace of a product of separable spaces such that $Y \times Y$ is normal, then $Y$ is collectionwise normal. Another partial positive answer: assuming $2^\omega<2^{\omega_1}$ , any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). Another partial positive answer to Arkhangelskii’s question is the theorem due to Reznichenko: If $C_p(X)$ , which is a dense subspace of the product space $\mathbb{R}^X$ , is normal, then it is collectionwise normal (see Theorem I.5.12 in [2]). In this post, we highlight another partial positive answer to the question posted in [2]. Specifically, we prove the following theorem:

Theorem 1

$X=\prod_{\alpha<\omega_1} X_\alpha$

$X_\alpha$

$Y$

$X$

$Y$

Since any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post), it suffices to prove the following theorem:

Theorem 1a

$X=\prod_{\alpha<\omega_1} X_\alpha$

$X_\alpha$

$Y$

$X$

$Y$

Arkhangelskii’s question was studied by the author of [3] and [4]. Theorem 1 as presented in this post is essentially the Theorem 1 found in [3]. The proof given in [3] is a beautiful proof. The proof in this post is modeled on the proof in [3] with the exception that all the crucial details are filled in. Theorem 1a (as stated above) is used in [1] to show that the function space $C_p(\omega_1+1)$ contains no dense normal subspace.

It is natural to wonder if Theorem 1 can be generalized to product space of $\tau$ many separable metric factors where $\tau$ is an arbitrary uncountable cardinal. The work of [4] shows that the question at the beginning of this post cannot be answered positively in ZFC. Recall the above mentioned result that assuming $2^\omega<2^{\omega_1}$ , any normal dense subspace of the product space of continuum many separable metric factors is collectionwise normal (see Corollary 4 in this previous post). A theorem in [4] implies that assuming $2^\omega=2^{\omega_1}$ , for any separable metric space $M$ with at least 2 points, the product of continuum many copies of $M$ contains a normal dense subspace $Y$ that is not collectionwise normal. A side note: for this normal subspace $Y$ , $Y \times Y$ is necessarily not normal (according to Corson’s theorem). Thus [3] and [4] collectively show that Arkhangelskii’s question stated here at the beginning of the post is answered positively (in ZFC) among product spaces of $\omega_1$ many separable metric factors and that outside of the $\omega_1$ case, it is impossible to answer the question positively in ZFC.

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Proving Theorem 1a

We use the following lemma. For a proof of this lemma, see the proof for Lemma 1 in this previous post.

Lemma 2

$X=\prod_{\alpha \in A} X_\alpha$

$Y$

$X$

$Y$ is normal.
For any pair of disjoint closed subsets $H$ and $K$ of $Y$ , there exists a countable $B \subset A$ such that $\overline{\pi_B(H)} \cap \overline{\pi_B(K)}=\varnothing$ .
For any pair of disjoint closed subsets $H$ and $K$ of $Y$ , there exists a countable $B \subset A$ such that $\pi_B(H)$ and $\pi_B(K)$ are separated in $\pi_B(Y)$ , meaning that $\overline{\pi_B(H)} \cap \pi_B(K)=\pi_B(H) \cap \overline{\pi_B(K)}=\varnothing$ .

For any $B \subset \omega_1$ , let $\pi_B$ be the natural projection from the product space $X=\prod_{\alpha<\omega_1} X_\alpha$ into the subproduct space $\prod_{\alpha \in B} X_\alpha$ .

Proof of Theorem 1a
Let $Y$ be a dense subspace of the product space $X=\prod_{\alpha<\omega_1} X_\alpha$ where each factor $X_\alpha$ has a countable base. Suppose that $D$ is an uncountable closed and discrete subset of $Y$ . We then construct a pair of disjoint closed subsets $H$ and $K$ of $Y$ such that for all countable $B \subset \omega_1$ , $\pi_B(H)$ and $\pi_B(K)$ are not separated, specifically $\pi_B(H) \cap \overline{\pi_B(K)}\ne \varnothing$ . Here the closure is taken in the space $\pi_B(Y)$ . By Lemma 2, the dense subspace $Y$ of $X$ is not normal.

For each $\alpha<\omega_1$ , let $\mathcal{B}_\alpha$ be a countable base for the space $X_\alpha$ . The standard basic open sets in the product space $X$ are of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that

each $O_\alpha$ is an open subset of $X_\alpha$ ,
if $O_\alpha \ne X_\alpha$ , then $O_\alpha \in \mathcal{B}_\alpha$ ,
$O_\alpha=X_\alpha$ for all but finitely many $\alpha<\omega_1$ .

We use $supp(O)$ to denote the finite set of $\alpha$ such that $O_\alpha \ne X_\alpha$ . Technically we should be working with standard basic open subsets of $Y$ , i.e., sets of the form $O \cap Y$ where $O$ is a standard basic open set as described above. Since $Y$ is dense in the product space, every standard open set contains points of $Y$ . Thus we can simply work with standard basic open sets in the product space as long as we are working with points of $Y$ in the construction.

Let $\mathcal{M}$ be the collection of all standard basic open sets as described above. Since there are only $\omega_1$ many factors in the product space, $\lvert \mathcal{M} \lvert=\omega_1$ . Recall that $D$ is an uncountable closed and discrete subset of $Y$ . Let $\mathcal{M}^*$ be the following:

$\mathcal{M}^*=\left\{U \in \mathcal{M}: U \cap D \text{ is uncountable } \right\}$

Claim 1. $\lvert \mathcal{M}^* \lvert=\omega_1$ .

First we show that $\mathcal{M}^* \ne \varnothing$ . Let $B \subset \omega_1$ be countable. Consider these two cases: Case 1. $\pi_B(D)$ is an uncountable subset of $\prod_{\alpha \in B} X_\alpha$ ; Case 2. $\pi_B(D)$ is countable.

Suppose Case 1 is true. Since $\prod_{\alpha \in B} X_\alpha$ is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point $y \in \pi_B(D)$ such that every open neighborhood of $y$ (open in $\prod_{\alpha \in B} X_\alpha$ ) contains uncountably many points of $\pi_B(D)$ . Thus every standard basic open set $U=\prod_{\alpha \in B} U_\alpha$ , with $y \in U$ , contains uncountably many points of $\pi_B(D)$ . Suppose Case 2 is true. There exists one point $y \in \pi_B(D)$ such that $y=\pi_B(t)$ for uncountably many $t \in D$ . Then in either case, every standard basic open set $V=\prod_{\alpha<\omega_1} V_\alpha$ , with $supp(V) \subset B$ and $y \in \pi_B(V)$ , contains uncountably many points of $D$ . Any one such $V$ is a member of $\mathcal{M}^*$ .

We can partition the index set $\omega_1$ into $\omega_1$ many disjoint countable sets $B$ . Then for each such $B$ , obtain a $V \in \mathcal{M}^*$ in either Case 1 or Case 2. Since $supp(V) \subset B$ , all such open sets $V$ are distinct. Thus Claim 1 is established.

Claim 2.
There exists an uncountable $H \subset D$ such that for each $U \in \mathcal{M}^*$ , $U \cap H \ne \varnothing$ and $U \cap (D-H) \ne \varnothing$ .

Enumerate $\mathcal{M}^*=\left\{U_\gamma: \gamma<\omega_1 \right\}$ . Choose $h_0,k_0 \in U_0 \cap D$ with $h_0 \ne k_0$ . Suppose that for all $\beta<\gamma$ , two points $h_\beta,k_\beta$ are chosen such that $h_\beta,k_\beta \in U_\beta \cap D$ , $h_\beta \ne k_\beta$ and such that $h_\beta \notin L_\beta$ and $k_\beta \notin L_\beta$ where $L_\beta=\left\{h_\rho: \rho<\beta \right\} \cup \left\{k_\rho: \rho<\beta \right\}$ . Then choose $h_\gamma,k_\gamma$ with $h_\gamma \ne k_\gamma$ such that $h_\gamma,k_\gamma \in U_\gamma \cap D$ and $h_\gamma \notin L_\gamma$ and $k_\gamma \notin L_\gamma$ where $L_\gamma=\left\{h_\rho: \rho<\gamma \right\} \cup \left\{k_\rho: \rho<\gamma \right\}$ .

Let $H=\left\{h_\gamma: \gamma<\omega_1 \right\}$ and let $K=D-H$ . Note that $K_0=\left\{k_\gamma: \gamma<\omega_1 \right\} \subset K$ . Based on the inductive process that is used to obtain $H$ and $K_0$ , it is clear that $H$ satisfies Claim 2.

Claim 3.
For each countable $B \subset \omega_1$ , the sets $\pi_B(H)$ and $\pi_B(K)$ are not separated in the space $\pi_B(Y)$ .

Let $B \subset \omega_1$ be countable. Consider the two cases: Case 1. $\pi_B(H)$ is uncountable; Case 2. $\pi_B(H)$ is countable. Suppose Case 1 is true. Since $\prod_{\alpha \in B} X_\alpha$ is a product of countably many separable metric spaces, it is hereditarily Lindelof. Then there exists a point $p \in \pi_B(H)$ such that every open neighborhood of $p$ (open in $\prod_{\alpha \in B} X_\alpha$ ) contains uncountably many points of $\pi_B(H)$ . Choose $h \in H$ such that $p=\pi_B(h)$ . Then the following statement holds:

For every basic open set $U=\prod_{\alpha<\omega_1} U_\alpha$ with $h \in U$ such that $supp(U) \subset B$ , the open set $U$ contains uncountably many points of $H$ .

Suppose Case 2 is true. There exists some $p \in \pi_B(H)$ such that $p=\pi_B(t)$ for uncountably many $t \in H$ . Choose $h \in H$ such that $p=\pi_B(h)$ . Then statement 1 also holds.

In either case, there exists $h \in H$ such that statement 1 holds. The open sets $U$ described in statement 1 are members of $\mathcal{M}^*$ . By Claim 2, the open sets described in statement 1 also contain points of $K$ . Since the open sets described in statement 1 have supports $\subset B$ , the following statement holds:

For every basic open set $V=\prod_{\alpha \in B} V_\alpha$ with $\pi_B(h) \in V$ , the open set $V$ contains points of $\pi_B(K)$ .

Statement 2 indicates that $\pi_B(h) \in \overline{\pi_B(K)}$ . Thus $\pi_B(h) \in \pi_B(H) \cap \overline{\pi_B(K)}$ . The closure here can be taken in either $\prod_{\alpha \in B} X_\alpha$ or $\pi_B(Y)$ (to apply Lemma 2, we only need the latter). Thus Claim 3 is established.

Claim 3 is the negation of condition 3 of Lemma 2. Therefore $Y$ is not normal. $\blacksquare$

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Remark

The proof of Theorem 1a, though a proof in ZFC only, clearly relies on the fact that the product space is a product of $\omega_1$ many factors. For example, in the inductive step in the proof of Claim 2, it is always possible to pick a pair of points not chosen previously. This is because the previously chosen points form a countable set and each open set in $\mathcal{M}^*$ contains $\omega_1$ many points of the closed and discrete set $D$ . With the “ $\omega$ versus $\omega_1$ ” situation, at each step, there are always points not previously chosen. When more than $\omega_1$ many factors are involved, there may be no such guarantee in the inductive process.

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Reference

Arkhangelskii, A. V., Normality and dense subspaces, Proc. Amer. Math. Soc., 130 (1), 283-291, 2001.
Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.
Baturov, D. P., On perfectly normal dense subspaces of products, Topology Appl., 154, 374-383, 2007.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

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.

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Some Non-Pseudonormal Examples

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

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

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

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

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

$O_x=V_x \cup D_x$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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An Example of Pseudonormal but not Normal

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

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

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

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

$V_H=O_H$

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

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

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Some Comments about Examples 1 – 3

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

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

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

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

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

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

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

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Reference

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.

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One way to find collectionwise normal spaces

Posted on December 5, 2012 by Dan Ma

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

$\text{ }$

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

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Collectionwise Normal Spaces

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

Definition

$X$

$\mathcal{A}$

$X$

$\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$

$X$

$A \subset U_A$

$A \in \mathcal{A}$

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

$\text{ }$

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

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

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When Normal implies Collectionwise Normal

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

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

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

We have the following propositions.

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

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

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

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

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Paracompact Spaces

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

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

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

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

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

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Reference

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

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Topological Spaces with Caliber Omega 1

Posted on November 17, 2012 by Dan Ma

Let $\omega_1$ be the first uncountable ordinal. A separable space is one that has a countable dense subset. Any separable space $X$ satisfies this property:

$\left\{U_\alpha: \alpha < \omega_1 \right\}$

$X$

$A \subset \omega_1$

$\bigcap \limits_{\alpha \in A} U_\alpha \ne \varnothing$

Any space that has this property is said to have caliber $\omega_1$ . Spaces that have caliber $\omega_1$ have the countable chain condition (abbreviated by CCC, which means that there is no uncountable pairwise disjoint collection of open subsets of the space). So we have the following implications:

$\text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}$

The chain condition “caliber $\omega_1$ ” is an interesting one. Some of the spaces that have the CCC actually have caliber $\omega_1$ . For example, separable spaces and products of separable spaces have caliber $\omega_1$ . Thus the product space $\left\{0,1 \right\}^{\mathcal{K}}$ (for any uncountable cardinal $\mathcal{K}$ ) not only has the countable chain condition. It has the stronger property of having caliber $\omega_1$ , which may make certain proof easier to do. In this post we also provide examples to show that none of the above implications is reversible.

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Delta System Lemma

In proving that product of separable spaces has caliber $\omega_1$ , the Delta system lemma is used. A collection $\mathcal{D}$ of sets is said to be a Delta-system (or $\Delta$ -system) if there is a set $D$ such that for every $A,B \in \mathcal{D}$ with $A \ne B$ , we have $A \cap B = D$ . When such set $D$ exists, it is called the root of the Delta-system $\mathcal{D}$ . The following is the statement of Delta-system lemma.

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

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

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Products of Spaces with Caliber $\omega_1$

Theorem 2 below shows that whenever “caliber $\omega_1$ ” is preserved by taking product with any finite number of factors, “caliber $\omega_1$ ” is preserved by taking the product of any number of factors. As a corollary, we have the result that the product of any number of separable spaces has caliber $\omega_1$ . The proof of Theorem 2 is similar to the one stating that whenever CCC is preserved by taking product with any finite number of factors, CCC is preserved by taking the product of any number of factors (see the previous post Product of Spaces with Countable Chain Condition).

Theorem 2
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of spaces such that $\prod \limits_{\alpha \in F} X_\alpha$ has caliber $\omega_1$ for every finite $F \subset T$ . Then $\prod \limits_{\alpha \in T} X_\alpha$ has caliber $\omega_1$ .

Proof
In proving the product space $\prod \limits_{\alpha \in T} X_\alpha$ having caliber $\omega_1$ , it suffices to work with basic open sets of the form $\prod \limits_{\alpha \in T} O_\alpha$ where $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in T$ . Let $\mathcal{U}=\left\{U_\beta: \beta < \omega_1 \right\}$ be an uncountable collection of such non-empty open sets. Our plan is to find an uncountable $W_0 \subset \omega_1$ such that $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ .

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For each $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ , let $F_\beta \subset T$ be the finite set such that $\alpha \in F_\beta$ if and only if $U_{\beta, \alpha} \ne X_\alpha$ . Consider $\mathcal{A}=\left\{F_\beta: \beta < \omega_1 \right\}$ . By the Delta-system lemma, there is an uncountable $W \subset \omega_1$ such that $\mathcal{F}=\left\{F_\beta: \beta \in W \right\}$ is a Delta-system. Let $F$ be the root of this Delta-system.

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Consider the case that the root of the Delta-system is empty. Then for any $F_{\beta_1} \in \mathcal{F}$ and $F_{\beta_2} \in \mathcal{F}$ where $F_{\beta_1} \ne F_{\beta_2}$ , we have $F_{\beta_1} \cap F_{\beta_2}=\varnothing$ . For each $\beta \in W$ , we have $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ and we choose $h_\beta$ in $\prod \limits_{\alpha \in F_\beta} U_{\beta,\alpha}$ . Then we can define an $h$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $h$ extends $h_\beta$ for all $\beta \in W$ . Then in this case let $W_0=W$ and we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ . So we move onto the case that $F \ne \varnothing$ .

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Now assume $F \ne \varnothing$ . Let $\mathcal{U}^*=\left\{U_\beta: \beta \in W \right\}$ . For each $U_\beta \in \mathcal{U}^*$ where $U_\beta=\prod \limits_{\alpha \in T} U_{\beta,\alpha}$ , let $p(U_\beta)$ be $\prod \limits_{\alpha \in F} U_{\beta,\alpha}$ (i.e. $p$ is a projection map). Let $\mathcal{U}^{**}=\left\{p(U_\beta): \beta \in W \right\}$ .

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Consider two cases. One is that $\mathcal{U}^{**}$ is countable. The second is that $\mathcal{U}^{**}$ is uncountable. Suppose $\mathcal{U}^{**}$ is countable. Then there is an uncountable $W_0 \subset W$ such that $p(U_\gamma)=p(U_\mu)$ for all $\gamma,\mu \in W_0$ . Then fix $\gamma \in W_0$ and choose $g^* \in p(U_\gamma)$ . Then $g^*$ is extendable to some $g$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $g \in U_\beta$ for all $\beta \in W_0$ . Thus we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ .

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Now assume $\mathcal{U}^{**}$ is uncountable. By assumption $\prod \limits_{\alpha \in F} X_\alpha$ has caliber $\omega_1$ . Then there is an uncountable $W_0 \subset W$ such that $\bigcap \limits_{\beta \in W_0} p(U_\beta) \ne \varnothing$ . Choose $g^* \in \bigcap \limits_{\beta \in W_0} p(U_\beta)$ . As in the previous case, $g^*$ is extendable to some $g$ in $\prod \limits_{\alpha \in T} X_\alpha$ such that $g \in U_\beta$ for all $\beta \in W_0$ . Thus we have $\bigcap \limits_{\beta \in W_0} U_\beta \ne \varnothing$ . $\blacksquare$

Corollary 3
Suppose that $\left\{X_\alpha: \alpha \in T \right\}$ is a family of separable spaces. Then $\prod \limits_{\alpha \in T} X_\alpha$ has caliber $\omega_1$ .

Proof
This follows directly from Theorem 2. Note that the product of finitely many separable is separable (hence has caliber $\omega_1$ ). $\blacksquare$

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Examples

We now show that the following implications are not reversible.

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$\text{separable} \Longrightarrow \text{caliber } \omega_1 \Longrightarrow \text{CCC}$

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To get a space with caliber $\omega_1$ that is not separable, consider the product of $\mathcal{K}$ many copies of $\left\{0,1 \right\}$ where $\mathcal{K}$ is a cardinal number greater than continuum. Since $\left\{0,1 \right\}^{\mathcal{K}}$ is a product of separable spaces, it has caliber $\omega_1$ according to Corollary 3. It is well known that the product of more than continuum many separable spaces is not separable (see Product of Separable Spaces).

To get a space with the CCC that does not have caliber $\omega_1$ . Consider the subspace $S$ of the product space $H=\left\{0,1 \right\}^{\omega_1}$ ( $\omega_1$ many copies of $\left\{0,1 \right\}$ ) where $S$ is the set of all $h \in H$ such that $h(\alpha) \ne 0$ for at most countably many $\alpha < \omega_1$ . Note that $H$ is a space with the CCC since it is a product of separable spaces. Furthermore $S$ is a dense subspace of $H$ . The property of CCC is hereditary with respect to dense subsets. Thus $S$ has the countable chain condition. Here's a discussion of the countable chain condition.

To see that $S$ does not have caliber $\omega_1$ , look at the collection of open sets $\left\{V_\alpha: \alpha < \omega_1 \right\}$ where each $V_\alpha=\left\{h \in S: h(\alpha)=1 \right\}$ . Note that each $h \in S$ belongs to at most countably many $V_\alpha$ . Thus for any uncountable $A \subset \omega_1$ , $\bigcap \limits_{\alpha \in A} V_\alpha = \varnothing$ .

The example $S$ shows that dense subspace of a space with caliber $\omega_1$ need not have caliber $\omega_1$ .

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Reference

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

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Dan Ma's Topology Blog

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