# Introducing Menger and Hurewicz spaces

This is an introduction of two classical properties in selection principles – Menger property and Hurewicz property. Both properties imply the Lindelof property and generalize $\sigma$-compactness.

A space $X$ is a Menger space (or has the Menger property) if for every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that $\bigcup_{n \in \omega} \mathcal{V}_n$ is an open cover of $X$.

A space $X$ is a Hurewicz space (or has the Hurewicz property) if for every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that $\{ \bigcup \mathcal{V}_n: n \in \omega \}$ is a $\gamma$-cover of $X$, i.e. each point of $X$ belongs to $\bigcup \mathcal{V}_n$ for all but finitely many $n$.

A space $X$ is a Rothberger space (or has the Rothberger property) if for every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, $V_n \in \mathcal{U}_n$ such that $\{V_n: n \in \omega \}$ is an open cover of $X$.

The Rothberger property, though not discussed here, is added for contrast. Based on the definitions, it is clear that all three properties imply that the space is Lindelof. By definition, it is straightforward to show that any $\sigma$-compact space is a Hurewicz space. In turn, the property of Menger follows from the property of Hurewicz. The definition of a Rothberger space is a special case of the definition of Menger spaces. The following diagram summarizes these implications.

Figure 1

\displaystyle \begin{aligned} & \sigma \text{-} \bold C \bold o \bold m \bold p \bold a \bold c \bold t \\&\ \ \ \ \ \downarrow \\& \bold H \bold u \bold r \bold e \bold w \bold i \bold c \bold z \\&\ \ \ \ \ \downarrow \\& \bold M \bold e \bold n \bold g \bold e \bold r \ \ \ \ \ \leftarrow \ \ \ \bold R \bold o \bold t \bold h \bold b \bold e \bold r \bold g \bold e \bold r \\&\ \ \ \ \ \downarrow \\& \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \end{aligned}

K. Menger [8] in 1924 defined a basis covering property in metric spaces. He conjectured that among metric spaces, this basis covering property is equivalent to $\sigma$-compactness. In 1925, Hurewicz [5] introduced a selection principle which led to the notion of Menger space as defined above. He showed that a metric space has Menger’s basis covering property if, and only if, it is a Menger space. However, Hurewicz did not settle Menger’s conjecture. Instead, Hurewicz formulated a related selection principle equivalent to the notion of Hurewicz space as defined above and conjectured that a non-compact metric space is $\sigma$-compact if, and only if, it is a Hurewicz space.

This is a brief introduction of the notions of Menger and Hurewicz spaces, focusing on basic facts. The goal is to give the reader a sense of what these spaces are like, especially among subsets of the real line. Another characterization of Menger spaces is discussed here. For subsets of the real line, another characterization of Hurewicz spaces is discussed here.

One comment about the term Menger space or Menger property. In the literature, the Menger space was at one time called the Hurewicz space. For example, the Hurewicz spaces discussed in [1], [7] and [15] are actually the Menger spaces as defined above. In this article, we use the modern terminology of Menger spaces (a notion based on a selection principle of Hurewicz that was proven to be equivalent to Menger’s basis covering property). Then the Hurewicz spaces discussed here is the notion derived from another selection principle of Hurewicz.

The articles [6] and [12] are very in-depth coverage of Menger, Hurewicz and Rothberger spaces where these spaces are discussed in the context of selection principles. The articles [13] and [14] are excellent survey articles for selection principles and covering properties.

The symbols $\mathbb{R}$, $\mathbb{P}$ and $\omega$ denote the real line, the set of all irrational numbers and the first infinite ordinal, respectively. The set $\omega$ is regarded as the set of all non-negative integers, i.e. $\omega=\{0,1,2,3,\cdots \}$. The set $\omega^\omega$ is the set of all functions $f:\omega \rightarrow \omega$. The set $\omega^\omega$ is endowed with the product topology. The set $\mathbb{P}$ has the subspace topology inherited from the real line. We use $\mathbb{P}$ and $\omega^\omega$ interchangeably since they are topologically equivalent (see here).

The first basic fact is that all these three properties are preserved by continuous maps and by taking closed subspaces.

Theorem 1
Suppose that $X$ has any one of the properties: Menger, Hurewicz or Rothberger. Then the following holds.

1. Any continuous image of $X$ has the same property.
2. Any closed subset of $X$ has the same property.

Equivalent Formulations

To make some of the results easier to do, we use equivalent formulations of Menger and Rothberger spaces.

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

1. The space $X$ is a Menger space.
2. For every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that for each $x \in X$, $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$.

Theorem 3
Let $X$ be a space. The following are equivalent.

1. The space $X$ is a Rothberger space.
2. For every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, $V_n \in \mathcal{U}_n$ such that for each $x \in X$, $x \in V_n$ for infinitely many $n$.

Dominating Sets and Unbounded Sets

To characterize the Menger property and the Hurewicz property, it is useful to use dominating subsets and unbounded subsets of $\omega^\omega$, the space of irrationals. Define the order $\le^*$ on $\omega^\omega$ as follows. For $f, g \in \omega^\omega$, we say $f \le^* g$ if $f(n) \le g(n)$ for all but finitely many $n$. The negation of $f \le^* g$ is denoted by $f \not \le^* g$. It follows that $f \not \le^* g$ if $g(n) < f(n)$ for infinitely many $n$. The order $\le^*$ is a reflexive and transitive relation.

Let $H$ be a subset of $\omega^\omega$. We say $H$ is a bounded set if $H$ has an upper bound with respect to $\le^*$, i.e. there exists $f \in \omega^\omega$ such that for each $g \in H$, we have $g \le^* f$. The set $H$ is an unbounded set if it is not a bounded set. To spell it out, $H$ is an unbounded set if for each $f \in \omega^\omega$, there exists $g \in H$ such that $g \not \le^* f$, i.e. $f(n) < g(n)$ for infinitely many $n$.

Let $D$ be a subset of $\omega^\omega$. We say $D$ is a dominating set if for each $f \in \omega^\omega$, there exists $g \in D$ such that $f \le^* g$. The set $D$ is not a dominating set would mean: there exists $f \in \omega^\omega$ such that for each $g \in D$, we have $f \not \le^* g$, i.e. $g(n) for infinitely many $n$.

More about dominating sets and unbounded sets in a later section.

Menger Basics

The first result is on the subsets of the space of irrationals that are Menger.

Theorem 4
Let $X \subset \omega^\omega$. Then if $X$ is a Menger space, then $X$ is not a dominating set.

Theorem 5
Let $X$ be a space. Then if $X$ is a Menger space, then every continuous image of $X$ in $\omega^\omega$ is not a dominating set.

Theorem 6
Let $X$ be a Lindelof and zero-dimensional space. Then $X$ is a Menger space if, and only if, every continuous image of $X$ in $\omega^\omega$ is not a dominating set.

Non-Menger Examples

We are in a position to look at some examples. In particular, we look at examples that are Lindelof but not Menger.

Example 1
Any dominating subset of $\omega^\omega$ is not a Menger space. This follows from Theorem 4. In particular the space of the irrational numbers $\mathbb{P}$ is not a Menger space since it is clearly a dominating set.

Example 2
A non-Menger subset of the Cantor space $2^\omega$.

The Cantor space $2^\omega$ is compact and is thus a Menger space. Furthermore it is a bounded subset of $\omega^\omega$. Thus it is not a dominating set. Any subset of $2^\omega$ is also not a dominating set. This example shows that the converse of Theorem 4 is not true.

Let $F: 2^\omega \rightarrow [0,1]$ be a continuous surjection. For example, $F(x)=\sum_{n=0}^\infty \frac{x(n)}{2^{n+1}}$. Let $Y=\{x \in 2^\omega: F(x) \in \mathbb{P} \}$. The map $F$ restricted to $Y$ is still a continuous map. It maps $Y$ onto the set of irrational numbers in $(0,1)$, which is homeomorphic to $\mathbb{P}$. By Theorem 5, $Y$ is not Menger. This example shows that “not dominating” cannot be a characterization of the Menger subsets of $\mathbb{P} \cong \omega^\omega$.

Example 3
The Sorgenfrey line $S$ is not a Menger space. There is a continuous map from $S$ onto $\omega^\omega$. See [11] for more information about Menger subsets of the Sorgenfrey line.

To see that it is not Menger, we define a sequence of open covers of $S$ that witnesses the non-Menger property. For any interval $(a,b)$ in the real line, we fix a scheme to obtain an increasing sequence $t_0=a,t_1,t_2,\cdots$ of real numbers such that $t_n \rightarrow b$ from the right. Let $t_0=a$, $t_1$ is the midpoint of $t_0$ and $b$, $t_2$ is the midpoint of $t_1$ and $b$, and so on.

Let $\mathcal{U}_0=\{U_0^n: n \in \omega \}$ where $U_0^0=[0,2)$, $U_0^2=[2,4)$, $U_0^4=[4,6)$ and so on, and that $U_0^1=[-2,0)$, $U_0^3=[-4,-2)$, $U_0^5=[-6,-4)$ and so on. In other words, the even indexed intervals are on the right of the origin and the odd indexed intervals are on the left of the origin. They are all half closed and half open intervals of length 2.

To define $\mathcal{U}_1$, take each interval $[a,b)$ in $\mathcal{U}_0$, and use the scheme stated above to obtain the intervals $[t_0,t_1)$, $[t_1,t_2)$, $[t_2,t_3)$ and so on. Then $\mathcal{U}_1$ consists of all these intervals for each $[a,b) \in \mathcal{U}_0$. The intervals in $\mathcal{U}_2$ are obtained in the same way using intervals in $\mathcal{U}_1$ (for each $[a,b) \in \mathcal{U}_1$, generate the intervals using the above scheme). Thus the sequence of open covers $\{ \mathcal{U}_n: n \in \omega \}$ is defined in this recursive fashion.

With the sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers defined, choose for each $n$ any finite $\mathcal{V}_n \subset \mathcal{U}_n$. We show that $\cup \{ \mathcal{V}_n: n \in \omega \}$ is not an open cover. Since $\mathcal{V}_0$ is finite, choose one $[a_0, b_0) \in \mathcal{U}_0$ such that $[a_0, b_0) \notin \mathcal{V}_0$. There are infinitely many intervals in $\mathcal{U}_1$ that are subsets of $[a_0, b_0)$. Choose one $[a_1,b_1) \in \mathcal{U}_1$ such that $[a_1,b_1) \subset [a_0,b_0)$ and $[a_1,b_1) \notin \mathcal{V}_1$. Continuing the inductive process, we obtain a sequence of intervals $[a_0,b_0) \supset [a_1,b_1) \supset [a_2,b_2) \supset \cdots [a_n,b_n) \supset \cdots$ with a single point $p$ in the intersection. Note that the lengths of the intervals go to zero and that $[a_{n+1},b_{n+1}] \subset [a_n,b_b)$ for each $n$. Then the point $p$ is not in any set in any $\mathcal{V}_n$.

The proof of Theorem 6 shows that given a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of a space that witnesses the non-Menger property of that space, we can define a continuous map from that space into a dominating subset of $\omega^\omega$. Applying that procedure to the sequence of open covers in this example, we would get a continuous map from the Sorgenfrey line onto $\omega^\omega$.

Hurewicz Basics

The three results in this section are parallel to Theorems 4 to 6 in the above Menger section. The proofs are similar but using different definitions and properties.

Theorem 7
Let $X \subset \omega^\omega$. Then if $X$ is a Hurewicz space, then $X$ is a bounded set.

Theorem 8
Let $X$ be a space. Then if $X$ is a Hurewicz space, then every continuous image of $X$ in $\omega^\omega$ is a bounded set.

Theorem 9
Let $X$ be a Lindelof and zero-dimensional space. Then $X$ is a Hurewicz space if, and only if, every continuous image of $X$ in $\omega^\omega$ is a bounded set.

Dominating Number and Bounding Number

We now continue the discussion on dominating sets and unbounded sets. First, one observation. If $H$ is a bounded set, then $H$ is not a dominating set. In other words, if $H$ is a dominating set, then $H$ is an unbounded set.

We now define two cardinals using the order $\le^*$ as follows:

$\mathfrak{b}=\text{min} \{ \ \lvert H \lvert \ : H \subset \omega^\omega \text{ and H is an unbounded set} \}$

$\mathfrak{d}=\text{min} \{ \ \lvert H \lvert \ : H \subset \omega^\omega \text{ and H is a dominating set} \}$

The cardinal $\mathfrak{b}$ is called the bounding number. It is the least cardinality of an unbounded subset of $\omega^\omega$. The cardinal $\mathfrak{d}$ is called the dominating number. It is the least cardinality of a dominating subset of $\omega^\omega$.

Based on the observation made at the beginning of this section, $\mathfrak{b} \le \lvert H \lvert$ for every dominating set $H$. Thus $\mathfrak{b} \le \mathfrak{d}$. Both of these cardinals are upper bounded by $\mathfrak{c}$, the cardinality of the continuum. Both of these cardinalys must be uncountable. This is because any countable subset of $\omega^\omega$ cannot be unbounded (using a diagonal argument). We always have: $\omega <\mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}$ or $\omega_1 \le \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}$.

The values of $\mathfrak{b}$ and $\mathfrak{d}$ are quite sensitive to set theoretic assumptions. For example, if continuum hypothesis holds, $\omega_1=\mathfrak{b}=\mathfrak{d}=\mathfrak{c}$. On the other hand, it is consistent that $\omega_1 \le \mathfrak{b} < \mathfrak{d} \le \mathfrak{c}$. We will see below that the non-property of Menger spaces is characterized by $\mathfrak{d}$ and the non-property of Hurewicz spaces is characterized by $\mathfrak{b}$. This fact shows that the notions discussed here are also set-theoretically sensitive. See [17] for more information on the bounding number $\mathfrak{b}$, the dominating number $\mathfrak{d}$ and other cardinal characteristics of the continuum.

More on Menger and Hurewicz

The dominating number $\mathfrak{d}$ and the bounding number $\mathfrak{b}$ are “cutoff points” for the Menger property and Hurewicz property, respectively. Any space with cardinality below the cutoff point have the respective property. Equivalently, any space that does not have the given property must be at or above the respective cutoff point. The goal of the theorems in this section is to make the cutoff points precise.

Theorem 10
Let $X$ be a Lindelof space. If the cardinality of $X$ is less than $\mathfrak{d}$, then $X$ is Menger space.

Theorem 11
Let $X$ be a Lindelof space. If the cardinality of $X$ is less than $\mathfrak{b}$, then $X$ is Hurewicz space.

To make the cutoff points precise, we define a cardinal of non-property. For any property $\mathcal{P}$, the cardinal $\text{non}(\mathcal{P})$ is defined as follows:

$\text{non}(\mathcal{P})=\text{min} \{ \ \lvert X \lvert \ : X \subset \mathbb{R} \text{ and } \mathcal{P} \text{ does not hold} \}$

Thus $\text{non}(\mathcal{P})$ is the least cardinality of a subset of the real line that does not have the property $\mathcal{P}$. This means that any set with cardinality less than $\text{non}(\mathcal{P})$ has property $\mathcal{P}$ while any set that does not have property $\mathcal{P}$ must have cardinality $\text{non}(\mathcal{P})$ or higher. We are interested in knowing more about the numbers $\text{non}(\mathcal{P})$ when $\mathcal{P}$ is the Menger property or the Hurewicz property. We denote these two cardinals by $\text{non}(\text{Menger})$ and $\text{non}(\text{Hurewicz})$.

Theorem 10 and Theorem 11 work in Lindelof spaces. Now we shift the focus to subsets of the real line. Theorem 10 can be restated: if $X$ does not have the Menger property, then $\mathfrak{d} \le \lvert X \lvert$. Theorem 11 can be restated: if $X$ does not have the Hurewicz property, then $\mathfrak{b} \le \lvert X \lvert$. The next two theorems show that $\mathfrak{d}$ and $\mathfrak{b}$ are precisely $\text{non}(\text{Menger})$ and $\text{non}(\text{Hurewicz})$, respectively.

Theorem 12
$\mathfrak{d}=\text{non}(\text{Menger})$

Theorem 13
$\mathfrak{b}=\text{non}(\text{Hurewicz})$

Theorem 12 indicates that the dominating number $\mathfrak{d}$ is the least cardinality of a subset of the real line that is not a Menger space. Theorem 13 indicates that the bounding number $\mathfrak{b}$ is the least cardinality of a subset of the real line that is not a Hurewicz space.

The Conjecture of Menger

In the section above on non-Menger examples, we give three examples of Lindelof spaces that are not Menger. Examples that are even more interesting would be spaces that have the Menger property but are not $\sigma$-compact. K. Menger conjectured that such examples do not exist. Counterexamples to Menger’s conjecture do exist.

Example 4
One example is the so called Lusin sets. A Lusin set (also called Luzin set) is a subset $X$ of the real line such that the intersection of $X$ and any meager (i.e. first category) subset is countable, i.e. if $F$ is a first category subset of the real line, then $X \cap F$ is countable. Any such set is a Menger space. In fact, any such set is a Rothberger space. Using CH, a Lusin set can be constructed. The existence of Lusin sets is independent of ZFC. Under MA and not CH, Lusin sets cannot exist. This example is further discussed in the next post.

Example 5
Miller and Fremlin [9] gave the first counterexample to Menger’s conjecture without using any axioms beyond ZFC. However, their proof is not constructive. It is instead a dichotomic argument. It looked at two cases – when a certain set-theoretic statement is true and when it is not true. In either case, there is a subset of the real line that is a Menger space and not $\sigma$-compact. This example is further discussed in the next post.

Example 6
Bartoszyński and Tsaban [3] gave a counterexample to Menger’s conjecture in ZFC and without using any dichotomic argument.

The Proof Section

This begins the section of proofs of the theorems stated above.

Theorem 1
Suppose that $X$ has any one of the properties: Menger, Hurewicz or Rothberger. Then the following holds.

1. Any continuous image of $X$ has the same property.
2. Any closed subset of $X$ has the same property.

Theorem 1 is a set of three theorems, one for each property: Menger, Hurewicz and Rothberger. The proofs are straightforward. We show the proof for the Menger proeprty.

Proof of Theorem 1
Let $X$ be a Menger space. Let $f: X \rightarrow Y$ be a continuous map such that $Y=f(X)$. We show that $Y$ is a Menger space. To this end, let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $Y$. For each $n$, let $\mathcal{W}_n=\{ f^{-1}(U): U \in \mathcal{U}_n \}$. Then $\{ \mathcal{W}_n: n \in \omega \}$ is a sequence of open covers of $X$. Since $X$ is Menger, we can choose, for each $n$, a finite $\mathcal{E}_n \subset \mathcal{W}_n$ such that $\bigcup_{n \in \omega} \mathcal{E}_n$ is an open cover of $X$. Then for each $n$, let $\mathcal{V}_n=\{ U \in \mathcal{U}_n: f^{-1}(U) \in \mathcal{E}_n \}$. Each $\mathcal{V}_n$ is finite. It is clear that $\bigcup_{n \in \omega} \mathcal{V}_n$ is an open cover of $Y$. This completes the proof that every continuous image of a Menger space is a Menger space.

Let $X$ be a Menger space. Let $Z$ be a closed subset of $X$. We show that $Z$ is a Menger space. To this end, let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $Z$. For each $n$, let $\mathcal{W}_n=\mathcal{U}_n^* \cup \{ X \backslash Z \}$, where each $A \in \mathcal{U}_n^*$ is an open subset of $X$ such that $A \cap Z \in \mathcal{U}_n$. Then $\{ \mathcal{W}_n: n \in \omega \}$ is a sequence of open covers of $X$. Since $X$ is Menger, there exists, for each $n$, a finite $\mathcal{D}_n \subset \mathcal{W}_n$ such that $\bigcup_{n \in \omega} \mathcal{D}_n$ is an open cover of $X$. In particular, $\bigcup_{n \in \omega} \mathcal{D}_n$ is an open cover of $Z$. The open sets in $\mathcal{D}_n$ that covers $Z$ are not $X \backslash Z$. For each $n$, let $\mathcal{V}_n=\{ A \cap Z: A \in \mathcal{D}_n \}$. Then $\bigcup_{n \in \omega} \mathcal{V}_n$ is an open cover of $Z$. $\square$

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

1. The space $X$ is a Menger space.
2. For every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that for each $x \in X$, $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$.

Proof of Theorem 2
$2 \rightarrow 1$ is clear. If for each $x \in X$, $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$, then $\bigcup_{n \in \omega} \mathcal{V}_n$ must be an open cover.

$1 \rightarrow 2$
Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. Break up $\omega$ in infinitely many infinite subsets. Say, $\omega=\bigcup_{n \in \omega} A_n$ where each $A_n$ is infinite and $A_i \cap A_j=\varnothing$ for any $i. For each $n$, let $\mathcal{O}_n=\{ \mathcal{U}_j: j \in A_n \}$. Each $\mathcal{O}_n$ is a sequence of open covers of $X$. By condition 1, for each $n$, we can do the following: for each $j \in A_n$, we can choose finite $\mathcal{V}_j \subset \mathcal{U}_j$ such that $\bigcup_{j \in A_n} \mathcal{V}_j$ is an open cover of $X$. Then for each $x \in X$, and for each $n$, $x$ belongs to some element of $\mathcal{V}_{k_n}$ where $k_n \in A_n$. Thus we can say that for each $x \in X$, $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$. $\square$

Theorem 3
Let $X$ be a space. The following are equivalent.

1. The space $X$ is a Rothberger space.
2. For every sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$, there exists, for each $n$, $V_n \in \mathcal{U}_n$ such that for each $x \in X$, $x \in V_n$ for infinitely many $n$.

Proof of Theorem 3
$2 \rightarrow 1$ is clear.

$1 \rightarrow 2$
Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. Break up $\omega$ in infinitely many infinite subsets. Say, $\omega=\bigcup_{n \in \omega} A_n$ where each $A_n$ is infinite and $A_i \cap A_j=\varnothing$ for any $i. For each $n$, let $\mathcal{O}_n=\{ \mathcal{U}_j: j \in A_n \}$. Each $\mathcal{O}_n$ is a sequence of open covers of $X$. By condition 1, for each $n$, we can do the following: for each $j \in A_n$, we can choose $V_j \in \mathcal{U}_j$ such that $\{ V_j: j \in A_n \}$ is an open cover of $X$. Then for each $x \in X$, and for each $n$, $x$ belongs to $V_{k_n}$ where $k_n \in A_n$. Thus we can say that for each $x \in X$, $x \in V_n$ for infinitely many $n$. $\square$

Theorem 4
Let $X \subset \omega^\omega$. Then if $X$ is a Menger space, then $X$ is not a dominating set.

Proof of Theorem 4
Let $X \subset \omega^\omega$. Suppose that $X$ is a dominating set. For each $n$, and for each $x \in X$, let $U_n^x=\{ h \in X: h(j)=x(j) \ \forall \ j \le n \}$. For each $n$, $\mathcal{U}_n=\{ U_n^x: x \in X \}$ is an open cover of $X$. Then $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. For each $n$, choose any finite $\mathcal{V}_n \subset \mathcal{U}_n$. Define the function $f \in \omega^\omega$ as follows:

$f(n)=\text{max} \{x(n): x \in U_n^x \text{ where } U_n^x \in \mathcal{V}_n \}+1$

Since $X$ is a dominating set, for the $f$ just defined, there exists $g \in X$ such that $f \le^* g$, i.e. $f(n) \le g(n)$ for all but finitely many $n$. This means that $g \notin U_n^x$ for each $U_n^x \in \mathcal{V}_n$ for all but finitely many $n$. Thus $g \notin \bigcup \mathcal{V}_n$ for all but finitely many $n$.

From a dominating set $X$, we can derive a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$ that witnesses the non-Menger property of $X$ (according to condition 2 in Theorem 2). It follows that if $X$ has the Menger property, then $X$ is not a dominating set. $\square$

Theorem 5
Let $X$ be a space. Then if $X$ is a Menger space, then every continuous image of $X$ in $\omega^\omega$ is not a dominating set.

Proof of Theorem 5
Let $X$ be a Menger space. Let $F:X \rightarrow \omega^\omega$ be a continuous map. Let $D=F(X)$. By Theorem 1, $D$ is a Menger space. By Theorem 4, $D$ is not a dominating set. $\square$

Theorem 6
Let $X$ be a Lindelof and zero-dimensional space. Then $X$ is a Menger space if, and only if, every continuous image of $X$ in $\omega^\omega$ is not a dominating set.

Proof of Theorem 6
The direction $\longrightarrow$ is Theorem 5. The assumption of Lindelof and zero-dimensional is not needed.

$\longleftarrow$
Let $X$ be a Lindelof and zero-dimensional space. Suppose $X$ is not a Menger space. There is a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$ that witnesses the non-Menger property of $X$ (according to condition 2 in Theorem 2). Since $X$ is Lindelof, we can assume each $\mathcal{U}_n$ is countable. For each $n$, arrange $\mathcal{U}_n$ as $\mathcal{U}_n=\{U_n^0, U_n^1, U_n^2, \cdots \}$. Since $X$ is zero-dimensional, we can assume each $U_n^j$ is both closed and open in $X$.

For each $x \in X$, define $f_x \in \omega^\omega$ as follows:

$f_x(n)=\text{min} \{j: x \in U_n^j \}$

Thus $f_x(n)$ is the least integer $j$ such that $x \in U_n^j$. Let $D=\{ f_x: x \in X \}$. We claim that $D$ is a dominating set. Suppose $D$ is not dominating. Then there exists $f \in \omega^\omega$ such that for each $f_x \in D$, $f_x(n) < f(n)$ for infinitely many $n$. For each $n$, let $\mathcal{V}_n=\{ U_n^j: j \le f(n) \}$. This means that for each $x \in X$, we have $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$. This contradicts the fact that the sequence $\{ \mathcal{U}_n: n \in \omega \}$ is to witness the non-Menger property of $X$. Thus $D$ is a dominating set.

Consider the map $F: X \rightarrow D$ by $F(x)=f_x$ for all $x \in X$. Clearly this is a surjection. We show that this is a continuous map. Let $x \in X$. Let $O$ be open in $D$ such that $f_x \in O$. Assume that $O=\{ f \in D: f(j)=f_x(j) \ \forall \ j \le n \}$ for some integer $n$. For each $j \le n$, let $G_j=U_j^{f_x(j)} \backslash \bigcup_{i. Note that each $G_j$ is open and $x \in G_j$. Let $W_x=\bigcap_{j \le n} G_j$. The set $W_x$ is open in $X$ and $x \in W_x$. We show that $F(W_x) \subset O$. To this end, let $y \in W_x$. Note that $y \in U_j^{f_x(j)}$ and $y \notin \bigcup_{i. Thus $f_y(j)=f_x(j)$ for all $j \le n$. This means $f_y \in O$. Thus $F(W_x) \subset O$. This shows that $F$ is a continuous map. Assuming that $X$ is not Menger, we can build a continuous map that maps $X$ onto a dominating set. This concludes the proof of the direction $\longleftarrow$. $\square$

Theorem 7
Let $X \subset \omega^\omega$. Then if $X$ is a Hurewicz space, then $X$ is a bounded set.

Proof of Theorem 7
Let $X \subset \omega^\omega$. Suppose that $X$ is an unbounded set. For each $n$, and for each $x \in X$, let $U_n^x=\{ h \in X: h(j)=x(j) \ \forall \ j \le n \}$. For each $n$, $\mathcal{U}_n=\{ U_n^x: x \in X \}$ is an open cover of $X$. Then $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. For each $n$, choose any finite $\mathcal{V}_n \subset \mathcal{U}_n$. Define the function $f \in \omega^\omega$ as follows:

$f(n)=\text{max} \{x(n): x \in U_n^x \text{ where } U_n^x \in \mathcal{V}_n \}+1$

Since $X$ is an unbounded set, for the $f$ just defined, there exists $g \in X$ such that $g \not \le^* f$, i.e. $f(n) < g(n)$ for infinitely many $n$. This means that $g \notin U_n^x$ for each $U_n^x \in \mathcal{V}_n$ for infinitely many $n$. Thus $g \notin \bigcup \mathcal{V}_n$ for infinitely many $n$.

From an unbounded set $X$, we can derive a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$ that witnesses the non-Hurewicz property of $X$. It follows that if $X$ has the Hurewicz property, then $X$ is a bounded set. $\square$

Theorem 8
Let $X$ be a space. Then if $X$ is a Hurewicz space, then every continuous image of $X$ in $\omega^\omega$ is a bounded set.

Proof of Theorem 8
Let $X$ be a Hurewicz space. Let $F:X \rightarrow \omega^\omega$ be a continuous map. Let $D=F(X)$. By Theorem 1, $D$ is a Hurewicz space. By Theorem 7, $D$ is a bounded set. $\square$

Theorem 9
Let $X$ be a Lindelof and zero-dimensional space. Then $X$ is a Hurewicz space if, and only if, every continuous image of $X$ in $\omega^\omega$ is a bounded set.

Proof of Theorem 9
The direction $\longrightarrow$ is Theorem 8. The assumption of Lindelof and zero-dimensional is not needed.

$\longleftarrow$
Let $X$ be a Lindelof and zero-dimensional space. Suppose $X$ is not a Hurewicz space. There is a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $X$ that witnesses the non-Hurewicz property of $X$. Since $X$ is Lindelof, we can assume each $\mathcal{U}_n$ is countable. For each $n$, arrange $\mathcal{U}_n$ as $\mathcal{U}_n=\{U_n^0, U_n^1, U_n^2, \cdots \}$. Since $X$ is zero-dimensional, we can assume each $U_n^j$ is both closed and open in $X$.

For each $x \in X$, define $f_x \in \omega^\omega$ as follows:

$f_x(n)=\text{min} \{j: x \in U_n^j \}$

Thus $f_x(n)$ is the least integer $j$ such that $x \in U_n^j$. Let $D=\{ f_x: x \in X \}$. We claim that $D$ is an unbounded set. Suppose $D$ is bounded. Then there exists $f \in \omega^\omega$ such that for each $f_x \in D$, $f_x \le^* f$, i.e. $f_x(n) \le f(n)$ for all but finitely many $n$. For each $n$, let $\mathcal{V}_n=\{ U_n^j: j \le f(n) \}$. This means that for each $x \in X$, we have $x \in \bigcup \mathcal{V}_n$ for all but finitely many $n$. This contradicts the fact that the sequence $\{ \mathcal{U}_n: n \in \omega \}$ is to witness the non-Hurewicz property of $X$. Thus $D$ is an unbounded set.

Consider the map $F: X \rightarrow D$ by $F(x)=f_x$ for all $x \in X$. Clearly this is a surjection. The function $F$ is also a continuous map. The proof is identical to the one in the proof of Theorem 6. Assuming that $X$ is not Hurewicz, we can build a continuous map that maps $X$ onto an unbounded set. This concludes the proof of the direction $\longleftarrow$. $\square$

Theorem 10
Let $X$ be a Lindelof space. If the cardinality of $X$ is less than $\mathfrak{d}$, then $X$ is Menger space.

Proof of Theorem 10
Suppose that the cardinality of $X$ is less than $\mathfrak{d}$. Suppose that $X$ is not a Menger space. Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$ such that for for each $n$ and for any finite $\mathcal{V}_n \subset \mathcal{U}_n$, there exists some $x \in X$ such that $x \notin \cup \mathcal{V}_n$ for all but finitely many $n$. Since $X$ is Lindelof, assume that for each $n$, $\mathcal{U}_n=\{ U_n^j: j \in \omega \}$. For each $x \in X$, define $f_x \in \omega^\omega$ by: $f_x(n)=\text{min} \{i: x \in U_n^i \}$ for each $n \in \omega$. Let $D=\{ f_x: x \in X \}$. Based on the proof in Theorem 6, the set $D$ is a dominating set. Based on the definition of the dominating number $\mathfrak{d}$, we have $\mathfrak{d} \le \lvert D \lvert$. Here we have a situation where a set $X$ with cardinality less than $\mathfrak{d}$ is mapped onto a set $D$ with $\mathfrak{d} \le \lvert D \lvert$. This is a contradiction. Thus $X$ must be a Menger space. $\square$

Theorem 11
Let $X$ be a Lindelof space. If the cardinality of $X$ is less than $\mathfrak{b}$, then $X$ is Hurewicz space.

Proof of Theorem 11
Suppose that the cardinality of $X$ is less than $\mathfrak{b}$. Suppose that $X$ is not a Hurewicz space. Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$ such that for for each $n$ and for any finite $\mathcal{V}_n \subset \mathcal{U}_n$, there exists some $x \in X$ such that $x \notin \cup \mathcal{V}_n$ for infinitely many $n$. Since $X$ is Lindelof, assume that for each $n$, $\mathcal{U}_n=\{ U_n^j: j \in \omega \}$. For each $x \in X$, define $f_x \in \omega^\omega$ by: $f_x(n)=\text{min} \{i: x \in U_n^i \}$ for each $n \in \omega$. Let $D=\{ f_x: x \in X \}$. Based on the proof in Theorem 9, the set $D$ is an unbounded set. Based on the definition of the bounding number $\mathfrak{b}$, we have $\mathfrak{b} \le \lvert D \lvert$. Here we have a situation where a set $X$ with cardinality less than $\mathfrak{b}$ is mapped onto a set $D$ with $\mathfrak{b} \le \lvert D \lvert$. This is a contradiction. Thus $X$ must be a Hurewicz space. $\square$

Theorem 12
$\mathfrak{d}=\text{non}(\text{Menger})$

Proof of Theorem 12
As a result of Theorem 10 (if $X$ is a subset of the real line and $\lvert X \lvert < \mathfrak{d}$, then $X$ is Menger), it follows that $\mathfrak{d} \le \text{non}(\text{Menger})$. We claim that $\mathfrak{d} \ge \text{non}(\text{Menger})$. Suppose that $\mathfrak{d} < \text{non}(\text{Menger})$. Choose $X \subset \omega^\omega$ such that $\lvert X \lvert=\mathfrak{d}$ and $X$ is a dominating set. Since $\lvert X \lvert<\text{non}(\text{Menger})$, $X$ is Menger. On the other hand, since $X$ is dominating, $X$ is not Menger (Theorem 4). Thus we have $\mathfrak{d} \ge \text{non}(\text{Menger})$, leading to the conclusion $\mathfrak{d}=\text{non}(\text{Menger})$. $\square$

Theorem 13
$\mathfrak{b}=\text{non}(\text{Hurewicz})$

Proof of Theorem 13
As a result of Theorem 11 (if $X$ is a subset of the real line and $\lvert X \lvert < \mathfrak{b}$, then $X$ is Hurewicz), it follows that $\mathfrak{b} \le \text{non}(\text{Hurewicz})$. We claim that $\mathfrak{b} \ge \text{non}(\text{Hurewicz})$. Suppose that $\mathfrak{b} < \text{non}(\text{Hurewicz})$. Choose $X \subset \omega^\omega$ such that $\lvert X \lvert=\mathfrak{b}$ and $X$ is an unbounded set. Since $\lvert X \lvert<\text{non}(\text{Hurewicz})$, $X$ is Hurewicz. On the other hand, since $X$ is unbounded, $X$ is not Hurewicz (Theorem 7). Thus we have $\mathfrak{b} \ge \text{non}(\text{Hurewicz})$, leading to the conclusion $\mathfrak{b}=\text{non}(\text{Hurewicz})$. $\square$

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Aurichi L., D-spaces, topological games, and selection principles, Topology Proc., 36, 107-122, 2010.
3. Bartoszyński T., Tsaban B., Hereditary topological diagonalizations and the Menger–Hurewicz Conjectures, Proc. Amer. Math. Soc., 134 (2), 605-615, 2005.
4. Blass, A., Combinatorial Cardinal Characteristics of the Continuum, Handbook of Set Theory (M. Foreman, A. Kanamori, eds), Springer Science+Business Media B. V., Netherlands, 395-489, 2010.
5. Hurewicz W., Uber die Verallgemeinerung des Borelschen Theorems, Mathematische
Zeitschrift, 24, 401-425, 1925.
6. Just W., Miller A. W., Scheepers M., Szeptycki P. J.,Combinatorics of
open covers (II)
, Topology Appl., 73, 241–266, 1996.
7. Lelek A., Some cover properties of spaces, Fund. Math., 64, 209-218, 1969.
8. Menger K., Einige Uberdeckungssatze der Punltmengenlehre, Sitzungsberichte Abt. 2a, Mathematik, Astronomie, Physik, Meteorologie und Mechanik (Wiener Akademie, Wien), 133, 421-444, 1924.
9. Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 17-33, 1988.
10. Reclaw I., Every Luzin set is undetermined in point-open game, Fund. Math., 144, 43-43, 1994.
11. Sakai M., Menger Subsets of the Sorgenfrey Line, Proc. Amer. Math. Soc., 137 (9), 3129-3138, 2009.
12. Scheepers M., Combinatorics of open covers I: Ramsey theory, Topology Appl., 69, 31-62, 1996.
13. Scheepers M., Selection principles and covering properties in topology, Note di Matematica, 22 (2), 3-41, 2003.
14. Scheepers M., Selection principles in topology: New directions, Filomat, 15, 111-126, 2001.
15. Wingers L., Box Products and Hurewicz Spaces, Topology Appl., 64, 9-21, 1995.
16. Tsaban B., Zdomskyy L.Scales, fields, and a problem of Hurewicz, J. Eur. Math. Soc. 10, 837–866, 2008.
17. Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 111-167, 1984.

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# Lindelof Exercise 2

The preceding post is an exercise showing that the product of countably many $\sigma$-compact spaces is a Lindelof space. The result is an example of a situation where the Lindelof property is countably productive if each factor is a “nice” Lindelof space. In this case, “nice” means $\sigma$-compact. This post gives several exercises surrounding the notion of $\sigma$-compactness.

Exercise 2.A

According to the preceding exercise, the product of countably many $\sigma$-compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many $\sigma$-compact spaces. More specifically, give an example of a product of uncountably many $\sigma$-compact spaces such that the product space is not Lindelof.

Exercise 2.B

Any $\sigma$-compact space is Lindelof. Since $\mathbb{R}=\bigcup_{n=1}^\infty [-n,n]$, the real line with the usual Euclidean topology is $\sigma$-compact. This exercise is to find an example of “Lindelof does not imply $\sigma$-compact.” Find one such example among the subspaces of the real line. Note that as a subspace of the real line, the example would be a separable metric space, hence would be a Lindelof space.

Exercise 2.C

This exercise is also to look for an example of a space that is Lindelof and not $\sigma$-compact. The example sought is a non-metric one, preferably a space whose underlying set is the real line and whose topology is finer than the Euclidean topology.

Exercise 2.D

Show that the product of two Lindelof spaces is a Lindelof space whenever one of the factors is a $\sigma$-compact space.

Exercise 2.E

Prove that the product of finitely many $\sigma$-compact spaces is a $\sigma$-compact space. Give an example of a space showing that the product of countably and infinitely many $\sigma$-compact spaces does not have to be $\sigma$-compact. For example, show that $\mathbb{R}^\omega$, the product of countably many copies of the real line, is not $\sigma$-compact.

The Lindelof property and $\sigma$-compactness are basic topological notions. The above exercises are natural questions based on these two basic notions. One immediate purpose of these exercises is that they provide further interaction with the two basic notions. More importantly, working on these exercise give exposure to mathematics that is seemingly unrelated to the two basic notions. For example, finding $\sigma$-compactness on subspaces of the real line and subspaces of compact spaces naturally uses a Baire category argument, which is a deep and rich topic that finds uses in multiple areas of mathematics. For this reason, these exercises present excellent learning opportunities not only in topology but also in other useful mathematical topics.

If preferred, the exercises can be attacked head on. The exercises are also intended to be a guided tour. Hints are also provided below. Two sets of hints are given – Hints (blue dividers) and Further Hints (maroon dividers). The proofs of certain key facts are also given (orange dividers). Concluding remarks are given at the end of the post.

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Hints for Exercise 2.A

Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if $X$ is a Lindelof space, then every closed subspace of $X$ is also Lindelof.

Prove that if $X$ is a Lindelof space, then every closed and discrete subset of $X$ is countable (every space that has this property is said to have countable extent).

Show that the product of uncountably many copies of the real line does not have countable extent. Specifically, focus on either one of the following two examples.

• Show that the product space $\mathbb{R}^c$ has a closed and discrete subspace of cardinality continuum where $c$ is cardinality of continuum. Hence $\mathbb{R}^c$ is not Lindelof.
• Show that the product space $\mathbb{R}^{\omega_1}$ has a closed and discrete subspace of cardinality $\omega_1$ where $\omega_1$ is the first uncountable ordinal. Hence $\mathbb{R}^{\omega_1}$ is not Lindelof.

Hints for Exercise 2.B

Let $\mathbb{P}$ be the set of all irrational numbers. Show that $\mathbb{P}$ as a subspace of the real line is not $\sigma$-compact.

Hints for Exercise 2.C

Let $S$ be the real line with the topology generated by the half open and half closed intervals of the form $[a,b)=\{ x \in \mathbb{R}: a \le x < b \}$. The real line with this topology is called the Sorgenfrey line. Show that $S$ is Lindelof and is not $\sigma$-compact.

Hints for Exercise 2.D

It is helpful to first prove: the product of two Lindelof space is Lindelof if one of the factors is a compact space. The Tube lemma is helpful.

Tube Lemma
Let $X$ be a space. Let $Y$ be a compact space. Suppose that $U$ is an open subset of $X \times Y$ and suppose that $\{ x \} \times Y \subset U$ where $x \in X$. Then there exists an open subset $V$ of $X$ such that $\{ x \} \times Y \subset V \times Y \subset U$.

Hints for Exercise 2.E

Since the real line $\mathbb{R}$ is homeomorphic to the open interval $(0,1)$, $\mathbb{R}^\omega$ is homeomorphic to $(0,1)^\omega$. Show that $(0,1)^\omega$ is not $\sigma$-compact.

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Further Hints for Exercise 2.A

The hints here focus on the example $\mathbb{R}^c$.

Let $I=[0,1]$. Let $\omega$ be the first infinite ordinal. For convenience, consider $\omega$ the set $\{ 0,1,2,3,\cdots \}$, the set of all non-negative integers. Since $\omega^I$ is a closed subset of $\mathbb{R}^I$, any closed and discrete subset of $\omega^I$ is a closed and discrete subset of $\mathbb{R}^I$. The task at hand is to find a closed and discrete subset of $Y=\omega^I$. To this end, we define $W=\{W_x: x \in I \}$ after setting up background information.

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

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

Note. For each $t \in I-\{0,1 \}$, the open intervals $O_{t,j}$ are of the form $(a,b)$. For $t=0$, the open intervals $O_{t,j}$ are of the form $[0,b)$. For $t=1$, the open intervals $O_{t,j}$ are of the form $(a,1]$.

For each $t \in I$, define the map $f_t: I \rightarrow \omega$ as follows:

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

We are now ready to define $W=\{W_x: x \in I \}$. For each $x \in I$, $W_x$ is the mapping $W_x:I \rightarrow \omega$ defined by $W_x(t)=f_t(x)$ for each $t \in I$.

Show the following:

• The set $W=\{W_x: x \in I \}$ has cardinality continuum.
• The set $W$ is a discrete space.
• The set $W$ is a closed subspace of $Y$.

Further Hints for Exercise 2.B

A subset $A$ of the real line $\mathbb{R}$ is nowhere dense in $\mathbb{R}$ if for any nonempty open subset $U$ of $\mathbb{R}$, there is a nonempty open subset $V$ of $U$ such that $V \cap A=\varnothing$. If we replace open sets by open intervals, we have the same notion.

Show that the real line $\mathbb{R}$ with the usual Euclidean topology cannot be the union of countably many closed and nowhere dense sets.

Further Hints for Exercise 2.C

Prove that if $X$ and $Y$ are $\sigma$-compact, then the product $X \times Y$ is $\sigma$-compact, hence Lindelof.

Prove that $S$, the Sorgenfrey line, is Lindelof while its square $S \times S$ is not Lindelof.

Further Hints for Exercise 2.D

As suggested in the hints given earlier, prove that $X \times Y$ is Lindelof if $X$ is Lindelof and $Y$ is compact. As suggested, the Tube lemma is a useful tool.

Further Hints for Exercise 2.E

The product space $(0,1)^\omega$ is a subspace of the product space $[0,1]^\omega$. Since $[0,1]^\omega$ is compact, we can fall back on a Baire category theorem argument to show why $(0,1)^\omega$ cannot be $\sigma$-compact. To this end, we consider the notion of Baire space. A space $X$ is said to be a Baire space if for each countable family $\{ U_1,U_2,U_3,\cdots \}$ of open and dense subsets of $X$, the intersection $\bigcap_{i=1}^\infty U_i$ is a dense subset of $X$. Prove the following results.

Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that $\overline{O_{n+1}} \subset O_n$ for each $n$. Then the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Fact E.2
Any compact Hausdorff space is Baire space.

Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Then $Y$ is not a $\sigma$-compact space.

Since $X=[0,1]^\omega$ is compact, it follows from Fact E.2 that the product space $X=[0,1]^\omega$ is a Baire space.

Fact E.4
Let $X=[0,1]^\omega$ and $Y=(0,1)^\omega$. The product space $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X=[0,1]^\omega$. Furthermore, $X-Y$ is a dense subset of $X$.

It follows from the above facts that the product space $(0,1)^\omega$ cannot be a $\sigma$-compact space.

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Proofs of Key Steps for Exercise 2.A

The proof here focuses on the example $\mathbb{R}^c$.

To see that $W=\{W_x: x \in I \}$ has the same cardinality as that of $I$, show that $W_x \ne W_y$ for $x \ne y$. This follows from the definition of the mapping $W_x$.

To see that $W$ is discrete, for each $x \in I$, consider the open set $U_x=\{ b \in Y: b(x)=0 \}$. Note that $W_x \in U_x$. Further note that $W_y \notin U_x$ for all $y \ne x$.

To see that $W$ is a closed subset of $Y$, let $k: I \rightarrow \omega$ such that $k \notin W$. Consider two cases.

Case 1. $k(r) \ne 0$ for all $r \in I$.
Note that $\{ O_{t,k(t)}: t \in I \}$ is an open cover of $I$ (in the usual topology). There exists a finite $H \subset I$ such that $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. Consider the open set $G=\{ b \in Y: \forall \ h \in H, \ b(h)=k(h) \}$. Define the set $F$ as follows:

$F=\{ c \in I: W_c \in G \}$

The set $F$ can be further described as follows:

\displaystyle \begin{aligned} F&=\{ c \in I: W_c \in G \} \\&=\{ c \in I: \forall \ h \in H, \ W_c(h)=f_h(c)=k(h) \ne 0 \} \\&=\{ c \in I: \forall \ h \in H, \ c \in I-O_{h,k(h)} \} \\&=\bigcap_{h \in H} (I-O_{h,k(h)}) \\&=I-\bigcup_{h \in H} O_{h,k(h)}=I-I =\varnothing \end{aligned}

The last step is $\varnothing$ because $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. The fact that $F=\varnothing$ means that $G$ is an open subset of $Y$ containing the point $k$ such that $G$ contains no point of $W$.

Case 2. $k(r) = 0$ for some $r \in I$.
Since $k \notin W$, $k \ne W_x$ for all $x \in I$. In particular, $k \ne W_r$. This means that $k(t) \ne W_r(t)$ for some $t \in I$. Define the open set $G$ as follows:

$G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}$

Clearly $k \in G$. Observe that $W_r \notin G$ since $W_r(t) \ne k(t)$. For each $p \in I-\{ r \}$, $W_p \notin G$ since $W_p(r) \ne 0$. Thus $G$ is an open set containing $k$ such that $G \cap W=\varnothing$.

Both cases show that $W$ is a closed subset of $Y=\omega^I$.

Proofs of Key Steps for Exercise 2.B

Suppose that $\mathbb{P}$, the set of all irrational numbers, is $\sigma$-compact. That is, $\mathbb{P}=A_1 \cup A_2 \cup A_3 \cup \cdots$ where each $A_i$ is a compact space as a subspace of $\mathbb{P}$. Any compact subspace of $\mathbb{P}$ is also a compact subspace of $\mathbb{R}$. As a result, each $A_i$ is a closed subset of $\mathbb{R}$. Furthermore, prove the following:

Each $A_i$ is a nowhere dense subset of $\mathbb{R}$.

Each singleton set $\{ r \}$ where $r$ is any rational number is also a closed and nowhere dense subset of $\mathbb{R}$. This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus $\mathbb{P}$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.C

The Sorgenfrey line $S$ is a Lindelof space whose square $S \times S$ is not normal. This is a famous example of a Lindelof space whose square is not Lindelof (not even normal). For reference, a proof is found here. An alternative proof of the non-normality of $S \times S$ uses the Baire category theorem and is found here.

If the Sorgenfrey line is $\sigma$-compact, then $S \times S$ would be $\sigma$-compact and hence Lindelof. Thus $S$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.D

Suppose that $X$ is Lindelof and that $Y$ is compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\{ x \} \times Y$. Putting it another way, $\{ x \} \times Y \subset \cup \mathcal{U}_x$. By the Tube lemma, for each $x \in X$, there is an open $O_x$ such that $\{ x \} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is Lindelof, there exists a countable set $\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $\{ O_{x_1},O_{x_2},O_{x_3},\cdots \}$ is a cover of $X$. Then $\mathcal{U}_{x_1} \cup \mathcal{U}_{x_2} \cup \mathcal{U}_{x_3} \cup \cdots$ is a countable subcover of $\mathcal{U}$. This completes the proof that $X \times Y$ is Lindelof when $X$ is Lindelof and $Y$ is compact.

To complete the exercise, observe that if $X$ is Lindelof and $Y$ is $\sigma$-compact, then $X \times Y$ is the union of countably many Lindelof subspaces.

Proofs of Key Steps for Exercise 2.E

Proof of Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that \$latex $\overline{O_{n+1}} \subset O_n$ for each $n$. Show that the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Suppose that $\bigcap_{i=1}^\infty O_i=\varnothing$. Choose $x_1 \in O_1$. There must exist some $n_1$ such that $x_1 \notin O_{n_1}$. Choose $x_2 \in O_{n_1}$. There must exist some $n_2>n_1$ such that $x_2 \notin O_{n_2}$. Continue in this manner we can choose inductively an infinite set $A=\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $x_i \ne x_j$ for $i \ne j$. Since $X$ is compact, the infinite set $A$ has a limit point $p$. This means that every open set containing $p$ contains some $x_j$ (in fact for infinitely many $j$). The point $p$ cannot be in the intersection $\bigcap_{i=1}^\infty O_i$. Thus for some $n$, $p \notin O_n$. Thus $p \notin \overline{O_{n+1}}$. We can choose an open set $U$ such that $p \in U$ and $U \cap \overline{O_{n+1}}=\varnothing$. However, $U$ must contain some point $x_j$ where $j>n+1$. This is a contradiction since $O_j \subset \overline{O_{n+1}}$ for all $j>n+1$. Thus Fact E.1 is established.

Proof of Fact E.2
Let $X$ be a compact space. Let $U_1,U_2,U_3,\cdots$ be open subsets of $X$ such that each $U_i$ is also a dense subset of $X$. Let $V$ a non-empty open subset of $X$. We wish to show that $V$ contains a point that belongs to each $U_i$. Since $U_1$ is dense in $X$, $O_1=V \cap U_1$ is non-empty. Since $U_2$ is dense in $X$, choose non-empty open $O_2$ such that $\overline{O_2} \subset O_1$ and $O_2 \subset U_2$. Since $U_3$ is dense in $X$, choose non-empty open $O_3$ such that $\overline{O_3} \subset O_2$ and $O_3 \subset U_3$. Continue inductively in this manner and we have a sequence of open sets $O_1,O_2,O_3,\cdots$ just like in Fact E.1. Then the intersection of the open sets $O_n$ is non-empty. Points in the intersection are in $V$ and in all the $U_n$. This completes the proof of Fact E.2.

Proof of Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Show that $Y$ is not a $\sigma$-compact space.

Suppose $Y$ is $\sigma$-compact. Let $Y=\bigcup_{n=1}^\infty B_n$ where each $B_n$ is compact. Each $B_n$ is obviously a closed subset of $X$. We claim that each $B_n$ is a closed nowhere dense subset of $X$. To see this, let $U$ be a non-empty open subset of $X$. Since $X-Y$ is dense in $X$, $U$ contains a point $p$ where $p \notin Y$. Since $p \notin B_n$, there exists a non-empty open $V \subset U$ such that $V \cap B_n=\varnothing$. This shows that each $B_n$ is a nowhere dense subset of $X$.

Since $Y$ is a dense $G_\delta$-subset of $X$, $Y=\bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open and dense subset of $X$. Then each $A_n=X-O_n$ is a closed nowhere dense subset of $X$. This means that $X$ is the union of countably many closed and nowhere dense subsets of $X$. More specifically, we have the following.

(1)………$X= \biggl( \bigcup_{n=1}^\infty A_n \biggr) \cup \biggl( \bigcup_{n=1}^\infty B_n \biggr)$

Statement (1) contradicts the fact that $X$ is a Baire space. Note that all $X-A_n$ and $X-B_n$ are open and dense subsets of $X$. Further note that the intersection of all these countably many open and dense subsets of $X$ is empty according to (1). Thus $Y$ cannot not a $\sigma$-compact space.

Proof of Fact E.4
The space $X=[0,1]^\omega$ is compact since it is a product of compact spaces. To see that $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X$, note that $Y=\bigcap_{n=1}^\infty U_n$ where for each integer $n \ge 1$

(2)………$U_n=(0,1) \times \cdots \times (0,1) \times [0,1] \times [0,1] \times \cdots$

Note that the first $n$ factors of $U_n$ are the open interval $(0,1)$ and the remaining factors are the closed interval $[0,1]$. It is also clear that $X-Y$ is a dense subset of $X$. This completes the proof of Fact E.4.

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Concluding Remarks

Exercise 2.A
The exercise is to show that the product of uncountably many $\sigma$-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that $\mathbb{R}^{c}$ has uncountable extent where $c$ is continuum. Having uncountable extent (i.e. having an uncountable subset that is both closed and discrete) implies the space is not Lindelof. The uncountable extent of the product space $\mathbb{R}^{\omega_1}$ is discussed in this post.

For $\mathbb{R}^{c}$ and $\mathbb{R}^{\omega_1}$, there is another way to show non-Lindelof. For example, both product spaces are not normal. As a result, both product spaces cannot be Lindelof. Note that every regular Lindelof space is normal. Both product spaces contain the product $\omega^{\omega_1}$ as a closed subspace. The non-normality of $\omega^{\omega_1}$ is discussed here.

Exercise 2.B
The hints given above is to show that the set of all irrational numbers, $\mathbb{P}$, is not $\sigma$-compact (as a subspace of the real line). The same argument showing that $\mathbb{P}$ is not $\sigma$-compact can be generalized. Note that the complement of $\mathbb{P}$ is $\mathbb{Q}$, the set of all rational numbers (a countable set). In this case, $\mathbb{Q}$ is a dense subset of the real line and is the union of countably many singleton sets. Each singleton set is a closed and nowhere dense subset of the real line. In general, we can let $B$, the complement of a set $A$, be dense in the real line and be the union of countably many closed nowhere dense subsets of the real line (not necessarily singleton sets). The same argument will show that $A$ cannot be a $\sigma$-compact space. This argument is captured in Fact E.3 in Exercise 2.E. Thus both Exercise 2.B and Exercise 2.E use a Baire category argument.

Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not $\sigma$-compact. In this case, the suggested space is $\mathbb{R}^{\omega}$, the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product space $\mathbb{R}^{\omega}$ is embedded in the compact space $[0,1]^{\omega}$, which is a Baire space. As mentioned earlier, Fact E.3 is essentially the same argument used for Exercise 2.B.

Using the same Baire category argument, it can be shown that $\omega^{\omega}$, the product of countably many copies of the countably infinite discrete space, is not $\sigma$-compact. The space $\omega$ of the non-negative integers, as a subspace of the real line, is certainly $\sigma$-compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not $\sigma$-compact. With the product space $\omega^{\omega}$, there is a connection with Exercise 2.B. The product $\omega^{\omega}$ is homeomorphic to $\mathbb{P}$. The idea of the homeomorphism is discussed here. Thus the non-$\sigma$-compactness of $\omega^{\omega}$ can be achieved by mapping it to the irrationals. Of course, the same Baire category argument runs through both exercises.

Exercise 2.C
Even the non-$\sigma$-compactness of the Sorgenfrey line $S$ can be achieved by a Baire category argument. The non-normality of the Sorgenfrey plane $S \times S$ can be achieved by Jones’ lemma argument or by the fact that $\mathbb{P}$ is not a first category set. Links to both arguments are given in the Proof section above.

See here for another introduction to the Baire category theorem.

The Tube lemma is discussed here.

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# Lindelof Exercise 1

A space $X$ is called a $\sigma$-compact space if it is the union of countably many compact subspaces. Clearly, any $\sigma$-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. The most well known example is perhaps the square of the Sorgenfrey line. In certain cases, the Lindelof property can be productive. For example, the product of countably many $\sigma$-compact spaces is a Lindelof space. The discussion here centers on the following theorem.

Theorem 1
Let $X_1,X_2,X_3,\cdots$ be $\sigma$-compact spaces. Then the product space $\prod_{i=1}^\infty X_i$ is Lindelof.

Theorem 1 is Exercise 3.8G in page 195 of General Topology by Engelking [1]. The reference for Exercise 3.8G is [2]. But the theorem is not found in [2] (it is not stated directly and it does not seem to be an obvious corollary of a theorem discussed in that paper). However, a hint is provided in Engelking for Exercise 3.8G. In this post, we discuss Theorem 1 as an exercise by giving expanded hint. Solutions to some of the key steps in the expanded hint are given at the end of the post.

Expanded Hint

It is helpful to first prove the following theorem.

Theorem 2
For each integer $i \ge 1$, let $C_{i,1},C_{i,2},\cdots$ be compact spaces and let $C_i$ be the topological sum:

$C_i=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots=\oplus_{j=1}^\infty C_{i,j}$

Then the product $\prod_{i=1}^\infty C_i$ is Lindelof.

Note that in the topological sum $C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots$, the spaces $C_{i,1},C_{i,2},C_{i,3},\cdots$ are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way to view this topology: each of the $C_{i,j}$ is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many $\sigma$-compact spaces is Lindelof if each $\sigma$-compact space is the union of countably many disjoint compact spaces. The hint for Exercise 3.8G can be applied much more naturally on Theorem 2 than on Theorem 1. The following is Exercise 3.8F (a), which is the hint for Exercise 3.8G.

Lemma 3
Let $Z$ be a compact space. Let $X$ be a subspace of $Z$. Suppose that there exist $F_1,F_2,F_3,\cdots$, closed subsets of $Z$, such that for all $x$ and $y$ where $x \in X$ and $y \in Z-X$, there exists $F_i$ such that $x \in F_i$ and $y \notin F_i$. Then $X$ is a Lindelof space.

The following theorem connects the hint (Lemma 3) with Theorem 2.

Theorem 4
For each integer $i \ge 1$, let $Z_i$ be the one-point compactification of $C_i$ in Theorem 2. Then the product $Z=\prod_{i=1}^\infty Z_i$ is a compact space. Furthermore, $X=\prod_{i=1}^\infty C_i$ is a subspace of $Z$. Prove that $Z$ and $X$ satisfy Lemma 3.

Each $C_i$ in Theorem 2 is a locally compact space. To define the one-point compactifications, for each $i$, choose $p_i \notin C_i$. Make sure that $p_i \ne p_j$ for $i \ne j$. Then $Z_i$ is simply

$Z_i=C_i \cup \{ p_i \}=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots \cup \{ p_i \}$

with the topology defined as follows:

• Open subsets of $C_i$ continue to be open in $Z_i$.
• An open set containing $p_i$ is of the form $\{ p_i \} \cup (C_i - \overline{D})$ where $D$ is open in $C_i$ and $D$ is contained in the union of finitely many $C_{i,j}$.

For convenience, each point $p_i$ is called a point at infinity.

Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem 2 to the product space in Theorem 1.

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Proofs of Key Steps

Proof of Lemma 3
Let $Z$, $X$ and $F_1,F_2,F_3,\cdots$ be as described in the statement for Lemma 3. Let $\mathcal{U}$ be a collection of open subsets of $Z$ such that $\mathcal{U}$ covers $X$. We would like to show that a countable subcollection of $\mathcal{U}$ is also a cover of $X$. Let $O=\cup \mathcal{U}$. If $Z-O=\varnothing$, then $\mathcal{U}$ is an open cover of $Z$ and there is a finite subset of $\mathcal{U}$ that is a cover of $Z$ and thus a cover of $X$. Thus we can assume that $Z-O \ne \varnothing$.

Let $F=\{ F_1,F_2,F_3,\cdots \}$. Let $K=Z-O$, which is compact. We make the following claim.

Claim. Let $Y$ be the union of all possible $\cap G$ where $G \subset F$ is finite and $\cap G \subset O$. Then $X \subset Y \subset O$.

To establish the claim, let $x \in X$. For each $y \in K=Z-O$, there exists $F_{n(y)}$ such that $x \in F_{n(y)}$ and $y \notin F_{n(y)}$. This means that $\{ Z-F_{n(y)}: y \in K \}$ is an open cover of $K$. By the compactness of $K$, there are finitely many $F_{n(y_1)}, \cdots, F_{n(y_k)}$ such that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$ misses $K$, or equivalently $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset O$. Note that $x \in F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$. Further note that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset Y$. This establishes the claim that $X \subset Y$. The claim that $Y \subset O$ is clear from the definition of $Y$.

Each set $F_i$ is compact since it is closed in $Z$. The intersection of finitely many $F_i$ is also compact. Thus the $\cap G$ in the definition of $Y$ in the above claim is compact. There can be only countably many $\cap G$ in the definition of $Y$. Thus $Y$ is a $\sigma$-compact space that is covered by the open cover $\mathcal{U}$. Choose a countable $\mathcal{V} \subset \mathcal{U}$ such that $\mathcal{V}$ covers $Y$. Then $\mathcal{V}$ is a cover of $X$ too. This completes the proof that $X$ is Lindelof.

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Proof of Theorem 4
Recall that $Z=\prod_{i=1}^\infty Z_i$ and that $X=\prod_{i=1}^\infty C_i$. Each $Z_i$ is the one-point compactification of $C_i$, which is the topological sum of the disjoint compact spaces $C_{i,1},C_{i,2},\cdots$.

For integers $i,j \ge 1$, define $K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$. For integers $n,j \ge 1$, define the product $F_{n,j}$ as follows:

$F_{n,j}=K_{1,j} \times \cdots \times K_{n,j} \times Z_{n+1} \times Z_{n+2} \times \cdots$

Since $F_{n,j}$ is a product of compact spaces, $F_{n,j}$ is compact and thus closed in $Z$. There are only countably many $F_{n,j}$.

We claim that the countably many $F_{n,j}$ have the property indicated in Lemma 3. To this end, let $f \in X=\prod_{i=1}^\infty C_i$ and $g \in Z-X$. There exists an integer $n \ge 1$ such that $g(n) \notin C_{n}$. This means that $g(n) \notin C_{n,j}$ for all $j$, i.e. $g(n)=p_n$ (so $g(n)$ must be the point at infinity). Choose $j \ge 1$ large enough such that

$f(i) \in K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$

for all $i \le n$. It follows that $f \in F_{n,j}$ and $g \notin F_{n,j}$. Thus the sequence of closed sets $F_{n,j}$ satisfies Lemma 3. By Lemma 3, $X=\prod_{i=1}^\infty C_i$ is Lindelof.

Reference

1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.

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# Helly Space

This is a discussion on a compact space called Helly space. The discussion here builds on the facts presented in Counterexample in Topology [2]. Helly space is Example 107 in [2]. The space is named after Eduard Helly.

Let $I=[0,1]$ be the closed unit interval with the usual topology. Let $C$ be the set of all functions $f:I \rightarrow I$. The set $C$ is endowed with the product space topology. The usual product space notation is $I^I$ or $\prod_{t \in I} W_t$ where each $W_t=I$. As a product of compact spaces, $C=I^I$ is compact.

Any function $f:I \rightarrow I$ is said to be increasing if $f(x) \le f(y)$ for all $x (such a function is usually referred to as non-decreasing). Helly space is the subspace $X$ consisting of all increasing functions. This space is Example 107 in Counterexample in Topology [2]. The following facts are discussed in [2].

• The space $X$ is compact.
• The space $X$ is first countable (having a countable base at each point).
• The space $X$ is separable.
• The space $X$ has an uncountable discrete subspace.

From the last two facts, Helly space is a compact non-metrizable space. Any separable metric space would have countable spread (all discrete subspaces must be countable).

The compactness of $X$ stems from the fact that $X$ is a closed subspace of the compact space $C$.

Further Discussion

Additional facts concerning Helly space are discussed.

1. The product space $\omega_1 \times X$ is normal.
2. Helly space $X$ contains a copy of the Sorgenfrey line.
3. Helly space $X$ is not hereditarily normal.

The space $\omega_1$ is the space of all countable ordinals with the order topology. Recall $C$ is the product space $I^I$. The product space $\omega_1 \times C$ is Example 106 in [2]. This product is not normal. The non-normality of $\omega_1 \times C$ is based on this theorem: for any compact space $Y$, the product $\omega_1 \times Y$ is normal if and only if the compact space $Y$ is countably tight. The compact product space $C$ is not countably tight (discussed here). Thus $\omega_1 \times C$ is not normal. However, the product $\omega_1 \times X$ is normal since Helly space $X$ is first countable.

To see that $X$ contains a copy of the Sorgenfrey line, consider the functions $h_t:I \rightarrow I$ defined as follows:

$\displaystyle h_t(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le x \le t \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ t

for all $0. Let $S=\{ h_t: 0. Consider the mapping $\gamma: (0,1) \rightarrow S$ defined by $\gamma(t)=h_t$. With the domain $(0,1)$ having the Sorgenfrey topology and with the range $S$ being a subspace of Helly space, it can be shown that $\gamma$ is a homeomorphism.

With the Sorgenfrey line $S$ embedded in $X$, the square $X \times X$ contains a copy of the Sorgenfrey plane $S \times S$, which is non-normal (discussed here). Thus the square of Helly space is not hereditarily normal. A more interesting fact is that Helly space is not hereditarily normal. This is discussed in the next section.

Finding a Non-Normal Subspace of Helly Space

As before, $C$ is the product space $I^I$ where $I=[0,1]$ and $X$ is Helly space consisting of all increasing functions in $C$. Consider the following two subspaces of $X$.

$Y_{0,1}=\{ f \in X: f(I) \subset \{0, 1 \} \}$

$Y=X - Y_{0,1}$

The subspace $Y_{0,1}$ is a closed subset of $X$, hence compact. We claim that subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that the subspace $Y$ is not a normal space.

First, we define a discrete subspace. For each $x$ with $0, define $f_x: I \rightarrow I$ as follows:

$\displaystyle f_x(y) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le y < x \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{2} &\ \ \ \ \ y=x \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ x

Let $H=\{ f_x: 0. The set $H$ as a subspace of $X$ is discrete. Of course it is not discrete in $X$ since $X$ is compact. In fact, for any $f \in Y_{0,1}$, $f \in \overline{H}$ (closure taken in $X$). However, it can be shown that $H$ is closed and discrete as a subset of $Y$.

We now construct a countable dense subset of $Y$. To this end, let $\mathcal{B}$ be a countable base for the usual topology on the unit interval $I=[0,1]$. For example, we can let $\mathcal{B}$ be the set of all open intervals with rational endpoints. Furthermore, let $A$ be a countable dense subset of the open interval $(0,1)$ (in the usual topology). For convenience, we enumerate the elements of $A$ and $\mathcal{B}$.

$A=\{ a_1,a_2,a_3,\cdots \}$

$\mathcal{B}=\{B_1,B_2,B_3,\cdots \}$

We also need the following collections.

$\mathcal{G}=\{G \subset \mathcal{B}: G \text{ is finite and is pairwise disjoint} \}$

$\mathcal{A}=\{F \subset A: F \text{ is finite} \}$

For each $G \in \mathcal{G}$ and for each $F \in \mathcal{A}$ with $\lvert G \lvert=\lvert F \lvert=n$, we would like to arrange the elements in increasing order, notated as follow:

$F=\{t_1,t_2,\cdots,t_n \}$

$G=\{E_1,E_2,\cdots,E_n \}$

For the set $F$, we have $0. For the set $G$, $E_i$ is to the left of $E_j$ for $i. Note that elements of $G$ are pairwise disjoint. Furthermore, write $E_i=(p_i,q_i)$. If $0 \in E_1$, then $E_1=[p_1,q_1)=[0,q_1)$. If $1 \in E_n$, then $E_n=(p_n,q_n]=(p_n,1]$.

For each $F$ and $G$ as detailed above, we define a function $L(F,G):I \rightarrow I$ as follows:

$\displaystyle L(F,G)(x) = \left\{ \begin{array}{ll} \displaystyle t_1 &\ \ \ \ \ 0 \le x < q_1 \\ \text{ } & \text{ } \\ \displaystyle t_2 &\ \ \ \ \ q_1 \le x < q_2 \\ \text{ } & \text{ } \\ \displaystyle \vdots &\ \ \ \ \ \vdots \\ \text{ } & \text{ } \\ \displaystyle t_{n-1} &\ \ \ \ \ q_{n-2} \le x < q_{n-1} \\ \text{ } & \text{ } \\ \displaystyle t_n &\ \ \ \ \ q_{n-1} \le x \le 1 \\ \end{array} \right.$

The following diagram illustrates the definition of $L(F,G)$ when both $F$ and $G$ have 4 elements.

Figure 1 – Member of a countable dense set

Let $D$ be the set of $L(F,G)$ over all $F \in \mathcal{A}$ and $G \in \mathcal{G}$. The set $D$ is a countable set. It can be shown that $D$ is dense in the subspace $Y$. In fact $D$ is dense in the entire Helly space $X$.

To summarize, the subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that $Y$ is not normal. Hence Helly space $X$ is not hereditarily normal. According to Jones’ lemma, in any normal separable space, the cardinality of any closed and discrete subspace must be less than continuum (discussed here).

Remarks

The preceding discussion shows that both Helly space and the square of Helly space are not hereditarily normal. This is actually not surprising. According to a theorem of Katetov, for any compact non-metrizable space $V$, the cube $V^3$ is not hereditarily normal (see Theorem 3 in this post). Thus a non-normal subspace is found in $V$, $V \times V$ or $V \times V \times V$. In fact, for any compact non-metric space $V$, an excellent exercise is to find where a non-normal subspace can be found. Is it in $V$, the square of $V$ or the cube of $V$? In the case of Helly space $X$, a non-normal subspace can be found in $X$.

A natural question is: is there a compact non-metric space $V$ such that both $V$ and $V \times V$ are hereditarily normal and $V \times V \times V$ is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyond ZFC, any compact non-metric space $V$ is likely to fail hereditarily normality in either $V$ or $V \times V$. See here for a discussion of this set-theoretic question.

Reference

1. Kelly, J. L., General Topology, Springer-Verlag, New York, 1955.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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# A little corner in the world of set-theoretic topology

This post puts a spot light on a little corner in the world of set-theoretic topology. There lies in this corner a simple topological statement that opens a door to the esoteric world of independence results. In this post, we give a proof of this basic fact and discuss its ramifications. This basic result is an excellent entry point to the study of S and L spaces.

The following paragraph is found in the paper called Gently killing S-spaces by Todd Eisworth, Peter Nyikos and Saharon Shelah [1]. The basic fact in question is highlighted in blue.

A simultaneous generalization of hereditarily separable and hereditarily Lindelof spaces is the class of spaces of countable spread – those spaces in which every discrete subspace is countable. One of the basic facts in this little corner of set-theoretic topology is that if a regular space of countable spread is not hereditarily separable, it contains an L-space, and if it is not hereditarily Lindelof, it contains an S-space. [1]

The same basic fact is also mentioned in the paper called The spread of regular spaces by Judith Roitman [2].

It is also well known that a regular space of countable spread which is not hereditarily separable contains an L-space and a regular space of countable spread which is not hereditarily Lindelof contains an S-space. Thus an absolute example of a space satisfying (Statement) A would contain a proof of the existence of S and L space – a consummation which some may devoutly wish, but which this paper does not attempt. [2]

Statement A in [2] is: There exists a 0-dimensional Hausdorff space of countable spread that is not the union of a hereditarily separable and a hereditarily Lindelof space. Statement A would mean the existence of a regular space of countable spread that is not hereditarily separable and that is also not hereditarily Lindelof. By the well known fact just mentioned, statement A would imply the existence of a space that is simultaneously an S-space and an L-space!

Let’s unpack the preceding section. First some basic definitions. A space $X$ is of countable spread (has countable spread) if every discrete subspace of $X$ is countable. A space $X$ is hereditarily separable if every subspace of $X$ is separable. A space $X$ is hereditarily Lindelof if every subspace of $X$ is Lindelof. A space is an S-space if it is hereditarily separable but not Lindelof. A space is an L-space if it is hereditarily Lindelof but not separable. See [3] for a basic discussion of S and L spaces.

Hereditarily separable but not Lindelof spaces as well as hereditarily Lindelof but not separable spaces can be easily defined in ZFC [3]. However, such examples are not regular. For the notions of S and L-spaces to be interesting, the definitions must include regularity. Thus in the discussion that follows, all spaces are assumed to be Hausdorff and regular.

One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions that are not far from the definition of topological spaces. The same can be said about hereditarily separable and hereditarily Lindelof spaces. Out of these basic ingredients come the notion of S-spaces and L-spaces, the existence of which is one of the key motivating questions in set-theoretic topology in the twentieth century. The study of S and L-spaces is a body of mathematics that had been developed for nearly a century. It is a fruitful area of research at the boundary of topology and axiomatic set theory.

The existence of an S-space is independent of ZFC (as a result of the work by Todorcevic in early 1980s). This means that there is a model of set theory in which an S-space exists and there is also a model of set theory in which S-spaces cannot exist. One half of the basic result mentioned in the preceding section is intimately tied to the existence of S-spaces and thus has interesting set-theoretic implications. The other half of the basic result involves the existence of L-spaces, which are shown to exist without using extra set theory axioms beyond ZFC by Justin Moore in 2005, which went against the common expectation that the existence of L-spaces would be independent of ZFC as well.

Let’s examine the basic notions in a little more details. The following diagram shows the properties surrounding the notion of countable spread.

Diagram 1 – Properties surrounding countable spread

The implications (the arrows) in Diagram 1 can be verified easily. Central to the discussion at hand, both hereditarily separable and hereditarily Lindelof imply countable spread. The best way to see this is that if a space has an uncountable discrete subspace, that subspace is simultaneously a non-separable subspace and a non-Lindelof subspace. A natural question is whether these implications can be reversed. Another question is whether the properties in Diagram 1 can be related in other ways. The following diagram attempts to ask these questions.

Diagram 2 – Reverse implications surrounding countable spread

Not shown in Diagram 2 are these four facts: separable $\not \rightarrow$ hereditarily separable, Lindelof $\not \rightarrow$ hereditarily Lindelof, separable $\not \rightarrow$ countable spread and Lindelof $\not \rightarrow$ countable spread. The examples supporting these facts are not set-theoretic in nature and are not discussed here.

Let’s focus on each question mark in Diagram 2. The two horizontal arrows with question marks at the top are about S-space and L-space. If $X$ is hereditarily separable, then is $X$ hereditarily Lindelof? A “no” answer would mean there is an S-space. A “yes” answer would mean there exists no S-space. So the top arrow from left to right is independent of ZFC. Since an L-space can be constructed within ZFC, the question mark in the top arrow in Diagram 2 from right to left has a “no” answer.

Now focus on the arrows emanating from countable spread in Diagram 2. These arrows are about the basic fact discussed earlier. From Diagram 1, we know that hereditarily separable implies countable spread. Can the implication be reversed? Any L-space would be an example showing that the implication cannot be reversed. Note that any L-space is of countable spread and is not separable and hence not hereditarily separable. Since L-space exists in ZFC, the question mark in the arrow from countable spread to hereditarily separable has a “no” answer. The same is true for the question mark in the arrow from countable spread to separable

We know that hereditarily Lindelof implies countable spread. Can the implication be reversed? According to the basic fact mentioned earlier, if the implication cannot be reversed, there exists an S-space. Thus if S-space does not exist, the implication can be reversed. Any S-space is an example showing that the implication cannot be reversed. Thus the question in the arrow from countable spread to hereditarily Lindelof cannot be answered without assuming axioms beyond ZFC. The same is true for the question mark for the arrow from countable spread to Lindelf.

Diagram 2 is set-theoretic in nature. The diagram is remarkable in that the properties in the diagram are basic notions that are only brief steps away from the definition of a topological space. Thus the basic highlighted here is a quick route to the world of independence results.

We now give a proof of the basic result, which is stated in the following theorem.

Theorem 1
Let $X$ is regular and Hausdorff space. Then the following is true.

• If $X$ is of countable spread and is not a hereditarily separable space, then $X$ contains an L-space.
• If $X$ is of countable spread and is not a hereditarily Lindelof space, then $X$ contains an S-space.

To that end, we use the concepts of right separated space and left separated space. Recall that an initial segment of a well-ordered set $(X,<)$ is a set of the form $\{y \in X: y where $x \in X$. A space $X$ is a right separated space if $X$ can be well-ordered in such a way that every initial segment is open. A right separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. A space $X$ is a left separated space if $X$ can be well-ordered in such a way that every initial segment is closed. A left separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. The following results are used in proving Theorem 1.

Theorem A
Let $X$ is regular and Hausdorff space. Then the following is true.

• The space $X$ is hereditarily separable space if and only if $X$ has no uncountable left separated subspace.
• The space $X$ is hereditarily Lindelof space if and only if $X$ has no uncountable right separated subspace.

Proof of Theorem A
$\Longrightarrow$ of the first bullet point.
Suppose $Y \subset X$ is an uncountable left separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $C_\alpha=\{ x_\beta: \beta<\alpha \}$ is a closed subset of $Y$. Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that for any countable $A \subset Y$, $A \subset C_\alpha$ for some $\alpha<\omega_1$. It follows that $Y$ is not separable. This means that $X$ is not hereditarily separable.

$\Longleftarrow$ of the first bullet point.
Suppose that $X$ is not hereditarily separable. Let $Y \subset X$ be a subspace that is not separable. We now inductively derive an uncountable left separated subspace of $Y$. Choose $y_0 \in Y$. For each $\alpha<\omega_1$, let $A_\alpha=\{ y_\beta \in Y: \beta <\alpha \}$. The set $A_\alpha$ is the set of all the points of $Y$ chosen before the step at $\alpha<\omega_1$. Since $A_\alpha$ is countable, its closure in $Y$ is not the entire space $Y$. Choose $y_\alpha \in Y-\overline{A_\alpha}=O_\alpha$.

Let $Y_L=\{ y_\alpha: \alpha<\omega_1 \}$. We claim that $Y_L$ is a left separated space. To this end, we need to show that each initial segment $A_\alpha$ is a closed subset of $Y_L$. Note that for each $\gamma \ge \alpha$, $O_\gamma=Y-\overline{A_\gamma}$ is an open subset of $Y$ with $y_\gamma \in O_\gamma$ such that $O_\gamma \cap \overline{A_\gamma}=\varnothing$ and thus $O_\gamma \cap \overline{A_\alpha}=\varnothing$ (closure in $Y$). Then $U_\gamma=O_\gamma \cap Y_L$ is an open subset of $Y_L$ containing $y_\gamma$ such that $U_\gamma \cap A_\alpha=\varnothing$. It follows that $Y-A_\alpha$ is open in $Y_L$ and that $A_\alpha$ is a closed subset of $Y_L$.

$\Longrightarrow$ of the second bullet point.
Suppose $Y \subset X$ is an uncountable right separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $U_\alpha=\{ x_\beta: \beta<\alpha \}$ is an open subset of $Y$.

Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that $\{ U_\alpha: \alpha<\omega_1 \}$ is an open cover of $Y$ that has no countable subcover. It follows that $Y$ is not Lindelof. This means that $X$ is not hereditarily Lindelof.

$\Longleftarrow$ of the second bullet point.
Suppose that $X$ is not hereditarily Lindelof. Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be an open cover of $Y$ that has no countable subcover. We now inductively derive a right separated subspace of $Y$ of type $\omega_1$.

Choose $U_0 \in \mathcal{U}$ and choose $y_0 \in U_0$. Choose $y_1 \in Y-U_0$ and choose $U_1 \in \mathcal{U}$ such that $y_1 \in U_1$. Let $\alpha<\omega_1$. Suppose that points $y_\beta$ and open sets $U_\beta$, $\beta<\alpha$, have been chosen such that $y_\beta \in Y-\bigcup_{\delta<\beta} U_\delta$ and $y_\beta \in U_\beta$. The countably many chosen open sets $U_\beta$, $\beta<\alpha$, cannot cover $Y$. Choose $y_\alpha \in Y-\bigcup_{\beta<\alpha} U_\beta$. Choose $U_\alpha \in \mathcal{U}$ such that $y_\alpha \in U_\alpha$.

Let $Y_R=\{ y_\alpha: \alpha<\omega_1 \}$. It follows that $Y_R$ is a right separated space. Note that for each $\alpha<\omega_1$, $\{ y_\beta: \beta<\alpha \} \subset \bigcup_{\beta<\alpha} U_\beta$ and the open set $\bigcup_{\beta<\alpha} U_\beta$ does not contain $y_\gamma$ for any $\gamma \ge \alpha$. This means that the initial segment $\{ y_\beta: \beta<\alpha \}$ is open in $Y_L$. $\square$

Lemma B
Let $X$ be a space that is a right separated space and also a left separated space based on the same well ordering. Then $X$ is a discrete space.

Proof of Lemma B
Let $X=\{ w_\alpha: \alpha<\kappa \}$ such that the well-ordering is given by the ordinals in the subscripts, i.e. $w_\beta if and only if $\beta<\gamma$. Suppose that $X$ with this well-ordering is both a right separated space and a left separated space. We claim that every point is a discrete point, i.e. $\{ x_\alpha \}$ is open for any $\alpha<\kappa$.

To see this, fix $\alpha<\kappa$. The initial segment $A_\alpha=\{ w_\beta: \beta<\alpha \}$ is closed in $X$ since $X$ is a left separated space. On the other hand, the initial segment $\{ w_\beta: \beta < \alpha+1 \}$ is open in $X$ since $X$ is a right separated space. Then $B_{\alpha}=\{ w_\beta: \beta \ge \alpha+1 \}$ is closed in $X$. It follows that $\{ x_\alpha \}$ must be open since $X=A_\alpha \cup B_\alpha \cup \{ w_\alpha \}$. $\square$

Theorem C
Let $X$ is regular and Hausdorff space. Then the following is true.

• Suppose the space $X$ is right separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an S-space or $X$ contains an S-space.
• Suppose the space $X$ is left separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an L-space or $X$ contains an L-space.

Proof of Theorem C
For the first bullet point, suppose the space $X$ is right separated space of type $\omega_1$. Then by Theorem A, $X$ is not hereditarily Lindelof. If $X$ is hereditarily separable, then $X$ is an S-space (if $X$ is not Lindelof) or $X$ contains an S-space (a non-Lindelof subspace of $X$). Suppose $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace of type $\omega_1$.

Let $X=\{ x_\alpha: \alpha<\omega_1 \}$ such that the well-ordering represented by the ordinals in the subscripts is a right separated space. Let $<_R$ be the symbol for the right separated well-ordering, i.e. $x_\beta <_R \ x_\delta$ if and only if $\beta<\delta$. As indicated in the preceding paragraph, $X$ has an uncountable left separated subspace. Let $Y=\{ y_\alpha \in X: \alpha<\omega_1 \}$ be this left separated subspace. Let $<_L$ be the symbol for the left separated well-ordering. The well-ordering $<_R$ may be different from the well-ordering $<_L$. However, we can obtain an uncountable subset of $Y$ such that the two well-orderings coincide on this subset.

To start, pick any $y_\gamma$ in $Y$ and relabel it $t_0$. The final segment $\{y_\beta \in Y: t_0 <_L \ y_\beta \}$ must intersect the final segment $\{x_\beta \in X: t_0 <_R \ x_\beta \}$ in uncountably many points. Choose the least such point (according to $<_R$) and call it $t_1$. It is clear how $t_{\delta+1}$ is chosen if $t_\delta$ has been chosen.

Suppose $\alpha<\omega_1$ is a limit ordinal and that $t_\beta$ has been chosen for all $\beta<\alpha$. Then the set $\{y_\tau: \forall \ \beta<\alpha, t_\beta <_L \ y_\tau \}$ and the set $\{x_\tau: \forall \ \beta<\alpha, t_\beta <_R \ x_\tau \}$ must intersect in uncountably many points. Choose the least such point and call it $t_\alpha$ (according to $<_R$). As a result, we have obtained $T=\{ t_\alpha: \alpha<\omega_1 \}$. It follows that T with the well-ordering represented by the ordinals in the subscript is a subset of $(X,<_R)$ and a subset of $(Y,<_L)$. Thus $T$ is both right separated and left separated.

By Lemma B, $T$ is a discrete subspace of $X$. However, $X$ is assumed to have no uncountable discrete subspace. Thus if $X$ has no uncountable discrete subspace, then $X$ must be hereditarily separable and as a result, must be an S-space or must contain an S-space.

The proof for the second bullet point is analogous to that of the first bullet point. $\square$

We are now ready to prove Theorem 1.

Proof of Theorem 1
Suppose that $X$ is of countable spread and that $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace $Y$ (assume it is of type $\omega_1$). The property of countable spread is hereditary. So $Y$ is of countable spread. By Theorem C, $Y$ is an L-space or $Y$ contains an L-space. In either way, $X$ contains an L-space.

Suppose that $X$ is of countable spread and that $X$ is not hereditarily Lindelof. By Theorem A, $X$ has an uncountable right separated subspace $Y$ (assume it is of type $\omega_1$). By Theorem C, $Y$ is an S-space or $Y$ contains an S-space. In either way, $X$ contains an S-space.

Reference

1. Eisworth T., Nyikos P., Shelah S., Gently killing S-spaces, Israel Journal of Mathmatics, 136, 189-220, 2003.
2. Roitman J., The spread of regular spaces, General Topology and Its Applications, 8, 85-91, 1978.
3. Roitman, J., Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 295-326, 1984.
4. Tatch-Moore J., A solution to the L space problem, Journal of the American Mathematical Society, 19, 717-736, 2006.

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Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

# Every space is star discrete

The statement in the title is a folklore fact, though the term star discrete is usually not used whenever this well known fact is invoked in the literature. We present a proof to this well known fact. We also discuss some related concepts.

All spaces are assumed to be Hausdorff and regular.

First, let’s define the star notation. Let $X$ be a space. Let $\mathcal{U}$ be a collection of subsets of $X$. Let $A \subset X$. Define $\text{St}(A,\mathcal{U})$ to be the set $\bigcup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$. In other words, the set $\text{St}(A,\mathcal{U})$ is simply the union of all elements of $\mathcal{U}$ that contains points of the set $A$. The set $\text{St}(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the collection $\mathcal{U}$. If $A=\{ x \}$, we use the notation $\text{St}(x,\mathcal{U})$ instead of $\text{St}( \{ x \},\mathcal{U})$. The following is the well known result in question.

Lemma 1
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $X=\text{St}(A,\mathcal{U})$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.

Any space that satisfies the condition in Lemma 1 is said to be a star discrete space. The proof shown below will work for any topological space. Hence every space is star discrete. We come across three references in which the lemma is stated or is used – Lemma IV.2.20 in page 135 of [3], page 137 of [2] and [1]. The first two references do not use the term star discrete. Star discrete is mentioned in [1] since that paper focuses on star properties. This property that is present in every topological space is at heart a covering property. Here’s a rewording of the lemma that makes it look like a covering property.

Lemma 1a
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.

Lemma 1a is clearly identical to Lemma 1. However, Lemma 1a makes it extra clear that this is a covering property. For every open cover of a space, instead of finding a sub cover or an open refinement, we find a discrete subspace so that the stars of the points of the discrete subspace with respect to the given open cover also cover the space.

Lemma 1a naturally leads to other star covering properties. For example, a space $X$ is said to be a star countable space if for any open cover $\mathcal{U}$ of $X$, there exists a countable subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. A space $X$ is said to be a star Lindelof space if for any open cover $\mathcal{U}$ of $X$, there exists a Lindelof subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. In general, for any topological property $\mathcal{P}$, a space $X$ is a star $\mathcal{P}$ space if for any open cover $\mathcal{U}$ of $X$, there exists a subspace $A$ of $X$ with property $\mathcal{P}$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$.

It follows that every Lindelof space is a star countable space. It is also clear that every star countable space is a star Lindelof space.

Lemma 1 or Lemma 1a, at first glance, may seem like a surprising result. However, one can argue that it is not a strong result at all since the property is possessed by every space. Indeed, the lemma has nothing to say about the size of the discrete set. It only says that there exists a star cover based on a discrete set for a given open cover. To derive more information about the given space, we may need to work with more information on the space in question.

Consider spaces such that every discrete subspace is countable (such a space is said to have countable spread or a space of countable spread). Also consider spaces such that every closed and discrete subspace is countable (such a space is said to have countable extent or a space of countable extent). Any space that has countable spread is also a space that has countable extent for the simple reason that if every discrete subspace is countable, then every closed and discrete subspace is countable.

Then it follows from Lemma 1 that any space $X$ that has countable extent is star countable. Any star countable space is obviously a star Lindelof space. The following diagram displays these relationships.

According to the diagram, the star countable and star Lindelof are both downstream from the countable spread property and the Lindelof property. The star properties being downstream from the Lindelof property is not surprising. What is interesting is that if a space has countable spread, then it is star countable and hence star Lindelof.

Do “countable spread” and “Lindelof” relate to each other? Lindelof spaces do not have to have countable spread. The simplest example is the one-point compactification of an uncountable discrete space. More specifically, let $X$ be an uncountable discrete space. Let $p$ be a point not in $X$. Then $Y=X \cup \{ p \}$ is a compact space (hence Lindelof) where $X$ is discrete and an open neighborhood of $p$ is of the form $\{ p \} \cup U$ where $X-U$ is a finite subset of $X$. The space $Y$ is not of countable spread since $X$ is an uncountable discrete subspace.

Does “countable spread” imply “Lindelof”? Is there a non-Lindelof space that has countable spread? It turns out that the answers are independent of ZFC. The next post has more details.

We now give a proof to Lemma 1. Suppose that $X$ is an infinite space (if it is finite, the lemma is true since the space is Hausdorff). Let $\kappa=\lvert X \lvert$. Let $\kappa^+$ be the next cardinal greater than $\kappa$. Let $\mathcal{U}$ be an open cover of the space $X$. Choose $x_0 \in X$. We choose a sequence of points $x_0,x_1,\cdots,x_\alpha,\cdots$ inductively. If $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) \ne X$, we can choose a point $x_\alpha \in X$ such that $x_\alpha \notin \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U})$.

We claim that the induction process must stop at some $\alpha<\kappa^+$. In other words, at some $\alpha<\kappa^+$, the star of the previous points must be the entire space and we run out of points to choose. Otherwise, we would have obtained a subset of $X$ with cardinality $\kappa^+$, a contradiction. Choose the least $\alpha<\kappa^+$ such that $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) = X$. Let $A=\{x_\beta: \beta<\alpha \}$.

Then it can be verified that the set $A$ is a discrete subspace of $X$ and that $A$ is a closed subset of $X$. Note that $x_\beta \in \text{St}(x_\beta, \mathcal{U})$ while $x_\gamma \notin \text{St}(x_\beta, \mathcal{U})$ for all $\gamma \ne \beta$. This follows from the way the points are chosen in the induction process. On the other hand, for any $x \in X-A$, $x \in \text{St}(x_\beta, \mathcal{U})$ for some $\beta<\alpha$. As discussed, the open set $\text{St}(x_\beta, \mathcal{U})$ contains only one point of $A$, namely $x_\beta$.

Reference

1. Alas O., Jumqueira L., van Mill J., Tkachuk V., Wilson R.On the extent of star countable spaces, Cent. Eur. J. Math., 9 (3), 603-615, 2011.
2. Alster, K., Pol, R.,On function spaces of compact subspaces of $\Sigma$-products of the real line, Fund. Math., 107, 35-46, 1980.
3. Arkhangelskii, A. V.,Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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Dan Ma math

Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

# Michael line and Morita’s conjectures

This post discusses Michael line from the point of view of the three conjectures of Kiiti Morita.

K. Morita defined the notion of P-spaces in [7]. The definition of P-spaces is discussed here in considerable details. K. Morita also proved that a space $X$ is a normal P-space if and only if the product $X \times Y$ is normal for every metrizable space $Y$. As a result of this characterization, the notion of normal P-space (a space that is a normal space and a P-space) is useful in the study of products of normal spaces. Just to be clear, we say a space is a non-normal P-space (i.e. a space that is not a normal P-space) if the space is a normal space that is not a P-space.

K. Morita formulated his three conjectures in 1976. The statements of the conjectures are given below. Here is a basic discussion of the three conjectures. The notion of normal P-spaces is a theme that runs through the three conjectures. The conjectures are actually theorems since 2001 [2].

Here’s where Michael line comes into the discussion. Based on the characterization of normal P-spaces mentioned above, to find a normal space that is not a P-space (a non-normal P-space), we would need to find a non-normal product $X \times Y$ such that one of the factors is a metric space and the other factor is a normal space. The first such example in ZFC is from an article by E. Michael in 1963 (found here and here). In this example, the normal space is $M$, which came be known as the Michael line, and the metric space is $\mathbb{P}$, the space of irrational numbers (as a subspace of the real line). Their product $M \times \mathbb{P}$ is not normal. A basic discussion of the Michael line is found here.

Because $M \times \mathbb{P}$ is not normal, the Michael line $M$ is not a normal P-space. Prior to E. Michael’s 1963 article, we have to reach back to 1955 to find an example of a non-normal product where one factor is a metric space. In 1955, M. E. Rudin used a Souslin line to construct a Dowker space, which is a normal space whose product with the closed unit interval is not normal. The existence of a Souslin line was shown to be independent of ZFC in the late 1960s. In 1971, Rudin constructed a Dowker space in ZFC. Thus finding a normal space that is not a normal P-space (finding a non-normal product $X \times Y$ where one factor is a metric space and the other factor is a normal space) is not a trivial matter.

Morita’s Three Conjectures

We show that the Michael line illustrates perfectly the three conjectures of K. Morita. Here’s the statements.

Morita’s Conjecture I. Let $X$ be a space. If the product $X \times Y$ is normal for every normal space $Y$ then $X$ is a discrete space.

Morita’s Conjecture II. Let $X$ be a space. If the product $X \times Y$ is normal for every normal P-space $Y$ then $X$ is a metrizable space.

Morita’s Conjecture III. Let $X$ be a space. If the product $X \times Y$ is normal for every normal countably paracompact space $Y$ then $X$ is a metrizable $\sigma$-locally compact space.

The contrapositive statement of Morita’s conjecture I is that for any non-discrete space $X$, there exists a normal space $Y$ such that $X \times Y$ is not normal. Thus any non-discrete space is paired with a normal space for forming a non-normal product. The Michael line $M$ is paired with the space of irrational numbers $\mathbb{P}$. Obviously, the space $\mathbb{P}$ is paired with the Michael line $M$.

The contrapositive statement of Morita’s conjecture II is that for any non-metrizable space $X$, there exists a normal P-space $Y$ such that $X \times Y$ is not normal. The pairing is more specific than for conjecture I. Any non-metrizable space is paired with a normal P-space to form a non-normal product. As illustration, the Michael line $M$ is not metrizable. The space $\mathbb{P}$ of irrational numbers is a metric space and hence a normal P-space. Here, $M$ is paired with $\mathbb{P}$ to form a non-normal product.

The contrapositive statement of Morita’s conjecture III is that for any space $X$ that is not both metrizable and $\sigma$-locally compact, there exists a normal countably paracompact space $Y$ such that $X \times Y$ is not normal. Note that the space $\mathbb{P}$ is not $\sigma$-locally compact (see Theorem 4 here). The Michael line $M$ is paracompact and hence normal and countably paracompact. Thus the metric non-$\sigma$-locally compact $\mathbb{P}$ is paired with normal countably paracompact $M$ to form a non-normal product. Here, the metric space $\mathbb{P}$ is paired with the non-normal P-space $M$.

In each conjecture, each space in a certain class of spaces is paired with one space in another class to form a non-normal product. For Morita’s conjecture I, each non-discrete space is paired with a normal space. For conjecture II, each non-metrizable space is paired with a normal P-space. For conjecture III, each metrizable but non-$\sigma$-locally compact is paired with a normal countably paracompact space to form a non-normal product. Note that the paired normal countably paracompact space would be a non-normal P-space.

Michael line as an example of a non-normal P-space is a great tool to help us walk through the three conjectures of Morita. Are there other examples of non-normal P-spaces? Dowker spaces mentioned above (normal spaces whose products with the closed unit interval are not normal) are non-normal P-spaces. Note that conjecture II guarantees a normal P-space to match every non-metric space for forming a non-normal product. Conjecture III guarantees a non-normal P-space to match every metrizable non-$\sigma$-locally compact space for forming a non-normal product. Based on the conjectures, examples of normal P-spaces and non-normal P-spaces, though may be hard to find, are guaranteed to exist.

We give more examples below to further illustrate the pairings for conjecture II and conjecture III. As indicated above, non-normal P-spaces are hard to come by. Some of the examples below are constructed using additional axioms beyond ZFC. The additional examples still give an impression that the availability of non-normal P-spaces, though guaranteed to exist, is limited.

Examples of Normal P-Spaces

One example is based on this classic theorem: for any normal space $X$, $X$ is paracompact if and only if the product $X \times \beta X$ is normal. Here $\beta X$ is the Stone-Cech compactification of the completely regular space $X$. Thus any normal but not paracompact space $X$ (a non-metrizable space) is paired with $\beta X$, a normal P-space, to form a non-normal product.

Naturally, the next class of non-metrizable spaces to be discussed should be the paracompact spaces that are not metrizable. If there is a readily available theorem to provide a normal P-space for each non-metrizable paracompact space, then there would be a simple proof of Morita’s conjecture II. The eventual solution of conjecture II is far from simple [2]. We narrow the focus to the non-metrizable compact spaces.

Consider this well known result: for any infinite compact space $X$, the product $\omega_1 \times X$ is normal if and only if the space $X$ has countable tightness (see Theorem 1 here). Thus any compact space with uncountable tightness is paired with $\omega_1$, the space of all countable ordinals, to form a non-normal product. The space $\omega_1$, being a countably compact space, is a normal P-space. A proof that normal countably compact space is a normal P-space is given here.

We now handle the case for non-metrizable compact spaces with countable tightness. In this case, compactness is not needed. For spaces with countable tightness, consider this result: every space with countable tightness, whose products with all perfectly normal spaces are normal, must be metrizable [3] (see Corollary 7). Thus any non-metrizable space with countable tightness is paired with some perfectly normal space to form a non-normal product. Any reader interested in what these perfectly normal spaces are can consult [3]. Note that perfectly normal spaces are normal P-spaces (see here for a proof).

Examples of Non-Normal P-Spaces

Another non-normal product is $X_B \times B$ where $B \subset \mathbb{R}$ is a Bernstein set and $X_B$ is the space with the real line as the underlying set such that points in $B$ are isolated and points in $\mathbb{R}-B$ retain the usual open sets. The set $B \subset \mathbb{R}$ is said to be a Bernstein set if every uncountable closed subset of the real line contains a point in B and contains a point in the complement of B. Such a set can be constructed using transfinite induction as shown here. The product $X_B \times B$ is not normal where $B$ is considered a subspace of the real line. The proof is essentially the same proof that shows $M \times \mathbb{P}$ is not normal (see here). The space $X_B$ is a Lindelof space. It is not a normal P-space since its product with $B$, a separable metric space, is not normal. However, this example is essentially the same example as the Michael line since the same technique and proof are used. On the one hand, the $X_B \times B$ example seems like an improvement over Michael line example since the first factor $X_B$ is Lindelof. On the other hand, it is inferior than the Michael line example since the second factor $B$ is not completely metrizable.

Moving away from the idea of Michael, there exist a Lindelof space and a completely metrizable (but not separable) space whose product is of weight $\omega_1$ and is not normal [5]. This would be a Lindelof space that is a non-normal P-space. However, this example is not as elementary as the Michael line, making it not as effective as an illustration of Morita’s three conjectures.

The next set of non-normal P-spaces requires set theory. A Michael space is a Lindelof space whose product with $\mathbb{P}$, the space of irrational numbers, is not normal. Michael problem is the question: is there a Michael space in ZFC? It is known that a Michael space can be constructed using continuum hypothesis [6] or using Martin’s axiom [1]. The construction using continuum hypothesis has been discussed in this blog (see here). The question of whether there exists a Michael space in ZFC is still unsolved.

The existence of a Michael space is equivalent to the existence of a Lindelof space and a separable completely metrizable space whose product is non-normal [4]. A Michael space, in the context of the discussion in this post, is a non-normal P-space.

The discussion in this post shows that the example of the Michael line and other examples of non-normal P-spaces are useful tools to illustrate Morita’s three conjectures.

Reference

1. Alster K.,On the product of a Lindelof space and the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., Vol. 110, 543-547, 1990.
2. Balogh Z.,Normality of product spaces and Morita’s conjectures, Topology Appl., Vol. 115, 333-341, 2001.
3. Chiba K., Przymusinski T., Rudin M. E.Nonshrinking open covers and K. Morita’s duality conjectures, Topology Appl., Vol. 22, 19-32, 1986.
4. Lawrence L. B., The influence of a small cardinal on the product of a Lindelof space and the irrationals, Proc. Amer. Math. Soc., 110, 535-542, 1990.
5. Lawrence L. B., A ZFC Example (of Minimum Weight) of a Lindelof Space and a Completely Metrizable Space with a Nonnormal Product, Proc. Amer. Math. Soc., 124, No 2, 627-632, 1996.
6. Michael E., Paracompactness and the Lindelof property in nite and countable cartesian products, Compositio Math., 23, 199-214, 1971.
7. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.
8. Rudin M. E., A Normal Space $X$ for which $X \times I$ is not Normal, Fund. Math., 73, 179-186, 1971.

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$\copyright$ 2018 – Dan Ma

# Three conjectures of K Morita

This post discusses the three conjectures that were proposed by K. Morita in 1976. These conjectures concern normality in product spaces. To start the discussion, here’s the conjectures.

Morita’s Conjecture I. Let $X$ be a space. The product $X \times Y$ is normal for every normal space $Y$ if and only if $X$ is a discrete space.

Morita’s Conjecture II. Let $X$ be a space. The product $X \times Y$ is normal for every normal P-space $Y$ if and only if $X$ is a metrizable space.

Morita’s Conjecture III. Let $X$ be a space. The product $X \times Y$ is normal for every normal countably paracompact space $Y$ if and only if $X$ is a metrizable $\sigma$-locally compact space.

These statements are no longer conjectures. Partial results appeared after the conjectures were proposed in 1976. The complete resolution of the conjectures came in 2001 in a paper by Zoli Balogh [5]. Though it is more appropriate to call these statements theorems, it is still convenient to call them conjectures. Just know that they are now known results rather open problems to be solved. The focus here is not on the evolution of the solutions. Instead, we discuss the relations among the three conjectures and why they are amazing results in the study of normality in product spaces.

As discussed below, in each of these conjectures, one direction is true based on prior known theorems (see Theorem 1, Theorem 2 and Theorem 4 below). The conjectures can be stated as follows.

Morita’s Conjecture I. Let $X$ be a space. If the product $X \times Y$ is normal for every normal space $Y$ then $X$ is a discrete space.

Morita’s Conjecture II. Let $X$ be a space. If the product $X \times Y$ is normal for every normal P-space $Y$ then $X$ is a metrizable space.

Morita’s Conjecture III. Let $X$ be a space. If the product $X \times Y$ is normal for every normal countably paracompact space $Y$ then $X$ is a metrizable $\sigma$-locally compact space.

P-spaces are defined by K. Morita [11]. He proved that a space $X$ is a normal P-space if and only if the product $X \times Y$ is normal for every metrizable space $Y$ (see theorem 2 below). Normal P-spaces are also discussed here. A space $X$ is $\sigma$-locally compact space if $X$ is the union of countably many locally compact subspaces each of which is also closed subspace of $X$.

As we will see below, these conjectures are also called duality conjectures because they are duals of known results.

[2] is a survey of Morita’s conjecture.

Duality Conjectures

Here’s three theorems that are duals to the conjectures.

Theorem 1
Let $X$ be a space. The product space $X \times Y$ is normal for every discrete space $Y$ if and only if $X$ is normal.

Theorem 2
Let $X$ be a space. The product space $X \times Y$ is normal for every metrizable space $Y$ if and only if $X$ is a normal P-space.

Theorem 3
Let $X$ be a space. The product space $X \times Y$ is normal for every metrizable $\sigma$-locally compact space $Y$ if and only if $X$ is normal countably paracompact.

The key words in red are for emphasis. In each of these three theorems, if we switch the two key words in red, we would obtain the statements for the conjectures. In this sense, the conjectures are called duality conjectures since they are duals of known results.

Theorem 1 is actually not found in the literature. It is an easy theorem. Theorem 2, found in [11], is a characterization of normal P-space (discussed here). Theorem 3 is a well known result based on the following theorem by K. Morita [10].

Theorem 4
Let $Y$ be a metrizable space. Then the product $X \times Y$ is normal for every normal countably paracompact space $X$ if and only if $Y$ is a $\sigma$-locally compact space.

We now show that Theorem 3 can be established using Theorem 4. Theorem 4 is also Theorem 3.5 in p. 111 of [2]. A proof of Theorem 4 is found in Theorem 1.8 in p. 130 of [8].

Proof of Theorem 3
$\Longleftarrow$ Suppose $X$ is normal and countably paracompact. Let $Y$ be a metrizable $\sigma$-locally compact space. By Theorem 4, $X \times Y$ is normal.

$\Longrightarrow$ This direction uses Dowker’s theorem. We give a contrapositive proof. Suppose that $X$ is not both normal and countably paracompact. Case 1. $X$ is not normal. Then $X \times \{ y \}$ is not normal where $\{ y \}$ is any one-point discrete space. Case 2. $X$ is normal and not countably paracompact. This means that $X$ is a Dowker space. Then $X \times [0,1]$ is not normal. In either case, $X \times Y$ is not normal for some compact metric space. Thus $X \times Y$ is not normal for some $\sigma$-locally compact metric space. This completes the proof of Theorem 3. $\square$

The First and Third Conjectures

The first conjecture of Morita was proved by Atsuji [1] and Rudin [13] in 1978. The proof in [13] is a constructive proof. The key to that solution is to define a $\kappa$-Dowker space. Suppose $X$ is a non-discrete space. Let $\kappa$ be the least cardinal of a non-discrete subspace of $X$. Then construct a $\kappa$-Dowker space $Y$ as in [13]. It follows that $X \times Y$ is not normal. The proof that $X \times Y$ is not normal is discussed here.

Conjecture III was confirmed by Balogh in 1998 [4]. We show here that the first and third conjectures of Morita can be confirmed by assuming the second conjecture.

Conjecture II implies Conjecture I
We give a contrapositive proof of Conjecture I. Suppose that $X$ is not discrete. We wish to find a normal space $Y$ such that $X \times Y$ is not normal. Consider two cases for $X$. Case 1. $X$ is not metrizable. By Conjecture II, $X \times Y$ is not normal for some normal P-space $Y$. Case 2. $X$ is metrizable. Since $X$ is infinite and metric, $X$ would contain an infinite compact metric space $S$. For example, $X$ contains a non-trivial convergent sequence and let $S$ be a convergence sequence plus the limit point. Let $Y$ be a Dowker space. Then the product $S \times Y$ is not normal. It follows that $X \times Y$ is not normal. Thus there exists a normal space $Y$ such that $X \times Y$ is not normal in either case. $\square$

Conjecture II implies Conjecture III
Suppose that the product $X \times Y$ is normal for every normal and countably paracompact space $Y$. Since any normal P-space is a normal countably paracompact space, $X \times Y$ is normal for every normal and P-space $Y$. By Conjecture II, $X$ is metrizable. By Theorem 4, $X$ is $\sigma$-locally compact. $\square$

The Second Conjecture

The above discussion shows that a complete solution to the three conjectures hinges on the resolution of the second conjecture. A partial resolution came in 1986 [6]. In that paper, it was shown that under V = L, conjecture II is true.

The complete solution of the second conjecture is given in a paper of Balogh [5] in 2001. The path to Balogh’s proof is through a conjecture of M. E. Rudin identified as Conjecture 9.

Rudin’s Conjecture 9. There exists a normal P-space $X$ such that some uncountable increasing open cover of $X$ cannot be shrunk.

Conjecture 9 was part of a set of 14 conjectures stated in [14]. It is also discussed in [7]. In [6], conjecture 9 was shown to be equivalent to Morita’s second conjecture. In [5], Balogh used his technique for constructing a Dowker space of cardinality continuum to obtain a space as described in conjecture 9.

The resolution of conjecture II is considered to be one of Balogh greatest hits [3].

Abundance of Non-Normal Products

One immediate observation from Morita’s conjecture I is that existence of non-normal products is wide spread. Conjecture I indicates that every normal non-discrete space $X$ is paired with some normal space $Y$ such that their product is not normal. So every normal non-discrete space forms a non-normal product with some normal space. Given any normal non-discrete space (no matter how nice it is or how exotic it is), it can always be paired with another normal space (sometimes paired with itself) for a non-normal product.

Suppose we narrow the focus to spaces that are normal and non-metrizable. Then any such space $X$ is paired with some normal P-space $Y$ to form a non-normal product space (Morita’s conjecture II). By narrowing the focus on $X$ to the non-metrizable spaces, we obtain more clarity on the paired space to form non-normal product, namely a normal P-space. As an example, let $X$ be the Michael line (normal and non-metrizable). It is well known that $X$ in this case is paired with $\mathbb{P}$, the space of irrational numbers with the usual Euclidean topology, to form a non-normal product (discussed here).

Another example is $X$ being the Sorgenfrey line. It is well known that $X$ in this case is paired with itself to form a non-normal product (discussed here). Morita’s conjectures are powerful indication that these two non-normal products are not isolated phenomena.

Another interesting observation about conjecture II is that normal P-spaces are not productive with respect to normality. More specifically, for any non-metrizable normal P-space $X$, conjecture II tells us that there exists another normal P-space $Y$ such that $X \times Y$ is not normal.

Now we narrow the focus to spaces that are metrizable but not $\sigma$-locally compact. For any such space $X$, conjecture III tells us that $X$ is paired with a normal countably paracompact space $Y$ to form a non-normal product. Using the Michael line example, this time let $X=\mathbb{P}$, the space of irrational numbers, which is a metric space that is not $\sigma$-locally compact. The paired normal and countably paracompact space $Y$ is the Michael line.

Each conjecture is about existence of a normal $Y$ that is paired with a given $X$ to form a non-normal product. For Conjecture I, the given $X$ is from a wide class (normal non-discrete). As a result, there is not much specific information on the paired $Y$, other than that it is normal. For Conjectures II and III, the given space $X$ is from narrower classes. As a result, there is more information on the paired $Y$.

The concept of Dowker spaces runs through the three conjectures, especially the first conjecture. Dowker spaces and $\kappa$-Dowker spaces provide reliable pairing for non-normal products. In fact this is one way to prove conjecture I [13], also see here. For any normal space $X$ with a countable non-discrete subspace, the product of $X$ and any Dowker space is not normal (discussed here). For any normal space $X$ such that the least cardinality of a non-discrete subspace is an uncountable cardinal $\kappa$, the product $X \times Y$ is not normal where $Y$ is a $\kappa$-Dowker space as constructed in [13], also discussed here.

In finding a normal pair $Y$ for a normal space $X$, if we do not care about $Y$ having a high degree of normal productiveness (e.g. normal P or normal countably paracompact), we can always let $Y$ be a Dowker space or $\kappa$-Dowker space. In fact, if the starting space $X$ is a metric space, the normal pair for a non-normal product (by definition) has to be a Dowker space. For example, if $X=[0,1]$, then the normal space $Y$ such that $X \times Y$ is by definition a Dowker space. The search for a Dowker space spanned a period of 20 years. For the real line $\mathbb{R}$, the normal pair for a non-normal product is also a Dowker space. For “nice” spaces such as metric spaces, finding a normal space to form non-normal product is no trivial problem.

Reference

1. Atsuji M.,On normality of the product of two spaces, General Topology and Its Relation to Modern Analysis and Algebra (Proc. Fourth Prague Topology sympos., 1976), Part B, 25–27, 1977.
2. Atsuji M.,Normality of product spaces I, in: K. Morita, J. Nagata (Eds.), Topics in General
Topology, North-Holland, Amsterdam, 81–116, 1989.
3. Burke D., Gruenhage G.,Zoli, Top. Proc., Vol. 27, No 1, i-xxii, 2003.
4. Balogh Z.,Normality of product spaces and K. Morita’s third conjecture, Topology Appl., Vol. 84, 185-198, 1998.
5. Balogh Z.,Normality of product spaces and Morita’s conjectures, Topology Appl., Vol. 115, 333-341, 2001.
6. Chiba K., Przymusinski T., Rudin M. E.Nonshrinking open covers and K. Morita’s duality conjectures, Topology Appl., Vol. 22, 19-32, 1986.
7. Gruenhage G.,Mary Ellen’s Conjectures,, Special Issue honoring the memory of Mary Ellen Rudin, Topology Appl., Vol. 195, 15-25, 2015.
8. Hoshina T.,Normality of product spaces II, in: K. Morita, J. Nagata (Eds.), Topics in General Topology, North-Holland, Amsterdam, 121–158, 1989.
9. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
10. Morita K., Products of Normal Spaces with Metric Spaces II, Sci. Rep. Tokyo Kyoiku Dagaiku Sec A, 8, 87-92, 1963.
11. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.
12. Morita K., Nagata J., Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
13. Rudin M. E., $\kappa$-Dowker Spaces, Czechoslovak Mathematical Journal, 28, No.2, 324-326, 1978.
14. Rudin M. E., Some conjectures, in: Open Problems in Topology, J. van Mill and G.M. Reed,
eds., North Holland, 184–193, 1990.
15. Telgárski R., A characterization of P-spaces, Proc. Japan Acad., Vol. 51, 802–807, 1975.

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Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

# Morita’s normal P-space

In this post we discuss K. Morita’s notion of P-space, which is a useful and interesting concept in the study of normality of product spaces.

The Definition

In [1] and [2], Morita defined the notion of P-spaces. First some notations. Let $\kappa$ be a cardinal number such that $\kappa \ge 1$. Conveniently, $\kappa$ is identified by the set of all ordinals preceding $\kappa$. Let $\Gamma$ be the set of all finite sequences $(\alpha_1,\alpha_2,\cdots,\alpha_n)$ where $n=1,2,\cdots$ and all $\alpha_i < \kappa$. Let $X$ be a space. The collection $\left\{A_\sigma \subset X: \sigma \in \Gamma \right\}$ is said to be decreasing if this condition holds: for any $\sigma \in \Gamma$ and $\delta \in \Gamma$ with

$\sigma =(\alpha_1,\alpha_2,\cdots,\alpha_n)$

$\delta =(\beta_1,\beta_2,\cdots,\beta_n, \cdots, \beta_m)$

such that $n and such that $\alpha_i=\beta_i$ for all $i \le n$, we have $A_{\delta} \subset A_{\sigma}$. On the other hand, the collection $\left\{A_\sigma \subset X: \sigma \in \Gamma \right\}$ is said to be increasing if for any $\sigma \in \Gamma$ and $\delta \in \Gamma$ as described above, we have $A_{\sigma} \subset A_{\delta}$.

The space $X$ is a P-space if for any cardinal $\kappa \ge 1$ and for any decreasing collection $\left\{F_\sigma \subset X: \sigma \in \Gamma \right\}$ of closed subsets of $X$, there exists open set $U_\sigma$ for each $\sigma \in \Gamma$ with $F_\sigma \subset U_\sigma$ such that for any countably infinite sequence $(\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots)$ where each finite subsequence $\sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n)$ is an element of $\Gamma$, if $\bigcap_{n=1}^\infty F_{\sigma_n}=\varnothing$, then $\bigcap_{n=1}^\infty U_{\sigma_n}=\varnothing$.

By switching closed sets and open sets and by switching decreasing collection and increasing collection, the following is an alternative but equivalent definition of P-spaces.

The space $X$ is a P-space if for any cardinal $\kappa \ge 1$ and for any increasing collection $\left\{U_\sigma \subset X: \sigma \in \Gamma \right\}$ of open subsets of $X$, there exists closed set $F_\sigma$ for each $\sigma \in \Gamma$ with $F_\sigma \subset U_\sigma$ such that for any countably infinite sequence $(\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots)$ where each finite subsequence $\sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n)$ is an element of $\Gamma$, if $\bigcup_{n=1}^\infty U_{\sigma_n}=X$, then $\bigcup_{n=1}^\infty F_{\sigma_n}=X$.

Note that the definition is per cardinal number $\kappa \ge 1$. To bring out more precision, we say a space $X$ is a P($\kappa$)-space of it satisfies the definition for P-space for the cardinal $\kappa$. Of course if a space is a P($\kappa$)-space for all $\kappa \ge 1$, then it is a P-space.

There is also a game characterization of P-spaces [4].

A Specific Case

It is instructive to examine a specific case of the definition. Let $\kappa=1=\{ 0 \}$. In other words, let’s look what what a P(1)-space looks like. The elements of the index set $\Gamma$ are simply finite sequences of 0’s. The relevant information about an element of $\Gamma$ is its length (i.e. a positive integer). Thus the closed sets $F_\sigma$ in the definition are essentially indexed by integers. For the case of $\kappa=1$, the definition can be stated as follows:

For any decreasing sequence $F_1 \supset F_2 \supset F_3 \cdots$ of closed subsets of $X$, there exist $U_1,U_2,U_3,\cdots$, open subsets of $X$, such that $F_n \subset U_n$ for all $n$ and such that if $\bigcap_{n=1}^\infty F_n=\varnothing$ then $\bigcap_{n=1}^\infty U_n=\varnothing$.

The above condition implies the following condition.

For any decreasing sequence $F_1 \supset F_2 \supset F_3 \cdots$ of closed subsets of $X$ such that $\bigcap_{n=1}^\infty F_n=\varnothing$, there exist $U_1,U_2,U_3,\cdots$, open subsets of $X$, such that $F_n \subset U_n$ for all $n$ and such that $\bigcap_{n=1}^\infty U_n=\varnothing$.

The last condition is one of the conditions in Dowker’s Theorem (condition 6 in Theorem 1 in this post and condition 7 in Theorem 1 in this post). Recall that Dowker’s theorem states that a normal space $X$ is countably paracompact if and only if the last condition holds if and only of the product $X \times Y$ is normal for every infinite compact metric space $Y$. Thus if a normal space $X$ is a P(1)-space, it is countably paracompact. More importantly P(1) space is about normality in product spaces where one factor is a class of metric spaces, namely the compact metric spaces.

Based on the above discussion, any normal space $X$ that is a P-space is a normal countably paracompact space.

The definition for P(1)-space is identical to one combinatorial condition in Dowker’s theorem which says that any decreasing sequence of closed sets with empty intersection has an open expansion that also has empty intersection.

For P($\kappa$)-space where $\kappa>1$, the decreasing family of closed sets are no longer indexed by the integers. Instead the decreasing closed sets are indexed by finite sequences of elements of $\kappa$. The index set $\Gamma$ would be more like a tree structure. However the look and feel of P-space is like the combinatorial condition in Dowker’s theorem. The decreasing closed sets are expanded by open sets. For any “path in the tree” (an infinite sequence of elements of $\kappa$), if the closed sets along the path has empty intersection, then the corresponding open sets would have empty intersection.

Not surprisingly, the notion of P-spaces is about normality in product spaces where one factor is a metric space. In fact, this is precisely the characterization of P-spaces (see Theorem 1 and Theorem 2 below).

A Characterization of P-Space

Morita gave the following characterization of P-spaces among normal spaces. The following theorems are found in [2].

Theorem 1
Let $X$ be a space. The space $X$ is a normal P-space if and only if the product space $X \times Y$ is normal for every metrizable space $Y$.

Thus the combinatorial definition involving decreasing families of closed sets being expanded by open sets is equivalent to a statement that is much easier to understand. A space that is normal and a P-space is precisely a normal space that is productively normal with every metric space. The following theorem is Theorem 1 broken out for each cardinal $\kappa$.

Theorem 2
Let $X$ be a space and let $\kappa \ge \omega$. Then $X$ is a normal P($\kappa$)-space if and only if the product space $X \times Y$ is normal for every metric space $Y$ of weight $\kappa$.

Theorem 2 only covers the infinite cardinals $\kappa$ starting with the countably infinite cardinal. Where are the P($n$)-spaces placed where $n$ are the positive integers? The following theorem gives the answer.

Theorem 3
Let $X$ be a space. Then $X$ is a normal P(2)-space if and only if the product space $X \times Y$ is normal for every separable metric space $Y$.

According to Theorem 2, $X$ is a normal P($\omega$)-space if and only if the product space $X \times Y$ is normal for every separable metric space $Y$. Thus suggests that any P(2)-space is a P($\omega$)-space. It seems to say that P(2) is identical to P($\kappa$) where $\kappa$ is the countably infinite cardinal. The following theorem captures the idea.

Theorem 4
Let $\kappa$ be the positive integers $2,3,4,\cdots$ or $\kappa=\omega$, the countably infinite cardinal. Let $X$ be a space. Then $X$ is a P(2)-space if and only if $X$ is a P($\kappa$)-space.

To give a context for Theorem 4, note that if $X$ is a P($\kappa$)-space, then $X$ is a P($\tau$)-space for any cardinal $\tau$ less than $\kappa$. Thus if $X$ is a P(3)-space, then it is a P(2)-space and also a P(1)-space. In the definition of P($\kappa$)-space, the index set $\Gamma$ is the set of all finite sequences of elements of $\kappa$. If the definition for P($\kappa$)-space holds, it would also hold for the index set consisting of finite sequences of elements of $\tau$ where $\tau<\kappa$. Thus if the definition for P($\omega$)-space holds, it would hold for P($n$)-space for all integers $n$.

Theorem 4 says that when the definition of P(2)-space holds, the definition would hold for all larger cardinals up to $\omega$.

In light of Theorem 1 and Dowker's theorem, we have the following corollary. If the product of a space $X$ with every metric space is normal, then the product of $X$ with every compact metric space is normal.

Corollary 5
Let $X$ be a space. If $X$ is a normal P-space, then $X$ is a normal and countably paracompact space.

Examples of Normal P-Space

Here’s several classes of spaces that are normal P-spaces.

• Metric spaces.
• $\sigma$-compact spaces (link).
• Paracompact locally compact spaces (link).
• Paracompact $\sigma$-locally compact spaces (link).
• Normal countably compact spaces (link).
• $\Sigma$-product of real lines.

Clearly any metric space is a normal P-space since the product of any two metric spaces is a metric space. Any compact space is a normal P-space since the product of a compact space and a paracompact space is paracompact, hence normal. For each of the classes of spaces listed above, the product with any metric space is normal. See the corresponding links for proofs of the key theorems.

The $\Sigma$-product of real lines $\Sigma_{\alpha<\tau} \mathbb{R}$ is a normal P-space. For any metric space $Y$, the product $(\Sigma_{\alpha<\tau} \mathbb{R}) \times Y$ is a $\Sigma$-product of metric spaces. By a well known result, the $\Sigma$-product of metric spaces is normal.

Examples of Non-Normal P-Spaces

Paracompact $\sigma$-locally compact spaces are normal P-spaces since the product of such a space with any paracompact space is paracompact. However, the product of paracompact spaces in general is not normal. The product of Michael line (a hereditarily paracompact space) and the space of irrational numbers (a metric space) is not normal (discussed here). Thus the Michael line is not a normal P-space. More specifically the Michael line fails to be a normal P(2)-space. However, it is a normal P(1)-space (i.e. normal and countably paracompact space).

The Michael line is obtained from the usual real line topology by making the irrational points isolated. Instead of using the irrational numbers, we can obtain a similar space by making points in a Bernstein set isolated. The resulting space $X$ is a Michael line-like space. The product of $X$ with the starting Bernstein set (a subset of the real line with the usual topology) is not normal. Thus this is another example of a normal space that is not a P(2)-space. See here for the details of how this space is constructed.

To look for more examples, look for non-normal product $X \times Y$ where one factor is normal and the other is a metric space.

More Examples

Based on the characterization theorem of Morita, normal P-spaces are very productively normal. Normal P-spaces are well behaved when taking product with metrizable spaces. However, they are not well behaved when taking product with non-metrizable spaces. Let’s look at several examples.

Consider the Sorgenfrey line. It is perfectly normal. Thus the product of the Sorgenfrey line with any metric space is also perfectly normal, hence normal. It is well known that the square of the Sorgenfrey line is not normal.

The space $\omega_1$ of all countable ordinals is a normal and countably compact space, hence a normal P-space. However, the product of $\omega_1$ and some compact spaces are not normal. For example, $\omega_1 \times (\omega_1 +1)$ is not normal. Another example: $\omega_1 \times I^I$ is not normal where $I=[0,1]$. The idea here is that the product of $\omega_1$ and any compact space with uncountable tightness is not normal (see here).

Compact spaces are normal P-spaces. As discussed in the preceding paragraph, the product of any compact space with uncountable tightness and the space $\omega_1$ is not normal.

Even as nice a space as the unit interval $[0,1]$, it is not always productive. The product of $[0,1]$ with a Dowker space is not normal (see here).

In general, normality is not preserved in the product space operation. the best we can ask for is that normal spaces be productively normal with respect to a narrow class of spaces. For normal P-spaces, that narrow class of spaces is the class of metric spaces. However, normal product is not a guarantee outside of the productive class in question.

Reference

1. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
2. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.
3. Morita K., Nagata J., Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
4. Telgárski R., A characterization of P-spaces, Proc. Japan Acad., Vol. 51, 802–807, 1975.

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# In between G-delta diagonal and submetrizable

This post discusses the property of having a $G_\delta$-diagonal and related diagonal properties. The focus is on the diagonal properties in between $G_\delta$-diagonal and submetrizability. The discussion is followed by a diagram displaying the relative strengths of these properties. Some examples and questions are discussed.

G-delta Diagonal

In any space $Y$, a subset $A$ is said to be a $G_\delta$-set in the space $Y$ (or $A$ is a $G_\delta$-subset of $Y$) if $A$ is the intersection of countably many open subsets of $Y$. A subset $A$ of $Y$ is an $F_\sigma$-set in $Y$ (or $A$ is an $F_\sigma$-subset of $Y$) if $A$ is the union of countably closed subsets of the space $Y$. Of course, the set $A$ is a $G_\delta$-set if and only if $Y-A$, the complement of $A$, is an $F_\sigma$-set.

The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$, which is a subset of the square $X \times X$. When the set $\Delta$ is a $G_\delta$-set in the space $X \times X$, we say that the space $X$ has a $G_\delta$-diagonal.

It is straightforward to verify that the space $X$ is a Hausdorff space if and only if the diagonal $\Delta$ is a closed subset of $X \times X$. As a result, if $X$ is a Hausdorff space such that $X \times X$ is perfectly normal, then the diagonal would be a closed set and thus a $G_\delta$-set. Such spaces, including metric spaces, would have a $G_\delta$-diagonal. Thus any metric space has a $G_\delta$-diagonal.

A space $X$ is submetrizable if there is a metrizable topology that is weaker than the topology for $X$. Then the diagonal $\Delta$ would be a $G_\delta$-set with respect to the weaker metrizable topology of $X \times X$ and thus with respect to the orginal topology of $X$. This means that the class of spaces having $G_\delta$-diagonals also include the submetrizable spaces. As a result, Sorgenfrey line and Michael line have $G_\delta$-diagonals since the Euclidean topology are weaker than both topologies.

A space having a $G_\delta$-diagonal is a simple topological property. Such spaces form a wide class of spaces containing many familiar spaces. According to the authors in [2], the property of having a $G_\delta$-diagonal is an important ingredient of submetrizability and metrizability. For example, any compact space with a $G_\delta$-diagonal is metrizable (see this blog post). Any paracompact or Lindelof space with a $G_\delta$-diagonal is submetrizable. Spaces with $G_\delta$-diagonals are also interesting in their own right. It is a property that had been research extensively. It is also a current research topic; see [7].

A Closer Look

To make the discussion more interesting, let’s point out a few essential definitions and notations. Let $X$ be a space. Let $\mathcal{U}$ be a collection of subsets of $X$. Let $A \subset X$. The notation $St(A, \mathcal{U})$ refers to the set $St(A, \mathcal{U})=\cup \{U \in \mathcal{U}: A \cap U \ne \varnothing \}$. In other words, $St(A, \mathcal{U})$ is the union of all the sets in $\mathcal{U}$ that intersect the set $A$. The set $St(A, \mathcal{U})$ is also called the star of the set $A$ with respect to the collection $\mathcal{U}$.

If $A=\{ x \}$, we write $St(x, \mathcal{U})$ instead of $St(\{ x \}, \mathcal{U})$. Then $St(x, \mathcal{U})$ refers to the union of all sets in $\mathcal{U}$ that contain the point $x$. The set $St(x, \mathcal{U})$ is then called the star of the point $x$ with respect to the collection $\mathcal{U}$.

Note that the statement of $X$ having a $G_\delta$-diagonal is defined by a statement about the product $X \times X$. It is desirable to have a translation that is a statement about the space $X$.

Theorem 1
Let $X$ be a space. Then the following statements are equivalent.

1. The space $X$ has a $G_\delta$-diagonal.
2. There exists a sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ such that for each $x \in X$, $\{ x \}=\bigcap \{ St(x, \mathcal{U}_n): n=0,1,2,\cdots \}$.

The sequence of open covers in condition 2 is called a $G_\delta$-diagonal sequence for the space $X$. According to condition 2, at any given point, the stars of the point with respect to the open covers in the sequence collapse to the given point.

One advantage of a $G_\delta$-diagonal sequence is that it is entirely about points of the space $X$. Thus we can work with such sequences of open covers of $X$ instead of the $G_\delta$-set $\Delta$ in $X \times X$. Theorem 1 is not a word for word translation. However, the proof is quote natural.

Suppose that $\Delta=\cap \{U_n: n=0,1,2,\cdots \}$ where each $U_n$ is an open subset of $X \times X$. Then let $\mathcal{U}_n=\{U \subset X: U \text{ open and } U \times U \subset U_n \}$. It can be verify that $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a $G_\delta$-diagonal sequence for $X$.

Suppose that $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a $G_\delta$-diagonal sequence for $X$. For each $n$, let $U_n=\cup \{ U \times U: U \in \mathcal{U}_n \}$. It follows that $\Delta=\bigcap_{n=0}^\infty U_n$. $\square$

It is informative to compare the property of $G_\delta$-diagonal with the definition of Moore spaces. A development for the space $X$ is a sequence $\mathcal{D}_0,\mathcal{D}_1,\mathcal{D}_2,\cdots$ of open covers of $X$ such that for each $x \in X$, $\{ St(x, \mathcal{D}_n): n=0,1,2,\cdots \}$ is a local base at the point $x$. A space is said to be developable if it has a development. The space $X$ is said to be a Moore space if $X$ is a Hausdorff and regular space that has a development.

The stars of a given point with respect to the open covers of a development form a local base at the given point, and thus collapse to the given point. Thus a development is also a $G_\delta$-diagonal sequence. It then follows that any Moore space has a $G_\delta$-diagonal.

A point in a space is a $G_\delta$-point if the point is the intersection of countably many open sets. Then having a $G_\delta$-diagonal sequence implies that that every point of the space is a $G_\delta$-point since every point is the intersection of the stars of that point with respect to a $G_\delta$-diagonal sequence. In contrast, any Moore space is necessarily a first countable space since the stars of any given point with respect to the development is a countable local base at the given point. The parallel suggests that spaces with $G_\delta$-diagonals can be thought of as a weak form of Moore spaces (at least a weak form of developable spaces).

Regular G-delta Diagonal

We discuss other diagonal properties. The space $X$ is said to have a regular $G_\delta$-diagonal if $\Delta=\cap \{\overline{U_n}:n=0,1,2,\cdots \}$ where each $U_n$ is an open subset of $X \times X$ such that $\Delta \subset U_n$. This diagonal property also has an equivalent condition in terms of a diagonal sequence.

Theorem 2
Let $X$ be a space. Then the following statements are equivalent.

1. The space $X$ has a regular $G_\delta$-diagonal.
2. There exists a sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ such that for every two distinct points $x,y \in X$, there exist open sets $U$ and $V$ with $x \in U$ and $y \in V$ and there also exists an $n$ such that no member of $\mathcal{U}_n$ intersects both $U$ and $V$.

For convenience, we call the sequence described in Theorem 2 a regular $G_\delta$-diagonal sequence. It is clear that if the diagonal of a space is a regular $G_\delta$-diagonal, then it is a $G_\delta$-diagonal. It can also be verified that a regular $G_\delta$-diagonal sequence is also a $G_\delta$-diagonal sequence. To see this, let $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ be a regular $G_\delta$-diagonal sequence for $X$. Suppose that $y \ne x$ and $y \in \bigcap_k St(x, \mathcal{U}_k)$. Choose open sets $U$ and $V$ and an integer $n$ guaranteed by the regular $G_\delta$-diagonal sequence. Since $y \in St(x, \mathcal{U}_n)$, choose $B \in \mathcal{U}_n$ such that $x,y \in B$. Then $B$ would be an element of $\mathcal{U}_n$ that meets both $U$ and $V$, a contradiction. Then $\{ x \}= \bigcap_k St(x, \mathcal{U}_k)$ for all $x \in X$.

To proof Theorem 2, suppose that $X$ has a regular $G_\delta$-diagonal. Let $\Delta=\bigcap_{k=0}^\infty \overline{U_k}$ where each $U_k$ is open in $X \times X$ and $\Delta \subset U_k$. For each $k$, let $\mathcal{U}_k$ be the collection of all open subsets $U$ of $X$ such that $U \times U \subset U_k$. It can be verified that $\{ \mathcal{U}_k \}$ is a regular $G_\delta$-diagonal sequence for $X$.

On the other hand, suppose that $\{ \mathcal{U}_k \}$ is a regular $G_\delta$-diagonal sequence for $X$. For each $k$, let $U_k=\cup \{U \times U: U \in \mathcal{U}_k \}$. It can be verified that $\Delta=\bigcap_{k=0}^\infty \overline{U_k}$. $\square$

Rank-k Diagonals

Metric spaces and submetrizable spaces have regular $G_\delta$-diagonals. We discuss this fact after introducing another set of diagonal properties. First some notations. For any family $\mathcal{U}$ of subsets of the space $X$ and for any $x \in X$, define $St^1(x, \mathcal{U})=St(x, \mathcal{U})$. For any integer $k \ge 2$, let $St^k(x, \mathcal{U})=St^{k-1}(St(x, \mathcal{U}))$. Thus $St^{2}(x, \mathcal{U})$ is the star of the star $St(x, \mathcal{U})$ with respect to $\mathcal{U}$ and $St^{3}(x, \mathcal{U})$ is the star of $St^{2}(x, \mathcal{U})$ and so on.

Let $X$ be a space. A sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ is said to be a rank-$k$ diagonal sequence of $X$ if for each $x \in X$, we have $\{ x \}=\bigcap_{j=0}^\infty St^k(x,\mathcal{U}_j)$. When the space $X$ has a rank-$k$ diagonal sequence, the space is said to have a rank-$k$ diagonal. Clearly a rank-1 diagonal sequence is simply a $G_\delta$-diagonal sequence as defined in Theorem 1. Thus having a rank-1 diagonal is the same as having a $G_\delta$-diagonal.

It is also clear that having a higher rank diagonal implies having a lower rank diagonal. This follows from the fact that a rank $k+1$ diagonal sequence is also a rank $k$ diagonal sequence.

The following lemma builds intuition of the rank-$k$ diagonal sequence. For any two distinct points $x$ and $y$ of a space $X$, and for any integer $d \ge 2$, a $d$-link path from $x$ to $y$ is a set of open sets $W_1,W_2,\cdots,W_d$ such that $x \in W_1$, $y \in W_d$ and $W_t \cap W_{t+1} \ne \varnothing$ for all $t=1,2,\cdots,d-1$. By default, a single open set $W$ containing both $x$ and $y$ is a d-link path from $x$ to $y$ for any integer $d \ge 1$.

Lemma 3
Let $X$ be a space. Let $k$ be a positive integer. Let $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ be a sequence of open covers of $X$. Then the following statements are equivalent.

1. The sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a rank-$k$ diagonal sequence for the space $X$.
2. For any two distinct points $x$ and $y$ of $X$, there is an integer $n$ such that $y \notin St^k(x,\mathcal{U}_n)$.
3. For any two distinct points $x$ and $y$ of $X$, there is an integer $n$ such that there is no $k$-link path from $x$ to $y$ consisting of elements of $\mathcal{U}_n$.

It can be seen directly from definition that Condition 1 and Condition 2 are equivalent. For Condition 3, observe that the set $St^k(x,\mathcal{U}_n)$ is the union of $k$ types of open sets – open sets in $\mathcal{U}_n$ containing $x$, open sets in $\mathcal{U}_n$ that intersect the first type, open sets in $\mathcal{U}_n$ that intersect the second type and so on down to the open sets in $\mathcal{U}_n$ that intersect $St^{k-1}(x,\mathcal{U}_n)$. A path is formed by taking one open set from each type.

We now show a few basic results that provide further insight on the rank-$k$ diagonal.

Theorem 4
Let $X$ be a space. If the space $X$ has a rank-2 diagonal, then $X$ is a Hausdorff space.

Theorem 5
Let $X$ be a Moore space. Then $X$ has a rank-2 diagonal.

Theorem 6
Let $X$ be a space. If $X$ has a rank-3 diagonal, then $X$ has a regular $G_\delta$-diagonal.

Once Lemma 3 is understood, Theorem 4 is also easily understood. If a space $X$ has a rank-2 diagonal sequence $\{ \mathcal{U}_n \}$, then for any two distinct points $x$ and $y$, we can always find an $n$ where there is no 2-link path from $x$ to $y$. Then $x$ and $y$ can be separated by open sets in $\mathcal{U}_n$. Thus these diagonal ranking properties confer separation axioms. We usually start off a topology discussion by assuming a reasonable separation axiom (usually implicitly). The fact that the diagonal ranking gives a bonus makes it even more interesting. Apparently many authors agree since $G_\delta$-diagonal and related topics had been researched extensively over decades.

To prove Theorem 5, let $\{ \mathcal{U}_n \}$ be a development for the space $X$. Let $x$ and $y$ be two distinct points of $X$. We claim that there exists some $n$ such that $y \notin St^2(x,\mathcal{U}_n)$. Suppose not. This means that for each $n$, $y \in St^2(x,\mathcal{U}_n)$. This also means that $St(x,\mathcal{U}_n) \cap St(y,\mathcal{U}_n) \ne \varnothing$ for each $n$. Choose $x_n \in St(x,\mathcal{U}_n) \cap St(y,\mathcal{U}_n)$ for each $n$. Since $X$ is a Moore space, $\{ St(x,\mathcal{U}_n) \}$ is a local base at $x$. Then $\{ x_n \}$ converges to $x$. Since $\{ St(y,\mathcal{U}_n) \}$ is a local base at $y$, $\{ x_n \}$ converges to $y$, a contradiction. Thus the claim that there exists some $n$ such that $y \notin St^2(x,\mathcal{U}_n)$ is true. By Lemma 3, a development for a Moore space is a rank-2 diagonal sequence.

To prove Theorem 6, let $\{ \mathcal{U}_n \}$ be a rank-3 diagonal sequence for the space $X$. We show that $\{ \mathcal{U}_n \}$ is also a regular $G_\delta$-diagonal sequence for $X$. Suppose $x$ and $y$ are two distinct points of $X$. By Lemma 3, there exists an $n$ such that there is no 3-link path consisting of open sets in $\mathcal{U}_n$ that goes from $x$ to $y$. Choose $U \in \mathcal{U}_n$ with $x \in U$. Choose $V \in \mathcal{U}_n$ with $y \in V$. Then it follows that no member of $\mathcal{U}_n$ can intersect both $U$ and $V$ (otherwise there would be a 3-link path from $x$ to $y$). Thus $\{ \mathcal{U}_n \}$ is also a regular $G_\delta$-diagonal sequence for $X$.

We now show that metric spaces have rank-$k$ diagonal for all integer $k \ge 1$.

Theorem 7
Let $X$ be a metrizable space. Then $X$ has rank-$k$ diagonal for all integers $k \ge 1$.

If $d$ is a metric that generates the topology of $X$, and if $\mathcal{U}_n$ is the collection of all open subsets with diameters $\le 2^{-n}$ with respect to the metrix $d$ then $\{ \mathcal{U}_n \}$ is a rank-$k$ diagonal sequence for $X$ for any integer $k \ge 1$.

We instead prove Theorem 7 topologically. To this end, we use an appropriate metrization theorem. The following theorem is a good candidate.

Alexandrov-Urysohn Metrization Theorem. A space $X$ is metrizable if and only if the space $X$ has a development $\{ \mathcal{U}_n \}$ such that for any $U_1,U_2 \in \mathcal{U}_{n+1}$ with $U_1 \cap U_2 \ne \varnothing$, the set $U_1 \cup U_2$ is contained in some element of $\mathcal{U}_n$. See Theorem 1.5 in p. 427 of [5].

Let $\{ \mathcal{U}_n \}$ be the development from Alexandrov-Urysohn Metrization Theorem. It is a development with a strong property. Each open cover in the development refines the preceding open cover in a special way. This refinement property allows us to show that it is a rank-$k$ diagonal sequence for $X$ for any integer $k \ge 1$.

First, we make a few observations about $\{ \mathcal{U}_n \}$. From the statement of the theorem, each $\mathcal{U}_{n+1}$ is a refinement of $\mathcal{U}_n$. As a result of this observation, $\mathcal{U}_{m}$ is a refinement of $\mathcal{U}_n$ for any $m>n$. Furthermore, for each $x \in X$, $\text{St}(x,\mathcal{U}_m) \subset \text{St}(x,\mathcal{U}_n)$ for any $m>n$.

Let $x, y \in X$ with $x \ne y$. Based on the preceding observations, it follows that there exists some $m$ such that $\text{St}(x,\mathcal{U}_m) \cap \text{St}(y,\mathcal{U}_m)=\varnothing$. We claim that there exists some integer $h>m$ such that there are no $k$-link path from $x$ to $y$ consisting of open sets from $\mathcal{U}_h$. Then $\{ \mathcal{U}_n \}$ is a rank-$k$ diagonal sequence for $X$ according to Lemma 3.

We show this claim is true for $k=2$. Observe that there cannot exist $U_1, U_2 \in \mathcal{U}_{m+1}$ such that $x \in U_1$, $y \in U_2$ and $U_1 \cap U_2 \ne \varnothing$. If there exists such a pair, then $U_1 \cup U_2$ would be contained in $\text{St}(x,\mathcal{U}_m)$ and $\text{St}(y,\mathcal{U}_m)$, a contradiction. Putting it in another way, there cannot be any 2-link path $U_1,U_2$ from $x$ to $y$ such that the open sets in the path are from $\mathcal{U}_{m+1}$. According to Lemma 3, the sequence $\{ \mathcal{U}_n \}$ is a rank-2 diagonal sequence for the space $X$.

In general for any $k \ge 2$, there cannot exist any $k$-link path $U_1,\cdots,U_k$ from $x$ to $y$ such that the open sets in the path are from $\mathcal{U}_{m+k-1}$. The argument goes just like the one for the case for $k=2$. Suppose the path $U_1,\cdots,U_k$ exists. Using the special property of $\{ \mathcal{U}_n \}$, the 2-link path $U_1,U_2$ is contained in some open set in $\mathcal{U}_{m+k-2}$. The path $U_1,\cdots,U_k$ is now contained in a $(k-1)$-link path consisting of elements from the open cover $\mathcal{U}_{m+k-2}$. Continuing the refinement process, the path $U_1,\cdots,U_k$ is contained in a 2-link path from $x$ to $y$ consisting of elements from $\mathcal{U}_{m+1}$. Like before this would lead to a contradiction. According to Lemma 3, $\{ \mathcal{U}_n \}$ is a rank-$k$ diagonal sequence for the space $X$ for any integer $k \ge 2$.

Of course, any metric space already has a $G_\delta$-diagonal. We conclude that any metrizable space has a rank-$k$ diagonal for any integer $k \ge 1$. $\square$

We have the following corollary.

Corollary 8
Let $X$ be a submetrizable space. Then $X$ has rank-$k$ diagonal for all integer $k \ge 1$.

In a submetrizable space, the weaker metrizable topology has a rank-$k$ diagonal sequence, which in turn is a rank-$k$ diagonal sequence in the original topology.

Examples and Questions

The preceding discussion focuses on properties that are in between $G_\delta$-diagonal and submetrizability. In fact, one of the properties has infinitely many levels (rank-$k$ diagonal for integers $k \ge 1$). We would like to have a diagram showing the relative strengths of these properties. Before we do so, consider one more diagonal property.

Let $X$ be a space. The set $A \subset X$ is said to be a zero-set in $X$ if there is a continuous $f:X \rightarrow [0,1]$ such that $A=f^{-1}(0)$. In other words, a zero-set is a set that is the inverse image of zero for some continuous real-valued function defined on the space in question.

A space $X$ has a zero-set diagonal if the diagonal $\Delta=\{ (x,x): x \in X \}$ is a zero-set in $X \times X$. The space $X$ having a zero-set diagonal implies that $X$ has a regular $G_\delta$-diagonal, and thus a $G_\delta$-diagonal. To see this, suppose that $\Delta=f^{-1}(0)$ where $f:X \times X \rightarrow [0,1]$ is continuous. Then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where $U_n=f^{-1}([0,1/n))$. Thus having a zero-set diagonal is a strong property.

We have the following diagram.

The diagram summarizes the preceding discussion. From top to bottom, the stronger properties are at the top. From left to right, the stronger properties are on the left. The diagram shows several properties in between $G_\delta$-diagonal at the bottom and submetrizability at the top.

Note that the statement at the very bottom is not explicitly a diagonal property. It is placed at the bottom because of the classic result that any compact space with a $G_\delta$-diagonal is metrizable.

In the diagram, “rank-k diagonal” means that the space has a rank-$k$ diagonal where $k \ge 1$ is an integer, which in terms means that the space has a rank-$k$ diagonal sequence as defined above. Thus rank-$k$ diagonal is not to be confused with the rank of a diagonal. The rank of the diagonal of a given space is the largest integer $k$ such that the space has a rank-$k$ diagonal. For example, for a space that has a rank-2 diagonal but has no rank-3 diagonal, the rank of the diagonal is 2.

To further make sense of the diagram, let’s examine examples.

The Mrowka space is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable (introduced here). Where is this space located in the diagram? The Mrowka space, also called Psi-space, is defined using a maximal almost disjoint family of subsets of $\omega$. We denote such a space by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$. It is a pseudocompact Moore space that is not submetrizable. As a Moore space, it has a rank-2 diagonal sequence. A well known result states that any pseudocompact space with a regular $G_\delta$-diagonal is metrizable (see here). As a non-submetrizable space, the Mrowka space cannot have a regular $G_\delta$-diagonal. Thus $\Psi(\mathcal{A})$ is an example of a space with a rank-2 diagonal but not a rank-3 diagonal sequence.

Examples of non-submetrizable spaces with stronger diagonal properties are harder to come by. We discuss examples that are found in the literature.

Example 2.9 in [2] is a Tychonoff separable Moore space $Z$ that has a rank-3 diagonal but not of higher diagonal rank. As a result of not having a rank-4 diagonal, $Z$ is not submetrizable. Thus $Z$ is an example of a space with rank-3 diagonal (hence with a regular $G_\delta$-diagonal) that is not submetrizable. According to a result in [6], any separable space with a zero-set diagonal is submetrizable. Then the space $Z$ is an example of a space with a regular $G_\delta$-diagonal that does not have a zero-set diagonal. In fact, the authors of [2] indicated that this is the first such example.

Example 2.9 of [2] shows that having a rank-3 diagonal does not imply having a zero-set diagonal. If a space is strengthened to have a rank-4 diagonal, does it imply having a zero-set diagonal? This is essentially Problem 2.13 in [2].

On the other hand, having a rank-3 diagonal implies a rank-2 diagonal. If we weaken the hypothesis to just having a regular regular $G_\delta$-diagonal, does it imply having a rank-2 diagonal? This is essentially Problem 2.14 in [2].

The authors of [2] conjectured that for each $n$, there exists a space $X_n$ with a rank-$n$ diagonal but not having a rank-$(n+1)$ diagonal. This conjecture was answered affirmatively in [8] by constructing, for each integer $k \ge 4$, a Tychonoff space with a rank-$k$ diagonal but not having a rank-$(k+1)$ diagonal. Thus even for high $k$, a non-submetrizable space can be found with rank-$k$ diagonal.

One natural question is this. Is there a non-submetrizable space that has rank-$k$ diagonal for all $k \ge 1$? We have not seen this question stated in the literature. But it is clearly a natural question.

Example 2.17 in [2] is a non-submetrizable Moore space that has a zero-set diagonal and has rank-3 diagonal exactly (i.e. it does not have a higher rank diagonal). This example shows that having a zero-set diagonal does not imply having a rank-4 diagonal. A natural question is then this. Does having a zero-set diagonal imply having a rank-3 diagonal? This appears to be an open question. This is hinted by Problem 2.19 in [2]. It asks, if $X$ is a normal space with a zero-set diagonal, does $X$ have at least a rank-2 diagonal?

The property of having a $G_\delta$-diagonal and related properties is a topic that had been researched extensively over the decades. It is still an active topic of research. The discussion in this post only touches on the surface. There are many other diagonal properties not covered here. To further investigate, check with the papers listed below and also consult with information available in the literature.

Reference

1. Arhangelskii A. V., Burke D. K., Spaces with a regular $G_\delta$-diagonal, Topology and its Applications, Vol. 153, No. 11, 1917–1929, 2006.
2. Arhangelskii A. V., Buzyakova R. Z., The rank of the diagonal and submetrizability, Comment. Math. Univ. Carolinae, Vol. 47, No. 4, 585-597, 2006.
3. Buzyakova R. Z., Cardinalities of ccc-spaces with regular $G_\delta$-diagonals, Topology and its Applications, Vol. 153, 1696–1698, 2006.
4. Buzyakova R. Z., Observations on spaces with zeroset or regular $G_\delta$-diagonals, Comment. Math. Univ. Carolinae, Vol. 46, No. 3, 469-473, 2005.
5. Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
6. Martin H. W., Contractibility of topological spaces onto metric spaces, Pacific J. Math., Vol. 61, No. 1, 209-217, 1975.
7. Xuan Wei-Feng, Shi Wei-Xue, On spaces with rank k-diagonals or zeroset diagonals, Topology Proceddings, Vol. 51, 245{251, 2018.
8. Yu Zuoming, Yun Ziqiu, A note on the rank of diagonals, Topology and its Applications, Vol. 157, 1011–1014, 2010.

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