# 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 $\mathfrak{b}=\text{non}(\text{Hurewicz})$. $\square$

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