# The space of irrational numbers is not Menger

This post could very well be titled Killing Two Birds in One Stone. We present one proof that can show two results – the set $\mathbb{P}$ of all irrational numbers does not satisfy the Menger property and that $\mathbb{P}$ is homeomorphic to the product space $\omega^\omega$.

In this previous post, we introduce the notion of Menger spaces. 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 $X$. In the definition if the open covers $\mathcal{U}_n$ are made identical, then the definition becomes that of a Lindelof space. Thus Menger implies Lindelof. For anyone encountering the notion of Menger spaces for the first time, one natural question is: are there Lindelof spaces that are not Menger? Another natural question: are there subsets of the real line that are not Menger?

The most handy example of “Lindelof but not Menger” is probably $\mathbb{P}$, the set of all irrational numbers. The way we show this fact in this previous post is by working in the product space $\omega^\omega$. This approach requires an understanding that $\mathbb{P}$ and the product space $\omega^\omega$ are identical topologically as well as working with the eventual domination order $\le^*$ in $\omega^\omega$. In this post, we give a direct proof that $\mathbb{P}$ is not Menger with the irrational numbers lying on a line. This proof will open up an opportunity to see that $\mathbb{P}$ is homeomorphic to $\omega^\omega$. The product space $\omega^\omega$ is also called the Baire space in the literature.

The sets $\mathbb{P}$ and $\omega^\omega$ are two different (but topologically equivalent) ways of looking at the irrational numbers. The first way is to look at the numbers on a line. With the rational numbers removed, the line will have countably many holes (but the holes are dense). The open sets in the line are open intervals or unions of open intervals. The second way is to view the irrational numbers as a product space – the product of countably many copies of $\omega$ with $\omega$ being the set of all non-negative numbers with the discrete topology. In many ways the product space version is easier to work with. For example, the eventual domination order $\le^*$ on $\omega^\omega$ help describe the classical covering properties such as Menger property and Hurewicz property.

A Direct Proof of Non-Menger

As mentioned above, the set $\mathbb{P}$ is the set of all irrational numbers. We let $\mathbb{Q}$ denote the set of all rational numbers.

To see that $\mathbb{P}$ does not satisfy the Menger property, we need to exhibit a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $\mathbb{P}$ such that whenever we pick, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$, the set $\{ \mathcal{V}_n: n \in \omega \}$ cannot be a cover of $\mathbb{P}$, i.e. no matter how we pick the finite $\mathcal{V}_n$, there is always an irrational number $x$ that is missed by all $\mathcal{V}_n$. The open covers $\mathcal{U}_n$ are derived inductively, starting with $\mathcal{U}_0$.

To prepare for the inductive steps, we fix a scheme of choosing convergent sequences of rational numbers. For any interval $(a,b)$ where $a, b \in \mathbb{Q}$, we choose a decreasing sequence $\{ a_n \in (a,b) \cap \mathbb{Q}: n \in \omega \}$ such that $a_n \rightarrow a$ from the right with $a_0=b$ and an increasing sequence $\{ b_n \in (a,b) \cap \mathbb{Q}: n \in \omega \}$ such that $b_n \rightarrow b$ from the left with $b_0=a$. Enumerate $\mathbb{Q}$ in a countable sequence $\{ r_0, r_1, r_2, \cdots \}$ with $r_0=0$. Several points to keep in mind when choosing such sequences.

1. In each inductive step, either the sequences of the type $a_n$ or the sequences of the type $b_n$ are chosen but not both.
2. in the $j$-th stage of the inductive process, in picking the sequence $a_n$ or $b_n$ for the interval $(a,b)$, we make sure that the rational number $r_j$ is used in the sequence $a_n$ or $b_n$ if $r_j$ is in the interval $(a,b)$ and if $r_j$ is not previously chosen. So at the end, all rational numbers are used up as endpoints of intervals in all $\mathcal{U}_n$.
3. If sequence of the type $a_n$ is chosen out of the interval $(a,b)$, we derive the subintervals $\cdots (a_3,a_2),(a_2,a_1),(a_1,a_0)$ whose union is $(a,b)$. We want the sequence $a_n$ chosen in such a way that the length of each subinterval $(a_n,a_{n-1})$ is less than 1/2 of $b-a$.
4. If sequence of the type $b_n$ is chosen out of the interval $(a,b)$, we derive the subintervals $(b_0,b_1), (b_1,b_2),(b_2,b_3),\cdots$ whose union is $(a,b)$. We want the sequence $b_n$ chosen in such a way that the length of each subinterval $(b_n,b_{n+1})$ is less than 1/2 of $b-a$.

To start, let $\mathcal{U}_0=\{ (0,1),(1,2),(2,3),\cdots,(-1,0),(-2,-1),(-3,-2),\cdots \}$. In other words, $\mathcal{U}_0$ consists of all open intervals whose endpoints are the integers and whose lengths are 1. Label the elements of $\mathcal{U}_0$ as $\mathcal{U}_0=\{ O_0,O_1,O_2,\cdots \}$.

For each $O_i=(a,b) \in \mathcal{U}_0$, choose a sequence of rational numbers $b_n$ converging to $b$ from the left (according to the scheme described above). Make sure that the rational number $r_1$ is picked if $r_1 \in (a,b)$. From this, we obtain the intervals $(b_0,b_1), (b_1,b_2),(b_2,b_3),\cdots$ covering all the irrational numbers in $(a,b)$. Label these intervals as $O_{i,0},O_{i,1},O_{i,2},\cdots$. Then $\mathcal{U}_1$ consists of all such intervals obtained from each $O_i=(a,b) \in \mathcal{U}_0$.

Here’s how to define $\mathcal{U}_2$. For each $O_{i,j}=(a,b) \in \mathcal{U}_0$, choose a sequence of rational numbers $a_n$ converging to $a$ from the right (according to the scheme described above). Make sure that the rational number $r_2$ is picked if $r_2 \in (a,b)$ and if $r_2$ has not been chosen previously. From this, we obtain the intervals $\cdots (a_3,a_2),(a_2,a_1),(a_1,a_0)$ covering all the irrational numbers in $(a,b)$. Label these intervals as $O_{i,j,0},O_{i,j,1},O_{i,j,2},\cdots$. Then $\mathcal{U}_2$ consists of all such intervals obtained from each $O_{i,j}=(a,b) \in \mathcal{U}_1$.

From here on out, we continue the same induction steps to derive $\mathcal{U}_n$, making sure that when $n$ is odd, we choose sequence $b_n$ converging to the end point $b$ from the left and when $n$ is even, we choose sequence $a_n$ converging to the end point $a$ from the right for each $O_{i_0,i_2,\cdots,i_{n-1}}=(a,b) \in \mathcal{U}_{n-1}$, producing the intervals $O_{i_0,i_2,\cdots,i_{n-1},0}$, $O_{i_0,i_2,\cdots,i_{n-1},1}$, $O_{i_0,i_2,\cdots,i_{n-1},2}$, $O_{i_0,i_2,\cdots,i_{n-1},3}\cdots$.

Now the sequence $\{ \mathcal{U}_n: n \in \omega \}$ has been defined. Note that every rational number is used as the endpoint of some open interval in some $\mathcal{U}_n$. Each $\mathcal{U}_n$ is clearly an open cover of $\mathbb{P}$. Another useful observation is that

$\bigcap_{n \in \omega} U_n=\bigcap_{n \in \omega} \overline{U_n}=\{ x \}$

whenever $U_n \in \mathcal{U}_n$ for all $n$ and $U_{n+1} \subset U_n$ for all $n$. This is because each chosen interval has length less than 1/2 of the previous interval. Furthermore, the point $x$ in the intersection must be an irrational number since all rational numbers are used up as endpoints.

Let’s pick, for each $n$, a finite $\mathcal{V}_n \subset \mathcal{U}_n$. We now find an irrational number $x$ that is not in any interval of any $\mathcal{V}_n$.

Since $\mathcal{V}_0$ is finite, choose some $O_{m_0} \in \mathcal{U}_0$ such that $O_{m_0} \notin \mathcal{V}_0$. Since $\mathcal{V}_1$ is finite, choose some $O_{m_0,m_1} \in \mathcal{U}_1$ such that $O_{m_0,m_1} \notin \mathcal{V}_1$. In the same manner, choose $O_{m_0,m_1,m_2} \in \mathcal{U}_2$ such that $O_{m_0,m_1,m_2} \notin \mathcal{V}_2$.

Continuing the inductive process, we have open intervals $O_{m_0},O_{m_0,m_1},O_{m_0,m_1,m_2},\cdots$ from $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$, respectively. Furthermore, for each $n$, $O_{m_0,\cdots,m_n} \notin \mathcal{V}_n$. Then the intersection

$\bigcap_{n \in \omega} O_{m_0,\cdots,m_n}=\bigcap_{n \in \omega} \overline{O_{m_0,\cdots,m_n}}=\{ x \}$

must be an irrational number $x$. This is a real number that is not covered by $\mathcal{V}_n$ for all $n$. This shows that $\{ \mathcal{V}_n: n \in \omega \}$ cannot be an open cover of $\mathbb{P}$. This completes the proof that $\mathbb{P}$ does not satisfy the Menger property. $\square$

A Homeomorphism

In defining the sequence $\{ \mathcal{U}_n: n \in \omega \}$ of open covers of $\mathbb{P}$, we have also define a one-to-one correspondence between $\mathbb{P}$ and the product space $\omega^\omega$.

First, each $x \in \mathbb{P}$ is uniquely identified with a sequence of non-negative integers $f_0,f_1,f_2,\cdots$.

$x \mapsto (f_0,f_1,f_2,\cdots)=f \in \omega^\omega$

Observe that each irrational number $x$ belongs to a unique element in each $\mathcal{U}_n$. Thus the sequence $f_0,f_1,f_2,\cdots$ is simply the subscripts of the open intervals from the open covers $\mathcal{U}_n$ such that $x \in O_{f_0,\cdots,f_n}$ for each $n$. With the way the intervals $O_{f_0,\cdots,f_n}$ are chosen, it must be the case that $O_{f_0} \supset O_{f_1} \supset O_{f_2} \cdots$. Furthermore the intervals collapse to the point $x$.

$\bigcap_{n \in \omega} O_{f_0,\cdots,f_n}=\bigcap_{n \in \omega} \overline{O_{f_0,\cdots,f_n}}=\{ x \}$……………(1)

The mapping goes in the reverse direction too. For each $f=(f_0,f_1,f_2,\cdots) \in \omega^\omega$, we can use $f$ to obtain the open intervals $O_{f_0,\cdots,f_n}$. These intervals collapse to a single point, which is the irrational number that is associated with $f=(f_0,f_1,f_2,\cdots)$.

Denote this mapping by $H: \mathbb{P} \rightarrow \omega^\omega$. For each $x \in \mathbb{P}$, $H(x)=f$ is derived in the manner described above. The function $H$ maps $\mathbb{P}$ onto $\omega^\omega$. It follows that $H$ is a one-to-one map and that both $H$ and the inverse $H^{-1}$ are continuous. To see this, we need to focus on the open sets in both the domain and the range.

For the correspondence $H(x)=(f_0,f_1,f_2,\cdots)=f$, we consider the two sets:

$\mathcal{B}_x=\{ O_{f_0}, O_{f_0,f_1}, O_{f_0,f_1,f_2}, \cdots, O_{f_0,\cdots,f_n}, \cdots \}$

$\mathcal{V}_f=\{ E(f,n): n \in \omega \}$

where $E(f,n)=\{ g \in \omega^\omega: f_j=g_j \ \forall \ j \le n \}$ for each $n$. Note that $\mathcal{B}_x$ is a local base at $x \in \mathbb{P}$. This is because the intervals $O_{f_0,\cdots,f_n}$ have lengths converging to zero and they collapse to the point $x$ in the manner described in (1) above. On the other hand, $\mathcal{V}_f$ is a local base at $H(x)=f \in \omega^\omega$. Furthermore, there is a clear correspondence between $\mathcal{B}_x$ and $\mathcal{V}_f$. Note that $H(O_{f_0,\cdots,f_n})=E(f,n)$. This means that both $H$ and the inverse $H^{-1}$ are continuous. Since $\{ \mathcal{B}_x: x \in \mathbb{P} \}$ is a base for $\mathbb{P}$, it is clear the map $H$ is one-to-one.

This previous post has a shorter (but similar) derivation of the homeomorphism between $\mathbb{P}$ and $\omega^\omega$.

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# Hurewicz spaces are sigma-compact-like

Menger spaces and Hurewicz spaces are situated in between $\sigma$-compactness and the Lindelof property.

$\sigma \text{-} \bold C \bold o \bold m \bold p \bold a \bold c \bold t \Longrightarrow \bold H \bold u \bold r \bold e \bold w \bold i \bold c \bold z \Longrightarrow \bold M \bold e \bold n \bold g \bold e \bold r \Longrightarrow \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f$

In the preceding post, we discuss a characterization of Menger spaces that is a Lindelof-like property. The attention is now turned to Hurewicz spaces. We discuss a $\sigma$-compact-like characterization of Hurewicz spaces among subsets of the real line.

It seems there is a “symmetry” here. The property that is closer to Lindelof property is Lindelof-like, while the one closer to $\sigma$-compactness is $\sigma$-compact-like. In addition to proving the $\sigma$-compact-like characterization, we use several examples to demonstrate this property. One of the examples is the Lusin sets. We also briefly discuss the Hurewicz problem.

Let $X$ be a space. The space $X$ is a Menger space (has the Menger property) if for each 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 $X$. The space $X$ is a Hurewicz space (has the Hurewicz property) if for each 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 $\{ \cup \mathcal{V}_n: n \in \omega \}$ is a $\gamma$-cover of $X$, i.e. each $x \in X$ belongs to $\cup \mathcal{V}_n$ for all but finitely many $n$.

Two previous posts (here and here) discuss basic facts of Menger spaces and Hurewicz spaces. Consider the following theorems.

Theorem 1
Let $X \subset \mathbb{R}$. Then the following conditions are equivalent.

1. The set $X$ has the Hurewicz property.
2. For every $G_\delta$-set $G$ with $X \subset G$, there exists an $F_\sigma$-set $F$ such that $X \subset F \subset G$.

This theorem is Theorem 5.7 in [3]. In the real line, any $F_\sigma$-set is $\sigma$-compact. Hurewicz subsets of the real line may not be $\sigma$-compact but are “approximated” by $\sigma$-compact sets in the manner described in Theorem 1. Whenever a Hurewicz subset is situated in a $G_\delta$-set, we can always find a $\sigma$-compact set cushioned in between.

Theorem 2
Let $X \subset \mathbb{R}$. If $X$ has empty interior and has the Hurewicz property, then $X$ is of first category.

Theorem 2 is useful in showing that Lusin sets are not Hurewicz spaces.

Examples

In order to understand more about this $\sigma$-compact-like property of Hurewicz subsets of the real line, consider the following examples.

Example 1
Clearly, any $F_\sigma$-set in the real line satisfies condition 2 in Theorem 1.

Example 2
In the real line, any $G_\delta$-set $X$ that is not $F_\sigma$ does not satisfy condition 2 of Theorem 1.

For any $G_\delta$-set $X$ that is not an $F_\sigma$-set, there can be no $F_\sigma$ cushioned in between $X$ and $X$. An example of such a set is the set $\mathbb{P}$ of all irrational numbers. We already know that $\mathbb{P}$ does not have the Menger property (see Theorem 4 here) and hence does not not have the Hurewicz property.

Example 3
Not only $G_\delta$-sets $X$ that are not $F_\sigma$ do not have the $\sigma$-compact-like property of Theorem 1, all Borel sets that that are not $F_\sigma$ do not satisfy condition 2 of Theorem 1. This is because such sets are not Menger. This observation is made in Example 4 of the preceding post.

Example 4
Any Bernstein set in the real line does not satisfy condition 2 of Theorem 1.

Bernstein sets are not Menger. This is discussed in Example 5 of the preceding post. Thus they are not Hurewicz. As illustration, we demonstrate that $\sigma$-compact sets cannot be cushioned in between Bernstein sets and $G_\delta$-sets. Let $X \subset \mathbb{R}$ be a Bernstein set. The complement $\mathbb{R} \backslash X$ is also a Bernstein set and hence dense in the real line. Let $D \subset \mathbb{R} \backslash X$ be a countable dense subset. Then $\mathbb{R} \backslash D$ is a dense $G_\delta$-set containing $X$.

Suppose that $X \subset F \subset \mathbb{R} \backslash D$ for some $F_\sigma$-set $F$. Let $F=\bigcup_{n \in \omega} F_n$ where each $F_n$ is a closed set. Observe that each $F_n$ is nowhere dense in the real line. To see this, let $U$ be an open interval. Since $D$ is dense, choose $x \in U \cap D$. Since $x \notin F_n$, choose open interval $V$ such that $x \in V$, $V \cap F_n=\varnothing$ and $V \subset U$.

Consider $W=\mathbb{R} \backslash F$. The set $W$ is a $G_\delta$-set. If $W$ is countable, then the real line is the union of countably many nowhere dense sets $F_n$ and a countable set which means that the real line is a first category set, a contradiction. Thus $W$ must be uncountable. Since $W$ is an uncountable $G_\delta$-set, $W$ contains a Cantor set $C$. Since $X$ is a Bernstein set, $X \cap C \ne \varnothing$. But $C$ cannot contain points of $X$ since $C \subset W=\mathbb{R} \backslash F$. Thus there can be no $F_\sigma$-set cushioned between the Bernstein set $X$ and the $G_\delta$-set $\mathbb{R} \backslash D$.

Example 5
Any Lusin set does not satisfy condition 2 of Theorem 1 and hence is not Hurewicz.

Let $X \subset \mathbb{R}$ where $X$ is uncountable. The set $X$ is a Lusin set (alternative spelling: Luzin) if for each $F$ that is of first category in the real line, $X \cap F$ is countable.

Suppose $X$ is a Lusin set. We show that there can be no $F_\sigma$-set cushioned between $X$ and some $G_\delta$-set containing $X$. Note that the Lusin set $X$ has empty interior. Otherwise the Lusin set would contain an open interval. This is impossible. The reason is that that open interval would contain a Cantor set, which is nowhere dense. By Theorem 2, if $X$ is Hurewicz, then it would be of first category. But a Lusin set by definition cannot be of first category. Thus $X$ cannot be Hurewicz and thus does not satisfy the $\sigma$-compact-like property.

Example 6
Let $X \subset \mathbb{R}$ such that $X$ is a Sierpinski set. Then $X$ has the Hurewicz property hence satisfies the $\sigma$-compact-like property.

The set $X$ is a Sierpinski set if $X$ is uncountable and $X \cap L$ is countable for every Lebesgue measure zero set $L$. See Theorem 2.3 in [10] for a proof that any Sierpinski set is Hurewicz. My note. Source is Menger’s and Hurewicz’s problems: solutions from the book and refinements.

Hurewicz’s Problem

Example 5 concerns Lusin sets and is a good lead-in to the Hurewicz’s Problem.

In 1924, Karl Menger [4] conjectured that in metric spaces, the Menger property is equivalent to $\sigma$-compactness. In 1928 Sierpinski [9] pointed out that when continuum hypothesis (CH) holds, Menger’s conjecture is false. The counterexample was the Lusin sets, which were shown to exist under CH by N. Lusin in 1914. Thus Lusin sets are Menger spaces that are not $\sigma$-compact (the steps are worked out in Example 4 in this previous post). Lusin sets do not exist under Martin’s axiom and the negation of CH. Thus Lusin sets are only consistent counterexamples to Menger’s conjecture. The first ZFC counterexample to Menger’s conjecture was given by Miller and Fremlin [5] (the steps are worked out in Example 5 in this previous post).

Example 5 above shows that Lusin sets are not Hurewicz spaces. This shows that Lusin sets are solutions to another problem. The Hurewicz problem is the following question:

Is there a metric space that is Menger but not Hurewicz?

Lusin sets answer this question in the affirmative (first solved by Sierpinski). The first ZFC example of a Menger space that is not Hurewicz is found in Chaber and Pol, which is unpublished source. Also see Theorem 5.3 in [10] for a combinatorial proof that contains the essence of the proof of Chaber and Pol. The proof in Chaber and Pol’s unpublished note and Theorem 5.3 in [10] are dichotomic. So no explicit examples are provided in these proofs. These proofs exploit the opposing cases of a set-theoretic statement. For example, when $\mathfrak{b}<\mathfrak{d}$, an example of a Menger but not Hurewicz space can be derived. On the other hand, when $\mathfrak{b}=\mathfrak{d}$, another example of a Menger but not Hurewicz space can be derived. The first explicit ZFC example of a Menger but not Hurewicz space can be found in [11].

Note that the Hurewicz problem is not the same as the Hurewicz’s conjecture. The latter is a conjecture made by W. Hurewicz in 1925 that a non-compact metric space is $\sigma$-compact if, and only if, it is a Hurewicz space. Sierpinski sets pointed out in Example 6 provide consistent counterexample to this conjecture. The first ZFC counterexample is provided in [3]. The proof in [3] is a dichotomic proof. See [10] for a more detailed discussion of Hurewicz’s conjecture.

The Proof Section

We now prove Theorem 1 in the following form. Proof of Theorem 2 follows the proof of Theorem 1.

Theorem 1
Let $X \subset \mathbb{R}$. Then the following conditions are equivalent.

1. The set $X$ has the Hurewicz property.
2. For each sequence $\{ \mathcal{U}_n: n \in \omega \}$ of $\gamma$-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 $X$.
3. For every $G_\delta$-set $G$ with $X \subset G$, there exists an $F_\sigma$-set $F$ such that $X \subset F \subset G$.

The direction $1 \rightarrow 2$ is straightforward. Condition 2 is the Hurewicz property restricted to sequences of $\gamma$-covers. Note that any $\gamma$-cover is also an open cover. So if a space satisfies the full Hurewicz property, it will then satisfy the restricted property.

$2 \rightarrow 3$
Suppose that $X \subset G$ such that $G=\bigcap_{n \in \omega} O_n$ where each $O_n$ is an open subset of the real line. We can also require that $O_0 \supset O_1 \supset O_2 \supset \cdots$. From the $G_\delta$-set $G$, we construct a sequence $\{ \mathcal{U}_n: n \in \omega \}$ of $\gamma$-covers of $X$.

To define $\mathcal{U}_n$, do the following. For each $x \in X$, choose an open interval $E_x^n$ such that $x \in E_x^n$ and $\overline{E_x^n} \subset O_n$. For the open cover $\{E_x^n: x \in X \}$, we can find a countable subcover $\{ E_{x_{n,k}}^n: k \in \omega \}$. To simplify notation, we rename $E_{x_{n,k}}^n=F_{k}^n$. For each $k$, define $H_k^n=\bigcup_{j \le k} F_{j}^n$. Let $\mathcal{U}_n=\{ H_k^n: k \in \omega \}$. Note that $\mathcal{U}_n$ is a $\gamma$-cover of $X$ and that for each $k$, $\overline{H_k^n} \subset O_n$.

For each $\mathcal{U}_n$, note that the open sets $H_k^n$ are increasing, i.e. for $i, $H_i^n \subset H_j^n$. So picking a finite subset of $\mathcal{U}_n$ is the same as picking one element (the largest one in the finite set). Using condition 2, we can choose, for each $n$, $H_{t(n)}^n \in \mathcal{U}_n$ such that $\{ H_{t(n)}^n: n \in \omega \}$ is a $\gamma$-cover of $X$.

For each $n$, define $L_n=\bigcap_{m \ge n} \overline{H_{t(m)}^m}$. Note that each closure in the intersection is contained in the open set $O_m$. To see this,

$\overline{H_{t(m)}^m}=\overline{\bigcup_{j \le t(m)} F_j^m}=\bigcup_{j \le t(m)} \overline{F_j^m}=\bigcup_{j \le t(m)} \overline{E_{x_{m,j}}^m} \subset O_m$

Thus $L_n=\bigcap_{m \ge n} \overline{H_{t(m)}^m} \subset \bigcap_{m \ge n} O_m \subset \bigcap_{k \in \omega} O_k=G$. Observe there is now an $F_\sigma$-set cushioned between $X$ and $G$.

$X \subset \bigcup_{n \in \omega} L_n \subset G$

$3 \rightarrow 1$
Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$ such that elements of each $\mathcal{U}_n$ are open subsets of the real line. By condition 3, there is an $F_\sigma$-set $L$ such that

$X \subset L=\bigcup_{n \in \omega} L_n \subset \bigcap_{n \in \omega} (\bigcup \mathcal{U}_n)$

Since the real line is $\sigma$-compact, we can assume each $L_n$ is compact. For each $n$, we can choose finite $\mathcal{V}_n \subset \mathcal{U}_n$ such that $L_n \subset \bigcup \mathcal{V}_n$. It follows that $\bigcup_{n \in \omega} \mathcal{V}_n$ is an open cover of $X$. This completes the proof that $X$ is a Hurewicz space. $\square$

Theorem 2
Let $X \subset \mathbb{R}$. If $X$ has empty interior and has the Hurewicz property, then $X$ is of first category.

Suppose $X$ has empty interior and has the Hurewicz property. Since $X$ has empty interior, $\mathbb{R} \backslash X$ is a dense subset of the real line. Let $D$ be a countable dense subset of $\mathbb{R} \backslash X$. Then $\mathbb{R} \backslash D$ is a $G_\delta$-set and $X \subset \mathbb{R} \backslash D$. By Theorem 1, there exists an $F_\sigma$-set $F$ such that $X \subset F \subset \mathbb{R} \backslash D$. Let $F=\bigcup_{n \in \omega} F_n$ where each $F_n$ is a closed set. By the same argument in Example 4 and Example 5, each $F_n$ is a nowhere dense set. Then $X$ is a set of first category since it is a subset of $F$. $\square$

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9. Sierpinski W., Sur un probleme de M. Menger, Fund. Math., 8, 223-224, 1926.
10. Tsaban B., Menger’s and Hurewicz’s Problems: Sulutions From “The Book” and Refinements, Contemporary Math. 533, 211–226, 2011.
11. Tsaban B., Zdomskyy L., Scales, fields, and a problem of Hurewicz, J. Eur. Math. Soc. 10, 837–866, 2008.

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# Menger spaces are Lindelof-like

This post extends two previous posts on Menger spaces (here and here). We show that the Menger property, though stronger than the Lindelof property, is equivalent to a Lindelof-like property.

Let $X$ be a space. The space $X$ is a Menger space (has the Menger property) if for each 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 $X$. It is sometimes easier to work with an equivalent definition of Menger spaces (found in Theorem 2 here).

The space $X$ is a Hurewicz space (has the Hurewicz property) if for each 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 $\{ \cup \mathcal{V}_n: n \in \omega \}$ is a $\gamma$-cover of $X$, i.e. each $x \in X$ belongs to $\cup \mathcal{V}_n$ for all but finitely many $n$.

It is clear from definitions that

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

The Menger property is strictly stronger than Lindelof. See here for examples that are Lindelof but not Menger. One of them is the space $\mathbb{P}$ of all irrational numbers. These examples show that having the property that every open cover has a countable subcover does not quite rise to the Menger property. Consider the following theorem.

Theorem 1
Let $X$ be a space. Then the space $X$ is a Menger space, if, and only if, for every Menger subspace $M$ of $X$, and for every $G_\delta$-subset $G_M$ of $X$ such that $M \subset G_M$, the cover $\{ G_M: M \subset X \text{ and } M \text{ is Menger} \}$ has a countable subcover.

The proof of Theorem 1 is found at the end of the post. Given a cover described in Theorem 1, every Menger subset is contained in one element of the cover. The covers in Theorem 1 can be called $G_\delta$ covers since the elements of such covers are $G_\delta$ sets. This is a “Lindelof” property using $G_\delta$ covers. To get better intuition about this “Lindelof” characterization, we look at some examples.

Example 1
Any $\sigma$-compact space satisfies the covering property in Theorem 1.

In a $\sigma$-compact space, each of the countably many compact sets that makes up the space is a Menger subspace. Suppose that $X=\bigcup_{n \in \omega} X_n$ where each $X_n$ is compact. For any $G_\delta$ cover $\{ G_M: M \subset X \text{ and } M \text{ is Menger} \}$, $\{ G_{X_0}, G_{X_1}, G_{X_2}, \cdots \}$ would be a countable subcover. Similarly, any space that is the union of countably many Menger subspaces would satisfy the property in Theorem 1.

Example 2
Let $X \subset \mathbb{R}$. The set $X$ is $\sigma$-compact if and only if $X$ is an $F_\sigma$-set. This is because the real line is $\sigma$-compact. Thus any $F_\sigma$-subset of the real line satisfies the covering property in Theorem 1.

Example 3
The space $\mathbb{P}$ of all irrational numbers is not a Menger space. Thus it does not satisfy the covering property in Theorem 1.

The set $\mathbb{P}$ is homeomorphic to $\omega^\omega$, which is a dominating set and thus not Menger. See Theorem 4 in this post. The set of irrational numbers $\mathbb{P}$ is a $G_\delta$-set that is not $F_\sigma$. The next example discusses other Borel sets that are not $F_\sigma$.

Example 4
Among the Borel sets in the real line, $F_\sigma$ sets are the only ones that have the Menger property. Thus any Borel set that is not $F_\sigma$ is not Menger. The only Borel sets satisfying the covering property in Theorem 1 are precisely the $F_\sigma$ sets.

The Borel sets in the real line are constructed using the open sets and closed sets through the operations of countable unions and countable intersections. The following shows the Borel sets from the first few steps.

open sets, $\text{ }$ closed sets, $F_\sigma$, $\text{ }$ $G_\delta$, $\text{ }$ $G_{\delta \sigma}$, $\text{ }$ $F_{\sigma \delta}$, $\text{ }$ $F_{\sigma \delta \sigma}$, $\text{ }$ $G_{\delta \sigma \delta}$, $\text{ }$ $\cdots$

In the above steps, the subscript $\sigma$ denotes union and the subscript $\delta$ denotes intersection. After the steps for the finite ordinals are completed ($n < \omega$), the construction continues on through all the countable ordinals $\alpha<\omega_1$ (through taking countable unions and taking countable intersections of all the Borel sets that come before the $\alpha$th step). We claim that the only Borel sets that are Menger are the $F_\sigma$-sets, which of course include the open sets and the closed sets. This fact follows from the following theorem [3].

Theorem 2
Let $X \subset \mathbb{R}$. If the set $X$ is an analytic set, then either $X$ is $\sigma$-compact or $X$ contains a closed copy of the space $\mathbb{P}$ of all irrational numbers.

A subset $X$ of the real line is an analytic set if it is the continuous image of $\mathbb{P}$, the space of all irrational numbers. Any Borel set is an analytic set. Closed subsets of any Menger space are Menger. Thus for any analytic set $X$, if $X$ is Menger, it cannot have $\mathbb{P}$ as a closed subspace and thus must be $\sigma$-compact. It follows that Borel sets that are not $F_\sigma$ are not Menger spaces.

Example 5
Borel sets are “nice” sets. For example, they are measurable sets. Let’s look at a pathological example. We turn to the Bernstein sets. We show that any Bernstein set does not have the Menger property and hence does not satisfy the covering property in Theorem 1. A subset $B$ of the real line is a Bernstein set if both $B$ and the complement of $B$ intersect every uncountable closed subset of the real line. Such a set can be constructed using transfinite induction (see here for an illustration).

Let $B \subset \mathbb{R}$ be a Bernstein set. Note that the complement $\mathbb{R} \backslash B$ is also a Bernstein set. Any Bernstein set is dense in the real line. Choose a countable dense subset $D$ of $\mathbb{R} \backslash B$. Note that $\mathbb{R} \backslash D$ is a dense $G_\delta$-set and is homeomorphic to $\mathbb{P}$, the set of all irrational numbers. The set $\mathbb{P}$ is in turn homeomorphic to $\omega^\omega$. With $B \subset \mathbb{R} \backslash D$, we can consider $B$ as a subset of $\omega^\omega$.

Let $B \subset \omega^\omega$ be a Bernstein set. We show that it is a dominating set. By Theorem 4 in this post, it is not a Menger set. To this end, let $f \in \omega^\omega$. Consider the compact subset $C$ of $\omega^\omega$ where $C$ is the following set.

$C=\{ f(0)+1, f(0)+2 \} \times \{ f(1)+1, f(1)+2 \} \times \{ f(2)+1, f(2)+2 \} \times \cdots$

The set $C$ is a Cantor set and hence is an uncountable closed set. Thus $B \cap C \ne \varnothing$. Let $g \in B \cap C$. It is clear that $f \le^* g$. Thus $B$ is a dominating set and hence not Menger.

Remarks

There exist uncountable subsets of the real line that have the Menger property but are not $\sigma$-compact (see here). These are ZFC counterexamples to the conjecture of Menger. In 1924, Karl Menger conjectured that in metric spaces, Menger spaces are precisely the $\sigma$-compact spaces. Based on the discussion for the above examples, counterexamples to Menger’s conjecture are not to be found among Borel sets or analytic sets.

As an application of Theorem 1, consider a notion called Alster spaces. Let $X$ be a space. A collection $\mathcal{G}$ of $G_\delta$-subsets of $X$ is called an Alster cover if for any compact subset K of $X$, $K \subset G$ for some $G \in \mathcal{G}$. Such a collection of $G_\delta$-sets is necessarily a cover of the space $X$. A space $X$ is said to be an Alster space if for every Alster cover $\mathcal{G}$ of $X$, there is a countable $\mathcal{U} \subset \mathcal{G}$ such that $\mathcal{U}$ is a cover of $X$. In other words, such a space has the property that every Alster cover has a countable subcover. Alster spaces were introduced by K. Alster [1] to characterize the class of productively Lindelof spaces (a space $L$ is productively Lindelof if $L \times Y$ is Lindelof for every Lindelof space $Y$).

With the $G_\delta$ covering of Menger spaces (as described in Theorem 1), there is a clear connection between Alster spaces and Menger spaces. Observe that any $G_\delta$ covering as described in Theorem 1 is also an Alster cover. Thus it follows that every Alster space is a Menger space. Menger spaces are D-spaces [2]. Thus Alster spaces are D-spaces. Thus the notion of Menger spaces is connected to the notion of Alster spaces and other notions that are used in the study of productively Lindelof spaces.

The Proof Section

We now give a proof of Theorem 1.

Theorem 1
Let $X$ be a space. Then the space $X$ is a Menger space, if, and only if, for every Menger subspace $M$ of $X$, and for every $G_\delta$-subset $G_M$ of $X$ such that $M \subset G_M$, the cover $\{ G_M: M \subset X \text{ and } M \text{ is Menger} \}$ has a countable subcover.

The direction $\Rightarrow$ is clear. If $X$ is Menger, then $G_X$ belongs to any $G_\delta$ cover $\{ G_M: M \subset X \text{ and } M \text{ is Menger} \}$.

$\Leftarrow$
Let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. For each Menger $M \subset X$, each $\mathcal{U}_n$ is a open cover of $M$. Thus we can choose, for each $n$, a finite $\mathcal{V}_n^M \subset \mathcal{U}_n$ such that for each $x \in M$, $x \in \bigcup \mathcal{V}_n^M$ for infinitely many $n$. We are using a characterization of Menger space found here (see Theorem 2). Define $G_M=\bigcap_{n \in \omega} W_n^M$ where each $W_n^M$ is defined as follows:

$W_n^M=\bigcup_{j \ge n} (\bigcup \mathcal{V}_j^M)$

Note that for each $n$, $M \subset W_n^M$. Thus $M \subset G_M$. Consider the $G_\delta$ cover just defined.

$\{ G_M: M \subset X \text{ and } M \text{ is Menger} \}$

By assumption it has a countable subcover $\{ G_{M_0},G_{M_1},G_{M_2},\cdots \}$. For each $n$, define $\mathcal{V}_n$ by $\mathcal{V}_n=\bigcup_{k \le n} \mathcal{V}_n^{M_k}$. Note that each $\mathcal{V}_n$ is finite and $\mathcal{V}_n \subset \mathcal{U}_n$. We now show that for each $x \in X$, we have $x \in \bigcup \mathcal{V}_n$ for infinitely many $n$.

Let $x \in X$. Then $x \in G_{M_j}=\bigcap_{n \in \omega} W_n^{M_j}$ for some $j$. By the definition of $W_n^{M_j}$, we have $x \in \bigcup \mathcal{V}_m^{M_j}$ for infinitely many $m$. Then it follows that $x \in \bigcup \mathcal{V}_m^{M_j}$ for infinitely many $m$ where $m \ge j$. Note that $x \in \bigcup \mathcal{V}_m^{M_j}$ for $m \ge j$ means that $x \in \bigcup \mathcal{V}_m$. Therefore, $x \in \bigcup \mathcal{V}_m$ for infinitely many $m$. $\square$

Reference

1. Alster K., On the class of all spaces of weight not greater than $\omega_1$ whose cartesian product with every Lindelof space is Lindelof, Fund. Math., 129, 133–140, 1988.
2. Aurichi L., D-Spaces, Topological Games, and Selection Principles, Topology Proc., 36, 107–122, 2010.
3. Kechris A. S., Classical descriptive set theory, Graduate Texts in Mathematics, 156, Springer-Verlag, New York, 1995.

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# Counterexamples to the conjecture of Menger

This post is a continuation of the preceding post on Menger and Hurewicz spaces. We aim to provide the details on Example 4 and Example 5 in that post. These are counterexamples to Karl Menger’s conjecture concerning Menger spaces.

In 1924, Karl Menger introduced a basis covering property in metric spaces. He conjectured that among metric spaces this basis covering property is equivalent to $\sigma$-compactness. In 1925 Witold Hurewicz introduced a selection principle that led to a definition called Menger space as defined in the preceding post and below, which he showed to be equivalent to Menger’s basis covering property. In 1928, Sierpinski showed that Continuum Hypothesis (CH) implies that Menger’s conjecture is false. The counterexample is the Lusin sets. This is Example 4 in the preceding post, which is further discussed here. Lusin sets can be constructed using CH. On the other hand, Lusin sets do not exist under Martin’s axiom and the negation of CH. Thus Lusin sets, as counterexample to Menger’s conjecture, are only consistent counterexamples.

In a 1988 paper, Miller and Fremlin [3] gave the first ZFC counterexample to Menger’s conjecture. This is the subject of Example 5 in the preceding post. We plan to give a more detailed account of this example here.

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 $X$. If in the definition the $\mathcal{V}_n$ can always be made to consist of only one element, then the space is said to be a Rothberger space. It is clear from definition that $\sigma$-compact $\rightarrow$ Menger $\rightarrow$ Lindelof. On the other hand, Rothberger $\rightarrow$ Menger. The following diagram is from the preceding post.

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}

Example 4 – Lusin Sets

In this example, we show that any Lusin set is a Menger space that is not $\sigma$-compact. This fact is established through a series of claims listed below. The proofs are found at the end of the post.

Let $X \subset \mathbb{R}$ be uncountable. The set $X$ is said to be a Lusin set (alternative spelling: Luzin) if the intersection of $X$ with any first category subset of the real line is countable, i.e. for every first category subset $A$ of $\mathbb{R}$, $X \cap A$ is countable. In other words, an uncountable subset $X$ of the real line is a Lusin set if every first category subset of the real line can contain at most countably many points of $X$. Thus any Lusin set (if it exists) must necessarily be a subset of the real line that is of second category. Sets of first category are also called meager sets. For more information about Lusin sets, see [2].

Let $X \subset \mathbb{R}$. Let $B \subset \mathbb{R}$. The set $X$ is said to be concentrated on the set $B$ if for any open set $U$ with $B \subset U$, $X \backslash U$ is countable. In other words, $X$ is concentrated on the set $B$ if any open set containing the $B$ contains all but countably many points of $X$.

Claim 1
Let $X \subset \mathbb{R}$. If $X$ is a Lusin set, then $X$ is concentrated on any countable dense subset of $X$.

Claim 2
Let $X \subset \mathbb{R}$. Suppose that $D$ is a countable dense subset of $X$. Then if $X$ is concentrated on the set $D$, then $X$ is a Rothberger space and hence a Menger space.

Claim 3
Let $X \subset \mathbb{R}$. If $X$ is an uncountable $\sigma$-compact, then $X$ contains a Cantor set.

Note. By Cantor set here, we mean any subset of the real line that is homeomorphic to the middle third Cantor in the unit interval or the product space $\{0, 1 \}^\omega$.

Claim 4
Let $X \subset \mathbb{R}$. If $X$ is a Lusin set, then $X$ is a Menger space that is not $\sigma$-compact.

Lusin sets can be constructed under CH while they cannot exist under Martin’s axiom and the negation of CH. Thus any Lusin set is a consistent counterexample to Menger’s conjecture.

Example 5 (Miller and Fremlin)

The example makes use of two cardinals $\mathfrak{b}$ and $\mathfrak{d}$ that are associated with subsets of $\omega^\omega$, which is the set of all functions from $\omega$ into $\omega$. Define an order $\le^*$ on $\omega^\omega$ as follows. For any $f, g \in \omega^\omega$, we say $f \le^* g$ if $f(n) \le g(n)$ for all but finitely many $n$. A subset $A$ of $\omega^\omega$ is a bounded set if there exists some $f \in \omega^\omega$ such that for all $g \in A$, $g \le^* f$, i.e. $f$ is an upper bound of $A$ according to $\le^*$. The set $A$ is said to be unbounded if it is not bounded. The set $A$ is a dominating set if for each $f \in \omega^\omega$, there exists $g \in A$ such that $f \le^* g$.

The cardinal $\mathfrak{b}$ is called the bounding number and is the least cardinality of an unbounded subset of $\omega^\omega$. The cardinal $\mathfrak{d}$ is called the dominating number and is the least cardinality of a dominating subset of $\omega^\omega$. It is always the case that $\omega_1 \le \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}$ where $\mathfrak{c}$ is the cardinality of the continuum. For more information on bounded and dominating sets and the associated cardinals, see the preceding post or [4].

Furthermore, the cardinal $\text{non}(\text{Menger})$ is the least cardinality of a non-Menger subset of the real line. It is shown in the preceding post that $\mathfrak{d}=\text{non}(\text{Menger})$. As a result, any subset of the real line with cardinality less than $\mathfrak{d}$ must be a Menger space. It is consistent with ZFC that $\mathfrak{b} < \mathfrak{d}$. When this happens, any subset of the real line with cardinality $\omega_1$ is a Menger space that is not $\sigma$-compact. But this by itself is only a consistent example and not a ZFC example. We continue to fine tune until we get a ZFC example.

Another notion that is required is that of a scale. Let $X \subset \omega^\omega$. The set $X$ is said to be a scale if it is a dominating set and is well ordered by $\le^*$. The following claims will produce the desired example. In proving the following claims, we make of the fact that the set of irrational numbers, as a subset of the real line, is homeomorphic to the product space $\omega^\omega$. Thus in dealing with a set of irrational numbers, we sometimes think of it as a subset of $\omega^\omega$.

Claim 5
Let $X \subset \omega^\omega$. If $X$ is a dominating set of cardinality $\omega_1$, then there exists $Y \subset X$ such that $Y$ is a scale.

Claim 6
Let $Y \subset \omega^\omega$ be compact. Then $Y$ is a bounded set.

Claim 7
Let $X \subset \omega^\omega$. If $X$ is a scale of cardinality $\omega_1$, then $X$ is concentrated on $\mathbb{Q}$ where $\mathbb{Q}$ is the set of all rational numbers.

Claim 8
If $X$ is an uncountable $\sigma$-compact subset of the real line, then $X$ contains a Cantor set $C$ such that $C$ contains no rational number.

Note. By Cantor set here, we mean any subset of the real line that is homeomorphic to the middle third Cantor in the unit interval or the product space $\{0, 1 \}^\omega$.

Claim 9
There exists a subset $X$ of the real line with $\lvert X \lvert=\omega_1$ such that $X$ is a Menger space that is not $\sigma$-compact.

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The Proof Section

Proof of Claim 1
Let $X$ be a Lusin set. Let $D$ be any countable dense subset of $X$. Let $U$ be any open subset of the real line such that $D \subset U$. Let $W=U \cup (\mathbb{R} \backslash \overline{X})$. Note that $W$ is a dense open subset of the real line. Then $\mathbb{R} \backslash W$ is a closed nowhere dense subset of the real line. Since $X$ is a Lusin set, $\mathbb{R} \backslash W$ can contain at most countably many points of $X$. Note that $\mathbb{R} \backslash \overline{X}$ contains no points of $X$. The set $\mathbb{R} \backslash U$ can contain at most countably many points of $X$. It follows that $X \backslash U$ is countable. This shows that $X$ is concentrated on any countable dense subset of $X$. $\square$

Proof of Claim 2
Let $X \subset \mathbb{R}$. Suppose that $D$ is a countable subset of $X$ such that $D$ is a dense subset of $X$. Suppose $X$ is concentrated on the set $D$. We show that $X$ is a Rothberger space. To this end, let $\{ \mathcal{U}_n: n \in \omega \}$ be a sequence of open covers of $X$. Enumerate the set $D=\{ d_0, d_2, d_4, \cdots \}$ (indexed by the even integers). For each even $n$, choose $V_n \in \mathcal{U}_n$ such that $d_n \in V_n$. Each $V_n$ is an open subset of $X$. Find $W_n$, open in the real line, such that $V_n=W_n \cap X$. Let $W=\bigcup_{\text{even } n} W_n$. Note that $D \subset W$. Since $X$ is concentrated on $D$, there are at most countably many points of $X$ not in $W$. Say, the set of these points is labeled as $\{d_1, d_3, d_5, \cdots \}$ (indexed by the odd integers). Then for each odd $n$, choose $V_n \in \mathcal{U}_n$ such that $d_n \in V_n$. Overall, we can choose, for each $n$, one set $V_n \in \mathcal{U}_n$ such that $\{V_0,V_1,V_2,V_3,\cdots \}$ is a cover of $X$. This shows that $X$ is a Rothberger space, hence a Menger space. $\square$

Proof of Claim 3
Let $X$ be an uncountable $\sigma$-compact subset of the real line. Let $X=\bigcup_{n \in \omega} Y_n$ where each $Y_n$ is a compact subset of the real line. Then for at least one $n$, $Y_n$ is uncountable. Let $Y_n$ be one such. It is well known that any uncountable closed subset of the real line contains a Cantor set. For an algorithm of how to construct a Cantor set, see here. $\square$

Proof of Claim 4
This is where we put the preceding three claims together to prove that any Lusin set is a desired counterexample. Let $X$ be a Lusin set. By Claim 1, $X$ is concentrated on any countable dense subset of $X$. Since there is some countable subset $D \subset X$ such that $D$ is dense in $X$, by Claim 2, $X$ is a Menger space.

If $X$ is $\sigma$-compact, then by Claim 3 $X$ would contain a Cantor set $C$. The set $C$ is a nowhere dense subset of the real line. Because $X$ is a Lusin set, $X$ can have at most countably many points in $C$ (a contradiction). Thus $X$ cannot be $\sigma$-compact. $\square$

Proof of Claim 5
Let $X=\{ x_\alpha: \alpha<\omega_1 \}$ be a dominating set of size $\omega_1$. We now derive $Y=\{ y_\alpha: \alpha<\omega_1 \} \subset X$ such that $Y$ is also a dominating set and such that for any $\alpha<\beta$, we have $y_\alpha \le^* y_\beta$. We derive by induction. First let $y_0=x_0$. Suppose that for $\beta<\omega_1$, $Y_\beta=\{ y_\alpha \in X: \alpha<\beta \}$ has been chosen such that for any $\gamma<\tau<\beta$, we have $y_\gamma \le^* y_\tau$. The set $Y_\beta \cup \{ x_\beta \}$ is a countable subset of $\omega^\omega$. As a countable set, it is bounded. There exists $f \in \omega^\omega$ such that for each $g \in Y_\beta \cup \{ x_\beta \}$, we have $g \le^* f$. Since $X$ is dominating, there exists $x_\mu \in X$ such that $f \le^* x_\mu$. Let $y_\beta=x_\mu$. Now $Y=\{ y_\alpha: \alpha<\omega_1 \}$ has been inductively derived. It is clear that $Y$ is a scale. $\square$

Proof of Claim 6
Let $Y \subset \omega^\omega$ be compact. Let $B_n$ be the projection of $Y$ into the $n$th coordinate $\omega$. The set $B_n$ must be a finite set since $\omega$ is discrete. Note that $Y \subset \prod_{n \in \omega} B_n$. Furthermore, $\prod_{n \in \omega} B_n$ is a bounded set. $\square$

Proof of Claim 7
Let $X=\{ x_\alpha: \alpha<\omega_1 \}$ be a scale of size $\omega_1$. Let $U$ be an open subset of the real line such that $\mathbb{Q} \subset U$. For each positive integer $n$, consider the set $K_n=[-n,n] \backslash U$. The set $K_n$ is a compact subset of the real line consisting entirely irrational numbers. Thus we can consider each $K_n$ as a subset of $\omega^\omega$. By Claim 6, any compact subset of $\omega^\omega$ is bounded. Thus for each $n$, there exists $f_n \in \omega^\omega$ such that $f_n$ is an upper bound of $K_n$, i.e. for each $h \in K_n$, we have $h \le^* f_n$. For each $n$, let $g_n \in X$ such that $f_n \le^* g_n$ (using the fact that $X$ is dominating).

For each $n$, let $X_n=X \cap K_n$. Note that $X \backslash U=\bigcup_{n \ge 1} X_n$. Furthermore, for each $h \in X_n$, we have $h \le^* f_n \le^* g_n$. Since $X$ is a scale of size $\omega_1$, there are only countably many points in $X$ that are $\le^* g_n$. Thus each $X_n$ is countable. As a result, $X \backslash U$ is countable. This proves that the scale $X$ is concentrated on the set $\mathbb{Q}$ of rationals. $\square$

Proof of Claim 8
This is similar to Claim 3, except that this time we wish to obtain a Cantor set containing no rational numbers. Let enumerate the rational numbers in a sequence $\{ r_0, r_1, r_2, \cdots \}$. In the algorithm indicated here, at the $n$th step of the inductive process, we can make sure that the chosen intervals miss $r_n$. The selected intervals would then collapse in a Cantor set consisting entirely of irrational numbers. $\square$

Proof of Claim 9
This is the step where we put everything together to obtain ZFC example of Menger but not $\sigma$-compact. Recall the dominating number $\mathfrak{d}$. We consider two cases. Case 1. $\omega_2 \le \mathfrak{d}$. Case 2. $\mathfrak{d}=\omega_1$. The first case is that $\mathfrak{d}$ is at least $\omega_2$, the second uncountable cardinal. The second case is that $\mathfrak{d}$ is less than $\omega_2$, hence is $\omega_1$, the first uncountable cardinal. In either case, we can obtain a subset of the real line that is Menger but not $\sigma$-compact.

Case 1. $\omega_2 \le \mathfrak{d}$
Then any subset of the real line with cardinality $\omega_1$ must be Menger (see Theorem 10 in the preceding post). Such a set cannot be $\sigma$-compact. If it is, it would contain a Cantor set which has cardinality continuum (see Claim 3). In this scenario, $\omega_1$ is strictly less than continuum since $\omega_1 < \omega_2 \le \mathfrak{d} \le \mathfrak{c}$.

Case 2. $\mathfrak{d}=\omega_1$
Let $Y \subset \omega^\omega$ be a dominating set of cardinality $\omega_1$. This set is not a Menger space. If a subset of $\omega^\omega$ is a Menger space, it must be a non-dominating set (see Theorem 4 in the preceding post). By Claim 5, we can assume that $Y$ is a scale. Let $X=Y \cup \mathbb{Q}$. By Claim 7, $X$ is concentrated on $\mathbb{Q}$. By Claim 2, $X$ is a Menger space.

We claim that $X$ is not $\sigma$-compact. Suppose $X=\bigcup_{n \in \omega} H_n$ where each $H_n$ is compact. Then $H_m$ is uncountable for some $m$. By Claim 8, there is a Cantor set $C$ such that $C \subset H_n$ and that $C$ contains no rational numbers. Let $U=\mathbb{R} \backslash C$. Here $U$ is an open set containing the rational numbers but uncountably many points of $X$ are outside of $U$. This means that $X$ is not concentrated on $\mathbb{Q}$, going against Claim 7. Thus $X$ is not $\sigma$-compact. $\square$

Remarks on Example 5

At heart, the derivation of Example 5 is deeply set-theoretic. It relies heavily on set theory. Just that when the set theory goes one way, there is a counterexample and when the set theory goes the opposite way, there is also a counterexample. Taken altogether, it appears no extra set theory is needed. However, when each case is taken by itself, extra set theory is required. Case 2 says there is a scale of cardinality $\omega_1$. This is possible when $\mathfrak{b}=\mathfrak{d}=\omega_1$. This equality is consistent with ZFC. It is also consistent that $\mathfrak{b}=\mathfrak{d}=\kappa$ where $\kappa$ is a regular cardinal and $\kappa>\omega_1$. It is also consistent that $\omega_1=\mathfrak{b}<\mathfrak{d}$ or $\omega_1<\mathfrak{b}<\mathfrak{d}$. In general, a scale exists when $\mathfrak{b}=\mathfrak{d}$ (the scale is then of cardinality $\mathfrak{b}$). So Case 2 is possible because of set theory. For more information about the bounding number $\mathfrak{b}$ and the dominating number $\mathfrak{d}$, see [4].

Case 1, being the opposite of Case 2, is also deeply set-theoretically sensitive. However, the two cases taken together have the effect that set theory beyond ZFC is not used. This argument is a dichotomy. The dichotomic proof covers all bases. The manipulation of set theory to produce a ZFC result is clever. It is indeed a very nifty proof! Though the proof is not constructive, the example shows that Menger’s conjecture is false in ZFC. There are better counterexamples, e.g. [1]. However, the example of Miller and Fremlin shows that Menger’s conjecture is indeed false in ZFC.

Reference

1. Bartoszyński T., Tsaban B., Hereditary topological diagonalizations and the Menger–Hurewicz Conjectures, Proc. Amer. Math. Soc., 134 (2), 605-615, 2005.
2. Miller, A. W., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 201-233, 1984.
3. Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 17-33, 1988.
4. 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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# 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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