# 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 $F$. $\square$

Reference

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2. Hurewicz W., Uber die Verallgemeinerung des Borelschen Theorems, Mathematische
Zeitschrift, 24, 401-425, 1925.
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9. Sierpinski W., Sur un probleme de M. Menger, Fund. Math., 8, 223-224, 1926.
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