Menger spaces and Hurewicz spaces are situated in between compactness and the Lindelof property.
In the preceding post, we discuss a characterization of Menger spaces that is a Lindeloflike property. The attention is now turned to Hurewicz spaces. We discuss a compactlike 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 Lindeloflike, while the one closer to compactness is compactlike. In addition to proving the compactlike 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 be a space. The space is a Menger space (has the Menger property) if for each sequence of open covers of , there exists, for each , a finite such that is an open cover of . The space is a Hurewicz space (has the Hurewicz property) if for each sequence of open covers of , there exists, for each , a finite such that is a cover of , i.e. each belongs to for all but finitely many .
Two previous posts (here and here) discuss basic facts of Menger spaces and Hurewicz spaces. Consider the following theorems.
Theorem 1
Let . Then the following conditions are equivalent.
 The set has the Hurewicz property.
 For every set with , there exists an set such that .
This theorem is Theorem 5.7 in [3]. In the real line, any set is compact. Hurewicz subsets of the real line may not be compact but are “approximated” by compact sets in the manner described in Theorem 1. Whenever a Hurewicz subset is situated in a set, we can always find a compact set cushioned in between.
Theorem 2
Let . If has empty interior and has the Hurewicz property, then 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 compactlike property of Hurewicz subsets of the real line, consider the following examples.
Example 1
Clearly, any set in the real line satisfies condition 2 in Theorem 1.
Example 2
In the real line, any set that is not does not satisfy condition 2 of Theorem 1.
For any set that is not an set, there can be no cushioned in between and . An example of such a set is the set of all irrational numbers. We already know that does not have the Menger property (see Theorem 4 here) and hence does not not have the Hurewicz property.
Example 3
Not only sets that are not do not have the compactlike property of Theorem 1, all Borel sets that that are not 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 compact sets cannot be cushioned in between Bernstein sets and sets. Let be a Bernstein set. The complement is also a Bernstein set and hence dense in the real line. Let be a countable dense subset. Then is a dense set containing .
Suppose that for some set . Let where each is a closed set. Observe that each is nowhere dense in the real line. To see this, let be an open interval. Since is dense, choose . Since , choose open interval such that , and .
Consider . The set is a set. If is countable, then the real line is the union of countably many nowhere dense sets and a countable set which means that the real line is a first category set, a contradiction. Thus must be uncountable. Since is an uncountable set, contains a Cantor set . Since is a Bernstein set, . But cannot contain points of since . Thus there can be no set cushioned between the Bernstein set and the set .
Example 5
Any Lusin set does not satisfy condition 2 of Theorem 1 and hence is not Hurewicz.
Let where is uncountable. The set is a Lusin set (alternative spelling: Luzin) if for each that is of first category in the real line, is countable.
Suppose is a Lusin set. We show that there can be no set cushioned between and some set containing . Note that the Lusin set 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 is Hurewicz, then it would be of first category. But a Lusin set by definition cannot be of first category. Thus cannot be Hurewicz and thus does not satisfy the compactlike property.
Example 6
Let such that is a Sierpinski set. Then has the Hurewicz property hence satisfies the compactlike property.
The set is a Sierpinski set if is uncountable and is countable for every Lebesgue measure zero set . 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 leadin to the Hurewicz’s Problem.
In 1924, Karl Menger [4] conjectured that in metric spaces, the Menger property is equivalent to 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 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 settheoretic statement. For example, when , an example of a Menger but not Hurewicz space can be derived. On the other hand, when , 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 noncompact metric space is 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 . Then the following conditions are equivalent.
 The set has the Hurewicz property.
 For each sequence of covers of , there exists, for each , a finite such that is a cover of .
 For every set with , there exists an set such that .
The direction is straightforward. Condition 2 is the Hurewicz property restricted to sequences of covers. Note that any cover is also an open cover. So if a space satisfies the full Hurewicz property, it will then satisfy the restricted property.
Suppose that such that where each is an open subset of the real line. We can also require that . From the set , we construct a sequence of covers of .
To define , do the following. For each , choose an open interval such that and . For the open cover , we can find a countable subcover . To simplify notation, we rename . For each , define . Let . Note that is a cover of and that for each , .
For each , note that the open sets are increasing, i.e. for , . So picking a finite subset of is the same as picking one element (the largest one in the finite set). Using condition 2, we can choose, for each , such that is a cover of .
For each , define . Note that each closure in the intersection is contained in the open set . To see this,
Thus . Observe there is now an set cushioned between and .
Let be a sequence of open covers of such that elements of each are open subsets of the real line. By condition 3, there is an set such that
Since the real line is compact, we can assume each is compact. For each , we can choose finite such that . It follows that is an open cover of . This completes the proof that is a Hurewicz space.
Theorem 2
Let . If has empty interior and has the Hurewicz property, then is of first category.
Suppose has empty interior and has the Hurewicz property. Since has empty interior, is a dense subset of the real line. Let be a countable dense subset of . Then is a set and . By Theorem 1, there exists an set such that . Let where each is a closed set. By the same argument in Example 4 and Example 5, each is a nowhere dense set. Then is a set of first category since it is a subset of .
Reference
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 Hurewicz W., Uber die Verallgemeinerung des Borelschen Theorems, Mathematische
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 Miller A. W., Fremlin D. H., On some properties of Hurewicz, Menger, and Rothberger, Fund. Math., 129, 1733, 1988.
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