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 Lindeloflike property.
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 . It is sometimes easier to work with an equivalent definition of Menger spaces (found in Theorem 2 here).
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 .
It is clear from definitions that
The Menger property is strictly stronger than Lindelof. See here for examples that are Lindelof but not Menger. One of them is the space 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 be a space. Then the space is a Menger space, if, and only if, for every Menger subspace of , and for every subset of such that , the cover 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 covers since the elements of such covers are sets. This is a “Lindelof” property using covers. To get better intuition about this “Lindelof” characterization, we look at some examples.
Example 1
Any compact space satisfies the covering property in Theorem 1.
In a compact space, each of the countably many compact sets that makes up the space is a Menger subspace. Suppose that where each is compact. For any cover , 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 . The set is compact if and only if is an set. This is because the real line is compact. Thus any subset of the real line satisfies the covering property in Theorem 1.
Example 3
The space of all irrational numbers is not a Menger space. Thus it does not satisfy the covering property in Theorem 1.
The set is homeomorphic to , which is a dominating set and thus not Menger. See Theorem 4 in this post. The set of irrational numbers is a set that is not . The next example discusses other Borel sets that are not .
Example 4
Among the Borel sets in the real line, sets are the only ones that have the Menger property. Thus any Borel set that is not is not Menger. The only Borel sets satisfying the covering property in Theorem 1 are precisely the 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, closed sets, , , , , , ,
In the above steps, the subscript denotes union and the subscript denotes intersection. After the steps for the finite ordinals are completed (), the construction continues on through all the countable ordinals (through taking countable unions and taking countable intersections of all the Borel sets that come before the th step). We claim that the only Borel sets that are Menger are the sets, which of course include the open sets and the closed sets. This fact follows from the following theorem [3].

Theorem 2
Let . If the set is an analytic set, then either is compact or contains a closed copy of the space of all irrational numbers.
A subset of the real line is an analytic set if it is the continuous image of , 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 , if is Menger, it cannot have as a closed subspace and thus must be compact. It follows that Borel sets that are not 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 of the real line is a Bernstein set if both and the complement of intersect every uncountable closed subset of the real line. Such a set can be constructed using transfinite induction (see here for an illustration).
Let be a Bernstein set. Note that the complement is also a Bernstein set. Any Bernstein set is dense in the real line. Choose a countable dense subset of . Note that is a dense set and is homeomorphic to , the set of all irrational numbers. The set is in turn homeomorphic to . With , we can consider as a subset of .
Let 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 . Consider the compact subset of where is the following set.
The set is a Cantor set and hence is an uncountable closed set. Thus . Let . It is clear that . Thus 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 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 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 be a space. A collection of subsets of is called an Alster cover if for any compact subset K of , for some . Such a collection of sets is necessarily a cover of the space . A space is said to be an Alster space if for every Alster cover of , there is a countable such that is a cover of . 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 is productively Lindelof if is Lindelof for every Lindelof space ).
With the covering of Menger spaces (as described in Theorem 1), there is a clear connection between Alster spaces and Menger spaces. Observe that any 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 Dspaces [2]. Thus Alster spaces are Dspaces. 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 be a space. Then the space is a Menger space, if, and only if, for every Menger subspace of , and for every subset of such that , the cover has a countable subcover.
The direction is clear. If is Menger, then belongs to any cover .
Let be a sequence of open covers of . For each Menger , each is a open cover of . Thus we can choose, for each , a finite such that for each , for infinitely many . We are using a characterization of Menger space found here (see Theorem 2). Define where each is defined as follows:
Note that for each , . Thus . Consider the cover just defined.
By assumption it has a countable subcover . For each , define by . Note that each is finite and . We now show that for each , we have for infinitely many .
Let . Then for some . By the definition of , we have for infinitely many . Then it follows that for infinitely many where . Note that for means that . Therefore, for infinitely many .
Reference
 Alster K., On the class of all spaces of weight not greater than whose cartesian product with every Lindelof space is Lindelof, Fund. Math., 129, 133–140, 1988.
 Aurichi L., DSpaces, Topological Games, and Selection Principles, Topology Proc., 36, 107–122, 2010.
 Kechris A. S., Classical descriptive set theory, Graduate Texts in Mathematics, 156, SpringerVerlag, New York, 1995.
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