This post could very well be titled Killing Two Birds in One Stone. We present one proof that can show two results – the set of all irrational numbers does not satisfy the Menger property and that is homeomorphic to the product space .
In this previous post, we introduce the notion of Menger spaces. A space is a Menger space (or has the Menger property) if for every sequence of open covers of , there exists, for each , a finite such that is an open cover of . In the definition if the open covers 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 , the set of all irrational numbers. The way we show this fact in this previous post is by working in the product space . This approach requires an understanding that and the product space are identical topologically as well as working with the eventual domination order in . In this post, we give a direct proof that is not Menger with the irrational numbers lying on a line. This proof will open up an opportunity to see that is homeomorphic to . The product space is also called the Baire space in the literature.
The sets and 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 with being the set of all nonnegative numbers with the discrete topology. In many ways the product space version is easier to work with. For example, the eventual domination order on help describe the classical covering properties such as Menger property and Hurewicz property.
A Direct Proof of NonMenger
As mentioned above, the set is the set of all irrational numbers. We let denote the set of all rational numbers.
To see that does not satisfy the Menger property, we need to exhibit a sequence of open covers of such that whenever we pick, for each , a finite , the set cannot be a cover of , i.e. no matter how we pick the finite , there is always an irrational number that is missed by all . The open covers are derived inductively, starting with .
To prepare for the inductive steps, we fix a scheme of choosing convergent sequences of rational numbers. For any interval where , we choose a decreasing sequence such that from the right with and an increasing sequence such that from the left with . Enumerate in a countable sequence with . Several points to keep in mind when choosing such sequences.
 In each inductive step, either the sequences of the type or the sequences of the type are chosen but not both.
 in the th stage of the inductive process, in picking the sequence or for the interval , we make sure that the rational number is used in the sequence or if is in the interval and if is not previously chosen. So at the end, all rational numbers are used up as endpoints of intervals in all .
 If sequence of the type is chosen out of the interval , we derive the subintervals whose union is . We want the sequence chosen in such a way that the length of each subinterval is less than 1/2 of .
 If sequence of the type is chosen out of the interval , we derive the subintervals whose union is . We want the sequence chosen in such a way that the length of each subinterval is less than 1/2 of .
To start, let . In other words, consists of all open intervals whose endpoints are the integers and whose lengths are 1. Label the elements of as .
For each , choose a sequence of rational numbers converging to from the left (according to the scheme described above). Make sure that the rational number is picked if . From this, we obtain the intervals covering all the irrational numbers in . Label these intervals as . Then consists of all such intervals obtained from each .
Here’s how to define . For each , choose a sequence of rational numbers converging to from the right (according to the scheme described above). Make sure that the rational number is picked if and if has not been chosen previously. From this, we obtain the intervals covering all the irrational numbers in . Label these intervals as . Then consists of all such intervals obtained from each .
From here on out, we continue the same induction steps to derive , making sure that when is odd, we choose sequence converging to the end point from the left and when is even, we choose sequence converging to the end point from the right for each , producing the intervals , , , .
Now the sequence has been defined. Note that every rational number is used as the endpoint of some open interval in some . Each is clearly an open cover of . Another useful observation is that
whenever for all and for all . This is because each chosen interval has length less than 1/2 of the previous interval. Furthermore, the point in the intersection must be an irrational number since all rational numbers are used up as endpoints.
Let’s pick, for each , a finite . We now find an irrational number that is not in any interval of any .
Since is finite, choose some such that . Since is finite, choose some such that . In the same manner, choose such that .
Continuing the inductive process, we have open intervals from , respectively. Furthermore, for each , . Then the intersection
must be an irrational number . This is a real number that is not covered by for all . This shows that cannot be an open cover of . This completes the proof that does not satisfy the Menger property.
A Homeomorphism
In defining the sequence of open covers of , we have also define a onetoone correspondence between and the product space .
First, each is uniquely identified with a sequence of nonnegative integers .
Observe that each irrational number belongs to a unique element in each . Thus the sequence is simply the subscripts of the open intervals from the open covers such that for each . With the way the intervals are chosen, it must be the case that . Furthermore the intervals collapse to the point .

……………(1)
The mapping goes in the reverse direction too. For each , we can use to obtain the open intervals . These intervals collapse to a single point, which is the irrational number that is associated with .
Denote this mapping by . For each , is derived in the manner described above. The function maps onto . It follows that is a onetoone map and that both and the inverse are continuous. To see this, we need to focus on the open sets in both the domain and the range.
For the correspondence , we consider the two sets:
where for each . Note that is a local base at . This is because the intervals have lengths converging to zero and they collapse to the point in the manner described in (1) above. On the other hand, is a local base at . Furthermore, there is a clear correspondence between and . Note that . This means that both and the inverse are continuous. Since is a base for , it is clear the map is onetoone.
This previous post has a shorter (but similar) derivation of the homeomorphism between and .
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