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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\copyright 2020 – Dan Ma

4 thoughts on “The space of irrational numbers is not Menger

  1. Pingback: Topological meaning of bounded sets | Dan Ma's Topology Blog

  2. Pingback: Adding up to a non-meager set | Dan Ma's Topology Blog

  3. Pingback: The ideal of bounded sets | Dan Ma's Topology Blog

  4. Pingback: The ideal of meager sets | Dan Ma's Topology Blog

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