# 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 $\omega^\omega$.

$\text{ }$

$\text{ }$

$\text{ }$

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

$\copyright$ 2020 – Dan Ma