# Thinking about the Space of Irrationals Topologically

Let $\mathbb{R}$ denote the real number line and $\mathbb{P}$ denote the set of all irrational numbers. The irrational numbers and the set $\mathbb{P}$ occupy an important place in mathematics. The set $\mathbb{P}$ with the Euclidean topology inherited from the full real line is a topological space in its own right. Thus the space $\mathbb{P}$ has some of the same properties inherited from the Euclidean real line, e.g., just to name a few, it is hereditarily separable, hereditarily Lindelof, paracompact and metrizable. The space of the irrational numbers $\mathbb{P}$ has many properties apart from the full real line (e.g. $\mathbb{P}$ is totally disconnected). In this post, we look at a topological description of the space $\mathbb{P}$.

Let $\omega$ be the set of all nonnegative integers. Then the space $\mathbb{P}$ of irrational numbers is topologically equivalent (i.e. homeomorphic to) the product space $\prod \limits_{i=0}^\infty X_i$ where each $X_i=\omega$ has the discrete topology. We will also denote the product space $\prod \limits_{i=0}^\infty X_i$ by $\omega^\omega$. We have the following theorem.

Theorem
The space $\mathbb{P}$ of irrational numbers is homeomorphic to the product space $\omega^\omega$.

Because of this theorem, we can look at irrational numbers as sequences of nonnegative integers. Specifically each irrational number can be identified by a unique sequence of nonnegative integers. We can think of each such unique sequence as an address to locate an irrational number. In the remainder of the post, we describe at a high level how to define the unique addresses, which will also give us a homeomorphic map between the space $\mathbb{P}$ and the product space $\omega^\omega$.

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Step 0
Put the rational numbers in a sequence $r_0,r_1,r_2,r_3,\cdots$ such that $r_0=0$. We divide the real line into countably many non-overlapping intervals. Specifically, let $A_0=[0,1]$, $A_1=[-1,0]$, $A_2=[1,2]$, etc (see the following figure).

To facilitate the remaining construction, we denote the left endpoint of the interval $A_i$ by $L_i$ and denote the right endpoint by $R_i$.

Step 1
In each of the interval $A_i$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_i$ are $L_i$ and $R_i$, respectively. We choose a sequence $x_{i,0}, x_{i,1}, x_{i,2},\cdots$ of rational numbers converging to the right endpoint $R_i$. Then let $A_{i,0}=[L_i,x_{i,0}]$, $A_{i,1}=[x_{i,0},x_{i,1}]$, $A_{i,2}=[x_{i,1},x_{i,2}]$, etc (see the following figure).

Two important points to consider in Step $1$. One is that we make sure the rational number $r_1$ is chosen as an endpoint of some interval in Step 1. The second is that the length of each $A_{i,j}$ is less than $\displaystyle \frac{1}{2^1}$.

Step 2
In each of the interval $A_{i,j}$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_{i,j}$ are $L_{i,j}$ and $R_{i,j}$, respectively. We choose a sequence $x_{i,j,0}, x_{i,j,1}, x_{i,j,2},\cdots$ of rational numbers converging to the left endpoint $L_{i,j}$. Then let $A_{i,j,0}=[x_{i,j,0},R_{i,j}]$, $A_{i,j,1}=[x_{i,j,1},x_{i,j,0}]$, $A_{i,j,2}=[x_{i,j,2},x_{i,j,1}]$, etc (see the following figure).

As in the previous step, two important points to consider in Step $2$. One is that we make sure the rational number $r_2$ is chosen as an endpoint of some interval in Step 2. The second is that the length of each $A_{i,j,k}$ is less than $\displaystyle \frac{1}{2^2}$.

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Remark

In the process described above, the endpoints of the intervals $A_{f(0),\cdots,f(n)}$ are rational numbers and we make sure that all the rational numbers are used as endpoints. We also make sure that the intervals from the successive steps are nested closed intervals with lengths $\rightarrow 0$. The consequence of this point is that the nested decreasing closed intervals will collapse to one single point (since the real line is a complete metric space) and this single point must be an irrational number (since all the rational numbers are used up as endpoints of the nested closed intervals).

In Step $i$ where $i$ is an odd integer, we make the endpoints of the new intervals converge to the right. In Step $i$ where $i>1$ is an even integer, we make the endpoints of the new intervals converge to the left. This manipulation is to ensure that the nested closed intervals will never share the same endpoint from one step all the way to the end of the process.

Thus if we have $f \in \omega^\omega$, then $\bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)} \ne \varnothing$ and in fact has only one point that is an irrational number.

On the other hand, for each point $x \in \mathbb{P}$, we can locate inductively a sequence of intervals, $A_{f(0)}, A_{f(0),f(1)}, A_{f(0),f(1),f(2)}, \cdots$, containing the point $x$. This sequence of nested closed intervals must collapse to a single point and the single point must be the irrational number $x$.

The process described above gives us a one-to-one mapping from $\mathbb{P}$ onto the product space $\omega^\omega$. This mapping is also continuous in both directions, making it a homeomorphism. the nested intervals defined above form a base for the Euclidean topology on $\mathbb{P}$. These basic open sets have a natural correspondance with basic open sets in the product space $\omega^\omega$.

For example, for $f \in \omega^\omega$, $\left\{ A_{f(0),\cdots,f(n)} \cap \mathbb{P}: n \in \omega \right\}$ is a local base at the point $x \in \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)}$. One the other hand, each $A_{f(0),\cdots,f(n)} \cap \mathbb{P}$ has a natural counterpart in a basic open set in the product space, namely the following set:

$\displaystyle . \ \ \ \ \left\{g \in \omega^\omega: g \restriction n = f \restriction n \right\}$

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The above process establishes that the countable product of the integers, $\omega^\omega$, is equivalent topologically to the Euclidean space $\mathbb{P}$.

# When is a Lindelof Space Normal?

When is a Lindelof space normal? This question may seem strange since it is a well known basic result that Lindelof spaces are normal (see Theorem 3.8.2 in [1] and Theorem 16.8 in [6]). The proofs in both of these theorems require that the Lindelof space is also a regular space. We present an example showing that the requirement of regularity cannot be removed. The example defined below is a space that is Hausdorff, hereditarily Lindelof, not separable, not regular and not normal.

The example defined below makes it clear that we need to be careful in making the leap from the Lindelof property to normality of a space. We need to make sure that the space in question is also a regular space. The Lindelof property is a powerful property; it also implies paracompactness. But for this to happen, some separation axiom is required (it needs to be a regular space).

Another motivation for looking at examples such as the one defined here is that they are background information to the theory of S-spaces and L-spaces (see the remark given at the end of the post).

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The example is the Example 14.7 in [2]. The underlying set is the real number line $\mathbb{R}$. Let $\tau$ be the usual topology on $\mathbb{R}$. Let $X=\mathbb{R}$. We use the following collection of sets as a base for a new topology on $X$:

$\displaystyle \mathbb{B}=\left\{O-C: O \in \tau, C \subset \mathbb{R}, \lvert C \vert \le \omega \right\}$

The collection $\mathbb{B}$ needs to satisfy two conditions for being a base for a topology. One is that it covers the underlying set. The second is that whenever $B_1, B_2 \in \mathbb{B}$ with $x \in B_1 \cap B_2$, there is some $B_3 \in \mathbb{B}$ such that $x \in B_3 \subset B_1 \cap B_2$. Note that $\tau \subset \mathbb{B}$. This implies that $\mathbb{B}$ is a cover of $X$. It is also clear that the second condition is satisfied.

Let $\tau^{*}$ be the topology generated by the base $\mathbb{B}$. Another consequence of $\tau \subset \mathbb{B}$ is that the new topology on $X$ is finer than the usual topology, i.e., $\tau \subset \tau^{*}$. It is then clear that $(X, \tau^{*})$ is a Hausdorff space.

We denote the real line with the new topology $\tau^{*}$ by $X$ rather than $(X, \tau^{*})$ since the new topology $\tau^{*}$ is the primary focus.

In $X$, the complement of every countable set is open. So no countable set can be dense in $X$, making it not separable. Equivalently, every countable subset of $X$ is closed.

To see that $X$ is not regular, we first make the following observation.

Observation
Let $W \subset \mathbb{R}$ be a Euclidean open set. Let $A \subset \mathbb{R}$ be a countable set that is dense in $W$ (dense with respect to the Euclidean topology). Then for each $x \in A \cap W$, $x \in \overline{W-A}$ (with respect to the topology $\tau^*$).

To see that this observation is correct, let $x \in A \cap W$ and let $V-B$ be an open set containing $x$, where $V$ is a Euclidean open set and $B$ is a countable set. We can assume that $V \subset W$. Let $A_0=A-\left\{x \right\}$. Note that $x \in V-(A_0 \cup B) \subset V-B$ and that $V-(A_0 \cup B) \subset W-A_0$. Furthermore, for any $y \in V-(A_0 \cup B)$ with $y \ne x$, $y \in W-A$. Thus any open set containing $x$ contains many points of $W-A$. Hence, we have $x \in \overline{W-A}$.

To see that $X$ is not regular, we show that there is a closed set $C \subset X$ such that for any open set $O \subset X-C$, $\overline{O} \cap C \ne \varnothing$.

Let $C=\mathbb{Q}$ (the set of all rational numbers). Let $O$ be open in $X$ such that $O \subset X-C$. We assume that $O=W-A$ where $W$ is open in the Euclidean topology and $A$ is countable. Note that $O=W-A$ consisting of entirely of irrational numbers. Then $\mathbb{Q} \cap W \subset A$. Thus $A$ is dense in $W$ (with respect to the Euclidean topology). By the above observation, any point of $A$ is a member of $\overline{W-A}$ (with respect to the topology $\tau^*$). In particular, for each $x \in C=\mathbb{Q}$, $x \in \overline{W-A}$. Thus $X$ is not regular.

Any space that is not regular is also not normal. Thus $X$ is not normal. But we can also see this directly. Let $H=\mathbb{Q}$ and $K=\left\{x \right\}$ where $x$ is any irrational number. Then $H$ and $K$ are two disjoint closed sets that cannot be separated by disjoint open sets in $X$.

The space $X$ is Lindelof and also hereditarily Lindelof (both properties are piggy backed on the same properties of the Euclidean real line). To see that $X$ is Lindelof, suppose that a collection of open sets $O_a-A_a \in \mathbb{B}$ is a cover of $X$. The Euclidean open sets $O_a$ also form a cover of $X=\mathbb{R}$. Then pick countably many $O_a$ that also cover the real line (using the Lindelof property of the Euclidean real line). Then these countably many $O_a-A_a$ will cover $X$ except for countably many points. A countable subcover will result once we use open sets in the original cover to cover these countably many points. Using the same proof, it can be shown that every subspace of $X$ is also Lindelof.

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Remark

An S-space is any regular topological space that is hereditarily separable but not Lindelof. An L-space is any regular topological space that is hereditarily Lindelof but not separable. Elementary examples of hereditarily separable space that is not Lindelof and hereditarily Lindelof space that is not separable are quite easy to define (but they are not regular). The example we define above is an example of non-regular hereditarily Lindelof space that is not separable. Thus S-space and L-space are typically defined with the requirement of regularity.

It was shown in the early 1980’s that the existence of S-space is independent of the usual axioms of ZFC. This means that to prove the existence of an S-space or to prove the non-existence of S-space, one needs to assume axioms beyond those of ZFC. The L-space problem (whether an L-space can exist without assuming additional set-theoretic assumptions beyond those of ZFC) was not resolved until quite recently. For quite a long time, it was believed the L-space problem would have a similar solution (that its existence would be independent of ZFC). In 2005, Justin Tatch-Moore solved the L-space problem by constructing an L-space without assuming additional axioms.

For back ground information about S-space and L-space, see [3] and [4]. For Justin Moore’s proof, see [5].

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Hodel, R.,Cardinal Functions I, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 1-61.
3. Mastros, S.,S and L Spaces, Master’s Thesis, 2009, University of Pittsburgh. Link
4. Roitman, J.,Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 295-326.
5. Tatch-Moore, J.,A solution to the L space problem, Journal of the American Mathematical Society, 19 (2006), 3, 717—736. Link
6. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.