A Question About The Rational Numbers

Let $\mathbb{R}$ be the real line and $\mathbb{Q}$ be the set of all rational numbers. Consider the following question:

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Question

• For each nonnegative integer $n$, let $U_n$ be an open subset of $\mathbb{R}$ such that that $\mathbb{Q} \subset U_n$. The intersection $\bigcap \limits_{n=0}^\infty U_n$ is certainly nonempty since it contains $\mathbb{Q}$. Does this intersection necessarily contain some irrational numbers?

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While taking a real analysis course, the above question was posted to the author of this blog by the professor. Indeed, the question is an excellent opening of the subject of category. We first discuss the Baire category theorem and then discuss the above question. A discussion of Baire spaces follow. For any notions not defined here and for detailed discussion of any terms discussed here, see [1] and [2].

In the above question, the set $\bigcap \limits_{n=0}^\infty U_n$ is a $G_\delta$ set since it is the intersection of countably many open sets. It is also dense in the real line $\mathbb{R}$ since it contains the rational numbers. So the question can be rephrased as: is the set of rational numbers $\mathbb{Q}$ a $G_\delta$ set? Can a dense $G_\delta$ set in the real line $\mathbb{R}$ be a “small” set such as $\mathbb{Q}$? The discussion below shows that $\mathbb{Q}$ is too “thin” to be a dense $G_\delta$ set. Put it another way, a dense $G_\delta$ subset of the real line is a “thick” set. First we present the Baire category theorem.
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Baire Category Theorem

Let $X$ be a complete metric space. For each nonnegative integer $n$, let $O_n$ be an open subset of $X$ that is also dense in $X$. Then $\bigcap \limits_{n=0}^\infty O_n$ is dense in $X$.

Proof
Let $A=\bigcap \limits_{n=0}^\infty O_n$. Let $V_0$ be any nonempty open subset of $X$. We show that $V_0$ contains some point of $A$.

Since $O_0$ is dense in $X$, $V_0$ contains some point of $O_0$. Let $x_0$ be one such point and choose open set $V_1$ such that $x_0 \in V_1$ and $\overline{V_1} \subset V_0 \cap O_0 \subset V_0$ with the additional condition that the diameter of $\overline{V_1}$ is less than $\displaystyle \frac{1}{2^1}$.

Since $O_1$ is dense in $X$, $V_1$ contains some point of $O_1$. Let $x_1$ be one such point and choose open set $V_2$ such that $x_1 \in V_2$ and $\overline{V_2} \subset V_1 \cap O_1 \subset V_1$ with the additional condition that the diameter of $\overline{V_2}$ is less than $\displaystyle \frac{1}{2^2}$.

By continuing this inductive process, we obtain a nested sequence of open sets $V_n$ and a sequence of points $x_n$ such that $x_n \in V_n \subset \overline{V_n} \subset V_{n-1} \cap O_{n-1} \subset V_0$ for each $n$ and that the diameters of $\overline{V_n}$ converge to zero (according to some complete metric on $X$). Then the sequence of points $x_n$ is a Cauchy sequence. Since $X$ is a complete metric space, the sequence $x_n$ converges to a point $x \in X$.

We claim that $x \in V_0 \cap A$. To see this, note that for each $n$, $x_j \in \overline{V_n}$ for each $j \ge n$. Since $x$ is the sequential limit of $x_j$, $x \in \overline{V_n}$ for each $n$. It follows that $x \in O_n$ for each $n$ ($x \in A$) and $x \in V_0$. This completes the proof of Baire category theorem. $\blacksquare$

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Discussion of the Above Question

For each nonnegative integer $n$, let $U_n$ be an open subset of $\mathbb{R}$ such that that $\mathbb{Q} \subset U_n$. We claim that the intersection $\bigcap \limits_{n=0}^\infty U_n$ contain some irrational numbers.

Suppose the intersection contains no irrational numbers, that is, $\mathbb{Q}=\bigcap \limits_{n=0}^\infty U_n$.

Let $\mathbb{Q}$ be enumerated by $\left\{r_0,r_1,r_2,\cdots \right\}$. For each $n$, let $G_n=\mathbb{R}-\left\{ r_n \right\}$. Then each $G_n$ is an open and dense set in $\mathbb{R}$. Note that the set of irrational numbers $\mathbb{P}=\bigcap \limits_{n=0}^\infty G_n$.

We then have countably many open and dense sets $U_0,U_1,U_2,\cdots,G_0,G_1,G_2,\cdots$ whose intersection is empty. Note that any point that belongs to all $U_n$ has to be a rational number and any point that belongs to all $G_n$ has to be an irrational number. On the other hand, the real line $\mathbb{R}$ with the usual metric is a complete metric space. By the Baire category theorem, the intersection of all $U_n$ and $G_n$ must be nonempty. Thus the intersection $\bigcap \limits_{n=0}^\infty U_n$ must contain more than rational numbers.

It follows that the set of rational numbers $\mathbb{Q}$ cannot be a $G_\delta$ set in $\mathbb{R}$. In fact, the discussion below will show that the in a complete metric space such as the real line, any dense $G_\delta$ set must be a “thick” set (see Theorem 3 below).
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Baire Spaces

The version of the Baire category theorem discussed above involves complete metric spaces. However, the ideas behind the Baire category theorem are topological in nature. The following is the conclusion of the Baire category theorem:

$(*) \ \ \ \ X$ is a topological space such that for each countable family $\left\{U_0,U_1,U_2,\cdots \right\}$ of open and dense sets in $X$, the intersection $\bigcap \limits_{n=0}^\infty U_n$ is dense in $X$.

A Baire space is a topological space in which the condition $(*)$ holds. The Baire category theorem as stated above gives a sufficient condition for a topological space to be a Baire space. There are plenty of Baire spaces that are not complete metric spaces, in fact, not even metric spaces. The condition $(*)$ is a topological property. In order to delve deeper into this property, let’s look at some related notions.

Let $X$ be a topological space. A set $A \subset X$ is dense in $X$ if every open subset of $X$ contains a point of $A$ (i.e. $\overline{A}=X$). A set $A \subset X$ is nowhere dense in $X$ if for every open subset $U$ of $X$, there is some open set $V \subset U$ such that $V$ contains no point of $A$ (another way to describe this: $\overline{A}$ contains no interior point of $X$).

A set is dense if its points can be found in every nonempty open set. A set is nowhere dense if every nonempty open set has an open subset that misses it. For example, the set of integers $\mathbb{N}$ is nowhere dense in $\mathbb{R}$.

A set $A \subset X$ is of first category in $X$ if $A$ is the union of countably many nowhere dense sets in $X$. A set $A \subset X$ is of second category in $X$ if it is not of first category in $X$.

To make sense of these notions, the following observation is key:

$(**) \ \ \ \ F \subset X$ is nowhere dense in $X$ if and only if $X-\overline{F}$ is an open and dense set in $X$.

So in a Baire space, if you take away any countably many closed and nowhere dense sets (in other words, taking away a set of first category in $X$), there is a remainder (there are still points remaining) and the remainder is still dense in $X$. In thinking of sets of first category as “thin”, a Baire space is one that is considered “thick” or “fat” in that taking away a “thin” set still leaves a dense set.

A space $X$ is of second category in $X$ means that if you take away any countably many closed and nowhere dense sets in $X$, there are always points remaining. For a set $Y \subset X$, $Y$ is of second category in $X$ means that if you take away from $Y$ any countably many closed and nowhere dense sets in $X$, there are still points remaining in $Y$. A set of second category is “thick” in the sense that after taking away a “thin” set there are still points remaining.

For example, $\mathbb{N}$ is nowhere dense in $\mathbb{R}$ and thus of first category in $\mathbb{R}$. However, $\mathbb{N}$ is of second category in itself. In fact, $\mathbb{N}$ is a Baire space since it is a complete metric space (with the usual metric).

For example, $\mathbb{Q}$ is of first category in $\mathbb{R}$ since it is the union of countably many singleton sets ($\mathbb{Q}$ is also of first category in itself).

For example, let $T=[0,1] \cup (\mathbb{Q} \cap [2,3])$. The space $T$ is not a Baire space since after taking away the rational numbers in $[2,3]$, the remainder is no longer dense in $T$. However, $T$ is of second category in itself.

For example, any Cantor set defined in the real line is nowhere dense in $\mathbb{R}$. However, any Cantor set is of second category in itself (in fact a Baire space).

The following theorems summarize these concepts.

Theorem 1a
Let $X$ be a topological space. The following conditions are equivalent:

1. $X$ is of second category in itself.
2. The intersection of countably many dense open sets is nonempty.

Theorem 1b
Let $X$ be a topological space. Let $A \subset X$. The following conditions are equivalent:

1. The set $A$ is of second category in $X$.
2. The intersection of countably many dense open sets in $X$ must intersect $A$.

Theorem 2
Let $X$ be a topological space. The following conditions are equivalent:

1. $X$ is a Baire space, i.e., the intersection of countably many dense open sets is dense in $X$.
2. Every nonempty open subset of $X$ is of second category in $X$.

The above theorems can be verified by appealing to the relevant definitions, especially the observation $(**)$. Theorems 2 and 1a indicate that any Baire space is of second category in itself. The converse is not true (see the space $T=[0,1] \cup (\mathbb{Q} \cap [2,3])$ discussed above).

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Dense G delta Subsets of a Baire Space

In answering the question stated at the beginning, we have shown that $\mathbb{Q}$ cannot be a $G_\delta$ set. Being a set of first category, $\mathbb{Q}$ cannot be a dense $G_\delta$ set. In fact, it can be shown that in a Baire space, any dense $G_\delta$ subset is also a Baire space.

Theorem 3
Let $X$ be a Baire space. Then any dense $G_\delta$ subset of $X$ is also a Baire space.

Proof
Let $Y=\bigcap \limits_{n=0}^\infty U_n$ where each $U_n$ is open and dense in $X$. We show that $Y$ is a Baire space. In light of Theorem 2, we show that every nonempty open set of $Y$ is of second category in $Y$.

Soppuse that there is a nonempty open subset $U \subset Y$ such that $U$ is of first category in $Y$. Then $U=\bigcup \limits_{n=0}^\infty W_n$ where each $W_n$ is nowhere dense in $Y$. It can be shown that each $W_n$ is also nowhere dense in $X$.

Since $U$ is open in $Y$, there is an open set $U^* \subset X$ such that $U^* \cap Y=U$. Note that for each $n$, $F_n=X-U_n$ is closed and nowhere dense in $X$. Then we have:

$\displaystyle (1) \ \ \ \ \ U^*=\bigcup \limits_{n=0}^\infty (F_n \cap U^*) \cup \bigcup \limits_{n=0}^\infty W_n$

$(1)$ shows that $U^*$ is the union of countably many nowhere dense sets in $X$, contracting that every nonempty subset of $X$ is of second category in $X$. Thus we can conclude that every nonempty open subset of $Y$ is of second category in $Y$. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.