Every open set in $\mathbb{R}$ is the countable union of rational open intervals

Question

How would I prove that every open set in $\mathbb{R}$ is the countable union of open intervals with rational endpoints?

Context: I am trying to see whether the set of rational open subsets of $\mathbb{R}$ is countable.

Answer 1

Let $U$ be an open set in $\mathbb R$ we have to prove there is a countable family of open intervals $(a_i,b_i)$ with $a_i,b_i \in \mathbb R$ so that $U=\bigcup\limits_{i\in\mathbb N}(a_i,b_i)$.

To do this for each $x\in U$ we have $(a_x,b_x)$ so that $x\in(a_x,b_x) \subseteq U$. This is because $U$ is open. Of course, there is a rational number $a'_x$ between $a_x$ and $x$ and a rational number $b'_x$ between $b_x$ and $x$. This is because the rational numbers are dense in $\mathbb R$.

So we can consider the familiy of open intervals with rational endpoints. $(a'_x,b'_x)$ Why is this family countable? Because the number of intervals with rational coordinates is countable, since its cardinality does not exceed $|\mathbb Q\times \mathbb Q |=|\mathbb N|$.

Moreover $\bigcup\limits_{x\in U}(a'_x,b'_x)$ is clearly a subset of $U$ since every interval in the union is contained in $U$. But this union also contains $U$ since every element $x\in U$ is contained in the interval $(a'_x,b'_x)$ and hence is contained in the union. Hence $U$ is a countable union of rational open intervals.

Answer 2

The following link shows that opne set can is countable disjoint union of open intervals:

Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]

Let's asume $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)$,where $a_i,b_i\in \mathbb{R}$, then for each interval $(a_i,b_i)$, since rational number is dense in real number, we can always find rational sequence $\{b_{i,n}\}$ increasingly approaching $b_i$ and $\{a_{i,n}\}$ decreasingly approaching $a_i$, then $(a_i,b_i)=\bigcup_{n=1}^\infty (a_{i,n},b_{i,n})$, then $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)=\bigcup_{i=1}^\infty\bigcup_{n=1}^\infty(a_{i,n},b_{i,n})$, and countable union of countable union is countable union too.

As for can we find disjoint countable union of rational intervals, I didn't prove it.

Answer 3

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each $n\in\mathbb{Z}_{\geq 1}$, there exists $q_{n}\in\mathbb{Q}$ such that $|a-q_{n}|\leq\frac{1}{n}$. For each $n$, set $a_{n}:=q_{n}+\frac{1}{n}$. Note that $\{a_{n}\}_{n=1}^{\infty}$ is a decreasing sequence of rational numbers. Moreover, manipulating the inequality $|a-q_{n}|\leq\frac{1}{n}$ a little shows that $a\leq a_{n}\leq \frac{2}{n}+a$, and thus the Squeeze Theorem implies that $a_{n}\searrow a$. Similarly, the density of $\mathbb{Q}$ in $\mathbb{R}$ implies that, for each $n\in\mathbb{Z}_{\geq 1}$, there exists $p_{n}\in\mathbb{Q}$ such that $|b-p_{n}|\leq \frac{1}{n}$. For each $n$, set $b_{n}:=p_{n}-\frac{1}{n}$ so that $\{b_{n}\}_{n=1}^{\infty}$ is an increasing sequence of rational numbers and $b-\frac{2}{n}\leq b_{n}\leq b$, which implies that $b_{n}\nearrow b$. It then follows that \begin{equation*} (a,b)=\bigcup_{n=1}^{\infty}(a_{n},b_{n}). \end{equation*}