Proving Every open set in $\Bbb R$ is a countable union of open intervals.

Question

This question is from William R. Wade's Introduction to Analysis book:

Prove that every open set in $\Bbb R$ is a countable union of open intervals.

I have no ideas honestly. Thank you.

Answer 1

Let U the open set. Fix $x \in U$. Exist $\delta_x > 0$ such that $(x - \delta_x , x+ \delta_x ) \subset U$ (because $U$ is open). Then

$$ U = \displaystyle\bigcup_{x \in U} (x - \delta_x , x+ \delta_x )$$

Then by the Lindelöf covering theorem we can extract a enumerable subcovering. then exist $x_1, x_2 , ....$ elements of $U$ with :

$$ U = \displaystyle\bigcup_{i=1}^{\infty} (x_i - \delta_{x_i} , x+ \delta_{x_i} )$$

the affirmation is proved.

Answer 2

The complete argument can be divided into several steps. Let $U \subseteq \mathbb{R}$ be open.

1. $U$ is a disjoint union of open and connected sets. This is just decomposing $U$ into its connected components and noting that they are open too.
   
2. There can be at most countable such components. Hint as mentioned before: Rationals are dense in $\mathbb{R}$.
   
3. An open connected subset of $\mathbb{R}$ must be a (nonempty) open interval.

If you are convinced that this strategy works, then you can move on to justify why the individual steps are true. Maybe you already saw some of them before.

Answer 3

Hint: the rational numbers are countable, and they are dense in $\mathbb{R}^n$ (so every irrational number is contained in a neighborhood of a rational number or arbitrarily small radius).