Here is Prob. 29, Chap. 2, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Prove that every open set in $\mathbb{R}^1$ is the union of an at most countable collection of disjoint segments.
How to give this proof?
My effort:
Let $E$ be a non-empty open subset of $\mathbb{R}^1$, and let $p \in E$. Then we can find a real number $\delta > 0$ such that
$$(\, p-\delta \, , \, p+\delta \, ) \subset E.$$
Now we can find rational numbers $q_1$, $q_2$ such that
$$p-\delta < q_1 < p < q_2 < p+\delta.$$
So we have
$$ \left( q_1, q_2 \right) \subset (\, p-\delta \, , \, p+\delta \, ) \subset E.$$
In this way, we can show that
$$E = \cup (q_1, q_2),$$
and this union is of course that of a countable collection of segments.
Now how to show that these intervals are disjoint?
Is this line of reasoning going to lead to our desired conclusion?
Best Answer
The intervals as you've defined them won't generally be disjoint. You could try the following: for each $x \in E$ define $a_x = \inf\{a : [a,x] \subset E\}$ and $b_x = \sup\{b : [x,b] \subset E\}$. $a_x$ is possibly $-\infty$ and $b_x$ is possibly $\infty$. Show that 1) if $x \in E$ then $(a_x,b_x) \subset E$, and 2) if $x,y \in E$ then $(a_x,b_x) = (a_y,b_y)$ or $(a_x,b_x) \cap (a_y,b_y) = \emptyset$. Since each open interval contains a rational number there are at most countably many distinct intervals $(a_x,b_x)$, they are pairwise disjoint, and $\displaystyle E = \bigcup_{x \in E} (a_x,b_x)$.