Hi I see many people asked this question,but none of them proved this by considering all possible cases which my professor did.
Below is the solution that I have.
Show that a union of open intervals can be written as a disjoint union of open intervals.
Solution. The whole problem can be reduced to writing the union of two intervals as a union of disjoint ones (in
case we have several, or uncountably many, we do induction). We are given (a, b) and (c, d) and we want to write
(a, b) ∪ (c, d) as a union of disjoint intervals. There are three cases that can occur:
• a < b ≤ c < d. Then the two intervals are already disjoint, so we are done.
• a ≤ c < b ≤ d. Then (a, b) ∪ (c, d) = (a, d).
• a ≤ c < d ≤ b. Then (a, b) ∪ (c, d) = (a, b).
I am not sure why there are only three cases.
Can we have (c,d) and (c,b) as well? so I think there must be 5 cases that we have to consider. Also, I do not understand how I can apply this to show that any union of open intervals can be written as union of disjoint open intervals.
Can anyone explain?
Best Answer
Let $U \subset \mathbb{R}$ be open.
Define an equivalence relation $x \sim y $ iff $[\min(x,y), \max(x,y)] \subset U$.
Let $[x] \subset U$ denote the equivalence class containing $x$ (somewhat standard, but slightly ambiguous notation in this context).
Show that if $x \in U$, then $[x]$ is an open interval.
Let $\cal U = \{ [x] \}_{x \in U}$. Then $\cal U$ is a collection of disjoint open intervals such that $U = \cup_{A\in {\cal U}} A$.
Aside: By considering $U \cap \mathbb{Q}$, it is not difficult to show that the collection $\cal U$ is at most countable.