Union of 2 non-disjoint open intervals is open interval. I will not use word open later.
I want to obtain formal proof of this fact that is different from mine.
My attemp:
We know that union of two open sets is open set. Assume it is not an interval. Then it must be a union of intervals (because any open set is union of intervals). But union of more than one interval is not connected – we can just apply defintion of connectedness. And we know that union of nondisjoint connected sets is connected. Contradiction shows it must be an interval.
I am very unhappy with my attemp because I am trying to prove that any open set is union of disjoint intervals and in order to do I must prove that that union of non-dijsoint open intervals is open interval. So it is just cheating from my side, and in fact I do not understand two proofs instead of one proof. In fact I am not even sure that my attemp could be saved as a proof.
Is it possible to prove the fact using only properties of real numbers, including completeness and archimedean property without using connectedness?
Or maybe there is another way to deal with these 2 propositions at once without logic flaws?
Best Answer
Say $(a, b)$ and $(c, d)$ are your intervals. If there's an $x$ in both intervals, then that x is strictly between $c$ and $b$, so $c<b$, and so the union of these intervals is $(a, d)$ or $(a, b)$, depending on which of $b, d$ is larger.