I know that all open sets of $\mathbb{R}$ can be expressed as countable union of disjoint open intervals. However, I was hoping to restrict the case to finite union.
Can all bounded open sets of $\mathbb{R}$ be expressed as finite union of disjoint open intervals? If yes, how should the proof look like? If no, what are some counter-examples and is there anything we can say about the relationship between bounded open sets and finite union of disjoint open intervals?
Best Answer
Well, looking at the Cantor set, it's complement is
$$O:= \bigcup_{n=0}^\infty \bigcup_{k=0}^{3^n-1} \left(\frac{3k+1}{3^{n+1}},\frac{3k+2}{3^{n+1}}\right)$$ which is open and bounded and needs infinitely many intervals (the decomposition into open intervals is unique, as these are exactly the connected components of the open set).