# Bounded open sets of $\mathbb{R}$ as finite union of disjoint open intervals

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?

• Consider the union of all intervals of the form $(1/(2n+1), 1/(2n))$. Commented Oct 1, 2021 at 9:20
• @MartinR Thanks for the simple counter-example. Does it mean there is no way to relate open sets to finite union of disjoint intervals?
– Tham
Commented Oct 1, 2021 at 9:23
• None that I know of. Commented Oct 1, 2021 at 9:24
• @MartinR Got it, thank you!
– Tham
Commented Oct 1, 2021 at 9:25
• The complement of the Cantor set is another example. Commented Oct 1, 2021 at 9:26

$$O:= \bigcup_{n=0}^\infty \bigcup_{k=0}^{3^n-1} \left(\frac{3k+1}{3^{n+1}},\frac{3k+2}{3^{n+1}}\right)$$ 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).