1
$\begingroup$

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?

$\endgroup$
11
  • 4
    $\begingroup$ Consider the union of all intervals of the form $(1/(2n+1), 1/(2n))$. $\endgroup$
    – Martin R
    Oct 1, 2021 at 9:20
  • $\begingroup$ @MartinR Thanks for the simple counter-example. Does it mean there is no way to relate open sets to finite union of disjoint intervals? $\endgroup$
    – Tham
    Oct 1, 2021 at 9:23
  • 1
    $\begingroup$ None that I know of. $\endgroup$
    – Martin R
    Oct 1, 2021 at 9:24
  • $\begingroup$ @MartinR Got it, thank you! $\endgroup$
    – Tham
    Oct 1, 2021 at 9:25
  • $\begingroup$ The complement of the Cantor set is another example. $\endgroup$
    – Martin R
    Oct 1, 2021 at 9:26

1 Answer 1

1
$\begingroup$

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).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .