0
$\begingroup$

Could anyone please give me an example of a open set of $\mathbb{R}$ that can be written as a countable union of mutually disjoint open intervals?

I could not understand why the union of disjoint countable open sets could be an open subset of $\mathbb{R}$.

I mean, how could a union of countable sets equal an uncountable sets? As far as I understand, all infinite subsets of $\mathbb{R}$ are uncountable. I am not sure whether or not I misunderstood the theorem somehow.

Thank you so much, Riemann, Podiki, Soumik Mukherjee by saying countable union of sets, does it mean the union is a countable set, i.e. with the same cardinality of $\mathbb{N}$ (the set of natural numbers), or it just means a finite number of sets in a union?

$\endgroup$
7
  • $\begingroup$ It's not "countable open sets", but "countable family of open sets". There are countable number of sets in a family or collection, and each set in the family or collection is an open set. $\endgroup$
    – Riemann
    May 9 at 13:29
  • $\begingroup$ The union is countable (meaning it is the union of countably many sets), but each interval you consider will have uncountably many elements in it. $\endgroup$
    – podiki
    May 9 at 13:30
  • $\begingroup$ It is not union of countable sets, it is countable union of sets. That is, total number of sets is countable, the individual sets are open sets. $\endgroup$ May 9 at 13:31
  • $\begingroup$ Also, all infinite subsets of $\mathbb{R}$ need not be uncountable, $\mathbb{Q}$ or $\mathbb{Z}$ are infinite but countable. $\endgroup$ May 9 at 13:34
  • $\begingroup$ It means (at most) countable, i.e. it may have finite number of sets or countable number of sets. $\endgroup$ May 9 at 13:42

1 Answer 1

0
$\begingroup$

First we'll show that it cannot be an uncountable union of disjoint intervals. Given a family of disjoint open intervals, pick a rational number from each. These rationals are distinct and there are only countably-many, so the family must be countable.

To see that each open set is the union of disjoint intervals, take any point of the set and it belongs to an interval that's a subset of the set. One can prove that there exists an equivalence relation for points in this set for which two points are equivalent of there exists a chain of the intervals are pairwise intersect. You can then prove that the union of each equivalence class is an open interval.

$\endgroup$
0

You must log in to answer this question.