0

It is said that an open interval is a countable union of disjoint intervals.

I do not get a vision of how this is possible.

For example if we consider$(0,3)$.if we write $(0,1)U(1,2)U(2,3)$ then it will exclude the point $1$ and $2$.I mean this will happen always..

Can anybody represent $(0,4)$ as a union of disjoint open intervals

Rayees Ahmad
  • 1,325
  • Since $\emptyset$ is disjoint to every other set and an interval, you can take the union to be the interval itself and a bunch of empty sets. – Henricus V. Mar 23 '16 at 05:16
  • Many proofs have been given for this (e.g. here). Though I think this answer is the most helpful visually. – Jeremy Upsal Mar 23 '16 at 05:23
  • Then that is true for every set.What is the scope of saying this sir. – Rayees Ahmad Mar 23 '16 at 05:24
  • 3
    Perhaps you read that an open set, rather than an open interval, is a countable union of disjoint intervals. $\qquad$ – Michael Hardy Mar 23 '16 at 05:33
  • Do you want to say that an open interval is a countable union of disjoint *closed* intervals? – miracle173 Mar 23 '16 at 06:02
  • Well if the intervals can be clopen... Example (0,2) = union (1/(n+1),1/n] U (1,2). – fleablood Mar 23 '16 at 06:45
  • Also finite is countable, isn't it? Discounting the empty set (and dirty trick) I'm fairly certain an open interval is not the union of more than one disjoint open interval. Where have you heard it said? I've never heard it. I think you are mishearing every open set in R is an at most countable union of disjoint open intervals. – fleablood Mar 23 '16 at 07:54
  • I read it from the book (Real Analysis) of renowned Professor Ranbir sing Dhoot page 394

    He says if says if $G$ is an open set of $[a,b]$ then G can be written as countable union of mutually disjoint open intervals ${I_n}$ that is $$G=UI_n$$ where $I_n\intersection I_m=\phi$

     – Rayees Ahmad Mar 23 '16 at 08:41

0 Answers0