Writing $\mathbb{R}$ as union of more than one disjoint open sets

Question

Can we write $\mathbb{R}$ as union of more than one disjoint open sets (in $\mathbb{R}$) ? If no, then can you explain why it is not possible ?

You can write it as $\Bbb{R} = \Bbb{R} \cup \emptyset$. Otherwise no; this is a property known as connectedness. See the duplicate target for how to prove this. — Theo Bendit, Sep 03 '19 at 08:44
You also asked why not: it's because of the least upper bound property of R — Sort of Damocles, Sep 03 '19 at 12:33
@MeetR If you know that open sets in $\Bbb{R}$ are countable unions of disjoint open intervals, then consider any single open interval $I$. By the LUB property it has a least upper bound $z = \mathrm{lub} I$; since it is open we have $z \notin I$ (or else it would be the largest element). Thus $\Bbb{R} \setminus I$ cannot be open. Here's wikipedia on the LUB property: https://en.wikipedia.org/wiki/Least-upper-bound_property — Sort of Damocles, Sep 03 '19 at 14:09

score 2 · Accepted Answer · edited Sep 03 '19 at 08:45

2

$\mathbb R$ is connected. So you cannot write it as union of two non-empty disjoint open sets.

edited Sep 03 '19 at 08:45

answered Sep 03 '19 at 08:42

Kavi Rama Murthy

2

And if you write $\mathbb R$ as a union of any number (even an infinite number) of nonempty pairwise-disjoint open sets, then taking one of the sets and the union of all the rest gets you back to the union of two. – GEdgar Sep 03 '19 at 12:17

1 Answers1