# Explicit open cover of $\mathbb{R}$ by disjoint countable unions of open intervals

It is well-known that every open set in $$\mathbb{R}$$ is a disjoint countable union of open intervals; is there an explicit construction of an open cover of the whole real line consisting of collections of countable disjoint open intervals?

• I think trivially, the whole real line, $(-\infty, \infty)$ works for what you are requesting. May 18 at 0:47
• @N. Owad. Yes, that is what I thought first; but I think in the proof of the proposition that every open set in $\mathbb{R}$ is a disjoint countable union of open intervals, intervals are assumed to be of finite length; thus I am more interested in the non trivial case. May 18 at 0:59
• By the way, if you carefully read the proof of the theorem "every open set in $\mathbb{R}$ is a disjoint countable union of open intervals", you will find these disjoint intervals are unique, i.e. $(-\infty, +\infty)$ is also the unique decomposition of $\mathbb{R}$. May 18 at 1:00
• @Asigan. I see, thank you and I think that $(-\infty, \infty)$ might be the only option. May 18 at 1:02
• Yeah, the well known theorem isn't true if you only allow bounded open intervals. May 18 at 1:06

Suppose $$]a,b[$$ is an interval of the family covering $$R$$. The point $$a$$ is in $$R$$ but not in $$]a,b[$$. If $$a$$ were in another interval $$]c,d[$$, a nbrd of $$a$$ enclosed in $$]c,d[$$ should contain points of $$]a,b[$$, which is impossible. The family does't cover $$R$$.