Reference request: alternative proof for every open set in $\mathbb{R}^n$ can be expressed as countable disjoint union of open boxes

Question

A "box" is a cartesian product of intervals of the type $[a,b]$

I am using Terence Tao's introduction to measure theory and on page 24 a proof of title statement is given, however, it is quite difficult

I am aware that a lot of posts already exists in this direction, for example this one: Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs] But it is always about intervals on $\mathbb{R}$ and the proofs are all fairly tough.

Does anyone know if there exists a reference of this proof that is sufficiently easy for beginners in analysis?

score 2 · Accepted Answer · answered Apr 26 '16 at 02:25

2

Chapter 1 of Stein and Shakarchi's Real Analysis textbook has a similar result, and the proof includes a helpful picture. But I think it's worth taking the time to understand Tao's proof. Dyadic cubes are quite useful.

answered Apr 26 '16 at 02:25

carmichael561

53,688

Thanks, I will take a look at it. I'm using this book where literally every proof cites "every open set can be written..." yet I do not know how to prove this. By the way is the proof of this fact considered "natural" proof in the sense it can be proved on the spot and does not require some sort of gimmick? I want to assess whether if this is something that might show up on a test....:-) – Fraïssé Apr 26 '16 at 02:27
I suppose it is, though natural is of course a subjective term. It's important to note that only in $\mathbb{R}$ can the intervals be made disjoint; in higher dimensions "almost disjoint" is the best you can do. – carmichael561 Apr 26 '16 at 02:29
1

In higher dimensions, if you use intervals of the form $(a_1,b_1]\times\cdots(a_n,b_n]$ then they can be made disjoint. This is useful when applying countable additivity of measures. – Apr 26 '16 at 16:51
@Bungo: Those aren't open though. – carmichael561 Apr 26 '16 at 18:01
@carmichael561 Of course. I just wanted to point out the "workaround" that can be used if a countable disjoint union of intervals (of some kind) is desired. – Apr 26 '16 at 20:10

Reference request: alternative proof for every open set in $\mathbb{R}^n$ can be expressed as countable disjoint union of open boxes

1 Answers1