open set in $\mathbb{R}^n$ is a countable union of non-overlapping closed intervals in $\mathbb{R}^n$

Question

Every open set in $\mathbb{R}^n$ is a countable union of non-overlapping closed intervals in $\mathbb{R}^n$. I want to know interesting and different proofs for this.

Here closed interval in $\mathbb{R}^n$ means the set $[a_1,b_1]\times[a_2,b_2]\times...\times[a_n, b_n] $,where $[a_i , b_i]$ are real closed intervals for each $i$.

Two intervals are non-overlapping if the intersection of their interior is empty

Answer 1

Assume this is possible for $(0,1)\subset\Bbb R^1$. For the countable set $A$ of midpoints of the closed intervals, we have

• $A$ is not empty
• $A$ has no maximal and no minimal element
• if $a,b\in A$, $a<b$, then there exists $c\in A$, $a<c<b$.

It follows that $A$ is order-isomorphic to $\Bbb Q$, i.e., there exists a bijection $f\colon A\to \Bbb Q $ such that $a<b\to f(a)<f(b)$. Let $x=\sup\{\,a\in A\mid f(a)<\sqrt 2\,\}$. Then $x$ is not covered by the given intervals.