# Characterization of Open sets on $\mathbb{R}^m, m > 1$

I could get the proof for characterization of open sets in $\mathbb{R}$ from the book by NL Carothers (Theorem 4.6). However, I could not extend it to higher dimensions. Could any point me to a reference (a text book would be great) or answer this question?

Any help is much appreciated.

Thanks, Phanindra

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Would you be so kind and mention what characterization you have in mind? I for one don't have access to Carother's book... – t.b. Aug 4 '11 at 9:35
From what I could see in the Google books snippet views, you seem to ask: how can I extend the fact that every open subset of $\mathbb{R}$ is a union of countably many disjoint intervals to $\mathbb{R}^m$ for $m \gt 1$. Is that what you're asking? – t.b. Aug 4 '11 at 9:40
Theoream 4.6 says: If $U$ is an open subset of $\mathbb{R}$, then $U$ may be written as a countable union of disjoint open intervals. That is, $U=\bigcup_{n=1}^{\infty} I_n$, where $I_n=(a_n,b_n)$ (these may be unbounded) and $I_n \cap I_m=\oslash \ for \ n \neq m$ – ulead86 Aug 4 '11 at 10:04
Yes; $\mathbb R^n$ is 2nd-countable, meaning that any open set in it is the countable union of many intervals (tho you cannot guarantee that these sets will be disjoint), where here intervals are boxes $a_i<x<b_i$; for $i=1,2,..,n$. This is true because $\mathbb R^n$ is second-countable, and hasa basis of open intervals centered in points of $\mathbb Q^n$, with rational "length". So any proof of the 2nd countability should do. – gary Aug 4 '11 at 10:12
Theo: I should have included the google books link. Yes, I wanted know if we have that every open set in $\mathbb{R}^m, m > 1$ can be written as a countable union of disjoint open balls. – jpv Aug 4 '11 at 11:17

Yes; $\mathbb R^n$ is second-countable, meaning that any open set in it is the countable union of many intervals (although you cannot guarantee that these sets will be disjoint), where here intervals are boxes $a_i<x_i<b_i$; for $i=1,2,..,n$. This is true because $\mathbb R^n$ is second-countable, and has a basis of open intervals centered at points of $\mathbb Q^n$, with rational "length". So any proof of the second countability should do.
By definition of basis, every open subset of $\mathbb R^n$ is then the union of countably-many open basic open sets ; open boxes.
@Didier: What about the open ball of finite radius $r$ centered at a point $x$? (in dimension 2 or greater) It's connected, so it can't consist of more than one open interval. I believe the reason disjoint intervals characterize open sets of $\mathbb{R}^1$ is that any connected open set is an interval! – Shaun Ault Aug 4 '11 at 11:23