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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...2011-08-04
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    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?2011-08-04
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    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$2011-08-04
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    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; 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.2011-08-04
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    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.2011-08-04
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    I cannot access math.stackexchange.com (this website) from my work place. I am not sure where to post this problem. Any help?2011-08-06

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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; 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.

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    gary: Hope you do not mind the (small) edits. // I thought one could guarantee that the boxes are disjoint if one wanted to, am I wrong here?2011-08-04
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    Didier: I think there are cases where one cannot, like, e.g., for "non-standard" open sets, but let me think of an actual example.2011-08-04
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    My last comment is stupid, please forget it: I missed the condition that the boxes are *open*.2011-08-04
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    @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!2011-08-04
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    gary: Thanks. Theorem 5 (page 83) of this book http://books.google.com/books?id=z8IaHgZ9PwQC&printsec=frontcover&dq=real+analysis+kolmogorov&hl=en&ei=DYE6TqvhKaPz0gGQ4Ny4Aw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q=separable%20metric%20space&f=false shows that this is possible with non-disjoint open balls.2011-08-04
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    I apologize if the question was not clear. I wanted to know if every open set in $\mathbb{R}^m, m>1$ can be written as a union of countable disjoint open balls. Thanks all for replies.2011-08-04