It has been shown that every open set in $\mathbb R$ is the union of disjoint open intervals. Is this true for $\mathbb R^n$ in general?
It seems to me that it is not, but I cannot think of an example.
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1How are you defining interval in $\Bbb R^n$? – Brian M. Scott Aug 22 '16 at 17:59
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I am defining an open interval in R^n as a product on n open intervals in R. – user137957 Aug 22 '16 at 18:03
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It is not true in general for $\mathbb R^n$, but we do have the slightly weaker (but still very useful) result that every open set can be decomposed as a countable union of almost disjoint closed intervals – Aug 22 '16 at 18:04
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Looks like @BrianM.Scott has been answering this question for years. http://math.stackexchange.com/questions/435596/check-my-answer-prove-that-every-open-set-in-bbb-rn-is-a-countable-union-of – Alexis Olson Aug 22 '16 at 18:06
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1@AlexisOlson That's a different question, since there is no requirement of disjointness there. – Noah Schweber Aug 22 '16 at 18:08
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2I think the closest $\mathbb{R}^n$-counterpart would be: "every open set in $\mathbb{R}^n$ can be written (uniquely) as a countable union of disjoint connected open sets." It's just the case that connected open sets in $\mathbb{R}$ are open intervals. – Batominovski Aug 22 '16 at 18:36
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product of open intervals are rectangles. An open disc is not a union of disjoint rectangles. – fleablood Aug 22 '16 at 18:36
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@Batominovski that captures the spirit of the question. "intervals" was arbitrary and unimportant. In 1-dimension connected open sets are intervals. – fleablood Aug 22 '16 at 18:40
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It's unclear what you mean by "interval" in a higher-dimensional context. Reasonable candidates include boxes ( = Cartesian products of open intervals) and open balls. However, here's a general result showing that almost any such claim will fail even in $\mathbb{R}^2$:
Let $\mathcal{O}$ be some family of open sets (e.g., all open boxes), and let $U$ be a connected open set which is not an element of $\mathcal{O}$ (e.g., $U$ is an open ball). Then $U$ cannot be written as a disjoint union of elements of $\mathcal{O}$. Why? Well, suppose it could - as $U=\bigsqcup A_i$. Fix some nonempty $A_i$ in particular. Since $U\not\in\mathcal{O}$, we have $A_i\subsetneq U$; so (why? this is where connectedness of $U$ gets used!) we can find a point $x$ on the boundary of $A_i$, which is in $U$. Then any open set at all containing $x$ has to intersect $A_i$. But $x\in A_j$ for some $j$, since $x\in U$! So $A_i\cap A_j\not=\emptyset$, contradiction.
EDIT: Or, as Bungo points out below, $A_i$ and $\sqcup_{j\not=i} A_j$ partition $U$ into two disjoint open sets, contradicting the assumption that $U$ was connected.
The key difference between the two-(or higher-)dimensional case and the one-dimensional case? In one dimension, the connected open sets are easy to classify - they're exactly the open intervals! In two dimensions, though, there are lots of connected open sets. Really, the theorem that works for all dimensions is
Any open set can be written as a disjoint union of connected open sets;
it's just that for $n=1$ this can be phrased in a particularly snappy way.
Noah Schweber
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Maybe even simpler to note that that $A_i$ and $\cup_{n\neq i}A_n$ are disjoint nonempty open sets whose union is $U$, hence obtaining the contradiction that $U$ is disconnected? – Aug 22 '16 at 18:12
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