Every open subset of $\mathbb{R}$ can be expressed uniquely as a disjoint union of open intervals. Does this generalize to $\mathbb{R}^n$?

Question

I know that every open subset of $\mathbb{R}$ can be expressed uniquely as a disjoint union of open intervals. Further, only countably many intervals feature in any such decomposition.

Supposing we replace $\mathbb{R}$ with $\mathbb{R}^n$, and 'open intervals' with 'open connected subsets,' does the above result still hold?

score 2 · Accepted Answer · answered Aug 11 '13 at 05:59

2

Yes, of course. Let $U$ be an open subset of $\mathbb{R}^n$. The components of $U$ are open in $U$, and hence in $\mathbb{R}^n$, by the local connectedness of $\mathbb{R}^n$. (In fact, this condition is equivalent to the local connectedness of $\mathbb{R}^n$; local connectedness is the key word here!)

I hope this helps!

answered Aug 11 '13 at 05:59

Amitesh Datta

20,779

1

Ah thank you. So the result holds in any locally connected topological space? Edit. Apart from the countability requirement, of course. – goblin GONE Aug 11 '13 at 06:00
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Yes @user18921 and you're welcome! – Amitesh Datta Aug 11 '13 at 06:01

Every open subset of $\mathbb{R}$ can be expressed uniquely as a disjoint union of open intervals. Does this generalize to $\mathbb{R}^n$?

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