It seems that if a base $B$ of the euclidean topology on $\mathbb{R}^n$ has the property that any open $U \subset \mathbb{R}^n$ can be written as a disjoint union of members of $B$, then $B$ must contain all connected open subsets. However I don't have a proof. Is this choice of $B$ (i.e. all connected open subsets) the unique base of the topology on $\mathbb{R}^n$, such that any open $U$ is a disjoint union of members of $B$, and such that $B$ is minimal i.e. no subcollection of $B$ satisfies the same property?
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Take balls around rational vectors with rational radius. This is a (countable) base of $\mathbb{R}^n$ which is smaller than all connected open subsets. – archipelago Sep 26 '13 at 18:49
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Right but then I don't think arbitrary open $U$ can be written as a disjoint union of such balls. – user2566092 Sep 26 '13 at 18:50
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Oh, sorry. I misread your question. – archipelago Sep 26 '13 at 18:51
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Let $B$ be a base as you wish and $O$ a connected subset. Suppose $O\not\in B$, then you can write $O$ as a union of elements of $B$ (which are open), so $O$ isn't connected. Contradiction. – archipelago Sep 26 '13 at 18:54
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So $B$ contains all open connected subsets. And as you mentioned, the family of all open connected subset is a base as you whish. – archipelago Sep 26 '13 at 18:56
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So yes, it is the unique base with these properties. – archipelago Sep 26 '13 at 18:57
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Thanks, that almost answers the question and I'll upvote if you post as an answer. The only remaining question is whether arbitrary open $U$ can be written as a disjoint union of connected open sets. – user2566092 Sep 26 '13 at 19:00
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I'll give you a proof in my answer. Just a few minutes. – archipelago Sep 26 '13 at 19:03
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The connected components of an open set in $\mathbb R^n$ are open sets. – Etienne Sep 26 '13 at 19:04
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Yes, because it's locally connected. – archipelago Sep 26 '13 at 19:07
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As my comments suggested: Yes, the collection of all connected open subsets of $\mathbb{R}^n$ is the unique base, such that any open subset is a disjoint union of members of this base.
Let $C$ be the collection of all open connected subsets and $O$ a open subset. As open sets of $\mathbb{R}^n$ are locally connected the connected components of $O$ are open (and connected), so they are in $O$ and $C$ is the disjoint union of those. (See A space $X$ is locally connected if and only if every component of every open set of $X$ is open?) Hence $C$ serves as a base with your mentioned properties.
Let $B$ another base of that type and $U$ a connected open subset. Suppose $U\not\in B$, then you can write $U$ as a union of elements of $B$ (which are open), so $U$ isn't connected. Contradiction. This shows, that $B$ contains all open connected subsets.
archipelago
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