# Minimal bases $B$ of topology on $\mathbb{R}^n$ such that any open $U$ is a disjoint union of members of $B$

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?

• 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
• 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
• Oh, sorry. I misread your question. – archipelago Sep 26 '13 at 18:51
• 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
• 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

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.