0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – archipelago Sep 26 '13 at 18:49
  • $\begingroup$ Right but then I don't think arbitrary open $U$ can be written as a disjoint union of such balls. $\endgroup$ – user2566092 Sep 26 '13 at 18:50
  • $\begingroup$ Oh, sorry. I misread your question. $\endgroup$ – archipelago Sep 26 '13 at 18:51
  • $\begingroup$ 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. $\endgroup$ – archipelago Sep 26 '13 at 18:54
  • $\begingroup$ So $B$ contains all open connected subsets. And as you mentioned, the family of all open connected subset is a base as you whish. $\endgroup$ – archipelago Sep 26 '13 at 18:56
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.