Decompositions of open sets in $\mathbb{R}^n$ This may be a basic question. It's well known that open sets in $\mathbb{R}$ are the union of disjoint open intervals. Does it similarly hold that open sets in $\mathbb{R}^n$ are the union of disjoint connected open sets? Moreover, is a connected open set the union of disjoint connected convex sets? Thanks.
 A: The first property you ask for is fairly trivial to show; in particular, for any open set $O$ define an equivalence relation $$a\sim b$$ defined by the property that, if $S$ is a set of disjoint open sets covering $O$, then $a\in s\in S$ implies $b\in s$ - that is, $a$ and $b$ are in the same connected component. The equivalence classes of this can easily be shown to be disjoint open sets, whose union is $O$.
The second property is somewhat more difficult, but still doable. One approach is essentially to build an octree as a partition. In particular, let $S$ be the set of intervals of the form
$$\left[\frac{x_1}{2^n},\frac{x_1+1}{2^n}\right)\times \left[\frac{x_2}{2^n},\frac{x_2+1}{2^n}\right)\times \left[\frac{x_3}{2^n},\frac{x_3+1}{2^n}\right)$$
along with the set $\mathbb R^2$ itself. That is, we have axis aligned cubes whose corners are at "adjacent" dyadic rationals - if we put $8$ of these cubes together, we get a bigger cube which is also in the set. Then, we can quickly come up with a partition $P\subseteq S$ of $O$ - in particular, let $P$ be the set of $s\in S$ such that $s\subseteq O$ but for which there does not exist any $s'\in S$ which (strictly) contains $s$ but is contained in $O$ - that is, $P$ is the set of cubes of maximal size fitting in $O$.
This is a partition since, for any $p\in O$, there is some ball around $p$ contained in $O$ - and therefore, as every point is contained in a cube of arbitrarily small radius, there is some $s\in S$ such that $p\in s\subseteq O$. We can then argue that if two cubes $s_1,s_2\in S$ contain $p$, then either $s_1\subseteq s_2$ or $s_2\subseteq s_1$ (because of how the cubes nest) - and thus, the set of cubes containing $p$ forms a chain when ordered by inclusion - and our partition has the property that the union over such a chain is still in $S$ - therefore, we conclude that there is a maximal element in $S$ containing $p$ and that this is in $P$, thus $P$ is a partition.
We can generalize this argument to say that, if $S$ is a set satisfying:


*

*If $s_1,s_2\in S$ then $s_1\cap s_2$ is either $s_1$, $s_2$, or the empty set.

*Every subset $S'\subseteq S$ which is totally ordered by inclusion (i.e. a chain) has that the union of all $s\in S'$ is a member of $S$. (Equivalently, every subset of $S'$ has a least upper bound when ordered by inclusion) 

*For any open set $O$ and $p\in O$, there exists a $s\in S$ which contains $p$ and which is contained wholly within $O$.


then there is a partition $P$ of any open set satisfying $P\subseteq S$. In our particular example, we chose a suitable $S$ composed of convex sets with non-empty interior. 
A: The open sets of $\mathbb{R}^n$ in the standard Euclidean topology are generated by the open balls $B(x,\epsilon) = \{ y \in \mathbb{R}^n | ||x-y||<\epsilon \}$ where $\epsilon >0$ and $x \in \mathbb{R}^n$. Notably, for $n=1$ the $||x-y||$ (Euclidean norm) is just $|x-y|$, the absolute value.
It is true that in $\mathbb{R}$, the open sets are disjoint unions of intervals. For instance, the set $(-\infty,6) \cup (14,\infty)$ is a disjoint union of intervals, but if I write something like $(-\infty,6) \cup (4,\infty) = (-\infty, \infty)$, that is still an open set, even though I've written it as the union of non-disjoint open intervals.
After simplification of set union wherever it may occur, every open set in $\mathbb{R}^n$ will be the disjoint union of open balls, but it may not necessarily be written that way. Once we do all simplification, we should end up with a collection of disjoint open sets.
