Not every open set in $\mathbb{R}^n$, for $n\geq2$, is a disjoint union of countable open balls.

Question

I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help.

Answer 1

If any connected open set is written as a countbale disjoint union of open bals then all but one of them must be empty (because of conncetedness) so the original open set must itself be an open ball. So an open rectangle would be a counter-example.

Answer 2

Any open subset $O$ of $\Bbb R^n$ is a finite or countable disjoint union of open connected sets and that’s all that can be said, really. This is true in any separable locally connected space. In $\Bbb R$ those open connected sets can only be intervals or segments. In $\Bbb R^n, n>1$ there can be more variation (open rectangles minus finitely many points, interiors of closed curves in all sorts of shapes etc) but you do know the connected open sets are path-connected too.