I saw a proof of the following theorem.
Every open subset $\mathcal{O}$ of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals.
The proof was convincing, but can anyone help me writing out explicitly such a representation of the interval $(0,1)$? Or maybe $(0,1)$ itself is already such a representation? Can I write it as countable union of "smaller" disjoint open intervals?
Best Answer
$\{(0,1)\}$ is indeed already a countable set of vacuously pairwise disjoint open intervals whose union is $(0,1)$. (Countable means finite or countably infinite.) It is the only countable set of pairwise disjoint non-empty open intervals whose union is $(0,1)$. To see this, suppose that $(0,1)=\bigcup\mathscr{U}$ for some countable family $\mathscr{U}$ of pairwise disjoint non-empty open intervals. Let $U\in\mathscr{U}$, and let $V=\bigcup(\mathscr{U}\setminus\{U\})$. Then $\{U,V\}$ is a partition of $(0,1)$ into disjoint non-empty open sets, contradicting the fact that $(0,1)$ is connected.