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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?

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    $\begingroup$ $(0,1)$ is already such a representation, and no other representation is possible. $\endgroup$ Commented Nov 13, 2016 at 3:16

4 Answers 4

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$\{(0,1)\}$ 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.

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As noted by Dr. Scott, the partition $\{(0, 1)\}$ is the only possible partition of $(0, 1)$ into countable open sets. If it were possible to write $(0, 1)$ as the union of two or more nonempty disjoint open sets, then it would be possible to write $(0, 1)$ as the union of exactly two nonempty disjoint open subsets (since the union of any arbitrary collection of open sets is open). Suppose $(0,1) = V \cup W$, where $V$ and $W$ are nonempty disjoint open sets. If $x \in \overline{V} \setminus V$, then we cannot have $x \in W$.

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  • $\begingroup$ Umm... So I guess the "power" of the theorem is that, no matter how strange an open set in $\mathbb{R}$ is, we can always decompose it into unions of open set. But using the theorem on things like $(0,1)$ would be an overkill? $\endgroup$
    – user34183
    Commented Nov 13, 2016 at 3:23
  • $\begingroup$ The theorem works fine on $(0, 1)$ as long as you're willing to accept that $\{(0, 1) \}$ is the only possible such partition. $\endgroup$
    – Kaj Hansen
    Commented Nov 13, 2016 at 3:25
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Yes $(0,1)$ is itself a representation, and there are no others: Suppose $(0,1)$ is written as a countable disjoint union of more than one open interval. Call one of those open intervals $I$. By hypothesis $\inf\{x\in I\}\neq 0$ or $\sup\{x\in I\}\neq 1$. Either way there must exist $x_0\in\partial I$ such that $x_0\in(0,1)$. Therefore there must exist another open interval $J$ that contains $x_0$. As $J$ is an open set and $x_0\in J$ there exists an open ball $(x_0-\delta,x_0+\delta)\subset J$. As $x_0\in\partial I$ this leads us to the contradiction $$I\cap J\supset I\cap(x_0-\delta,x_0+\delta)\neq\phi.$$

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After unioning the first two disjoint intervals, the complement will clearly have a component that's a closed interval (possibly trivial). You certainly can't write that as any union of open sets.

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