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Theorem: Every non-empty open set S in $\mathbb{R^1}$ is the union of a countable collection of disjoint component intervals of S.

I think it is quite easy to prove that the component intervals is disjoint but i am not sure how to do for countable union.

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possible duplicate. – David Mitra Jul 4 '12 at 4:08

Every collection of pairwise disjoint non-empty open intervals of $\Bbb R$ is countable. If $\mathscr{I}$ is such a family of intervals, each $I\in\mathscr{I}$ contains some rational number $r(I)$. If $I,J\in\mathscr{I}$, and $I\ne J$, then $I\cap J=\varnothing$, so $r(I)\ne r(J)$. Thus, the map $r:\mathscr{I}\to\Bbb Q:I\mapsto r(I)$ is injective, and it follows that $|\mathscr{I}|\le|\Bbb Q|=\omega$, i.e., that $\mathscr{I}$ is countable. This is the easier part of the theorem.

To finish proving the theorem you must show that every open $S\subseteq\Bbb R$ is a union of pairwise disjoint non-empty open intervals. The easiest way to do this is to define an equivalence relation $\sim$ on $S$ as follows: if $x,y\in S$, then $x\sim y$ iff either $x\le y$ and $[x,y]\subseteq S$, or $y\le x$ and $[y,x]\subseteq S$. In other words, $x\sim y$ iff the entire closed interval between $x$ and $y$ is contained in $S$. To finish the proof you must do two things:

  1. Prove that $\sim$ actually is an equivalence relation on $S$.
  2. Prove that each $\sim$-equivalence class is an open interval in $\Bbb R$.

I’ll let you try; if you get stuck, I can add to the answer.

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i don't quite understood why we need the equivalence relation or how can it helps to deduce the conclusion? – Mathematics Jul 4 '12 at 7:56

Question: Have you shown that the components are in fact open intervals?

As for countability, here's an (unavoidably big) hint: consider a countable dense subset.

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Let $\Omega \subset \mathbb{R}$ be a non-empty open set. For all $q \in \mathbb{Q}$, define $r(q)= \sup\{ r \geq 0 | B(q,r) \subset \Omega \}$ (so $r(q)=0$ iff $q \notin \Omega$). You can show that $\Omega = \bigcup\limits_{q \in \mathbb{Q}} B(q,r(q))$.

Then, you can identify subcovers on each components of $\Omega$.

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