Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
possible duplicate. –  David Mitra Jul 4 '12 at 4:08
add comment

3 Answers

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

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 $\Omega$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.