# Partitioning an open interval

• Oct 31st 2008, 07:25 PM
Jes
Partitioning an open interval
I need to prove that every bounded open interval in $\displaystyle \mathbb{R}$ is a union of disjoint open sets. Not sure, but here's what I got:

Let $\displaystyle (a,b)$ be an open interval and let $\displaystyle x,y \in (a,b)$ with $\displaystyle x \not = y$ and assume $\displaystyle x < y$. Put $\displaystyle \epsilon = \min\{d(x,y), d(x,a), d(y,b)\}$ and define $\displaystyle S = \cup \{ B(x;\epsilon) : x \in (a,b)\}$. Then $\displaystyle S=(a,b)$ and it is a union of open sets.

Now we need to show they are pairwise disjoint. Using the previous definition of $\displaystyle \epsilon$, there is an open ball $\displaystyle B(x; \epsilon /3)$ and the ball $\displaystyle B(y; \epsilon /3)$ with $\displaystyle B(x; \epsilon /3), B(y; \epsilon /3) \subset (a,b)$. Suppose some $\displaystyle z \in B(x; \epsilon / 3) \cap B(y; \epsilon /3)$. Then $\displaystyle \epsilon \leq d(x,y) \leq d(x,z) + d(z,y) \leq \frac {\epsilon} {3} + \frac {\epsilon} {3} = \frac{2\epsilon}{3}$ gives a contradiction.
• Nov 1st 2008, 01:48 AM
Opalg
Quote:

Originally Posted by Jes
I need to prove that every bounded open interval in $\displaystyle \mathbb{R}$ is a union of disjoint open sets.

A bounded open interval in $\displaystyle \mathbb{R}$ is an open set all by itself. It is not possible to express it as the disjoint union of more than one (nonempty) open set.

You seem to have the wording of the problem the wrong way round. It should surely read: "Every bounded open set in $\displaystyle \mathbb{R}$ is a union of disjoint open intervals." You can find a neat proof of that here, using an equivalence relation.
• Nov 1st 2008, 04:53 AM
Jes
Good call, Opalg. You are right and thanks for the link.