# Understanding the concept of every open set being a disjoint union

1. Aug 25, 2013

### jdinatale

of a countable collection of open intervals.

I'm having a hard time seeing how this could be true. For instance, take the open set (0, 10). I'm having a hard time seeing how one could make this into a union of countable open intervals.

For instance, (0,1) U (1, 10) or (0, 3) U (3, 6) U (6, 10) wouldn't work because those open intervals miss some points. There are "gaps" missing from the initial open set (0, 10). It seems like any union of DISJOINT intervals would have "gaps" missing from the initial open set. And if any of the open sets overlap to fill those gaps, then they are no longer disjoint.

I've read several proofs of this theorem, and they don't clear up my confusion.

2. Aug 25, 2013

### rubi

An open set in $\mathbb R$ is by definition a (not necessarily disjoint) union of open intervals. You can show that $\mathbb R$ is second countable, so an at most countable number of open intervals suffices. Now a union of (not necessarily disjoint) open intervals is either an open interval again (for example $(0,2) \cup (1,3) = (0,3)$) or it is already a disjoint union of open intervals. So given an open set as a coutable union of not necessarily disjoint open intervals, you can always make it into a disjoint union of open intervals by joining the sets that have nonzero intersection.

In your case, $(0, 10)$ is already an open interval, so it is already a countable union of open intervals. Just take $A_0 = (0,10)$ and $A_n = \varnothing$ for $n\neq 0$.

3. Aug 28, 2013

### Erland

Countable includes finite, so a finite set is also countable. So {(0,10)} is countable set and its union is (0,10), which is therefore a countable union of open intervals.

4. Sep 13, 2013

### A David

You might be interested to know then that every open subset of the reals can be written as a countable union of almost disjoint closed intervals.