# Bounded open sets

Gear300
I am asked to prove that any bounded open subset of R is the union of disjoint open intervals.
If S = open interval (a,b), I don't really see how this could be the case (there will always be points in S that are not in the union of the disjoint sets).

## Answers and Replies

TylerH
The first thing that comes to mind is $$\left(a, \frac{a+b}{2} \right) \cup \left( \frac{a+b}{2}, b \right)$$, but like you said, what about $$\frac{a+b}{2}$$?

Gear300
The first thing that comes to mind is $$\left(a, \frac{a+b}{2} \right) \cup \left( \frac{a+b}{2}, b \right)$$, but like you said, what about $$\frac{a+b}{2}$$?

I don't know. Perhaps its a trivial instance in this case: S U {}, a union with a null set, in which the null set is both open and closed. Though, I'm not sure if this is what they have in mind.

Homework Helper
You are interpreting the phrase "disjoint union of open intervals" incorrectly. The union may be of any number of open intervals including 1. The example you are asking about, (a, b), is an open interval and so is the union of 1 open set- itself.

The "meat" of the theorem is that open sets that are NOT intervals can be written as unions of open intervals.

Gear300
You are interpreting the phrase "disjoint union of open intervals" incorrectly. The union may be of any number of open intervals including 1. The example you are asking about, (a, b), is an open interval and so is the union of 1 open set- itself.

The "meat" of the theorem is that open sets that are NOT intervals can be written as unions of open intervals.

That makes things more clear. Thanks.