# Bounded Open Subset as Open Intervals

1. Nov 17, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Prove the any bounded open subset of R is the union of disjoint open intervals.

The attempt at a solution
I've seen a proof of this using equivalence classes, which is fine, but I want an unsophisticated solution, e.g. one relying on just the definitions of "bounded", "open" and some properties of the reals. I have an approache in mind, but I don't think it works:

Let S be a bounded open subset of R. Then S is contained in contained in some open interval (-M, M). If we remove a point x from (-M, M), we get a disjoint union of open intervals, namley (-M, x) union (x, M). If we remove another point, we still get a disjoint union of open intervals. Thus, if we remove all the points of (-M, M) - S from (-M, M), which leaves S, we obtain S as the disjoint union of open intervals. The problem with this approach is that I'm removing an uncountably infinite amount of points one by one, which seems like a dubious process to me.

2. Nov 17, 2008

### Dick

Take the open interval (-1,2). Remove all of the points which are not in [0,1], 'one-by-one'. You wind up with [0,1] a closed interval. In fact, you can turn (-1,2) into ANY of it's subsets by doing that. No, removing an uncountable number of points 'one-by-one' isn't a very good idea. Your argument works fine for the removal of any finite number of points. But what makes you think the limit set is much like the finite steps? What's so 'sophisticated' about equivalence classes?

3. Nov 17, 2008

### e(ho0n3

It's the best argument I could think of and I had some hope it could somehow be made to work.

Nothing much really. I just want to solve this without them if it is possible.

4. Nov 17, 2008

### Dick

The argument is that the complement of the open set S in a closed interval is closed (call it C). If x is in the open set then the open interval between the limits defined by sup{y:y<x and y in C} and inf{y:y>x and y in C} is contained in S. To show they can be chosen disjoint, I don't see any way around some kind of an 'equivalence class' type argument. Two points will lead to the same open interval if they are contained in the same open interval.

5. Nov 19, 2008

### e(ho0n3

Yeah...I came up with another approach similar to yours which has an 'equivalence class' flavor to it. Anywho, thanks a lot.