# On open set on the line can be written as countable union of disjoint segments

1. Sep 26, 2007

### SiddharthM

Again, my method of proof is far too messy. Let E be a open subset of R.

Any collection of disjoint segments in R must be at most countable - pick one rational in each segment and from there we see that the collection of sets can be mapped to a subset of a countable set. In other words any collection of disjoint segments in R cannot be uncountable.

Where it gets messy is when I go from the collection of neighborhoods that equals E being an uncountable and NOT-pairwise disjoint. My construction of a new collection goes by first taking all nhbds that intersect finitely many other nhbds - and then each finite union of nhbds of this sort are also segments. Add these segments to our new collection. The nhbds that don't intersect anything but themselves should also be added. The nhbds that intersect infinitely many other nhbds will have a union that is either bounded or unbounded. If it's bounded then their union is a segment and these such unions should be added to our new collection of sets. The class of unions of nhbds that are unbounded - well I'm not even sure what to do with them.

There are tons of problems with this construction. There has to be an easier way to show that every open set in R can be written as a union of disjoint segments (the fact that this collection must not be uncountable is simple).

HELP PLEASE!!! I hate it when things get messy like this.

FYI :segment is of the form (a,b), a and b real.

2. Sep 26, 2007

### EnumaElish

Assume, without loss of generality, a = 0 and b = 1.

First segment = (0, 1/2)
Second seg. = [1/2, 1/2 + 1/3)
...
k'th seg = [1/2 + ... + 1/k, 1/2 + ... + 1/k + 1/{k+1})

3. Sep 26, 2007

### StatusX

What are the connected components of an open subset of R?

4. Sep 26, 2007

### SiddharthM

Enuma: segments are of the form (a,b) NOT [a,b).

Status X: The connected components of an open subset of R are neighborhoods, or more generally, segments.

5. Sep 26, 2007

### EnumaElish

Your title says "open set on the line can be written as countable union of disjoint segments." I don't see "open segments."

6. Sep 26, 2007

### SiddharthM

look at the last line of my post - the FYI.

7. Sep 26, 2007

### EnumaElish

Then I don't see how they can be open and disjoint, yet add up to (A,B).

8. Sep 26, 2007

### SiddharthM

sorry, I have not been clear. Every open set E can be written by a AT MOST countable union of disjoint segments. So for a basic open set (a,b), just simply (a,b) would suffice.

I can't come up with a countable collection of disjoint open segments that 'adds up' to (a,infinity).

9. Sep 26, 2007

### EnumaElish

I don't see how they can be open and disjoint, yet "add up" to (A, +infinity).

10. Sep 26, 2007

### SiddharthM

gives a solution.

The trick is to define the equivalence relation x ~y iff x and y are in some (a,b) which is contained in our open set E. Then the equivalence classes form disjoint open intervals which are at most countable.

The proof is unconvincing but is given as such in textbook. It's unconvincing b/c I CAN'T FIND A COUNTABLE UNION OF DISJOINT INTERVALS that equals (a, infinity).

I'm beginning to think Rudin considers (a, infinity) as a segment.

cheerio!

11. Sep 26, 2007

### EnumaElish

That's my guess, as well.