# Real analysis

1. Sep 29, 2010

### mynameisfunk

Hey guys, doin another rudin-related question. Here Goes:

Show that if E $$\subseteq$$ $$\Re$$ is open, then E can be written as an at most countable union of disjoint open intervals, i.e., E=$$\bigcup$$n(an,bn). (It's possible that an=-$$\infty$$ bn=+$$\infty$$ for some n.)

My attempt:
Take the set of all Neighborhoods of all of the rationals of a rational radius in R to be A. Now all members of E intersect A make up E. Take the union of all of the neighborhoods in this set E intersect A and this is a countable union of disjoint sets.

Is there a problem with this?

Last edited: Sep 30, 2010
2. Sep 30, 2010

### mynameisfunk

I now realize this doesnt create a disjoint union. I had the idea of taking an arbitrarily large neighborhood of a point in E that is still a proper subset of E and then filling up the gaps on either side with more neighborhoods, always still contained in E, until I have a countably dense set of neighborhoods contained in E...

Would this do?

3. Oct 2, 2010

### kntsy

Here is my suggestion:
First let E be a union of disjoint open sets F. Then E is an open set. We know that by Lindelof there is a countable subset F' of F. However, since F is disjoint, F=F'. So F is countable.

Accurately, F is a collection of sets and E is the union of sets taken from F.

I guess that your first suggestion proves the lindelof that countable subets exist. Although they are "not disjoint", you have proven a lindelof and you can apply this property to the open set which is a union "disjoint" open sets.

4. Oct 3, 2010

### Landau

This is a familiar exercise/result. See e.g. here, here, and your own (!!) thread here.