Real analysis help(countable union)

[mynameisfunk]

Homework Statement

Show that if E \subseteq R is open, then E can be written as an at most countable union of disjoint intervals, i.e., E=\bigcup_n(a_n,b_n). (It's possible that a_n=-\inf or b_n=+\inf for some n.) Hint: One way to do this is to put open intervals around each rational point in E in such a way that every point of E and only points of E are contained somewhere in these intervals. Then combine the intervals that intersect.

The Attempt at a Solution

OK, intuitively, I get this, but what confuses me about the method they suggest is when I'm told to combine the intervals that intersect. Doesn't that imply that they aren't disjoint?! At firstI had the idea to take the set of all Neigborhoods of all the rationals in E of rational radius that do not intersect the complement of E, but these are not disjoint. Then I decided to take an arbitrarily large neighborhood in E and then take neighborhoods of the space that is left over and keep filling in the gaps with more and more neighborhoods until I have an at most countable amount of neighborhoods that are dense in E, but I had trouble getting that down and I also figured I should probably utilize the hint... But the hint confuses me more than the problem statement.

 

[ystael]

The hint actually seems kind of backwards to me (to see why it's backwards, consider trying to cover all of \mathbb{R} by covering rational q < \pi with (-\infty, \pi) and rational q > \pi with (\pi, \infty)).

Try following this outline instead.

1. Express E = \bigcup_{x \in E} U_x where U_x is an open interval containing x.

2. Figure out some way to group the U_x together to produce E = \bigcup_{\alpha\in A} V_\alpha, where \{V_\alpha\}_{\alpha \in A} is a pairwise disjoint family of open intervals, and each V_\alpha is a union of some of the U_x.

3. Prove that the disjoint family \{V_\alpha\}_{\alpha\in A} is at most countable. (Here is where you need to think about rational numbers.)