Can every open subset of R be written as a countable union of open intervals?

[Buri]

I'm currently reading Real Analysis by Stein and Sharkarchi and they state the following theorem:

Every open subset O of R can be written uniquely as as a countable union of disjoint open intervals.

They prove it and I understand the proof. I was just playing around with open sets, but it seems like there's no punch to it, so to speak. If I take (0,1) the 'countable union of disjoint open intervals' is the set itself and if I take the union of two open intervals then again its the set itself. I guess the open set would have to be something weird to be more interesting. Is there maybe any more interesting examples of an open set being written as a countable union of disjoint open intervals?

Thanks

 

[micromass]

Perhaps the complement of $$\{0\}\cup\{1/n~\vert~n\in \mathbb{N}\}$$, but that's also trivially a union of open intervals. But it's more interesting...

Or take the complement of the Cantor set...

 

[Petek]

I think you're looking for an example of an open subset of R whose description as a union of disjoint open intervals is non-trivial. How about this: Define the set F by

$$F = \{\frac{1}{p^n} : p\ is\ a\ prime\ and\ n \in \mathbb{N}\} \cup \{0\}$$

Then F is closed (since it contains its only limit point 0), so R - F is open. However, it seems to me that writing R - F as a union of disjoint open intervals is complicated.

 

[Buri]

Thanks to both of you! For some reason I hadn't thought of taking closed sets and looking at their compliments. So thanks!

 

[Petek]

Just to clarify my definition of the set F: p is supposed to run over the entire set of primes, not a single fixed value. Also, the complement of the Cantor set is an excellent example.