# Open Sets of R^n: Countable Union of Open Rectangles/Balls?

• qspeechc
So if you want to cover an interval with rectangles, you can do so without having to overlap any of the intervals. It's a bit of a stretch, but that's what they use to prove their theorem. In summary, this proof relies on the fact that open sets in the real line can be expressed as the countable union of open rectangles.

#### qspeechc

Hi

Ok, so I know open sets of the real line are the countable union of disjoint open intervals (or open balls). Does this in any way extent to R^n? Say, any open set in R^n is the countable union of open rectangles or balls? I ask because I was reading some proof, and at a crucial step they use the fact that open sets in R^n can be expressed as the countable union of open rectangles, and I have no idea where this comes from! It doesn't even seem plausible to me, if one considers the open ball in the plane-- how would you describe that as the union of countably many open rectangles?
I know every open set is the union of open balls, and maybe the Heine-Borel theorem comes in somewhere...but I'm just lost.
Any help?
Thanks.

qspeechc said:
I ask because I was reading some proof, and at a crucial step they use the fact that open sets in R^n can be expressed as the countable union of open rectangles, and I have no idea where this comes from! It doesn't even seem plausible to me, if one considers the open ball in the plane-- how would you describe that as the union of countably many open rectangles?

Hi qspeechc!

What's uncountable about it?

n times countable is still countable.

For your open ball, just keep halving the size of the rectangles … that'll do the job, won't it?

Errr...I'm not sure I follow you...? Are you saying I can go from the case of the real line to R^n with no difficulty?

qspeechc said:
Errr...I'm not sure I follow you...? Are you saying I can go from the case of the real line to R^n with no difficulty?

Yes.

Hhmm, actually I found a proof, and it has nothing to do with the case of the real line, lol. It has to do with the fact that the rationals are dense in the reals.