Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that every open set in $\mathbb{R}$ is a disjoint union of at most countable segments.

But how do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles?

Moreover, is it true for $\mathbb{R}^k$ if open rectangle is replaced to open $k$-dimensional open rectangle?

share|improve this question
1  
Do you ask for a disjoint union or any countable union of open rectangles? –  Thomas E. Jan 6 '13 at 8:39
3  
All rectangles (or higher-dimensional analogues) with "all-rational" corners form a countable base for the topology of Euclidean spaces, so any open set can be written as a union of those. As the base is countable, so is the union. Disjoint unions cannot be achieved in general (due to connectedness arguments). –  Henno Brandsma Jan 6 '13 at 8:44
    
@Thomas: No, that is obviously false Henno: I didn't know that $\mathbb{Q}^k\cap V$ is dense in any open set $V$ in $\mathbb{R}^k$. Now it's done thank you –  Katlus Jan 6 '13 at 9:00
    
@Katlus. I think it is less obvious than the other property. Showing that rectangles form a basis for the topology of $\mathbb{R}^{n}$ is just a straight-forward use of triangle inequality. –  Thomas E. Jan 6 '13 at 15:25

1 Answer 1

up vote 6 down vote accepted

Let $E$ be open in $\mathbb R^k$. For each $x\in E$, there is an open ball centred at $x$ that is contained in $E$. Now one can construct an open $k$-cell containing $x$ and contained in this ball of the form $(a_1,b_1)\times ...\times (a_k,b_k)$ where $a_1,b_1,...,a_k,b_k$ are all rational, and then $E$ would be the union of these cells. This union is at most countable (after discarding identical cells) because $\{(a_1,b_1)\times ...\times (a_k,b_k):a_1,b_1,...,a_k,b_k\in\mathbb Q\}$ is countable.

share|improve this answer
    
Thank you. It is off the question (somewhat related to the question), but is it also true that every infinite closed set in $\mathbb{R}^k$ is separable? –  Katlus Jan 6 '13 at 9:04
    
How do you know the union is a countable union ? –  Katlus Jan 6 '13 at 10:28
    
@Katlus: in $\Bbb R^k$ every open cover contains a countable subcover since $\Bbb R^k$ is metric separable and hence Lindelof –  Ilya Jan 6 '13 at 11:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.