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

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?

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Do you ask for a disjoint union or any countable union of open rectangles? –  Thomas E. Jan 6 '13 at 8:39
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

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.
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
@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