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Prove that the unit open ball in $\mathbb{R}^2$ cannot be expressed as a countable disjoint union of open rectangles. Open rectangles in $\mathbb{R}^2$ are subsets of the form $(a,b)\times(c,d)$.

Thanks a lot!

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Mitt, as Arthur said you should show us what you have done and we can help you –  Nameless Nov 18 '12 at 11:26
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Suppose that it is possible to cover unit ball the by a sequence of disjoint open sets $U_1,U_2,U_3,U_4,U_5...$. Let U be $U_1$ and V be the union of $U_2,U_3,U_4,U_5...$. It follows easily that U,V are disjoint open sets and the unit open ball is contained in the union of U,V. This contradicts the fact that the unit open ball in $R^2$ is connected

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This is a great approach since it shows the problem has nothing to do with hving rectangles (except for the need to have more than one open set.) In fact this approach works for any cover with more than one open set, by letting $U$ be one of them and $V$ the union of all the others. +1 –  coffeemath Nov 18 '12 at 11:56
Yes it works for any disjoint collection of open sets that form an open cover of the unit ball. –  Amr Nov 18 '12 at 11:59
@coffeemath I edited my answer according to your suggestion –  Amr Nov 18 '12 at 12:07
A solution much more elegant than I expected! Thanks again. –  mitt Nov 18 '12 at 12:16
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Any specific rectangle $R$ has the property that for any point $P$ interior to the rectangle there is $\epsilon>0$ so that the $\epsilon$ ball about $P$ is contained in the interior of $R$. Now in the proposed cover, consider any rectangle $R_1$. Then the only points on the boundary of $R_1$ which might be on the unit circle are its four corners; choose some point $P$ on one of the edges of $R_1$ other than at a corner of $R_1$.

Since the union of the open rectangles is to cover the ball, and $P_1$ is not in the open rectangle $R_1$, there must be some other rectangle $R_2$ for which $P$ is interior to $R_2$. But then there is $\epsilon >0$ so that $R_2$ contains every point within $\epsilon$ of $P$, which would cause some interior points near $P$ in the first rectangle $R_1$ to also be in the interior of rectangle $R_2$, forcing an overlap between the two rectangles.

Note that this approach shows that even an uncountable union cannot do the job.

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