Prove that the unit open ball in $\mathbb{R}^2$ cannot be expressed as a countable disjoint union of open rectangles. 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!
 A: 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.
A: 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
