Rectangle as a Union of Cubes 
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*Can any closed rectangle in R^n be written as a countable union of pairwise disjoint
closed cubes ? Justify .

*Can any closed rectangle in R^n be written as a finite union of pairwise disjoint 
closed cubes ? Justify .


I can prove that it can be covered by cubes , but having problem the the other part .
Thanks for any help .
 A: The answer is ‘no’.
Take a longest side of a non-cube closed rectangle in $ \mathbb{R}^{n} $. If the rectangle were the union of a countable collection of disjoint closed cubes, then this line segment, which is homeomorphic to $ [0,1] $, would be the union of a countable collection of disjoint closed intervals (which includes single points). This is not possible; Professor Terence Tao has a nice proof of this fact on his research blog (as he offers the option of leaving the proof as an exercise for the reader, you need to highlight the missing text in order to see it). :)
The idea is as follows. Suppose that $ [0,1] $ is the union of a countable collection of disjoint closed intervals. Choose $ a,b \in [0,1] $ such that $ a < b $ and $ a $ and $ b $ belong to different closed sub-intervals on $ [0,1] $. Then there is a point $ c $ between $ a $ and $ b $ that is contained in a third closed sub-interval of $ [0,1] $. Repeat the argument for the pairs $ (a,c) $ and $ (c,b) $ to obtain more closed sub-intervals. Continuing the process ad infinitum, the collection of such intervals forms something like the complement of the Cantor set. As the Cantor set is uncountable, we get uncountably many points left that are not yet covered. However, these points can only be covered by uncountably many closed sub-intervals, which contradicts the fact that we started off with only countably many such closed sub-intervals in the beginning.
