$B(0,1)$ can not be expressed as countable collection of disjoint open cell

Question

how to prove $ b(0,1)$={$x\in \mathbb R^n:|x|<1$} can not be expressed as countable collection of disjoint open cells.where open cells are open rectangles.As the rectangles are open and disjoint therefore whenever I am trying to express the $b(0,1)$ as the union of open and disjoint rectangles,the boundary of the rectangles that intersect with $b(0,1)$ can not be covered with other open rectangles,as the collection of open rectangles are disjoint, that is why it seems to me that$b(0,1)$ can not be covered by any collection of open and disjoint cells(rectangles), but how can I prove it formally?

Answer 1

@Paul Sinclair, sometimes while solving a mathematical problem one can think a rough solution by intuition but can't able to write it formally, that's why I was unable to write anything about what I have tried..my intention was not to get my home work done by others.Now after thinking I have written roughly what I have thought please have a look..

If possible let $b(0,1)$ can be expressed as union of a countable collection of disjoint cells (open rectangles) ${R_i:i\in \mathbb{N}}$ Hence, $b(0,1)=\cup_{i\in\mathbb{N}}R_i$ Hence each $R_i$ is a proper subset of $b(0,1)$.Now if a rectangle contained in a circle then the boundary of the rectangle must contained in the circle. clearly boundary of a each $R_i$ is not contained in $R_i$,as $R_i$ is open. Hence boundary of$R_i$ $ \subset R_j(where j\neq i)$ let $x\in$ boundary of $R_i$,then $R_j$ is a neighborhood of x and hence must intersect with $R_i$, a contradiction as $R_i$'s are disjoint.this completes the proof.