# Writing an open set as a union of boxes, showing that the open unit ball can't be written as a disjoint union of sets of the form $(a,b)\times(c,d)$

I have the following exercise that I wish to solve:

Let $U\subseteq\mathbb{R}^{n}$ be an open set, $T\subseteq\mathbb{R}^{n}$ will be called a box if its of the form $[a_{1},b_{1}]\times\ldots[a_{n},b_{n}]$ where $a_{i},b_{i}\in\mathbb{R}$.

1. Show that there exist a sequence $\{T_{k}\}_{k=1}^{\infty}$if boxes with disjoint interior (they intersect, at most, at their sides) s.t $U=\cup_{k=1}^{\infty}T_{k}$

2. Explain why $U$ can not be a partition of a finite number of boxes

3. Show that the open unit ball in $\mathbb{R}^{2}$ can not be written as a disjoint countable union if sets of the form $(a_{1},b_{1})\times(a_{2},b_{2})$.

My work:

For the first part of the question I tried to partition $\mathbb{R}^{n}$ into boxes of sizes $r\times r$ and take those boxes that are contained in $U$. I hoped that for a small enough $r$ this set will be $U$, since $U$ is open for any $u\in U$ there is some $r'$ s.t $B(u,r')\subseteq U$ so I wanted $r<r'$. The problem with going this way is that there is a sequence of $u_{i}$ with corresponding $r'_{i}$ with $Inf_{i}r_{i}'=0$ and then Those won't be boxes anymore.

For $2$ I think that such a partitioning means $U$ is also closed hence each $[a_{i},b_{i}]=(-\infty,\infty)$ but its clear that you can't partition $(-\infty,\infty)$ in the described way.

I have no clue on how to start $3$.

I would appriciate any help on doing this exercise!

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1. There's no need for a single $r$ to produce all boxes. You can use boxes of different sizes. For example $1$, $1/2$, $1/4$, $\dotsc$ (refining each box into smaller ones). Then add boxes when they're small enough to fit into $U$.
3. Show that if $(a_1, b_1) \times (a_2, b_2)$ is contained in the unit ball then no point on the boundary of that box can be covered anymore.
Thanks for the answer! I think I understand what you mean with $2,3$ ($2$ is what I also said, right ?) but I don't completely follow you on $1$: Given the original partition to size $r$ I take all those boxes that are not contained in $U$ and I make them smaller each step and take them to the union when I can ? It seems like a good idea but how do I prove that each point in $U$ is in one of the boxes ? – Belgi Dec 11 '12 at 19:33
The argument for (2) works precisely because $\mathbb{R}^{2}$ is connected, so closed sets are not open sets other than $\mathbb{R}^{2}$ itself and $\varnothing$. This is to prevent the OP and beginning topologists from thinking that nontrivial open sets can never be closed sets --- an assertion that is not true in some topological spaces. – Haskell Curry Dec 11 '12 at 19:33
A point in an open set in the product space $\mathbb{R}^n$ is by definition an element of some (open) box. Alternatively, what is the diameter of a box with sides $r$? – WimC Dec 11 '12 at 19:36