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Question:

Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles, $G= \bigcup_{k=1}^\infty I_k$.

Attempt: My idea is to tile $\mathbb{R}^n$ with cubes of side 1. Then discard the cubes that do not intersect with $G$. Then repeat. However, I'm not sure if this is the right approach, since it may only approximate $G$. I also would be unsure of how to formalize such an approach. Alternative ideas would be appreciated.

Here's some additional context since there's quite a bit of discussion on the problem. The problem as originally written in Chapter 2 of Lebesgue Integration on Euclidean Space by Frank Jones reads: enter image description here

Of course, we need to know what a special rectangle is. This is defined as follows (essentially just a closed rectangle): enter image description here

mathjacks
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  • Must the rectangles not overlap? – Unit Dec 23 '14 at 19:22
  • Are rectangles $2$-dimensional or $n$-dimensional. – Nishant Dec 23 '14 at 19:26
  • A cartesian product of intervals is commonly called a rectangle regardless of the ambient dimension. – Umberto P. Dec 23 '14 at 19:27
  • For this questions, the rectangles must not overlap and they are $n$-dimensional. They could also be cubes. – mathjacks Dec 23 '14 at 19:31
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    I'm having trouble even covering the interval $(a,b)$ with non-overlapping rectangles. Any closed rectangle $[a_1,b_1]$ contained in $(a,b)$ leaves $(a,a_1)$ and $(b_1,b)$ uncovered; any two closed rectangles contained in these (one in one, the other in the other) gives us four still-uncovered intervals, and it just gets messier. If they are allowed to overlap on "small" sets, then it can be done: for $(a, b)$, pick two sequences $a_n\searrow a$ and $b_n\nearrow b$ with $a_1<b_1$. Then $(a,b)=\dotsb\cup [a_3,a_2]\cup [a_2,a_1]\cup [a_1,b_1]\cup [b_1,b_2]\cup [b_2,b_3]\cup\dotsb$. – Unit Dec 23 '14 at 19:32
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    @Unit: An open interval $(a,b)$ on the real line can not be expressed as a union of disjoint closed intervals. – bof Dec 23 '14 at 20:36
  • Are you sure you quoted that problem right? It's not "prove or disporove"? What book is that from? – bof Dec 23 '14 at 20:37
  • Maybe you meant almost disjoint, i.e. their intersections have measure zero. As bof said this is impossible for $(a,b)$. You may see this link for a proof http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/ – clark Dec 23 '14 at 21:39
  • @bof, I suppose the problem could be wrong, but I am quite sure that I am quoting the problem correctly. It is from the book Lebesgue Integration on Euclidean Space by Frank Jones. It is Chapter 2 Section A Problem 9. – mathjacks Dec 24 '14 at 01:32
  • he defines rectangles to be nonoverlapping if their $interiors$ are disjoint (see page 28) – Alain Aug 13 '17 at 03:43

3 Answers3

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The statement is false. A proof by Terence Tao is given in his blog post ‘Covering a non-closed interval by disjoint closed intervals’. To summarize:

Consider the one dimensional case. Suppose $\left\{I_k = \left[a_{k},b_{k}\right]\right\}_{k=1}^{\infty}$ is a collection of disjoint closed intervals such that $\mathbb{R} = \bigcup_{k=1}^{\infty}I_k$. Now consider the set

$$E = \bigcup_{k=1}^{\infty}\left\{a_k,b_k\right\}$$

of endpoints of the intervals. Clearly $E$ is countably infinite. Now $E = \mathbb{R}\setminus \bigcup_{k=1}^{\infty}(a_k,b_k)$, and is thus closed. We claim that $E$ is also perfect. That is $E$ is closed with no isolated points. Let $x \in E$. Then $E$ is either a right end point or a left end point but not both. If it is a right endpoint, then it's easy to see that $x$ is the limit of the left endpoints to the right of $x$. There is a theorem in analysis that says that every perfect subset of a complete metric space is uncountably infinite. Several proofs of this theorem are given here: Proof that a perfect set is uncountable

The Amplitwist
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Albert Steppi
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    Why does the density of ${x_n}$ mean that all points in $G$ must have been covered? Aren't the points in $G$ uncountable? Maybe I have some trouble with the terminology "countably dense" which I have not encountered before. – mathjacks Dec 23 '14 at 20:22
  • You're right, there's a gap in my proof. We might not have covered every point. I'm working on patching it now. – Albert Steppi Dec 23 '14 at 20:25
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Your idea seems to work, maybe with slight changes.

You should discard cubes that are not contained in $G$, rather than do not intersect $G$.

Why does it work? Call the cubes $(R_i)_{i=1}^\infty$. Because of the above line, no doubt that $$\bigcup_iR_i\subset G.$$On the other hand, since $G$ is open, every $x\in G$ is contained in a ball which is contained in $G$. As the cubes get small enough, one of them will contain $x$ and be contained in the ball.

Amitai Yuval
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1

Here's a somewhat different approach:

For each $x \in G$, there exists $r > 0$ such that $B(x,r) \subset G$. For each $x$, construct a closed rectangle with rational vertices containing $x$ and contained in $B(x,r)$. Since there are only a countable number of such rectangles and the union of all the rectangles equals $G$, we can write $G = \cup_{k=1}^\infty I_k$ where each $I_k$ is a rectangle with rational vertices.

Now define $P_1 = I_1$ and $P_k = I_k \setminus \text{int}(\cup_{j=1}^{k-1} I_j)$ for $k = 2, 3, \cdots$. Then the $P_k$ are non-overlapping and each $P_k$ can be written as the finite union of non-overlapping special rectangles, so the result follows.

David
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