# Disjoint multirectangles in an open set

It is well known that any open set in $$\mathbb R$$ can be written as the union of disjoint open intervals, why in $$\mathbb R^n$$ we cannot write any open set as union of disjoint balls. So I'm asking if one of these weaker propositions holds  1) Every open set in $$\mathbb R^n$$ can be written as disjoint union of multi-rectangles  2) for every $$A\subset \mathbb R^n$$ open, $$\exists C_n$$, sequence of disjoint open hypercubes in $$\mathbb R^n$$ such that $$\cup_n C_n\subseteq A,\ \ A\subseteq \cup_n \bar C_n$$

• (1) is impossible for the same reason that a disjoint union of open balls is not always possible: if the open set is connected and not a multi-rectangle, then by definition it is not a disjoint union of open sets. Dec 6, 2019 at 11:19

Hope this isn't too late to be of help!

1) As noted in the comments by @almagest, it is impossible to create a connected set by taking the disjoint union of any two open sets. Similarly if you require the multi-rectangles to be closed. If you do not require the multi-rectangles to be open or closed, but optionally containing points on their boundaries, then the problem reduces to part 2, since given a set of open multirectangles satisfying those conditions, we can assign each point in A to either the interior of a multi-rectangle or a point on the boundary of one of the multi-rectangles, and include or exclude that point accordingly.

2) I think that the answer is yes, although apologies in advance for the tedious and perhaps overly complicated proof.

From the definition of the product topology, and the fact the $$\mathbb{R}^n$$ is second countable, we have that any arbitrary open set $$U \subseteq \mathbb{R}^n$$ is the countable (not disjoint!) union $$U = \cup_m B_m$$ of basis sets, where the basis sets are open multi-rectangles.

We can then note that

$$\cup_m (B_m \setminus \cup_{k \lt m} \bar{B}_k) \subseteq U \subseteq \cup_m (\bar{B}_m \setminus \cup_{k \lt m} \bar{B}_k)$$

and note that $$\cup_m (B_m \setminus \cup_{k \lt m} \bar{B}_k)$$ is disjoint by construction.

We now claim that for each $$m$$ there is a finite set of of disjoint open multirectangles $$C_i$$ such that

$$\cup_i C_i \subseteq B_m \setminus \cup_{k \lt m} \bar{B}_k$$

$$\bar{B}_m \setminus \cup_{k \lt m} \bar{B}_k \subseteq \cup_i \bar{C}_i$$

Intuitively, this corresponds to the observation that a open (multi-)rectangle with closed rectangular holes cut out can be decomposed into disjoint open rectangles and parts of their boundaries. I collected this as the lemma at the end, since the proof isn't very enlightening.

Once we have this, than we can collect together all such open multi-rectangles $$C_i$$ for each $$m$$, noting that they are pairwise disjoint for distinct m, since they are contained in the sets $$B \setminus \cup_k \bar{B}_k$$, disjoint for distinct $$m$$.

Lemma: Given an open multirectangle $$B$$ and a finite set of open multirectangles $$B_k, k \in S$$ in $$\mathbb{R}^n$$, there exists a countable set of disjoint open multirectangles $$C_i$$ such that

$$\cup_i C_i \subseteq (B \setminus \cup_k \bar{B}_k) \subseteq (\bar{B} \setminus \cup_k \bar{B}_k)\subseteq \cup_i \bar{C_i}$$.

proof:

By induction on the dimension $$n$$.

Base step For $$n=1$$, this reduces to the observation that the set difference of an open interval by a finite set of closed intervals is the finite union of open intervals.

Induction step:

Write $$B = I \times C$$ and $$B_k = I_k \times C_k$$, where $$I,I_k$$ are open intervals and $$C,C_k$$ are open multi-rectangles in $$\mathbb{R}^{n-1}$$. Then we can pick intervals $$J_i$$ so that $$\cup_i J_i \subseteq I \subseteq \cup_i \bar{J}_i$$ and each $$J_i$$ is either contained or disjoint from each of the $$I_k$$, and write:

$$\cup_i [J_i \times (C \setminus \cap_{k:I_k \cap J_i \neq 0} \bar{C}_k)] \subseteq (B \setminus \cup_k \bar{B}_k) \subseteq (\bar{B} \setminus \cup_k \bar{B}_k) \subseteq \cup_i [\bar{J}_i \times (\bar{C} \setminus \cap_{k:I_k \cap J_i \neq 0} \bar{C}_k)]$$.

Then we can apply the inductive hypothesis to each $$C \setminus \cap_{k:I_k \cap J_i \neq 0} \bar{C}_k$$ to obtain some union $$\cup_\ell V_{i\ell}$$ of disjoint open multirectangles and obtain

$$\cup_{i\ell} J_i \times V_{i\ell} \subseteq (B \setminus \cup_k \bar{B}_k) \subseteq (\bar{B} \setminus \cup_k \bar{B}_k) \subseteq \cup_{i\ell} \bar{J}_i \times \bar{V_{i\ell}} = \cup_{i\ell} \overline{J_i \times V_{i\ell}}$$.

The multirectangles $$J_i \times V_{i\ell}$$ are disjoint, since the $$J_i$$ are disjoint for each $$i$$ and the $$V_{i\ell}$$ are disjoint for each $$\ell$$.

• thanks a lot for your work, I started reading and it seems right Dec 20, 2019 at 15:52