Disjoint multirectangles in an open set

Question

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$$

Answer 1

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 $\ell$.