An exact cover of an open set with in $\mathbb{R}^d$ with cubes

Question

Consider an open set $\mathcal{U} \subseteq \mathbb{R}^d$ being completely covered by a countable collection of boxes defined using the $d$-dimensional unit cube $Q = [0,1]^d$ as follows: $$ Q_{n,k} = 2^{-n}Q + 2^{-n} k, \text{ for some } n \in \mathbb{N}, k \in \mathbb{Z}^d $$

Take the set, $I_0$ to be: $$ I_0 = \{k \in \mathbb{Z}^d: Q_{0, k} \subseteq \mathcal{U} \} $$

Then we define $I_1$ to be the set of all indices $k \in \mathbb{Z}^d$, such that $Q_{1, k} \subseteq \mathcal{U}$, but $Q_{1, k}$ are not contained in any of the boxes $Q_{0, k}$ and so on. In this fashion, we construct smaller and smaller boxes such that are all contained in $\mathcal{U}$. So we have the inclusion, $$ \bigcup_{n \geq 0} \bigcup_{k \in I_n} Q_{n, k} \subseteq \mathcal{U} $$ What I am stuck in is the reverse inclusion, that is each $x \in \mathcal{U}$ is inside at least one of the constructed $Q_{n,k}$, that is,

$$ \bigcup_{n \geq 0} \bigcup_{k \in I_n} Q_{n, k} = \mathcal{U} $$

is eluding me. I have tried to use boxes that are contained in some open ball surrounding the point $x$, but having difficulty in seeing why we are able to find such a box $Q_{n,k}$ with $k \in I_n$ for some $n$ satisfying $2^{-n} < \sqrt{\frac{2}{d}} \delta \leq \sqrt{2}\delta$, where $\delta$ is the radius of the open ball surrounding $x$.

Edit: This "decomposition" of $\mathcal{U}$ concerns a lemma in my text that says that any open set in $\mathbb{R}^d$ can be covered exactly by at most countable collection of non-overlapping closed boxes.

Answer 1

You practically got this.

For each $n$ and $k$ the diameter of $Q_{n,k}$ is equal to $2^{-n} \sqrt{d}$, which converges to $0$ as $n \to \infty$.

Now let $x \in \mathcal U$. All we have to do is to prove that there exists $n,k$ such that $x \in Q_{n,k} \subset \mathcal U$, for once that is proved then one simply chooses $n_1 \ge 0,k_1 \in \mathbb Z^d$ to satisfy two properties:

1. $x \in Q_{n_1,k_1} \subset \mathcal U$
2. $n_1 \ge 0$ is minimal with respect to the previous property.

It follows that $Q_{n_1,k_1}$ is in your collection, that is, $k _1\in I_{n_1}$.

To prove the existence of $n,k$, let $\delta > 0$ be any radius such that $B(x,\delta) \subset \mathcal U$. Since $2^{-n} \sqrt{d}$ converges to $0$, there exists $n$ such that $2^{-n} \sqrt{d} < \delta$. Now choose any $k$ such that $x \in Q_{n,k}$; this $k$ exists because for fixed $n$ the boxes $Q_{n,k}$ for $k \in \mathbb Z^d$ cover $\mathbb R^n$. It follows that $Q_{n,k} \subset B(x,\delta)$, because if $y \in Q_{n,k}$ then $d(x,y) \le 2^{-n} \sqrt{d} < \delta$.