Cover an open set of $\mathbb{R}^n.$

Question

I have difficulty understanding the following result.

Theorem. Let $A\subseteq\mathbb{R}^n$ an open subset. Then for each $\delta>0$ the set $A$ is countable union of close cube having disjointed interiors and diagonal smaller than $\delta.$

Proof. We consider the family of close cube $$C_k:=\bigg[0,\frac{1}{2^k}\bigg]^n$$ and the family of translates $$K_k:=\bigg\{C_k+\frac{q}{2^k}\bigg\}\quad k\in\mathbb{N},$$ where $q=(q_1,\dots, q_n)$, $q_k\in\mathbb{Z}$ for all $k=1,\dots, n.$

First Problem. I can't visualize these translated sets, could you give me a concrete example?

We note that if $k\le l$, $C\in K_k$ and $\hat{C}\in K_l$, then either $\hat{C}\in C$ or $\text{Int}(C)\cap\text{Int}(\hat{C})=\emptyset.$

Second Problem. Why is the above statement true?

Let $\delta >0$; let $k_0$ the smallest integer such that $\frac{\sqrt{n}}{2^k}<\delta.$ We define $$\mathcal{C}_{k_0}:=\{C\in K_{k_0}\;|\; C\subseteq A\};$$

for induction we define for all $k\ge k_0$ $$ P_k:=\bigcup_{l=k_0}^k\bigg[\bigcup_{C\in\mathcal{C}_l} C\bigg];$$ $$H_k:= A\setminus\text{Int}(P_k),$$ $$\mathcal{C}_{k+1}:=\{C\in K_{k+1}\;|\; C\subseteq H_k\}.$$

Third problem How do I prove at this point that: $$A=\bigcup_{k=k_0}^\infty P_k$$

Could someone explain the details of the proof? Thank you in advance.

Answer 1

First problem (visualization): Probably the easiest is for $n=2, k=0$, that means tesselating the Euclidean plane with unit squares, like a chessboard, infinitely extended:

Somehwere is the origin point $(0,0)$ and $C_0$ with vertices $(0,0), (1,0), (1,1), (0,1)$, shown in black. All the other squares in $K_0$ are of the form $(q_1,q_2),(q_1+1,q_2),(q_1+1,q_2+1),(q_1,q_2+1), q_1,q_2 \in \mathbb Z.$

If $k$ becomes greater, nothing changes in the form, just the scale, as the squares get smaller and smaller. For $k=1$, $C_1$ is only a fourth of $C_0$, it has vertices $(0,0), (\frac12,0), (\frac12,\frac12)$ and $(0,\frac12)$. But since now the translate values are also allowed to talk half integer values, the remaining parts of $C_0$ are also covered by some translates in $K_1$.

Now if $n=3$, those squares becomes cubes, and it should be easy to understand how they fill the 3d space. For higher $n$, the $C_k$ are hypercubes, but that I can't visualize.

Second problem (inclusion or no common interior): With the above description, think about what happens when you move from $k$ to $k+1$ for some $n$. The hypercube $ C_k$ is divided into $2^n$ hypercubes of half it's side length. The same goes for all the translates in $K_k$. An element $\hat{C} \in K_{k+1}$ is contained in the element $C\in K_k$ that it was divided from. You could say $C$ is the parent of $\hat{C}$, and $\hat{C}$ the child of $C$.

So in this case a parent contains (in the subset sense) all of it's children and grandchildren, grandgrandchildren, etc.

Note also that different elements of the same 'generation' (in the same $K_k$) have no common interior!

So to get at the statement from the proof:

$$C \in K_k, \hat{C} \in K_l, k \le l$$

Go up from $\hat{C}$ to its parent, grandparent etc. until it reaches it's ancestor in the generation $k$. If it is $C$, we know from above that $\hat{C} \subseteq C$. If it is not $C$, but $C'$, then $\hat{C} \subseteq C'$, but since $C$ and $C'$ have no common interior, so have $C$ and $\hat{C}$.

Third problem (A is the union):

First note that each $C_k$ consists of cubes that are subsets of $A$. It's true for the first $C_{k_0}$ (see the definition) and also for each new generation, as $H_k \subseteq A$ (see the defintion). Each $P_k$ is the union of elements in some $C_l$, so we get

$$P_k \subseteq A$$

and thus

$$\bigcup_{k=k_0}^\infty P_k \subseteq A.$$

Now for the other direction, we need the condition that $A$ is open for the first time. Let $p\in A$ be any point from $A$. We will show $p \in P_k$ for some $k$. Because of the openness of $A$, there is a small ball of radius $\epsilon$ around $p$ that completely lies in $A$.

Now let's set $k_p$ equal the some value that fullfilles $\frac{\sqrt{n}}{2^{k_p}} < \epsilon$ and $k_p > k_0$ (so we can consider $P_{k_p}$ a.s.o. as defined in the problem statement). As we've seen in the consideration for the first problem, each $K_k$ covers the whole $R^n$. So there is a $C \in K_{k_p}$ with $p \in C$. That $C$ is a hypercube of sidelength $\frac1{2^{k_p}}$, so it's main diagonal has length

$$\sqrt{n}\frac1{2^{k_p}} = \frac{\sqrt{n}}{2^{k_p}} < \epsilon.$$

Now the main diagonal of such a hypercube is the largest distance between any 2 points of the hypercube. That means that the distance from $p$ to any other point in $C$ is at most $\frac{\sqrt{n}}{2^{k_p}}$ and thus less than $\epsilon$. By the construction of $\epsilon$, that means each point of $C$ is in $A$!.

That means $C \subseteq A$. From the second problem we know that either $C \subseteq P_{k_p-1}$ or $C$ and $P_{k_p-1}$ have no common interior.

(This is a bit handwaving, as $P_{k_p-1}$ may contain points that were not interior to any hypercubes it is made of. But those points form $n-1$ dimensional manifolds, so the $n$-dimensional cube $C$ cannot 'hide' in them without also containing interior points of any building block of $P_{k_p-1}$.)

In the first case we are done, we have found $p \in C \subseteq P_{k_p-1}$. In the second case, we have $C \cap \text{Int}(P_{k_p-1}) = \emptyset$. That means $C \subseteq H_{k_p-1}$ and hence $C \in \mathcal{C}_{k_p}$ and thus $p \in C \subset P_{k_p}$ by the definition.

So in all cases we found that the arbitrary point $p \in A$ is in some $P_{k_p}$, so we finally get

$$\bigcup_{k=k_0}^\infty P_k = A.$$