Can we almost cover any shape in the plane by disjoint/tangent disks of prescribed radii?

Question

This is a cross-post.

Let $(a_n)_{n \in \mathbb{Z}}$ be some given, strictly increasing sequence of positive numbers, such that $\lim_{n \to -\infty} a_n=0,\lim_{n \to +\infty} a_n=+\infty$.

Let $\Omega \subseteq \mathbb{R}^2$ be a bounded, connected, open subset, with Lipschitz boundary.

Let $\epsilon >0$. Does there exist a countable collection of closed disks $B(x_k,r_k)$ with the following properties:

1. For each $k$, $B(x_k,r_k) \subseteq \Omega$ and $r_k \in \{a_n\}$.
2. For every two distinct disks, either their intersection is empty, or they are tangent to each other. (The interiors of the disks do not intersect in any case).
3. $\Omega \setminus\cup_{k} B(x_k,r_k)$ has Lebesgue measure less then $\epsilon$.

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Here $B(x_k,r_k)$ denotes the closed Euclidean disk with radius $r_k$, centered around $x_k$.

It's possible that we only need the assumption $\lim_{n \to -\infty} a_n=0$.

I am fine with assuming higher regularity of the boundary $\partial \Omega$, say $C^1$.

Comment:

I am not sure if this cane be done for any such sequence $a_n$. One could suppose e.g. that perhaps certain bounds on the ratios between the different $a_n$ are needed.

Perhaps we could start with an easier question: If there was no restriction on the radii at all, could such a disjoint\tangent cover be constructed?

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This question resembles in a sense Apollonian gaskets.

Answer 1

The answer is yes, in an arbitrary dimension $d$. Here is a short proof.

Suppose, contrary to the above claim, that for some Jordan measurable $\Omega$ the infimum of what is left is equal to $m > 0$: $$ m = \inf \biggl\{\biggl|\Omega \setminus \bigcup_{k = 1}^K B(x_k, r_k)\biggr| : x_k, r_k \text{ as in the question}\biggr\} > 0 .$$ Note that we only consider finite collections of balls.

Fix a small $\epsilon > 0$ and choose a collection of balls $B(x_k, r_k)$ such that the measure of $\Omega' := \Omega \setminus \bigcup_{k = 1}^K B(x_k, r_k)$ is at most $m + \epsilon$. Since $\Omega'$ is Jordan-measurable, we can find a small $\delta = 2 a_n > 0$ such that if $Q_j$ is the collection of all elementary cubes of a $\delta$-lattice that are contained in $\Omega'$, then the measure of $\Omega' \setminus \bigcup_{j = 1}^J Q_j$ is less than $\epsilon$. Every cube $Q_j$ contains a ball $B(y_j, a_n)$, and $|B(y_j, a_n)| = C_d |Q_j|$ for some constant $C_d > 0$. It follows that $$ \begin{aligned}\biggl|\Omega \setminus \biggl(\bigcup_{k = 1}^K B(x_k, r_k) \cup \bigcup_{j = 1}^J B(y_j, a_n)\biggr)\biggr| & = \biggl|\Omega' \setminus \bigcup_{j = 1}^J B(y_j, a_n)\biggr| \\ & = |\Omega'| - \sum_{j = 1}^J |B(y_j, a_n)| \\ & = |\Omega'| - C_d |Q_j| \\ & = (1 - C_d) |\Omega'| + C_d \biggl|\Omega' \setminus \bigcup_{j = 1}^J Q_j\biggr| \\ & \leqslant (1 - C_d) (m + \epsilon) + C_d \epsilon \\ & = (1 - C_d) m + \epsilon . \end{aligned} $$ If $\epsilon$ is chosen to be less than $C_d m$, we arrive at a contradiction.

Answer 2

There is a stronger result. Suppose that $(b_n)$ is a sequence with the properties $b_n>0,\; b_n\to 0$ and $\sum b_n^2=\infty$. Then for any region $D$, and for every $\epsilon>0$ there are disjoint disks $B(z_n,r_n)\subset D$, such that the area of their union is at least area of $D$ minus $\epsilon$, $(r_n)$ is a subsequence of $(a_n)$ (that is each $r_n$ occurs among $a_n$ only once).

To obtain your result, first remove all terms with positive index from your sequence $(a_n)$, and second, if $\sum a_{-n}^2=\infty$, just take $b_n=a_{-n}$, and if this sum is finite, repeat each $a_n$ several times in the sequence $(b_n)$ to achieve the condition $\sum b_n^2=\infty$.

By the way, a similar result holds in any dimension $d$, if we replace the condition $\sum b_n^2=\infty$ by $\sum b_n^d=\infty$.

(This problem was offered on a student olimpiad in Kharkiv some time in the early 1970s).

Answer 3

Let $\Omega \subset \mathbb R^d$ be a given open set of finite Lebesgue measure. Here is an argument that shows that we can in fact cover $\Omega$ up to a set of Lebesgue measure zero by pairwise disjoint closed balls contained in $\Omega$, with radii forming a subsequence of a given decreasing sequence $r_n$ such that $\lim r_n = 0$ and $\sum r_n^d = \infty$ (as in Alexandre Eremenko's answer).

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Step 1. We use the greedy strategy: in step $n$, we choose any ball $B_n$ with radius $r_n$ contained in $$\Omega_n' := \Omega \setminus (B_1 \cup \ldots \cup B_{n-1})$$ if such a ball exists, and we define $B_n = \varnothing$ otherwise. Our goal is to prove that $$\Omega' := \Omega \setminus (B_1 \cup B_2 \cup \ldots)$$ has zero Lebesgue measure.

Step 2. We do this by showing the following claim: for every $N > 0$, $\Omega$ is contained in the union of slightly larger balls: $$ \Omega \subseteq (B_1 \cup B_2 \cup \ldots \cup B_{N - 1}) \cup (2 B_N \cup 2 B_{N+1} \cup \ldots) $$ (where, of course, the balls $B_n$ and $2 B_n$ have the same centre, but the radius of $2 B_n$ is equal to $2 r_n$). Once this claim is proved, we conclude the argument as follows. We have $$ |\Omega'| \leqslant \biggl| \bigcup_{n = N}^\infty ((2 B_n) \setminus B_n) \biggr| \leqslant (2^d - 1) \sum_{n = N}^\infty |B_n| . $$ Since $\sum_{n = 1}^\infty |B_n| \leqslant |\Omega| < \infty$, the right-hand side goes to zero as $N \to \infty$, and, consequently, $|\Omega'| = 0$. Therefore, it remains to prove the claim.

Step 3. Suppose, contrary to our claim, that for some $N$ there is a point $$x \in \Omega \setminus \bigl((B_1 \cup \ldots \cup B_{N-1}) \cup (2B_N \cup 2B_{N+1} \cup \ldots)\bigr).$$ Recall that $\lim r_n = 0$, so that if $n$ is large enough, say $n \geqslant n_0$, the ball $B(x, r_n)$ is contained in $\Omega_N'$.

Suppose that $n \geqslant n_0$ and $B_n = \varnothing$. This means that $B(x, r_n)$ is not contained in $\Omega_n'$. However, $B(x, r_n)$ is a subset of $\Omega_N'$, and so $B(x, r_n)$ necessarily intersects some ball $B_k$ with $N \leqslant k < n$. Since $r_k \geqslant r_n$, this implies that $x \in 2 B_k$, contrary to the definition of $x$. It follows that $B_n \ne \varnothing$ for $n \geqslant n_0$.

But this leads to a contradiction: on one hand, $\sum_{n = n_0}^\infty |B_n| \leqslant |\Omega| < \infty$, and on the other one, $\sum_{n = n_0}^\infty |B_n| = C_d \sum_{n = n_0}^\infty r_n^d = \infty$. This completes the proof of our claim.

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The above argument is, of course, a variant of the proof of Vitali's covering theorem.

Answer 4

Here is a proof that we can pick the balls $B(x_k,r_k)$ in such a way that $\Omega \setminus\cup_{k} B(x_k,r_k)$ has measure 0.

It is enough to prove that given an open $\Omega$, we can pick finitely many $B(x_k,r_k)$ such that $|\Omega \setminus\cup_{k} B(x_k,r_k)|<\frac{1}{2}|\Omega|$. Indeed, then we can call $\Omega_1=\Omega \setminus\cup_{k} \overline{B(x_k,r_k)}$, then repeat the process with $\Omega_1$ to obtain some $\Omega_2$ with $|\Omega_2|<\frac{1}{2}|\Omega_1|$, and after repeating the process infinitely many times we are left with a set of measure 0.

To do that, given $\Omega$ we call $\Omega_\varepsilon=\{x\in\Omega;d(x,\mathbb{R}^2\setminus\Omega)>\varepsilon\}$. Pick $\varepsilon>0$ such that $|\Omega_\varepsilon|>0.9|\Omega|$ and consider an hexagonal tesellation of the plane with hexagons of apothem some $a_k<\frac{\varepsilon}{3}$ from your sequence. Now take the union of all the inscribed circles in hexagons which intersect $\Omega_\varepsilon$. These circles cover more than $0.6$ times the area of the hexagons touching $\Omega_\varepsilon$, so they cover more than half of the area of $\Omega$, as we wanted. The same argument can be used for higher dimensions.