This is a [cross-post](https://math.stackexchange.com/questions/4131916/can-we-almost-cover-any-shape-in-the-plane-by-disjoint-disks-of-prescribed-radii).


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](https://en.wikipedia.org/wiki/Apollonian_gasket).



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


