Theorems Bescovitch’s covering theorem Attributions

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Proof

Idea of the proof: We begin by considering the bounded case. We inductively select a maximal disjoint subcollection of balls in $\mathcal{F}$ whose radii are not too small compared to the supremum of available radii. Using geometric estimates, we prove that these balls cover $A$ and that only a finite number (bounded in terms of dimension $n$) can overlap any fixed ball. We then extend the result to unbounded sets by partitioning $A$ into annuli and applying the bounded case on each piece.

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**Step 1 (Construction of disjoint ball sequence)** Let $D := \sup\{\operatorname{diam} B \mid B \in \mathcal{F}\} < \infty$, since $A$ is bounded. Choose $B_1 = B(a_1, \tau_1) \in \mathcal{F}$ with $\tau_1 \geq \frac{3}{4} \frac{D}{2}$. Inductively, define $A_j := A \setminus \bigcup_{i=1}^{j-1} B_i$. If $A_j = \emptyset$, stop and set $J := j - 1$. Otherwise, select $B_j = B(a_j, r_j) \in \mathcal{F}$ with $a_j \in A_j$ and \begin{align*} r_j \geq \frac{3}{4} \sup\{r \mid B(a, r) \in \mathcal{F}, a \in A_j\}. \end{align*} If $A_j \neq \emptyset$ for all $j$, set $J := \infty$.

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**Step 2 (Estimate of radius decay)** [claim]If $j > i$, then $r_j \leq \frac{4}{3} r_i$.[/claim]

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[proof] Since $a_j \in A_i$, it follows that \begin{align*} r_i \geq \frac{3}{4} \sup\{r \mid B(a, r) \in \mathcal{F}, a \in A_i\} \geq \frac{3}{4} r_j, \end{align*} which implies $r_j \leq \frac{4}{3} r_i$. [/proof]

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**Step 3 (Disjointness of shrunk balls)** [claim]The balls $\{B(a_j, \frac{r_j}{3})\}_{j=1}^{J}$ are disjoint.[/claim]

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[proof] Assume $j > i$. Then $a_j \notin B_i$, hence $|a_j - a_i| > r_i$. Note: \begin{align*} r_i = \frac{r_i}{3} + \frac{2 r_i}{3} \geq \frac{r_i}{3} + \frac{2}{3} \cdot \frac{3}{4} r_j > \frac{r_i}{3} + \frac{r_j}{3}. \end{align*} Thus $B(a_j, \frac{r_j}{3}) \cap B(a_i, \frac{r_i}{3}) = \emptyset$. [/proof]

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**Step 4 (Vanishing radii if infinite)** [claim]If $J = \infty$, then $\lim_{j \to \infty} r_j = 0$.[/claim]

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[proof] By Claim 2, the balls $\{B(a_j, \frac{r_j}{3})\}_j$ are disjoint. Since $A$ is bounded and $a_j \in A$, the radii $r_j$ must converge to $0$; otherwise, the total measure of disjoint balls would be infinite in a bounded set. [/proof]

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**Step 5 (Covering property)** [claim]$A \subseteq \bigcup_{j=1}^{J} B_j$.[/claim]

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[proof] If $J < \infty$, then $A = \bigcup_{j=1}^{J} A_j \subseteq \bigcup_{j=1}^{J} B_j$.

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Suppose $J = \infty$ and let $a \in A$. Then there exists $r > 0$ such that $B(a, r) \in \mathcal{F}$. By Claim 4, $\lim r_j = 0$, so for some $j$, $r_j < \frac{3}{4} r$. But then $a \notin \bigcup_{i=1}^{j-1} B_i$, and $a_j$ is chosen from this remainder. The inequality $r_j \geq \frac{3}{4} \sup\{r \mid B(a, r) \in \mathcal{F}, a \in A_j\} \geq \frac{3}{4} r$ leads to a contradiction. [/proof]

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**Step 6 (Cardinality of intersecting balls)** Fix $k > 1$ and let $I := \{j \mid 1 \leq j < k, B_j \cap B_k \neq \emptyset\}$. Define $K := \{j \in I \mid r_j \leq 3 r_k\}$.

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[claim]$\operatorname{Card}(K) \leq 20^n$.[/claim]

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[proof] For $j \in K$, choose $x \in B(a_j, \frac{r_j}{3})$. Then \begin{align*} |x - a_k| \leq |x - a_j| + |a_j - a_k| \leq \frac{r_j}{3} + r_j + r_k = \frac{4 r_j}{3} + r_k \leq 4 r_k + r_k = 5 r_k. \end{align*} Hence $B(a_j, \frac{r_j}{3}) \subseteq B(a_k, 5 r_k)$.

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By Claim 2, these balls are disjoint. Denote by $\alpha(n)$ the volume of the unit ball in $\mathbb{R}^n$, then: \begin{align*} \alpha(n) 5^n r_k^n \geq \sum_{j \in K} \alpha(n) \left(\frac{r_j}{3}\right)^n \geq \sum_{j \in K} \alpha(n) \left(\frac{r_k}{4}\right)^n = \operatorname{Card}(K) \cdot \alpha(n) \cdot \left(\frac{r_k}{4}\right)^n, \end{align*} so \begin{align*} 5^n \geq \operatorname{Card}(K) \cdot \left(\frac{1}{4}\right)^n \Rightarrow \operatorname{Card}(K) \leq 20^n. \end{align*} [/proof]

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**Step 7 (Estimate for $I \setminus K$)** Let $i, j \in I \setminus K$, $i \neq j$. Assume $a_k = 0$ for simplicity. Then \begin{align*} 3 r_k < r_i < |a_i| \leq r_i + r_k, \quad 3 r_k < r_j < |a_j| \leq r_j + r_k. \end{align*} Assume $|a_i| \leq |a_j|$.

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[claim]If $\cos \theta > \frac{5}{6}$, then $a_i \in B_j$.[/claim]

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[proof] If $|a_i - a_j| \geq |a_j|$, then by the law of cosines: \begin{align*} \cos \theta \leq \frac{|a_i|^2}{2 |a_i| |a_j|} = \frac{|a_i|}{2 |a_j|} \leq \frac{1}{2} < \frac{5}{6}. \end{align*} Otherwise, if $a_i \notin B_j$, then $r_j < |a_i - a_j|$, and similarly we get \begin{align*} \cos \theta \leq \frac{1}{2} + \frac{r_k}{r_i} \leq \frac{5}{6}. \end{align*} [/proof]

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[claim]If $a_i \in B_j$, then \begin{align*} 0 \leq |a_i - a_j| + |a_i| - |a_j| \leq |a_j| \epsilon(\theta), \quad \epsilon(\theta) := \frac{8}{3}(1 - \cos \theta). \end{align*} [/claim]

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[proof] Using triangle and cosine estimates: \begin{align*} |a_i - a_j| + |a_i| - |a_j| \leq \frac{2 |a_i| (1 - \cos \theta)}{|a_i - a_j|} \leq \epsilon(\theta) |a_j|. \end{align*} [/proof]

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[claim]If $a_i \in B_j$, then $\cos \theta \leq \frac{61}{64}$.[/claim]

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[proof] From the previous bounds and assumptions on $r_i$, $r_j$, and $r_k$, we derive: \begin{align*} \frac{1}{8} |a_j| \leq |a_i - a_j| + |a_i| - |a_j| \leq |a_j| \epsilon(\theta), \end{align*} which gives $\epsilon(\theta) \geq \frac{1}{8}$, so $\cos \theta \leq 1 - \frac{3}{8} = \frac{61}{64}$. [/proof]

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[claim]For all $i \neq j \in I \setminus K$, the angle $\theta$ between $a_i - a_k$ and $a_j - a_k$ satisfies $\theta \geq \theta_0 := \arccos \frac{61}{64} > 0$.[/claim]

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[proof] Follows from Claims 6a–6c. [/proof]

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[claim]There exists $L_n$ such that $\operatorname{Card}(I \setminus K) \leq L_n$.[/claim]

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[proof] Fix $r_0 > 0$ such that no two vectors with angle $\geq \theta_0$ lie in a ball of radius $r_0$ on $\partial B(1)$. Then $\partial B_k$ can be covered with $L_n$ such balls of radius $r_0 r_k$, and thus $\operatorname{Card}(I \setminus K) \leq L_n$. [/proof]

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**Step 8 (Partition into disjoint subfamilies)** Set $M_n := 20^n + L_n + 1$.

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Define $\sigma : \mathbb{N} \to \{1, \dotsc, M_n\}$: - For $1 \leq i \leq M_n$, set $\sigma(i) = i$. - For $k \geq M_n$, define $\sigma(k+1)$ inductively: since fewer than $M_n$ balls intersect $B_{k+1}$, assign it to an unused color $l$ among $\{1, \dotsc, M_n\}$.

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Let $\mathcal{G}_j := \{B_i \mid \sigma(i) = j\}$ for $1 \leq j \leq M_n$.

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Then each $\mathcal{G}_j$ is disjoint, and $A \subseteq \bigcup_{j=1}^{M_n} \bigcup_{B \in \mathcal{G}_j} B$.

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**Step 9 (Extension to unbounded $A$)** Let $A_l := A \cap \{x \mid 3 D (l - 1) \leq |x| < 3 D l\}$ and $\mathcal{F}_l := \{B(a, r) \in \mathcal{F} \mid a \in A_l\}$.

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Apply Step 8 to each $A_l$, obtaining disjoint subfamilies $\mathcal{G}_1^l, \dotsc, \mathcal{G}_{M_n}^l$. Then define: \begin{align*} \mathcal{G}_j := \bigcup_{l=1}^{\infty} \mathcal{G}_j^{2l-1}, \quad \mathcal{G}_{j + M_n} := \bigcup_{l=1}^{\infty} \mathcal{G}_j^{2l} \quad \text{for } 1 \leq j \leq M_n. \end{align*}

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Then $A \subseteq \bigcup_{j=1}^{2 M_n} \bigcup_{B \in \mathcal{G}_j} B$. Set $N_n := 2 M_n$.

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