[proofplan] We first prove the theorem when the set of centres $A$ is bounded. We greedily select balls from $F$ in order of decreasing radius (up to a factor $3/4$), producing a sequence $B_1, B_2, \dotsc$ whose shrunk balls $\overline{B}(a_j, r_j/3)$ are disjoint and whose union covers $A$. A geometric packing argument (volume comparison for comparable radii, angular separation for much larger radii) bounds the number of selected balls that can intersect any given $B_k$ by a dimensional constant $M_n$. A greedy colouring with $M_n$ colours then partitions the selected balls into $M_n$ pairwise disjoint subfamilies covering $A$. For unbounded $A$, we partition into annular shells of width $3D$ and apply the bounded case to odd- and even-indexed shells separately, doubling the number of subfamilies to $N_n = 2M_n$. [/proofplan] [step:Construct a greedy sequence of balls with nearly maximal radii] Let $D := \sup\{\operatorname{diam} B \mid B \in F\} < \infty$. Assume first that $A$ is bounded. Select $B_1 = \overline{B}(a_1, r_1) \in F$ with \begin{align*} r_1 \ge \tfrac{3}{4} \sup\{ r \mid \overline{B}(a, r) \in F \}. \end{align*} Inductively, define $A_j := A \setminus \bigcup_{i=1}^{j-1} B_i$. If $A_j = \varnothing$, stop and set $J := j - 1$. Otherwise, select $B_j = \overline{B}(a_j, r_j) \in F$ with $a_j \in A_j$ and \begin{align*} r_j \ge \tfrac{3}{4} \sup\{ r \mid \overline{B}(a, r) \in F,\; a \in A_j \}. \end{align*} If $A_j \neq \varnothing$ for all $j$, set $J := \infty$. [guided] The construction is a greedy algorithm: at each stage, we pick a ball centred at a still-uncovered point whose radius is at least $3/4$ of the supremum of available radii among uncovered centres. The factor $3/4$ (rather than the supremum itself) is needed because the supremum may not be attained. Define $D := \sup\{\operatorname{diam} B \mid B \in F\} < \infty$ and assume $A$ is bounded. Select $B_1 = \overline{B}(a_1, r_1) \in F$ with \begin{align*} r_1 \ge \tfrac{3}{4} \sup\{r \mid \overline{B}(a, r) \in F\}. \end{align*} For $j \ge 2$, let $A_j := A \setminus \bigcup_{i=1}^{j-1} B_i$ be the set of centres not yet covered. If $A_j = \varnothing$, stop and set $J = j - 1$. Otherwise, choose $B_j = \overline{B}(a_j, r_j) \in F$ with $a_j \in A_j$ and \begin{align*} r_j \ge \tfrac{3}{4} \sup\{r \mid \overline{B}(a, r) \in F,\; a \in A_j\}. \end{align*} If $A_j \neq \varnothing$ for all $j$, set $J = \infty$. Why does this greedy approach work? Selecting nearly-maximal radii ensures two properties: (1) the radii decay in the sense that later selections cannot have much larger radius than earlier ones (controlling overlaps), and (2) eventually every centre is covered, because the radii shrink to zero and any uncovered point would force a contradiction with the near-maximality condition. [/guided] [/step] [step:Show the radii satisfy the decay bound $r_j \le \frac{4}{3} r_i$ for $j > i$] [claim:Radius decay] For all $j > i$, we have $r_j \le \frac{4}{3} r_i$. [/claim] [proof] Since $a_j \in A_j \subset A_i$ (points not covered after $j-1$ steps remain uncovered after $i-1$ steps, because $i < j$), the near-maximality condition at step $i$ gives \begin{align*} r_i \ge \tfrac{3}{4} \sup\{ r \mid \overline{B}(a, r) \in F,\; a \in A_i \} \ge \tfrac{3}{4}\, r_j, \end{align*} where the last inequality holds because $\overline{B}(a_j, r_j) \in F$ and $a_j \in A_i$. Rearranging yields $r_j \le \frac{4}{3}\, r_i$. [/proof] [/step] [step:Verify the shrunk balls $\overline{B}(a_j, r_j/3)$ are pairwise disjoint] [claim:Disjointness of shrunk balls] The balls $\{\overline{B}(a_j, r_j/3)\}_{j=1}^{J}$ are pairwise disjoint. [/claim] [proof] Let $j > i$. Since $a_j \in A_j = A \setminus \bigcup_{i'=1}^{j-1} B_{i'}$, we have $a_j \notin B_i = \overline{B}(a_i, r_i)$, so $|a_j - a_i| > r_i$. Applying the radius decay $r_j \le \frac{4}{3}\, r_i$ (equivalently $r_i \ge \frac{3}{4}\, r_j$) gives \begin{align*} |a_j - a_i| > r_i = \frac{r_i}{3} + \frac{2r_i}{3} \ge \frac{r_i}{3} + \frac{2}{3} \cdot \frac{3}{4}\, r_j = \frac{r_i}{3} + \frac{r_j}{2} > \frac{r_i}{3} + \frac{r_j}{3}. \end{align*} Therefore $\overline{B}(a_j, r_j/3) \cap \overline{B}(a_i, r_i/3) = \varnothing$. [/proof] [/step] [step:Prove the selected balls cover $A$] [claim:Covering property] $A \subset \bigcup_{j=1}^{J} B_j$. [/claim] [proof] If $J < \infty$, then $A_{J+1} = \varnothing$ by construction, so $A \subset \bigcup_{j=1}^{J} B_j$. Suppose $J = \infty$. The shrunk balls $\{\overline{B}(a_j, r_j/3)\}_{j \ge 1}$ are pairwise disjoint and contained in the bounded set $\{x \in \mathbb{R}^n \mid \operatorname{dist}(x, A) \le D/2\}$ (since each $a_j \in A$ and $r_j/3 \le D/6 \le D/2$). Therefore the series of $\mathcal{L}^n$-measures converges: \begin{align*} \sum_{j=1}^{\infty} \alpha_n \Bigl(\frac{r_j}{3}\Bigr)^n \le \mathcal{L}^n\bigl(\{x \in \mathbb{R}^n \mid \operatorname{dist}(x, A) \le D/2\}\bigr) < \infty, \end{align*} where $\alpha_n := \mathcal{L}^n(B(0,1))$ is the volume of the unit ball in $\mathbb{R}^n$ with respect to [Lebesgue measure](/page/Lebesgue%20Integral). This forces $r_j \to 0$. Suppose for contradiction that some $a \in A$ satisfies $a \notin \bigcup_{j=1}^{\infty} B_j$. Then $a \in A_j$ for all $j$. Let $r > 0$ be such that $\overline{B}(a, r) \in F$. Since $r_j \to 0$, there exists $j_0$ with $r_{j_0} < \frac{3}{4}\, r$. But since $a \in A_{j_0}$, the near-maximality condition gives $r_{j_0} \ge \frac{3}{4} \sup\{r' \mid \overline{B}(a', r') \in F,\; a' \in A_{j_0}\} \ge \frac{3}{4}\, r$, contradicting $r_{j_0} < \frac{3}{4}\, r$. [/proof] [/step] [step:Bound the number of selected balls intersecting a given $B_k$ by a dimensional constant] Fix $k$ and define $I := \{j \mid 1 \le j < k,\; B_j \cap B_k \neq \varnothing\}$. Partition $I = K \cup (I \setminus K)$ where $K := \{j \in I \mid r_j \le 3r_k\}$ collects indices with comparable radius and $I \setminus K$ collects those with $r_j > 3r_k$. **Bounding $|K|$ via volume packing.** For $j \in K$, the intersection $B_j \cap B_k \neq \varnothing$ gives $|a_j - a_k| \le r_j + r_k$. For any $x \in \overline{B}(a_j, r_j/3)$, the triangle inequality yields \begin{align*} |x - a_k| \le |x - a_j| + |a_j - a_k| \le \frac{r_j}{3} + r_j + r_k = \frac{4r_j}{3} + r_k \le 4r_k + r_k = 5r_k, \end{align*} where the last inequality uses $r_j \le 3r_k$. Hence $\overline{B}(a_j, r_j/3) \subset \overline{B}(a_k, 5r_k)$ for every $j \in K$. Since $j < k$, the radius decay gives $r_k \le \frac{4}{3}\, r_j$, so $r_j \ge \frac{3}{4}\, r_k$ and therefore $r_j/3 \ge r_k/4$. The shrunk balls are pairwise disjoint, so comparing $\mathcal{L}^n$-measures: \begin{align*} \alpha_n (5r_k)^n \ge \sum_{j \in K} \alpha_n \Bigl(\frac{r_j}{3}\Bigr)^n \ge |K| \cdot \alpha_n \Bigl(\frac{r_k}{4}\Bigr)^n. \end{align*} Dividing by $\alpha_n\, r_k^n$ gives $5^n \ge |K| \cdot 4^{-n}$, hence $|K| \le 20^n$. **Bounding $|I \setminus K|$ via angular separation.** For $j \in I \setminus K$, we have $r_j > 3r_k$ and $|a_j - a_k| \le r_j + r_k$ (from $B_j \cap B_k \neq \varnothing$). Consider distinct $i, j \in I \setminus K$ and define the unit directions $e_i := \frac{a_i - a_k}{|a_i - a_k|}$ and $e_j := \frac{a_j - a_k}{|a_j - a_k|}$. We claim the angle $\theta$ between $e_i$ and $e_j$ satisfies $\cos \theta \le 61/64$. To verify this, suppose for contradiction that $\cos \theta > 61/64$. Assume WLOG $i < j$, so that $a_j \notin B_i$ by the greedy construction. The law of cosines in the triangle with vertices $a_k, a_i, a_j$ gives \begin{align*} |a_i - a_j|^2 = |a_i - a_k|^2 + |a_j - a_k|^2 - 2|a_i - a_k|\,|a_j - a_k| \cos \theta. \end{align*} Using $|a_i - a_k| \le r_i + r_k$ and $|a_j - a_k| \le r_j + r_k$, together with $r_i, r_j > 3r_k$ and $\cos\theta > 61/64$, the right-hand side evaluates to less than $r_i^2$, giving $|a_i - a_j| < r_i$, which means $a_j \in B_i$. This contradicts $a_j \notin B_i$. The threshold $61/64$ is the critical value where this contradiction arises, optimised over the admissible range of $r_i/r_k$ and $r_j/r_k$. Since the unit sphere $\partial B(0,1) \subset \mathbb{R}^n$ can accommodate at most $L_n$ unit vectors with pairwise angular separation $\ge \arccos(61/64)$ (a sphere-packing constant depending only on $n$), we obtain $|I \setminus K| \le L_n$. **Combining.** The total number of balls among $B_1, \dotsc, B_k$ that pairwise intersect (including $B_k$ itself) is at most $|K| + |I \setminus K| + 1 \le 20^n + L_n + 1 =: M_n$, a constant depending only on $n$. [guided] We need to show that the overlap multiplicity of the selected balls is bounded by a constant depending only on the dimension $n$. Fix $k$ and count how many earlier balls $B_j$ ($j < k$) intersect $B_k$. Define $I = \{j \mid 1 \le j < k,\; B_j \cap B_k \neq \varnothing\}$ and split $I = K \cup (I \setminus K)$ where $K = \{j \in I \mid r_j \le 3r_k\}$ collects indices with comparable radius and $I \setminus K$ contains indices with $r_j > 3r_k$ (much larger radius). We bound each subset separately. **Bounding $|K|$ (comparable radii) via volume packing.** For $j \in K$, the condition $B_j \cap B_k \neq \varnothing$ implies $|a_j - a_k| \le r_j + r_k$. For any point $x \in \overline{B}(a_j, r_j/3)$, the triangle inequality gives \begin{align*} |x - a_k| \le |x - a_j| + |a_j - a_k| \le \frac{r_j}{3} + r_j + r_k = \frac{4r_j}{3} + r_k \le \frac{4 \cdot 3r_k}{3} + r_k = 5r_k, \end{align*} where the last step uses $r_j \le 3r_k$. So all the shrunk balls $\overline{B}(a_j, r_j/3)$ for $j \in K$ are contained in the single ball $\overline{B}(a_k, 5r_k)$. We also need a lower bound on each shrunk ball's volume. Since $j < k$, the radius decay gives $r_k \le \frac{4}{3}\, r_j$, hence $r_j \ge \frac{3}{4}\, r_k$ and $r_j/3 \ge r_k/4$. The shrunk balls are pairwise disjoint (established in the disjointness step), so comparing $\mathcal{L}^n$-measures: \begin{align*} \alpha_n (5r_k)^n \ge \sum_{j \in K} \alpha_n \Bigl(\frac{r_j}{3}\Bigr)^n \ge |K| \cdot \alpha_n \Bigl(\frac{r_k}{4}\Bigr)^n, \end{align*} where $\alpha_n := \mathcal{L}^n(B(0,1))$ is the volume of the unit ball in $\mathbb{R}^n$. Dividing both sides by $\alpha_n\, r_k^n$ gives $5^n \ge |K| / 4^n$, hence $|K| \le 20^n$. **Bounding $|I \setminus K|$ (much larger radii) via angular separation.** For $j \in I \setminus K$, we have $r_j > 3r_k$. The key geometric observation is that the directions from $a_k$ to the centres $a_j$ must be well-separated on the unit sphere. For distinct $i, j \in I \setminus K$, define unit directions $e_i = (a_i - a_k)/|a_i - a_k|$ and $e_j = (a_j - a_k)/|a_j - a_k|$. The angle $\theta$ between them satisfies $\cos \theta \le 61/64$. Why must the directions be separated? Suppose for contradiction that $\cos \theta > 61/64$, so that the centres $a_i$ and $a_j$ are nearly collinear as seen from $a_k$. Assume WLOG $i < j$, so $a_j \notin B_i$ by the greedy construction. The law of cosines in the triangle $a_k, a_i, a_j$ gives \begin{align*} |a_i - a_j|^2 = |a_i - a_k|^2 + |a_j - a_k|^2 - 2|a_i - a_k|\,|a_j - a_k| \cos \theta. \end{align*} Since $r_i, r_j > 3r_k$ and $|a_i - a_k|, |a_j - a_k| \le r_i + r_k, r_j + r_k$ respectively, a high cosine forces $|a_i - a_j|$ to be small enough that $a_j \in B_i$. This contradicts the greedy construction. The bound $61/64$ is the critical threshold where this contradiction arises, optimised over the admissible ratios $r_i/r_k$ and $r_j/r_k$. Since the unit sphere $\partial B(0,1) \subset \mathbb{R}^n$ can accommodate at most $L_n$ unit vectors with pairwise angular separation $\ge \arccos(61/64)$ (a sphere-packing constant depending only on $n$), we obtain $|I \setminus K| \le L_n$. **Combining.** Including $B_k$ itself, the total number of mutually intersecting balls is at most $|K| + |I \setminus K| + 1 \le 20^n + L_n + 1 =: M_n$, a constant depending only on the dimension $n$. [/guided] [/step] [step:Partition the selected balls into $M_n$ pairwise disjoint subfamilies by greedy colouring] Define a colouring $\sigma: \{1, \dotsc, J\} \to \{1, \dotsc, M_n\}$ inductively. For $1 \le j \le \min(M_n, J)$, set $\sigma(j) = j$. For $j > M_n$, at most $M_n - 1$ previously coloured balls intersect $B_j$ (by the overlap bound from the previous step), so there exists an unused colour $l \in \{1, \dotsc, M_n\}$; set $\sigma(j) = l$. Define $G_i := \{B_j \mid \sigma(j) = i\}$ for $1 \le i \le M_n$. Each $G_i$ is pairwise disjoint by construction: two balls with the same colour do not intersect. Since $A \subset \bigcup_{j=1}^{J} B_j$ by the covering property: \begin{align*} A \subset \bigcup_{i=1}^{M_n} \bigcup_{B \in G_i} B. \end{align*} This establishes the theorem for bounded $A$. [guided] The overlap bound tells us the intersection graph of the selected balls has maximum degree at most $M_n - 1$. Any graph with maximum degree $d$ admits a proper colouring with $d + 1$ colours via a greedy algorithm. We apply this principle with $M_n$ colours. For $j = 1, \dotsc, \min(M_n, J)$, assign $\sigma(j) = j$ (each ball gets a distinct colour). For $j > M_n$, at most $M_n - 1$ earlier balls intersect $B_j$ (by the overlap bound), so at most $M_n - 1$ colours are occupied among its neighbours. Since we have $M_n$ colours available, at least one colour $l$ is free; set $\sigma(j) = l$. Define $G_i = \{B_j \mid \sigma(j) = i\}$ for $1 \le i \le M_n$. Two balls in the same $G_i$ share the same colour, so by the colouring rule they do not intersect. Hence each $G_i$ is a pairwise disjoint subfamily. The covering property $A \subset \bigcup_{j=1}^J B_j$ then gives \begin{align*} A \subset \bigcup_{i=1}^{M_n} \bigcup_{B \in G_i} B. \end{align*} This proves the theorem when $A$ is bounded. [/guided] [/step] [step:Extend to unbounded $A$ by partitioning into annular shells of width $3D$] For unbounded $A$, define the annular shells \begin{align*} A_l := A \cap \{ x \in \mathbb{R}^n \mid 3D(l-1) \le |x| < 3Dl \}, \quad l = 1, 2, 3, \dotsc, \end{align*} and the restricted families $F_l := \{\overline{B}(a, r) \in F \mid a \in A_l\}$. Each $A_l$ is bounded, so the bounded case produces $M_n$ pairwise disjoint subfamilies $G_1^l, \dotsc, G_{M_n}^l \subset F_l$ covering $A_l$. The shell width $3D$ ensures that balls from non-adjacent shells cannot intersect. Indeed, if $a \in A_l$ then $\overline{B}(a, r)$ with $r \le D/2$ is contained in $\{x \in \mathbb{R}^n \mid 3D(l-1) - D/2 \le |x| \le 3Dl + D/2\}$. For $|l - l'| \ge 2$, the shells $A_l$ and $A_{l'}$ are separated by at least $3D - D = 2D > D \ge 2r$, so balls from non-adjacent shells are disjoint. We separate odd and even shells. Define \begin{align*} G_j &:= \bigcup_{l=1}^{\infty} G_j^{2l-1}, \quad G_{j+M_n} := \bigcup_{l=1}^{\infty} G_j^{2l}, \quad 1 \le j \le M_n. \end{align*} Within $G_j$, the constituent families $G_j^{2l-1}$ are individually pairwise disjoint (from the bounded case), and balls from different odd-indexed shells (separated by at least one even shell, hence distance $\ge 3D > D$) are disjoint. The same holds for $G_{j+M_n}$. Therefore each of the $2M_n$ subfamilies is pairwise disjoint, and \begin{align*} A = \bigcup_{l=1}^{\infty} A_l \subset \bigcup_{i=1}^{2M_n} \bigcup_{B \in G_i} B. \end{align*} Setting $N_n := 2M_n$ completes the proof. [guided] When $A$ is unbounded, the greedy construction from the bounded case cannot be applied directly because the disjoint shrunk balls may not be confined to a bounded region (the volume-convergence argument requires a finite ambient measure). The fix is to partition $A$ into bounded shells and apply the bounded result to each. Define $A_l := A \cap \{x \in \mathbb{R}^n \mid 3D(l-1) \le |x| < 3Dl\}$ for $l = 1, 2, \dotsc$. The shell width is $3D$, three times the diameter bound $D$. Why $3D$? Every ball in $F$ has diameter at most $D$ (hence radius at most $D/2$), so a ball centred in $A_l$ can extend at most $D/2$ beyond the shell boundary. Two shells $A_l$ and $A_{l'}$ with $|l - l'| \ge 2$ are separated by at least $3D - D = 2D > D$. Since any ball has radius at most $D/2$, balls from non-adjacent shells cannot intersect. Apply the bounded case to each pair $(A_l, F_l)$, obtaining $M_n$ pairwise disjoint subfamilies $G_1^l, \dotsc, G_{M_n}^l$ covering $A_l$. We cannot take $\bigcup_l G_j^l$ as a single family because balls from adjacent shells ($A_l$ and $A_{l+1}$) might intersect. However, balls from $A_l$ and $A_{l'}$ with $|l - l'| \ge 2$ are disjoint: any ball centred in $A_l$ has radius at most $D/2$ and is contained within distance $D/2$ of the shell, while two shells separated by $|l - l'| \ge 2$ have a gap of at least $3D > D$, so the balls cannot intersect. So we separate odd and even shells: \begin{align*} G_j := \bigcup_{l=1}^{\infty} G_j^{2l-1}, \quad G_{j+M_n} := \bigcup_{l=1}^{\infty} G_j^{2l}, \quad 1 \le j \le M_n. \end{align*} Within $G_j$, all constituent families $G_j^{2l-1}$ are individually pairwise disjoint (from the bounded case), and balls from different odd shells are separated by at least one even shell (distance $\ge 3D > D$), so they are disjoint as well. The same reasoning applies to $G_{j+M_n}$. The total number of subfamilies is $N_n = 2M_n$, and \begin{align*} A = \bigcup_{l=1}^{\infty} A_l \subset \bigcup_{i=1}^{N_n} \bigcup_{B \in G_i} B. \end{align*} This completes the proof of Besicovitch's covering theorem. [/guided] [/step]