[proofplan] The proof is a greedy selection argument organised by diameter scales. We partition $F$ into countably many layers $F_1, F_2, \ldots$ according to geometric scales of the diameter, then sweep through the layers in order, selecting at each stage a maximal disjoint subcollection from the balls that do not yet intersect anything previously chosen. Maximality within each layer forces every unchosen ball to meet a selected ball of comparable size, and the diameter comparison between adjacent scales converts this intersection into containment in the five-times dilation. [/proofplan] [step:Stratify $F$ by diameter scales and greedily select maximal disjoint subfamilies] Define the diameter bound \begin{align*} D := \sup\{\operatorname{diam} B \mid B \in F\}, \end{align*} and partition $F$ into layers indexed by $j \in \mathbb{N}$: \begin{align*} F_j := \{B \in F \mid D \cdot 2^{-j} < \operatorname{diam} B \leq D \cdot 2^{-j+1}\}. \end{align*} Since every $B \in F$ is non-degenerate ($\operatorname{diam} B > 0$) and $\operatorname{diam} B \leq D$, each $B$ belongs to exactly one $F_j$, so $\{F_j\}_{j \geq 1}$ is a partition of $F$. We construct $G$ inductively. Choose $G_1 \subseteq F_1$ to be a maximal subcollection with respect to pairwise disjointness (such a maximal subcollection exists by [Zorn's lemma](/page/1261) applied to the partially ordered set of disjoint subfamilies of $F_1$, ordered by inclusion). Having selected $G_1, \ldots, G_{k-1}$, define the candidate collection \begin{align*} \widetilde{F}_k := \Bigl\{B \in F_k \;\Big|\; B \cap B' = \varnothing \text{ for all } B' \in \bigcup_{j=1}^{k-1} G_j\Bigr\}, \end{align*} and choose $G_k \subseteq \widetilde{F}_k$ maximal with respect to pairwise disjointness. Set \begin{align*} G := \bigcup_{j=1}^{\infty} G_j. \end{align*} By construction, the balls in $G$ are pairwise disjoint: balls within each $G_j$ are disjoint by the maximality selection, and balls from different layers $G_j$, $G_k$ ($j < k$) are disjoint because every ball in $G_k$ was chosen from $\widetilde{F}_k$, which excludes balls intersecting $\bigcup_{i=1}^{k-1} G_i$. Moreover, $G \subseteq F$ since each $G_j \subseteq F_j \subseteq F$. Because each $G_j$ is a disjoint collection of balls with diameter at least $D \cdot 2^{-j}$, any bounded region of $\mathbb{R}^n$ meets only finitely many balls from $G_j$, whence $G$ is [countable](/page/1215). [guided] The key design choice is the geometric scaling factor $2$ in the partition. We define the diameter bound $D := \sup\{\operatorname{diam} B \mid B \in F\}$ and partition $F$ into layers \begin{align*} F_j := \{B \in F \mid D \cdot 2^{-j} < \operatorname{diam} B \leq D \cdot 2^{-j+1}\}, \qquad j \in \mathbb{N}. \end{align*} Why does every ball land in exactly one layer? Since each $B \in F$ is non-degenerate, $\operatorname{diam} B > 0$, and the supremum bound gives $\operatorname{diam} B \leq D$. The intervals $(D \cdot 2^{-j}, D \cdot 2^{-j+1}]$ for $j \geq 1$ are disjoint and their union is $(0, D]$, so each ball belongs to a unique $F_j$. This geometric scaling ensures that any two balls from layers $F_j$ and $F_k$ with $k \leq j$ satisfy a diameter comparison: \begin{align*} \operatorname{diam} B \leq D \cdot 2^{-j+1} \leq 2 \cdot D \cdot 2^{-k} < 2 \cdot \operatorname{diam} B', \end{align*} where $B \in F_j$ and $B' \in F_k$. This factor of $2$ is what ultimately produces the constant $5$ in the dilation $\hat{B}$. We now construct $G$ by sweeping from the largest balls downward. For the base case, choose $G_1 \subseteq F_1$ to be a maximal subcollection with respect to pairwise disjointness. Why does a maximal disjoint subcollection exist? Consider the partially ordered set of all disjoint subfamilies of $F_1$, ordered by inclusion. Every chain has an upper bound (the union of the chain is still a disjoint subfamily, since any two balls in the union belong to some common element of the chain). By Zorn's lemma, a maximal element exists. For the inductive step, having selected $G_1, \ldots, G_{k-1}$, we first discard all balls in $F_k$ that already intersect something previously chosen: \begin{align*} \widetilde{F}_k := \Bigl\{B \in F_k \;\Big|\; B \cap B' = \varnothing \text{ for all } B' \in \bigcup_{j=1}^{k-1} G_j\Bigr\}, \end{align*} and then choose $G_k \subseteq \widetilde{F}_k$ maximal with respect to pairwise disjointness (again by Zorn's lemma). Finally, set \begin{align*} G := \bigcup_{j=1}^{\infty} G_j. \end{align*} Why are the balls in $G$ pairwise disjoint? There are two cases. Balls within the same $G_j$ are disjoint by the maximality selection. Balls from different layers $G_j$ and $G_k$ with $j < k$ are disjoint because every ball in $G_k$ was chosen from $\widetilde{F}_k$, which by definition excludes balls intersecting any element of $\bigcup_{i=1}^{k-1} G_i \supseteq G_j$. Why is $G$ countable? Each $G_j$ is a disjoint collection of balls with diameter at least $D \cdot 2^{-j}$. Any bounded region of $\mathbb{R}^n$ can contain only finitely many disjoint balls of diameter exceeding a fixed positive threshold, so each $G_j$ is countable. A countable union of countable sets is countable, so $G$ is countable. Why greedily sweep from large to small? The greedy algorithm processes the largest balls first. When we later encounter a smaller ball $B \in F_j$, the maximality of $G_j$ guarantees that $B$ intersects some already-selected ball $B'$ from $G_1 \cup \cdots \cup G_j$. Because $B'$ was selected at a layer $k \leq j$, it has $\operatorname{diam} B' \geq D \cdot 2^{-k} \geq D \cdot 2^{-j}$, so $\operatorname{diam} B \leq 2 \cdot \operatorname{diam} B'$. This diameter bound is what makes the five-times dilation sufficient to absorb $B$. [/guided] [/step] [step:Show that every ball in $F$ is absorbed by the five-times dilation of some ball in $G$] [claim:Covering Property] For every $B \in F$ there exists $B' \in G$ such that $B \cap B' \neq \varnothing$ and $B \subseteq \hat{B}'$. [/claim] [proof] Fix $B \in F$ and let $j \in \mathbb{N}$ be the unique index with $B \in F_j$, so that \begin{align*} D \cdot 2^{-j} < \operatorname{diam} B \leq D \cdot 2^{-j+1}. \end{align*} **Case 1:** $B \in G_j$. Then $B' := B$ satisfies the claim since $B \subseteq \hat{B}$. **Case 2:** $B \notin G_j$. Then either $B \notin \widetilde{F}_j$ (meaning $B$ intersects some ball already selected in layers $1, \ldots, j-1$), or $B \in \widetilde{F}_j$ but was not chosen for $G_j$. In the latter case, the maximality of $G_j$ within $\widetilde{F}_j$ implies that $B$ intersects some ball in $G_j$ (otherwise $G_j \cup \{B\}$ would be a strictly larger disjoint subcollection of $\widetilde{F}_j$, contradicting maximality). In either case, there exists $B' \in G_k$ for some $k \leq j$ with $B \cap B' \neq \varnothing$. Write $B = \overline{B}(a, s)$ and $B' = \overline{B}(a', s')$ where $s, s' > 0$ denote the respective radii. Since $B'$ belongs to layer $k \leq j$, the diameter bounds give \begin{align*} \operatorname{diam} B' > D \cdot 2^{-k} \geq D \cdot 2^{-j}, \end{align*} hence $2s' = \operatorname{diam} B' > D \cdot 2^{-j}$, which yields $s' > D \cdot 2^{-j-1}$. Meanwhile, $2s = \operatorname{diam} B \leq D \cdot 2^{-j+1}$, so $s \leq D \cdot 2^{-j}$. Combining: \begin{align*} s \leq D \cdot 2^{-j} < 2s'. \end{align*} Since $B \cap B' \neq \varnothing$, there exists a point $z \in B \cap B'$, so $|a - a'| \leq |a - z| + |z - a'| \leq s + s'$. For any $x \in B$, the triangle inequality gives \begin{align*} |x - a'| &\leq |x - a| + |a - a'| \leq s + (s + s') = 2s + s' < 4s' + s' = 5s'. \end{align*} Therefore $x \in \overline{B}(a', 5s') = \hat{B}'$, and since $x \in B$ was arbitrary, $B \subseteq \hat{B}'$. [guided] The diameter comparison is the heart of the argument. Fix $B \in F$ with $B \in F_j$, so $D \cdot 2^{-j} < \operatorname{diam} B \leq D \cdot 2^{-j+1}$. If $B \in G_j$, then taking $B' := B$ gives $B \subseteq \hat{B}$ and we are done. So suppose $B \notin G_j$. Either $B$ was discarded when forming $\widetilde{F}_j$ (because it intersects some previously selected ball from layers $1, \ldots, j-1$), or $B$ survived into $\widetilde{F}_j$ but was not chosen for $G_j$. In the second sub-case, maximality of $G_j$ forces $B$ to intersect some ball in $G_j$: if $B$ were disjoint from every ball in $G_j$, then $G_j \cup \{B\}$ would be a strictly larger disjoint subcollection of $\widetilde{F}_j$, contradicting the maximality of $G_j$. In either case, there exists $B' \in G_k$ for some $k \leq j$ with $B \cap B' \neq \varnothing$. Write $B = \overline{B}(a, s)$ and $B' = \overline{B}(a', s')$. We now trace where the constant $5$ comes from. We need to show $|x - a'| \leq 5s'$ for every $x \in B$. First, the diameter comparison. Since $B' \in G_k \subseteq F_k$ with $k \leq j$: \begin{align*} 2s' = \operatorname{diam} B' > D \cdot 2^{-k} \geq D \cdot 2^{-j}, \qquad 2s = \operatorname{diam} B \leq D \cdot 2^{-j+1}, \end{align*} so $s' > D \cdot 2^{-j-1}$ and $s \leq D \cdot 2^{-j}$. In particular, $s < 2s'$. The triangle inequality decomposes the distance as: \begin{align*} |x - a'| \leq \underbrace{|x - a|}_{\leq\, s} + \underbrace{|a - a'|}_{\leq\, s + s'}. \end{align*} The first term is at most $s$ because $x \in B = \overline{B}(a, s)$. The second term is at most $s + s'$ because the intersection point $z \in B \cap B'$ satisfies $|a - z| \leq s$ and $|z - a'| \leq s'$. Adding these: $|x - a'| \leq 2s + s'$. Now we use the diameter comparison $s < 2s'$ (which came from the scale stratification: $B$ is in layer $j$ or later, while $B'$ is in layer $k \leq j$). Substituting: \begin{align*} 2s + s' < 2(2s') + s' = 5s'. \end{align*} So the constant $5 = 2 \cdot 2 + 1$ arises as follows: one factor of $2$ from the triangle inequality (we traverse the radius of $B$ twice -- once from $x$ to the center $a$, once from $a$ to the intersection point), and the other factor of $2$ from the diameter comparison between adjacent geometric scales. The final $+1$ accounts for the radius of $B'$ itself (from the intersection point to the center $a'$). This decomposition explains why $5$ is the sharp constant for this greedy algorithm: tighter stratification (e.g., ratio $3/2$ instead of $2$) would reduce the constant, but at the cost of more layers and a more intricate construction. The choice of factor $2$ gives the cleanest argument. [/guided] [/proof] [/step] [step:Conclude the covering property by combining the claim with the greedy selection] Let $x \in \bigcup_{B \in F} B$ be arbitrary. Choose $B \in F$ with $x \in B$. By the Covering Property claim, there exists $B' \in G$ with $B \subseteq \hat{B}'$. In particular, $x \in \hat{B}'$, so \begin{align*} x \in \hat{B}' \subseteq \bigcup_{B' \in G} \hat{B}'. \end{align*} Since $x$ was arbitrary, we conclude \begin{align*} \bigcup_{B \in F} B \subseteq \bigcup_{B' \in G} \hat{B}'. \end{align*} This, together with the disjointness and countability of $G$ established in the first step, completes the proof. [/step]