[proofplan] We prove the uniform covering estimate required by Gromov's compactness criterion for compact metric spaces. For each scale $\varepsilon > 0$, we choose a maximal $\varepsilon$-separated set; maximality makes it an $\varepsilon$-net, while separation makes the balls of radius $\varepsilon/2$ disjoint. Bishop-Gromov volume comparison converts the Ricci lower bound, the diameter bound, and the global noncollapse assumption into a uniform lower bound for each such small ball, and also gives a uniform upper bound for the total volume. These two estimates bound the size of every maximal separated set by a number depending only on $n,K,D,v,\varepsilon$, so Gromov compactness gives precompactness. [/proofplan] [step:Define the model volume constants used in the comparison estimates] Let $\mathbb{M}_{K}^{n}$ denote the complete simply connected $n$-dimensional Riemannian manifold of constant sectional curvature $K$. For each $r > 0$, define the model ball volume \begin{align*} V_{n,K}(r) := \operatorname{Vol}_{\mathbb{M}_{K}^{n}}\bigl(B_{\mathbb{M}_{K}^{n}}(o,r)\bigr), \end{align*} where $o \in \mathbb{M}_{K}^{n}$ is any point. Since $\mathbb{M}_{K}^{n}$ is homogeneous, this definition is independent of $o$. For each fixed $r > 0$, the number $V_{n,K}(r)$ is finite and positive. Fix $\varepsilon > 0$. Define \begin{align*} r_\varepsilon := \frac{\varepsilon}{2}, \qquad a_\varepsilon := \frac{V_{n,K}(r_\varepsilon)}{V_{n,K}(D)}, \qquad A := V_{n,K}(D). \end{align*} If $\varepsilon \ge 2D$, then every $(M,d_g)$ with $\operatorname{diam}(M,d_g) \le D$ is covered by one ball of radius $\varepsilon$, so the required covering bound is $1$. Hence assume $0 < \varepsilon < 2D$, so $0 < r_\varepsilon < D$ and $a_\varepsilon > 0$. [/step] [step:Choose a maximal separated set and identify it as a net] Let $(M,g) \in \mathcal{M}(n,K,D,v)$, and let $d_g: M \times M \to [0,\infty)$ be the Riemannian distance. Choose a maximal finite subset \begin{align*} P := \{p_1,\dots,p_N\} \subset M \end{align*} such that \begin{align*} d_g(p_i,p_j) \ge \varepsilon \end{align*} whenever $i \ne j$. Such a set exists because $M$ is compact. Maximality implies that $P$ is an $\varepsilon$-net: for every $x \in M$ there exists $i \in \{1,\dots,N\}$ such that $d_g(x,p_i) < \varepsilon$. Indeed, if no such $i$ existed, then $P \cup \{x\}$ would still be $\varepsilon$-separated, contradicting maximality. Therefore \begin{align*} M = \bigcup_{i=1}^{N} B_g(p_i,\varepsilon). \end{align*} The same separation condition implies that the open balls $B_g(p_i,r_\varepsilon)$ are pairwise disjoint. If $x \in B_g(p_i,r_\varepsilon) \cap B_g(p_j,r_\varepsilon)$ for $i \ne j$, then the triangle inequality gives \begin{align*} d_g(p_i,p_j) &\le d_g(p_i,x) + d_g(x,p_j) \\ &< r_\varepsilon + r_\varepsilon \\ &= \varepsilon, \end{align*} contradicting $d_g(p_i,p_j) \ge \varepsilon$. [guided] The point of choosing a maximal separated set is that it has two complementary properties. Separation gives disjoint small balls, which is useful for counting. Maximality gives covering by larger balls, which is exactly the covering estimate needed for Gromov compactness. Let \begin{align*} P := \{p_1,\dots,p_N\} \subset M \end{align*} be maximal among finite subsets satisfying $d_g(p_i,p_j) \ge \varepsilon$ for $i \ne j$. Compactness of $M$ ensures such a finite maximal set exists: an infinite $\varepsilon$-separated subset of a compact [metric space](/page/Metric%20Space) would have no Cauchy subsequence, contradicting [sequential compactness](/page/Sequential%20Compactness). We first prove that $P$ is an $\varepsilon$-net. Suppose that there were a point $x \in M$ such that $d_g(x,p_i) \ge \varepsilon$ for every $i \in \{1,\dots,N\}$. Then the enlarged set $P \cup \{x\}$ would still be $\varepsilon$-separated, because the old points are separated from one another and $x$ is separated from each old point. This contradicts the maximality of $P$. Hence every $x \in M$ satisfies $d_g(x,p_i) < \varepsilon$ for some $i$, and therefore \begin{align*} M = \bigcup_{i=1}^{N} B_g(p_i,\varepsilon). \end{align*} Next we prove that the smaller balls are disjoint. Define $r_\varepsilon := \varepsilon/2$. If $x$ belonged to both $B_g(p_i,r_\varepsilon)$ and $B_g(p_j,r_\varepsilon)$ with $i \ne j$, then the triangle inequality would give \begin{align*} d_g(p_i,p_j) &\le d_g(p_i,x) + d_g(x,p_j) \\ &< r_\varepsilon + r_\varepsilon \\ &= \varepsilon. \end{align*} This contradicts the defining separation condition $d_g(p_i,p_j) \ge \varepsilon$. Thus the balls $B_g(p_i,r_\varepsilon)$ are pairwise disjoint. [/guided] [/step] [step:Use Bishop-Gromov comparison to bound each small ball from below] We use Bishop-Gromov volume comparison (citing a result not yet in the wiki: [Bishop-Gromov Volume Comparison Theorem](/theorems/5371)). The theorem applies because $(M,g)$ is complete, since it is closed, and satisfies $\operatorname{Ric}_g \ge (n-1)K g$ by hypothesis. For each $i \in \{1,\dots,N\}$, the diameter bound gives $M \subset B_g(p_i,D)$. Since the reverse inclusion is automatic, up to the harmless distinction between open and closed balls at radius $D$, volume regularity gives \begin{align*} \operatorname{Vol}_g(B_g(p_i,D)) = \operatorname{Vol}_g(M). \end{align*} Bishop-Gromov monotonicity applied at the radii $r_\varepsilon < D$ gives \begin{align*} \frac{\operatorname{Vol}_g(B_g(p_i,r_\varepsilon))}{V_{n,K}(r_\varepsilon)} \ge \frac{\operatorname{Vol}_g(B_g(p_i,D))}{V_{n,K}(D)}. \end{align*} Therefore \begin{align*} \operatorname{Vol}_g(B_g(p_i,r_\varepsilon)) &\ge \frac{V_{n,K}(r_\varepsilon)}{V_{n,K}(D)} \operatorname{Vol}_g(M) \\ &\ge a_\varepsilon v. \end{align*} Thus every ball $B_g(p_i,r_\varepsilon)$ has volume at least $a_\varepsilon v > 0$, uniformly over all $(M,g) \in \mathcal{M}(n,K,D,v)$ and all $i$. [/step] [step:Use Bishop-Gromov comparison to bound the total volume from above] Fix any point $p \in M$. Since $\operatorname{diam}(M,d_g) \le D$, we have $M \subset B_g(p,D)$, and hence \begin{align*} \operatorname{Vol}_g(M) = \operatorname{Vol}_g(B_g(p,D)). \end{align*} Bishop-Gromov volume comparison at radius $D$ gives \begin{align*} \operatorname{Vol}_g(B_g(p,D)) \le V_{n,K}(D) = A. \end{align*} Consequently, \begin{align*} \operatorname{Vol}_g(M) \le A. \end{align*} [/step] [step:Convert the volume estimates into a uniform covering number] Because the balls $B_g(p_i,r_\varepsilon)$ are pairwise disjoint and contained in $M$, countable additivity of the Riemannian volume measure gives \begin{align*} \sum_{i=1}^{N} \operatorname{Vol}_g(B_g(p_i,r_\varepsilon)) \le \operatorname{Vol}_g(M). \end{align*} Using the lower bound from the small balls and the upper bound for the total volume, we obtain \begin{align*} N a_\varepsilon v &\le \sum_{i=1}^{N} \operatorname{Vol}_g(B_g(p_i,r_\varepsilon)) \\ &\le \operatorname{Vol}_g(M) \\ &\le A. \end{align*} Hence \begin{align*} N \le \frac{A}{a_\varepsilon v} = \frac{V_{n,K}(D)^2}{V_{n,K}(r_\varepsilon)v}. \end{align*} Define the integer \begin{align*} N_\varepsilon := \left\lceil \frac{V_{n,K}(D)^2}{V_{n,K}(\varepsilon/2)v} \right\rceil. \end{align*} Then every $(M,d_g)$ in the class can be covered by at most $N_\varepsilon$ open balls of radius $\varepsilon$. [/step] [step:Apply Gromov compactness to obtain precompactness] We apply Gromov's [compactness theorem](/theorems/2748) for compact metric spaces (citing a result not yet in the wiki: Gromov Compactness Theorem). Its hypotheses require a uniform diameter bound and, for every $\varepsilon > 0$, a uniform bound on the number of $\varepsilon$-balls needed to cover each space in the family. The diameter bound is part of the definition of $\mathcal{M}(n,K,D,v)$. The preceding step supplies, for each $\varepsilon > 0$, an integer $N_\varepsilon$ depending only on $n,K,D,v,\varepsilon$ such that every $(M,d_g)$ in the class is covered by at most $N_\varepsilon$ balls of radius $\varepsilon$. Therefore Gromov's compactness theorem implies that the class of compact metric spaces \begin{align*} \{(M,d_g) : (M,g) \in \mathcal{M}(n,K,D,v)\} \end{align*} is precompact in the Gromov-Hausdorff topology. Equivalently, every sequence $((M_j,g_j))_{j=1}^{\infty}$ in $\mathcal{M}(n,K,D,v)$ has a subsequence whose metric spaces $(M_j,d_{g_j})$ converge in the Gromov-Hausdorff topology to a compact metric space. This is precisely the asserted precompactness. [/step]