[proofplan] We prove both directions using finite approximations. If the Gromov-Hausdorff closure of $\mathcal{C}$ is compact, then diameter is uniformly bounded and finitely many Gromov-Hausdorff balls at a fixed scale transfer finite nets from their centers to all spaces in $\mathcal{C}$. Conversely, from the uniform diameter and covering bounds we approximate each space in a sequence by nested finite nets of controlled cardinality, pass to a diagonal subsequence so all finite distance matrices converge, and build a compact limit space as the completion of the resulting countable pseudometric space. The finite nets then give Gromov-Hausdorff approximations from the original spaces to the limit at scales tending to zero. [/proofplan]

[step:Transfer finite nets from a compact Gromov-Hausdorff closure]Assume that $\mathcal{C}$ is precompact. Let $\overline{\mathcal{C}}$ denote its closure in the Gromov-Hausdorff metric $d_{GH}$; by hypothesis $\overline{\mathcal{C}}$ is compact. Here $d_{GH}$ denotes the Gromov-Hausdorff metric on isometry classes of compact metric spaces. For a compact metric space $Y$ and a number $r > 0$, define the open Gromov-Hausdorff ball \begin{align*} B_{GH}(Y,r) := \{X : X \text{ is a compact metric space and } d_{GH}(X,Y) < r\}. \end{align*} First, the diameter function is continuous with respect to $d_{GH}$. Indeed, if compact metric spaces $X$ and $Y$ satisfy $d_{GH}(X,Y) < \eta$, then there is a metric space $(Z,d_Z)$ and isometric embeddings \begin{align*} \iota_X: X &\to Z, & \iota_Y: Y &\to Z \end{align*} such that the Hausdorff distance between $\iota_X(X)$ and $\iota_Y(Y)$ in $Z$ is less than $\eta$. For $x_1,x_2 \in X$, choose $y_1,y_2 \in Y$ with \begin{align*} d_Z(\iota_X(x_i),\iota_Y(y_i)) < \eta \end{align*} for $i \in \{1,2\}$. Then \begin{align*} d_X(x_1,x_2) &= d_Z(\iota_X(x_1),\iota_X(x_2)) \\ &\leq d_Z(\iota_X(x_1),\iota_Y(y_1)) + d_Z(\iota_Y(y_1),\iota_Y(y_2)) + d_Z(\iota_Y(y_2),\iota_X(x_2)) \\ &< \operatorname{diam}(Y) + 2\eta. \end{align*} Taking the supremum over $x_1,x_2 \in X$ gives $\operatorname{diam}(X) \leq \operatorname{diam}(Y) + 2\eta$. Reversing the roles of $X$ and $Y$ gives \begin{align*} |\operatorname{diam}(X)-\operatorname{diam}(Y)| \leq 2\eta. \end{align*} Thus $\operatorname{diam}: \overline{\mathcal{C}} \to [0,\infty)$ is continuous, so compactness of $\overline{\mathcal{C}}$ gives \begin{align*} D := \sup_{X \in \overline{\mathcal{C}}} \operatorname{diam}(X) < \infty. \end{align*} In particular, $\operatorname{diam}(X) \leq D$ for every $X \in \mathcal{C}$. Now fix $\varepsilon > 0$. Since $\overline{\mathcal{C}}$ is compact, there exist compact metric spaces \begin{align*} Y_1,\dots,Y_m \in \overline{\mathcal{C}} \end{align*} such that \begin{align*} \overline{\mathcal{C}} \subset \bigcup_{i=1}^m B_{GH}(Y_i,\varepsilon/4). \end{align*} For each $i \in \{1,\dots,m\}$, compactness of $Y_i$ gives a finite $\varepsilon/2$-net $B_i \subset Y_i$. Define \begin{align*} N(\varepsilon) := \max_{1 \leq i \leq m} |B_i|. \end{align*} Let $X \in \mathcal{C}$. Choose $i$ such that $d_{GH}(X,Y_i) < \varepsilon/4$. By the definition of $d_{GH}$, embed $X$ and $Y_i$ isometrically into a metric space $(Z,d_Z)$ so that their Hausdorff distance in $Z$ is less than $\varepsilon/4$. For every $b \in B_i$, choose $a_b \in X$ such that \begin{align*} d_Z(a_b,b) < \varepsilon/4. \end{align*} Let \begin{align*} A := \{a_b : b \in B_i\} \subset X. \end{align*} Then $|A| \leq |B_i| \leq N(\varepsilon)$. If $x \in X$, choose $y \in Y_i$ with $d_Z(x,y) < \varepsilon/4$, and then choose $b \in B_i$ with $d_{Y_i}(y,b) < \varepsilon/2$. Since the embedding of $Y_i$ is isometric, \begin{align*} d_X(x,a_b) &= d_Z(x,a_b) \\ &\leq d_Z(x,y) + d_Z(y,b) + d_Z(b,a_b) \\ &< \varepsilon/4 + \varepsilon/2 + \varepsilon/4 \\ &= \varepsilon. \end{align*} Thus $A$ is an $\varepsilon$-net in $X$ with at most $N(\varepsilon)$ points. This proves the uniform covering bound.[/step]

[guided]Assume that $\mathcal{C}$ is precompact, and write $\overline{\mathcal{C}}$ for its closure in the Gromov-Hausdorff metric $d_{GH}$. Here $d_{GH}$ denotes the Gromov-Hausdorff metric on isometry classes of compact metric spaces. For a compact metric space $Y$ and a number $r > 0$, define \begin{align*} B_{GH}(Y,r) := \{X : X \text{ is a compact metric space and } d_{GH}(X,Y) < r\}. \end{align*} Precompactness means precisely that $\overline{\mathcal{C}}$ is compact. We first prove the diameter bound. The point is that spaces close in Gromov-Hausdorff distance have almost the same diameter. Let $X$ and $Y$ be compact metric spaces with $d_{GH}(X,Y) < \eta$. By the definition of the Gromov-Hausdorff metric, we may place both spaces isometrically into a common metric space $(Z,d_Z)$ by maps \begin{align*} \iota_X: X &\to Z, & \iota_Y: Y &\to Z \end{align*} so that every point of $\iota_X(X)$ lies within $\eta$ of $\iota_Y(Y)$ and every point of $\iota_Y(Y)$ lies within $\eta$ of $\iota_X(X)$. Take arbitrary points $x_1,x_2 \in X$. Choose $y_1,y_2 \in Y$ such that \begin{align*} d_Z(\iota_X(x_i),\iota_Y(y_i)) < \eta \end{align*} for $i \in \{1,2\}$. The triangle inequality in $Z$ gives \begin{align*} d_X(x_1,x_2) &= d_Z(\iota_X(x_1),\iota_X(x_2)) \\ &\leq d_Z(\iota_X(x_1),\iota_Y(y_1)) + d_Z(\iota_Y(y_1),\iota_Y(y_2)) + d_Z(\iota_Y(y_2),\iota_X(x_2)) \\ &< \eta + \operatorname{diam}(Y) + \eta \\ &= \operatorname{diam}(Y) + 2\eta. \end{align*} Taking the supremum over all $x_1,x_2 \in X$ gives \begin{align*} \operatorname{diam}(X) \leq \operatorname{diam}(Y) + 2\eta. \end{align*} The same argument with $X$ and $Y$ interchanged gives \begin{align*} \operatorname{diam}(Y) \leq \operatorname{diam}(X) + 2\eta. \end{align*} Therefore \begin{align*} |\operatorname{diam}(X)-\operatorname{diam}(Y)| \leq 2\eta. \end{align*} This proves continuity of the diameter function on the Gromov-Hausdorff closure. Since a continuous real-valued function on a [compact space](/page/Compact%20Space) is bounded, the number \begin{align*} D := \sup_{X \in \overline{\mathcal{C}}} \operatorname{diam}(X) \end{align*} is finite. Hence $\operatorname{diam}(X) \leq D$ for every $X \in \mathcal{C}$. Now fix $\varepsilon > 0$. We must produce one integer $N(\varepsilon)$ that works for every $X \in \mathcal{C}$. Compactness of $\overline{\mathcal{C}}$ supplies finitely many Gromov-Hausdorff balls of radius $\varepsilon/4$: \begin{align*} \overline{\mathcal{C}} \subset \bigcup_{i=1}^m B_{GH}(Y_i,\varepsilon/4), \end{align*} where $Y_1,\dots,Y_m \in \overline{\mathcal{C}}$. Each $Y_i$ is compact, so it has a finite $\varepsilon/2$-net $B_i \subset Y_i$. Define \begin{align*} N(\varepsilon) := \max_{1 \leq i \leq m} |B_i|. \end{align*} Let $X \in \mathcal{C}$. Choose $i$ with $d_{GH}(X,Y_i) < \varepsilon/4$. Put $X$ and $Y_i$ into a common metric space $(Z,d_Z)$ so that their Hausdorff distance is less than $\varepsilon/4$. For each net point $b \in B_i$, choose a point $a_b \in X$ satisfying \begin{align*} d_Z(a_b,b) < \varepsilon/4. \end{align*} Set \begin{align*} A := \{a_b : b \in B_i\} \subset X. \end{align*} Then $|A| \leq |B_i| \leq N(\varepsilon)$. We verify that $A$ is an $\varepsilon$-net in $X$. Given $x \in X$, choose $y \in Y_i$ with $d_Z(x,y) < \varepsilon/4$, using Hausdorff closeness. Then choose $b \in B_i$ with $d_{Y_i}(y,b) < \varepsilon/2$, using the fact that $B_i$ is an $\varepsilon/2$-net. Since the embedding of $Y_i$ into $Z$ is isometric, $d_Z(y,b)=d_{Y_i}(y,b)$. Therefore \begin{align*} d_X(x,a_b) &= d_Z(x,a_b) \\ &\leq d_Z(x,y) + d_Z(y,b) + d_Z(b,a_b) \\ &< \varepsilon/4 + \varepsilon/2 + \varepsilon/4 \\ &= \varepsilon. \end{align*} Thus every point of $X$ lies within $\varepsilon$ of some point of $A$, and $A$ has at most $N(\varepsilon)$ points. This proves the uniform covering bound.[/guided]

[step:Choose nested finite nets for an arbitrary sequence] Assume conversely that the uniform diameter and covering bounds hold. Let $(X_j,d_j)_{j=1}^{\infty}$ be a sequence in $\mathcal{C}$. We prove that it has a Gromov-Hausdorff convergent subsequence. For each $k \in \mathbb{N}$, set \begin{align*} \varepsilon_k := 2^{-k}. \end{align*} By the covering hypothesis, choose $M_k \in \mathbb{N}$ such that every $X \in \mathcal{C}$ admits an $\varepsilon_k$-net with at most $M_k$ points. Replacing $M_k$ by the nondecreasing sequence \begin{align*} \widetilde{M}_k := \sum_{\ell=1}^k M_\ell, \end{align*} we may construct, for each $j$, finite subsets \begin{align*} A_{j,k} = \{x_{j,1},\dots,x_{j,\widetilde{M}_k}\} \subset X_j \end{align*} such that $A_{j,k} \subset A_{j,k+1}$ and $A_{j,k}$ is an $\varepsilon_k$-net in $X_j$. Repeated labels are allowed: if fewer than $\widetilde{M}_k$ distinct points are needed, we repeat already chosen points so that the index set is exactly $\{1,\dots,\widetilde{M}_k\}$. The diameter bound gives \begin{align*} 0 \leq d_j(x_{j,a},x_{j,b}) \leq D \end{align*} for all $j$, all $k$, and all $a,b \in \{1,\dots,\widetilde{M}_k\}$. [/step]

[step:Pass to a diagonal subsequence of all finite distance matrices]For each fixed $k \in \mathbb{N}$, the distance matrix \begin{align*} \left(d_j(x_{j,a},x_{j,b})\right)_{1 \leq a,b \leq \widetilde{M}_k} \end{align*} lies in the compact cube $[0,D]^{\widetilde{M}_k^2}$. Hence, by compactness of this finite-dimensional cube, the sequence has a subsequence along which this matrix converges. Apply this successively for $k=1,2,\dots$ and take the diagonal subsequence. Relabel it as $(X_j,d_j)$ again. Then, for every pair $a,b \in \mathbb{N}$, the limit \begin{align*} \rho(a,b) := \lim_{j \to \infty} d_j(x_{j,a},x_{j,b}) \end{align*} exists whenever $a$ and $b$ both occur among the labels of some $A_{j,k}$. The function \begin{align*} \rho: \mathbb{N} \times \mathbb{N} &\to [0,D] \end{align*} is a pseudometric. Symmetry and $\rho(a,a)=0$ follow by taking limits from the corresponding properties of $d_j$. For the triangle inequality, fix $a,b,c \in \mathbb{N}$. For all sufficiently large $k$, the labels $a,b,c$ occur in $A_{j,k}$, and for every $j$, \begin{align*} d_j(x_{j,a},x_{j,c}) \leq d_j(x_{j,a},x_{j,b}) + d_j(x_{j,b},x_{j,c}). \end{align*} Taking $j \to \infty$ gives \begin{align*} \rho(a,c) \leq \rho(a,b) + \rho(b,c). \end{align*} Thus $\rho$ is a pseudometric.[/step]

[guided]The goal of this step is to make all finite distance data converge at once. For a fixed scale $k \in \mathbb{N}$, the labelled finite net $A_{j,k}$ has exactly $\widetilde{M}_k$ labels, so its distance data form the matrix \begin{align*} \left(d_j(x_{j,a},x_{j,b})\right)_{1 \leq a,b \leq \widetilde{M}_k}. \end{align*} Each entry lies in $[0,D]$ by the uniform diameter bound, hence the whole matrix lies in the compact finite-dimensional cube $[0,D]^{\widetilde{M}_k^2}$. Compactness of this cube gives a subsequence along which the $k$-th matrix converges. We need convergence for every $k$ simultaneously, not just for one fixed $k$. Apply the preceding compactness argument successively for $k=1,2,3,\dots$: first choose a subsequence on which the $k=1$ matrices converge, then a further subsequence on which the $k=2$ matrices converge, and continue. Taking the diagonal subsequence preserves convergence for each fixed $k$, because after stage $k$ all later diagonal terms lie in the subsequence selected for that $k$. Relabel this diagonal subsequence as $(X_j,d_j)$ again. For labels $a,b \in \mathbb{N}$, choose $k \in \mathbb{N}$ so large that both $a$ and $b$ occur among the labels of $A_{j,k}$. Define \begin{align*} \rho(a,b) := \lim_{j \to \infty} d_j(x_{j,a},x_{j,b}). \end{align*} This limit exists by the diagonal construction. Moreover $0 \leq \rho(a,b) \leq D$ because each term in the convergent sequence lies in $[0,D]$. We now verify that \begin{align*} \rho: \mathbb{N} \times \mathbb{N} &\to [0,D] \end{align*} is a pseudometric. Since each $d_j$ is a metric, $d_j(x_{j,a},x_{j,a})=0$ for every $j$, and taking limits gives $\rho(a,a)=0$. Similarly, $d_j(x_{j,a},x_{j,b})=d_j(x_{j,b},x_{j,a})$ for every $j$, so $\rho(a,b)=\rho(b,a)$. It remains to check the triangle inequality. Fix $a,b,c \in \mathbb{N}$, and choose $k \in \mathbb{N}$ so large that all three labels occur in $A_{j,k}$. For every $j$, the metric triangle inequality in $X_j$ gives \begin{align*} d_j(x_{j,a},x_{j,c}) \leq d_j(x_{j,a},x_{j,b}) + d_j(x_{j,b},x_{j,c}). \end{align*} Taking limits as $j \to \infty$ preserves this inequality because all three distance sequences converge, so \begin{align*} \rho(a,c) \leq \rho(a,b) + \rho(b,c). \end{align*} Thus $\rho$ is a pseudometric on $\mathbb{N}$.[/guided]

[step:Complete the limiting pseudometric space and prove compactness]Define an [equivalence relation](/page/Equivalence%20Relation) $\sim$ on $\mathbb{N}$ by \begin{align*} a \sim b \quad \Longleftrightarrow \quad \rho(a,b)=0. \end{align*} Let \begin{align*} Q := \mathbb{N}/\sim \end{align*} be the quotient set, and define \begin{align*} d_Q: Q \times Q &\to [0,D] \\ ([a],[b]) &\mapsto \rho(a,b). \end{align*} The pseudometric properties of $\rho$ imply that $d_Q$ is a well-defined metric. Let $(Y,d_Y)$ be the metric completion of $(Q,d_Q)$. We show that $Y$ is compact. It suffices to show that $Y$ is complete and totally bounded. Completeness holds by construction. To prove [total boundedness](/page/Total%20Boundedness), fix $\varepsilon > 0$ and choose $k \in \mathbb{N}$ with $\varepsilon_k < \varepsilon/2$. We claim that \begin{align*} Q_k := \{[1],\dots,[\widetilde{M}_k]\} \subset Q \end{align*} is an $\varepsilon$-net in $Q$. Let $a \in \mathbb{N}$. Choose $\ell \geq k$ such that $a \leq \widetilde{M}_\ell$. Since $A_{j,k}$ is an $\varepsilon_k$-net in $X_j$ and $A_{j,k} \subset A_{j,\ell}$, for each $j$ there exists $b_j \in \{1,\dots,\widetilde{M}_k\}$ such that \begin{align*} d_j(x_{j,a},x_{j,b_j}) < \varepsilon_k. \end{align*} The set $\{1,\dots,\widetilde{M}_k\}$ is finite, so some $b \in \{1,\dots,\widetilde{M}_k\}$ occurs as $b_j$ along an infinite subsequence. Passing to that subsequence and taking limits gives \begin{align*} d_Q([a],[b]) = \rho(a,b) \leq \varepsilon_k < \varepsilon/2 < \varepsilon. \end{align*} Therefore $Q_k$ is an $\varepsilon$-net in $Q$. Since $Q$ is dense in $Y$, the same finite set is an $\varepsilon$-net in $Y$ after slightly increasing the radius, and because $\varepsilon > 0$ was arbitrary, $Y$ is totally bounded. Hence $Y$ is compact.[/step]

[guided]We now build the candidate Gromov-Hausdorff limit. The limiting distance function $\rho$ may assign distance zero to two different labels, so it is a pseudometric rather than necessarily a metric. We remove this defect by identifying labels at zero distance. Define $a \sim b$ exactly when $\rho(a,b)=0$, and let \begin{align*} Q := \mathbb{N}/\sim. \end{align*} For equivalence classes $[a],[b] \in Q$, define \begin{align*} d_Q([a],[b]) := \rho(a,b). \end{align*} This is well-defined: if $a \sim a'$ and $b \sim b'$, then the triangle inequality for $\rho$ gives \begin{align*} \rho(a,b) &\leq \rho(a,a') + \rho(a',b') + \rho(b',b) \\ &= \rho(a',b'), \end{align*} and the same argument with $(a,b)$ and $(a',b')$ interchanged gives equality. The function $d_Q$ is a metric because the only remaining zero distances occur between identical equivalence classes. Let $(Y,d_Y)$ be the metric completion of $(Q,d_Q)$. Completeness is built into the definition of completion, so compactness will follow once we prove total boundedness. Fix $\varepsilon > 0$. Choose $k \in \mathbb{N}$ such that \begin{align*} \varepsilon_k = 2^{-k} < \varepsilon/2. \end{align*} We claim that the finite set \begin{align*} Q_k := \{[1],\dots,[\widetilde{M}_k]\} \end{align*} is an $\varepsilon$-net in $Q$. Take any class $[a] \in Q$. Choose $\ell \geq k$ with $a \leq \widetilde{M}_\ell$, so the label $a$ appears in the finite net $A_{j,\ell}$ for every $j$. Since $A_{j,k}$ is an $\varepsilon_k$-net in $X_j$, for every $j$ there exists a label \begin{align*} b_j \in \{1,\dots,\widetilde{M}_k\} \end{align*} such that \begin{align*} d_j(x_{j,a},x_{j,b_j}) < \varepsilon_k. \end{align*} There are only finitely many possible values of $b_j$, so at least one value $b \in \{1,\dots,\widetilde{M}_k\}$ occurs for infinitely many $j$. Restricting to that infinite subsequence and using the already established convergence of distances, we get \begin{align*} d_Q([a],[b]) = \rho(a,b) = \lim d_j(x_{j,a},x_{j,b}) \leq \varepsilon_k < \varepsilon. \end{align*} Thus every point of $Q$ lies within $\varepsilon$ of the finite set $Q_k$. Because $Q$ is dense in its completion $Y$, total boundedness passes from $Q$ to $Y$: given any $y \in Y$, choose $q \in Q$ with $d_Y(y,q) < \varepsilon/2$, then choose $q_k \in Q_k$ with $d_Y(q,q_k) < \varepsilon/2$, and obtain \begin{align*} d_Y(y,q_k) \leq d_Y(y,q) + d_Y(q,q_k) < \varepsilon. \end{align*} Thus $Y$ is totally bounded. By the metric-space compactness criterion that a [complete metric space](/page/Complete%20Metric%20Space) is [compact](/page/Compactness) if and only if it is [totally bounded](/page/Totally%20Bounded%20Space), $Y$ is compact.[/guided]

[step:Compare each original space with the limiting finite model]For each $k \in \mathbb{N}$, define \begin{align*} r_k := \varepsilon_k \quad\text{and}\quad s_k := k+2. \end{align*} Then $\varepsilon_{s_k} < r_k/2$. By the total boundedness argument from the preceding step, the finite set \begin{align*} Q_{s_k} := \{[1],\dots,[\widetilde{M}_{s_k}]\} \subset Y \end{align*} is an $r_k$-net in $Y$. Also $A_{j,s_k}$ is an $\varepsilon_{s_k}$-net in $X_j$, hence an $r_k$-net in $X_j$. For fixed $k$, the convergence of the finite distance matrices gives \begin{align*} \delta_{j,k} := \max_{1 \leq a,b \leq \widetilde{M}_{s_k}} \left|d_j(x_{j,a},x_{j,b}) - d_Y([a],[b])\right| \longrightarrow 0 \end{align*} as $j \to \infty$. Define a relation \begin{align*} R_{j,k} \subset X_j \times Y \end{align*} by declaring $(x,y) \in R_{j,k}$ if and only if there exists $a \in \{1,\dots,\widetilde{M}_{s_k}\}$ such that \begin{align*} d_j(x,x_{j,a}) < r_k \quad\text{and}\quad d_Y(y,[a]) < r_k. \end{align*} This relation is a correspondence. Indeed, if $x \in X_j$, the fact that $A_{j,s_k}$ is an $r_k$-net gives $a \in \{1,\dots,\widetilde{M}_{s_k}\}$ with $d_j(x,x_{j,a}) < r_k$, and then $(x,[a]) \in R_{j,k}$. If $y \in Y$, the fact that $Q_{s_k}$ is an $r_k$-net gives $a \in \{1,\dots,\widetilde{M}_{s_k}\}$ with $d_Y(y,[a]) < r_k$, and then $(x_{j,a},y) \in R_{j,k}$. If $(x,y),(x',y') \in R_{j,k}$ are witnessed by labels $a,b \in \{1,\dots,\widetilde{M}_{s_k}\}$, then the triangle inequality in $X_j$ and $Y$ gives \begin{align*} \left|d_j(x,x') - d_Y(y,y')\right| &\leq d_j(x,x_{j,a}) + \left|d_j(x_{j,a},x_{j,b}) - d_Y([a],[b])\right| \\ &\quad + d_Y([a],y) + d_j(x',x_{j,b}) + d_Y([b],y') \\ &< 4r_k + \delta_{j,k}. \end{align*} Define the distortion of a correspondence $R \subset X_j \times Y$ by \begin{align*} \operatorname{dis}(R) := \sup\left\{\left|d_j(u,u') - d_Y(v,v')\right| : (u,v),(u',v') \in R\right\}. \end{align*} The preceding estimate shows that \begin{align*} \operatorname{dis}(R_{j,k}) \leq 4r_k+\delta_{j,k}. \end{align*} By the [correspondence-distortion characterization of the Gromov-Hausdorff metric](/page/Gromov-Hausdorff%20Metric), which states that \begin{align*} d_{GH}(X_j,Y) \leq \frac{1}{2}\operatorname{dis}(R) \end{align*} for every correspondence $R \subset X_j \times Y$, we obtain \begin{align*} d_{GH}(X_j,Y) \leq \frac{1}{2}\left(4r_k+\delta_{j,k}\right) = \frac{1}{2}\left(4\varepsilon_k+\delta_{j,k}\right). \end{align*}[/step]

[guided]We compare $X_j$ with the candidate limit $Y$ through finite labelled models. For each $k \in \mathbb{N}$, define \begin{align*} r_k := \varepsilon_k \quad\text{and}\quad s_k := k+2. \end{align*} Then $\varepsilon_{s_k}<r_k/2$. The finite set \begin{align*} Q_{s_k}:=\{[1],\dots,[\widetilde{M}_{s_k}]\}\subset Y \end{align*} is an $r_k$-net in $Y$ by the total boundedness argument, and $A_{j,s_k}$ is an $\varepsilon_{s_k}$-net in $X_j$, hence also an $r_k$-net in $X_j$. For this fixed $k$, define \begin{align*} \delta_{j,k} := \max_{1 \leq a,b \leq \widetilde{M}_{s_k}} \left|d_j(x_{j,a},x_{j,b}) - d_Y([a],[b])\right|. \end{align*} The finite distance matrices converge along the chosen subsequence, so $\delta_{j,k}\to 0$ as $j\to\infty$. Define a relation \begin{align*} R_{j,k}\subset X_j\times Y \end{align*} by declaring $(x,y)\in R_{j,k}$ exactly when there is a label $a\in\{1,\dots,\widetilde{M}_{s_k}\}$ such that \begin{align*} d_j(x,x_{j,a})<r_k \quad\text{and}\quad d_Y(y,[a])<r_k. \end{align*} This relation is a correspondence. If $x\in X_j$, the $r_k$-net property of $A_{j,s_k}$ gives a label $a$ with $d_j(x,x_{j,a})<r_k$, and then $(x,[a])\in R_{j,k}$. If $y\in Y$, the $r_k$-net property of $Q_{s_k}$ gives a label $a$ with $d_Y(y,[a])<r_k$, and then $(x_{j,a},y)\in R_{j,k}$. Now take two related pairs $(x,y),(x',y')\in R_{j,k}$, witnessed by labels $a,b\in\{1,\dots,\widetilde{M}_{s_k}\}$. The triangle inequality in $X_j$ and $Y$ gives \begin{align*} \left|d_j(x,x') - d_Y(y,y')\right| &\leq d_j(x,x_{j,a}) + \left|d_j(x_{j,a},x_{j,b}) - d_Y([a],[b])\right| \\ &\quad + d_Y([a],y) + d_j(x',x_{j,b}) + d_Y([b],y') \\ &< 4r_k + \delta_{j,k}. \end{align*} Define the distortion of a correspondence $R \subset X_j \times Y$ by \begin{align*} \operatorname{dis}(R) := \sup\left\{\left|d_j(u,u') - d_Y(v,v')\right| : (u,v),(u',v') \in R\right\}. \end{align*} The preceding inequality for arbitrary pairs in $R_{j,k}$ proves \begin{align*} \operatorname{dis}(R_{j,k}) \leq 4r_k+\delta_{j,k}. \end{align*} By the [correspondence-distortion characterization of the Gromov-Hausdorff metric](/page/Gromov-Hausdorff%20Metric), every correspondence $R\subset X_j\times Y$ satisfies \begin{align*} d_{GH}(X_j,Y)\leq \frac{1}{2}\operatorname{dis}(R). \end{align*} Applying this to $R_{j,k}$ yields \begin{align*} d_{GH}(X_j,Y) \leq \frac{1}{2}\left(4r_k+\delta_{j,k}\right) = \frac{1}{2}\left(4\varepsilon_k+\delta_{j,k}\right). \end{align*}[/guided]

[step:Conclude sequential precompactness and hence precompactness] Choose an increasing sequence of indices $j_k$ so large that \begin{align*} \delta_{j_k,k} < \varepsilon_k. \end{align*} The preceding estimate gives \begin{align*} d_{GH}(X_{j_k},Y) \leq \frac{1}{2}\left(4\varepsilon_k+\delta_{j_k,k}\right) < \frac{5}{2}\varepsilon_k. \end{align*} Since $\varepsilon_k = 2^{-k} \to 0$, we have \begin{align*} d_{GH}(X_{j_k},Y) \to 0. \end{align*} Thus every sequence in $\mathcal{C}$ has a Gromov-Hausdorff convergent subsequence with compact limit. In a [metric space](/page/Metric%20Space), precompactness is equivalent to every sequence having a Cauchy subsequence, equivalently a subsequence converging in the completion. Applying this criterion to the metric space of isometry classes of compact metric spaces with the Gromov-Hausdorff metric shows that $\mathcal{C}$ is precompact in the Gromov-Hausdorff topology. Combining this with the first direction proves the theorem. [/step]