[proofplan] The proof establishes the equivalence of three characterisations of compactness in metric spaces by a cyclic chain of implications: (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (1). Compactness implies sequential compactness via a covering argument: if a sequence has no convergent subsequence, no point is a limit point, so small balls around each point contain only finitely many terms, and a finite subcover contradicts the infinitude of the sequence. Sequential compactness implies completeness (Cauchy sequences have convergent subsequences) and total boundedness (a greedy construction produces a sequence with no convergent subsequence if a finite net fails). Finally, completeness plus total boundedness implies compactness via the Lebesgue number lemma and a finite net. [/proofplan] [step:Prove compact implies sequentially compact by contradiction with the open-cover property] Let $(x_n)_{n=1}^{\infty}$ be a [sequence](/page/Sequence) in $M$. Suppose for contradiction that $(x_n)$ has no convergent subsequence. Then no point $p \in M$ is a [limit](/page/Limit) point of the sequence: for each $p$, there exists $r_p > 0$ such that $B(p, r_p)$ contains only finitely many terms of $(x_n)$. (If every ball around $p$ contained infinitely many terms, we could extract a subsequence converging to $p$ by choosing terms in $B(p, 1/k)$ for $k = 1, 2, \ldots$.) The collection $\{B(p, r_p)\}_{p \in M}$ is an open cover of $M$. By compactness, finitely many suffice: \begin{align*} M = B(p_1, r_{p_1}) \cup \cdots \cup B(p_N, r_{p_N}). \end{align*} Each ball contains only finitely many terms of $(x_n)$, so $M$ contains only finitely many terms --- but $(x_n)$ has infinitely many indices $n = 1, 2, 3, \ldots$, each placing $x_n$ in some ball. This is a contradiction. [guided] The argument has a clean structure: assume no convergent subsequence exists, then use compactness to derive a contradiction. **No cluster points:** If $(x_n)$ has no convergent subsequence, then no point $p \in M$ is a cluster point. A cluster point would satisfy: for every $\varepsilon > 0$, infinitely many indices $n$ have $d(x_n, p) < \varepsilon$, from which we could extract a convergent subsequence. The negation gives: for each $p \in M$, there exists $r_p > 0$ with $B(p, r_p)$ containing only finitely many terms (finitely many indices $n$ with $x_n \in B(p, r_p)$). **Open cover:** The collection $\{B(p, r_p)\}_{p \in M}$ is an open cover of $M$. **Compactness gives finite subcover:** By compactness, $M = B(p_1, r_{p_1}) \cup \cdots \cup B(p_N, r_{p_N})$ for finitely many $p_1, \ldots, p_N$. **Contradiction:** Each ball contains finitely many terms of the sequence. The finite union of finitely many finite sets is finite, so only finitely many indices $n$ have $x_n \in M$. But $x_n \in M$ for every $n \in \mathbb{N}$ --- the sequence has infinitely many terms. This is a contradiction. Note: "finitely many terms" counts indices, not values. Even if $x_n = x_m$ for $n \neq m$, the indices $n$ and $m$ are distinct. [/guided] [/step] [step:Prove sequentially compact implies complete by extracting a convergent subsequence from a Cauchy sequence] Let $(x_n)$ be a [Cauchy sequence](/page/Cauchy%20Sequence) in $M$. By sequential compactness, $(x_n)$ has a subsequence $(x_{n_k})$ converging to some $x \in M$. We show $x_n \to x$. Fix $\varepsilon > 0$. Choose $N_1$ with $d(x_n, x_m) < \varepsilon/2$ for all $n, m \geq N_1$ (Cauchy property). Choose $K$ with $d(x_{n_k}, x) < \varepsilon/2$ for all $k \geq K$. Pick $k \geq K$ with $n_k \geq N_1$. For all $n \geq N_1$: \begin{align*} d(x_n, x) \leq d(x_n, x_{n_k}) + d(x_{n_k}, x) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $x_n \to x$, and $M$ is complete. [/step] [step:Prove sequentially compact implies totally bounded by a greedy construction] Suppose $M$ is not totally bounded. Then there exists $\varepsilon > 0$ such that no finite collection of balls of radius $\varepsilon$ covers $M$. Construct a sequence inductively: choose $x_1 \in M$ arbitrarily. Given $x_1, \ldots, x_n$, the balls $B(x_1, \varepsilon), \ldots, B(x_n, \varepsilon)$ do not cover $M$, so choose $x_{n+1} \in M \setminus \bigcup_{k=1}^{n} B(x_k, \varepsilon)$. Then $d(x_{n+1}, x_k) \geq \varepsilon$ for all $k \leq n$. The resulting sequence satisfies $d(x_n, x_m) \geq \varepsilon$ for all $n \neq m$. No subsequence can be Cauchy (all pairwise distances are at least $\varepsilon$), so no subsequence can converge. This contradicts sequential compactness. [guided] The greedy construction builds a sequence in which every pair of distinct terms is at least $\varepsilon$ apart. This is possible precisely because total boundedness fails. **Construction:** At each stage, the finitely many balls $B(x_1, \varepsilon), \ldots, B(x_n, \varepsilon)$ do not cover $M$ (by assumption that $M$ is not totally bounded). So there exists a point not covered; pick it as $x_{n+1}$. By construction, $d(x_{n+1}, x_k) \geq \varepsilon$ for all $k \leq n$. **No convergent subsequence:** Suppose for contradiction that $(x_{n_k})$ converges to some $x \in M$. Then for large enough $k$, $d(x_{n_k}, x) < \varepsilon/2$. For two such indices $k \neq \ell$, the triangle inequality gives $d(x_{n_k}, x_{n_\ell}) \leq d(x_{n_k}, x) + d(x, x_{n_\ell}) < \varepsilon/2 + \varepsilon/2 = \varepsilon$. But $d(x_{n_k}, x_{n_\ell}) \geq \varepsilon$ by construction --- a contradiction. **Contrapositive:** If $M$ is sequentially compact, then every sequence has a convergent subsequence. The greedy sequence would have one, contradicting the $\varepsilon$-separation. Therefore the assumption that $M$ is not totally bounded was false, and $M$ must be totally bounded. [/guided] [/step] [step:Establish the Lebesgue number lemma for sequentially compact metric spaces] [claim:Lebesgue Number Lemma] Let $(M, d)$ be a sequentially compact [metric space](/page/Metric%20Space) and $\mathcal{U}$ an open cover of $M$. There exists $\delta > 0$ such that every subset of $M$ with diameter less than $\delta$ is contained in some member of $\mathcal{U}$. [/claim] [proof] Suppose no such $\delta$ exists. Then for each $n \in \mathbb{N}$, there exists $A_n \subseteq M$ with $\operatorname{diam}(A_n) < 1/n$ not contained in any member of $\mathcal{U}$. Choose $x_n \in A_n$. By sequential compactness, $(x_n)$ has a subsequence $x_{n_k} \to x$ for some $x \in M$. Since $\mathcal{U}$ covers $M$, choose $U \in \mathcal{U}$ with $x \in U$. Since $U$ is open, $B(x, r) \subseteq U$ for some $r > 0$. Choose $K$ large enough that $d(x_{n_K}, x) < r/2$ and $1/n_K < r/2$. For any $y \in A_{n_K}$: \begin{align*} d(y, x) \leq d(y, x_{n_K}) + d(x_{n_K}, x) < \operatorname{diam}(A_{n_K}) + \frac{r}{2} < \frac{1}{n_K} + \frac{r}{2} < r. \end{align*} Therefore $A_{n_K} \subseteq B(x, r) \subseteq U$, contradicting the choice of $A_{n_K}$. [/proof] [guided] The Lebesgue number lemma converts an open cover into a uniform scale: there exists a single $\delta > 0$ such that any set of diameter $< \delta$ fits inside some cover element. This uniformity bridges the gap between a finite net and a finite subcover. **Proof by contradiction:** Suppose no Lebesgue number exists. For each $n \in \mathbb{N}$, there is a set $A_n \subseteq M$ with $\operatorname{diam}(A_n) < 1/n$ that is not contained in any member of $\mathcal{U}$. Choose $x_n \in A_n$. **Extracting a convergent subsequence:** By sequential compactness, $(x_n)$ has a subsequence $x_{n_k} \to x$ for some $x \in M$. Since $\mathcal{U}$ covers $M$, choose $U \in \mathcal{U}$ with $x \in U$. Since $U$ is open, $B(x, r) \subseteq U$ for some $r > 0$. **The triangle inequality delivers the contradiction:** Choose $K$ large enough that $d(x_{n_K}, x) < r/2$ and $1/n_K < r/2$. For any $y \in A_{n_K}$, the triangle inequality gives $d(y, x) \leq d(y, x_{n_K}) + d(x_{n_K}, x) < \operatorname{diam}(A_{n_K}) + r/2 < 1/n_K + r/2 < r$. Therefore $A_{n_K} \subseteq B(x, r) \subseteq U$, contradicting the choice of $A_{n_K}$ as not fitting in any cover element. [/guided] [/step] [step:Prove complete and totally bounded implies sequentially compact via a diagonal argument] Let $(x_n)$ be a sequence in $M$. Since $M$ is totally bounded, for each $k \in \mathbb{N}$, $M$ is covered by finitely many balls of radius $1/k$. Construct subsequences inductively: - $k = 1$: Cover $M$ by finitely many balls of radius $1$. At least one ball contains infinitely many terms of $(x_n)$; restrict to this subsequence, call it $(x_n^{(1)})$. - $k = 2$: Cover $M$ by finitely many balls of radius $1/2$. At least one contains infinitely many terms of $(x_n^{(1)})$; restrict to $(x_n^{(2)})$. - Continue: at stage $k$, $(x_n^{(k)})$ is a subsequence of $(x_n^{(k-1)})$ lying in a ball of radius $1/k$. The diagonal subsequence $y_n = x_n^{(n)}$ is Cauchy: for $n, m \geq k$, both $y_n$ and $y_m$ belong to (a subsequence of) a sequence lying in a ball of radius $1/k$, so $d(y_n, y_m) \leq 2/k$. Since $k$ is arbitrary, $(y_n)$ is Cauchy. By completeness, $(y_n)$ converges. Therefore $M$ is sequentially compact. [/step] [step:Prove complete and totally bounded implies compact using the Lebesgue number and a finite net] Let $\mathcal{U}$ be an open cover of $M$. By the previous step, $M$ is sequentially compact (since it is complete and totally bounded). By the Lebesgue number lemma, there exists $\delta > 0$ such that every subset of $M$ with diameter less than $\delta$ is contained in some member of $\mathcal{U}$. By total boundedness, there exists a finite $(\delta/2)$-net: points $y_1, \ldots, y_N \in M$ with \begin{align*} M = B(y_1, \delta/2) \cup \cdots \cup B(y_N, \delta/2). \end{align*} Each ball $B(y_i, \delta/2)$ has diameter at most $\delta$ (for any $a, b \in B(y_i, \delta/2)$, $d(a, b) \leq d(a, y_i) + d(y_i, b) < \delta$). By the Lebesgue number property, each $B(y_i, \delta/2)$ is contained in some $U_i \in \mathcal{U}$. Then $\{U_1, \ldots, U_N\}$ is a finite subcover of $\mathcal{U}$. [guided] This final step combines the Lebesgue number lemma with total boundedness to produce a finite subcover. **Lebesgue number:** The lemma provides $\delta > 0$ such that any subset of $M$ with diameter $< \delta$ is contained in some $U \in \mathcal{U}$. **Finite net:** Total boundedness provides points $y_1, \ldots, y_N \in M$ with $M = B(y_1, \delta/2) \cup \cdots \cup B(y_N, \delta/2)$. **Diameter check:** Each ball $B(y_i, \delta/2)$ has diameter at most $\delta$: for any $a, b \in B(y_i, \delta/2)$, the triangle inequality gives $d(a, b) \leq d(a, y_i) + d(y_i, b) < \delta/2 + \delta/2 = \delta$. **Assembling the subcover:** By the Lebesgue number property, each $B(y_i, \delta/2)$ fits inside some $U_i \in \mathcal{U}$. Then $\{U_1, \ldots, U_N\}$ is a finite subcover. The full logic chain is: complete + totally bounded $\Rightarrow$ sequentially compact (diagonal argument) $\Rightarrow$ Lebesgue number exists $\Rightarrow$ finite $(\delta/2)$-net gives finite subcover $\Rightarrow$ compact. [/guided] [/step]