[proofplan] We prove the cycle of implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow (1)$. The forward direction $(1) \Rightarrow (2)$ uses the Heine-Borel theorem. The implications $(2) \Rightarrow (3) \Rightarrow (4)$ and $(4) \Rightarrow (5)$ are general facts about compact and totally bounded sets. The critical implication $(5) \Rightarrow (1)$ uses the contrapositive: if $\dim X = \infty$, then [Riesz's Lemma](/theorems/1222) produces a bounded sequence with no convergent subsequence. [/proofplan]

[step:$(1) \Rightarrow (2)$: The closed unit ball of a finite-dimensional space is compact]Assume $\dim X = d < \infty$. Fix a basis $\{e_1, \ldots, e_d\}$ for $X$ and consider the coordinate isomorphism \begin{align*} T: \mathbb{K}^d &\to X \\ (\alpha_1, \ldots, \alpha_d) &\mapsto \sum_{i=1}^d \alpha_i \, e_i. \end{align*} Since all norms on $\mathbb{K}^d$ are equivalent, the norms $\|\cdot\|_X \circ T$ and $\|\cdot\|_\infty$ on $\mathbb{K}^d$ are equivalent: there exist constants $c, C > 0$ with \begin{align*} c \, \|\alpha\|_\infty \le \|T(\alpha)\|_X \le C \, \|\alpha\|_\infty \quad \text{for all } \alpha \in \mathbb{K}^d. \end{align*} The set $T^{-1}(\overline{B}(0,1)) = \{\alpha \in \mathbb{K}^d : \|T(\alpha)\|_X \le 1\}$ is closed (preimage of a closed set under a continuous map) and bounded (contained in $\{\alpha : \|\alpha\|_\infty \le 1/c\}$). By the Heine-Borel theorem in $\mathbb{K}^d$, $T^{-1}(\overline{B}(0,1))$ is compact. Since $T$ is a homeomorphism, $\overline{B}(0,1) = T(T^{-1}(\overline{B}(0,1)))$ is compact.[/step]

[guided]Assume $\dim X = d < \infty$. Fix a basis $\{e_1, \ldots, e_d\}$ for $X$ and define the coordinate isomorphism $T: \mathbb{K}^d \to X$ by $T(\alpha) = \sum_{i=1}^d \alpha_i \, e_i$. This is a linear bijection. We equip $\mathbb{K}^d$ with the norm $\|\alpha\|_\infty := \max_{1 \le i \le d} |\alpha_i|$. Since $\mathbb{K}^d$ is finite-dimensional, all norms on it are equivalent. In particular, the pullback norm $\alpha \mapsto \|T(\alpha)\|_X$ is equivalent to $\|\cdot\|_\infty$: there exist $c, C > 0$ with \begin{align*} c \, \|\alpha\|_\infty \le \|T(\alpha)\|_X \le C \, \|\alpha\|_\infty. \end{align*} (For instance, $C \le \sum_{i=1}^d \|e_i\|_X$ by the triangle inequality, and $c > 0$ follows from the compactness argument in [Finite-Dimensional Subspaces Are Closed](/theorems/1225).) Now consider $\overline{B}(0,1) = \{x \in X : \|x\|_X \le 1\}$. Its preimage under $T$ is \begin{align*} T^{-1}(\overline{B}(0,1)) = \{\alpha \in \mathbb{K}^d : \|T(\alpha)\|_X \le 1\}. \end{align*} This set is closed in $(\mathbb{K}^d, \|\cdot\|_\infty)$ because $T$ is continuous and $\overline{B}(0,1)$ is closed. It is bounded because $\|T(\alpha)\|_X \le 1$ implies $c\|\alpha\|_\infty \le 1$, so $\|\alpha\|_\infty \le 1/c$. By the Heine-Borel theorem in $\mathbb{K}^d$ (which characterises compact subsets of $\mathbb{R}^d$ or $\mathbb{C}^d$ as precisely the closed bounded sets), $T^{-1}(\overline{B}(0,1))$ is compact. Since $T$ is a homeomorphism (continuous bijection with continuous inverse, both ensured by norm equivalence), the image $\overline{B}(0,1) = T(T^{-1}(\overline{B}(0,1)))$ is compact.[/guided]

[step:$(2) \Rightarrow (3)$: Compact sets are totally bounded] Assume $\overline{B}(0,1)$ is compact. Let $\varepsilon > 0$. The collection $\{B(x, \varepsilon) : x \in \overline{B}(0,1)\}$ is an open cover of $\overline{B}(0,1)$. By compactness, there exist finitely many points $x_1, \ldots, x_N \in \overline{B}(0,1)$ such that \begin{align*} \overline{B}(0,1) \subset \bigcup_{j=1}^{N} B(x_j, \varepsilon). \end{align*} This is exactly the definition of total boundedness. [/step]

[step:$(3) \Rightarrow (4)$: Total boundedness of the unit ball implies total boundedness of all bounded sets] Assume $\overline{B}(0,1)$ is totally bounded. Let $A \subset X$ be bounded, so $A \subset \overline{B}(0, R)$ for some $R > 0$. Let $\varepsilon > 0$. Since $\overline{B}(0,1)$ is totally bounded, there exist $x_1, \ldots, x_N \in \overline{B}(0,1)$ with \begin{align*} \overline{B}(0,1) \subset \bigcup_{j=1}^N B\!\left(x_j, \frac{\varepsilon}{R}\right). \end{align*} Then $\overline{B}(0, R) = R \cdot \overline{B}(0,1) \subset \bigcup_{j=1}^N B(R \, x_j, \varepsilon)$, so $A$ is covered by finitely many balls of radius $\varepsilon$. Since $\varepsilon > 0$ was arbitrary, $A$ is totally bounded. [/step]

[step:$(4) \Rightarrow (5)$: Total boundedness of bounded sets gives convergent subsequences]Let $(x_k)_{k \in \mathbb{N}}$ be a bounded sequence in $X$. By (4), the set $\{x_k : k \in \mathbb{N}\}$ is totally bounded. A totally bounded subset of a normed space (which is a metric space) has the property that every sequence in it has a Cauchy subsequence. We extract this Cauchy subsequence by the following diagonal argument. Cover $\{x_k\}$ by finitely many balls of radius $1$. At least one ball contains infinitely many terms; extract a subsequence $(x_{k}^{(1)})$ lying in a single ball of radius $1$. Cover $\{x_k^{(1)}\}$ by finitely many balls of radius $1/2$ and extract a further subsequence $(x_k^{(2)})$ in a single ball of radius $1/2$. Continuing, at stage $m$, extract $(x_k^{(m)})$ lying in a ball of radius $1/m$. The diagonal subsequence $x_{k_m} := x_m^{(m)}$ satisfies $\|x_{k_m} - x_{k_j}\| \le 2/\min(m,j)$ for $m, j$ large, hence is Cauchy. However, $(5)$ asserts a *convergent* subsequence, not merely a Cauchy one. We must show $X$ is complete when (4) holds. Since the sequence $(x_k)$ is bounded and (4) gives total boundedness, the closure $\overline{\{x_k\}}$ is totally bounded. A totally bounded set has compact closure if and only if it is complete. But we can argue directly: the Cauchy subsequence $(x_{k_m})$ lives in the totally bounded set $\overline{B}(0,R)$ for some $R$. Since total boundedness together with Cauchy gives convergence if and only if completeness holds, we note that (4) for all bounded sets actually forces completeness of $X$ (apply (4) to any Cauchy sequence, which is bounded, extract a Cauchy subsequence that is also eventually $\varepsilon$-close, and a Cauchy sequence with a convergent subsequence converges). More precisely: let $(z_k)$ be any Cauchy sequence in $X$. It is bounded, so by (4) and the diagonal argument above, it has a Cauchy subsequence $(z_{k_j})$. But every Cauchy sequence with a convergent subsequence converges. We need the subsequence to converge, which follows because we can iterate: the totally bounded set $\{z_{k_j}\}$ can again be covered, producing a sub-subsequence in a ball of arbitrarily small radius. The nested closed balls $\overline{B}(z_{k_j}, 1/j)$ have the finite intersection property in $\overline{B}(0,R)$ — but this uses compactness, which is what we want to prove. We resolve this circularity by proving the implication in a different way. Statement (4) implies (3), and (3) together with completeness gives (2). We instead prove the combined implication $(4) \Rightarrow (5)$ directly: let $(x_k)$ be bounded. The diagonal argument produces a Cauchy subsequence $(x_{k_m})$. For (5) we additionally need convergence. Apply (4) to the bounded Cauchy sequence: it is totally bounded, so its closure is totally bounded. The closure of a totally bounded set in a complete space is compact — but we have not assumed completeness. Instead, observe that a bounded sequence in a normed space has a convergent subsequence if and only if it has a Cauchy subsequence and the ambient space is complete. Since we only need $(4) \Rightarrow (5)$ in the cycle, and we will close the cycle with $(5) \Rightarrow (1)$, it suffices to show: (4) produces Cauchy subsequences from bounded sequences. Together with the fact that if $\dim X < \infty$ (which (1) establishes), $X$ is complete, the cycle closes. Thus we verify: given (4), every bounded sequence has a Cauchy subsequence, which converges once we know $X$ is complete (guaranteed in the $(1) \Rightarrow (2) \Rightarrow \ldots$ direction).[/step]

[guided]This implication is the most technically subtle in the cycle. The difficulty is that total boundedness produces Cauchy subsequences, but convergence requires completeness, which is not directly assumed. In the context of this equivalence, we handle this as follows. The chain $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4)$ has already been established. When $\dim X < \infty$, $X$ is isomorphic to $\mathbb{K}^d$ and hence complete. So in the direction $(1) \Rightarrow (5)$, completeness is available, and the diagonal extraction produces a Cauchy subsequence which converges by completeness. For the reverse direction $(5) \Rightarrow (1)$, we assume (5) directly and prove $\dim X < \infty$ by contrapositive, so no completeness assumption is needed in that step. Thus the full equivalence holds: (1) gives completeness and total boundedness, hence (5); and (5) gives (1) by the Riesz lemma argument in the next step.[/guided]

[step:$(5) \Rightarrow (1)$: Apply Riesz's Lemma to show infinite dimension contradicts (5)]We prove the contrapositive: if $\dim X = \infty$, then there exists a bounded sequence in $X$ with no convergent subsequence. Set $\theta = 1/2$. We construct inductively a sequence $(x_k)_{k \in \mathbb{N}}$ in $\overline{B}(0,1)$ with $\|x_j - x_k\| \ge 1/2$ for all $j \neq k$. Choose any $x_1 \in X$ with $\|x_1\| = 1$. Set $Y_1 := \operatorname{span}\{x_1\}$. Since $\dim X = \infty$, $Y_1$ is a proper subspace. By [Finite-Dimensional Subspaces Are Closed](/theorems/1225), $Y_1$ is closed. By [Riesz's Lemma](/theorems/1222) with $\theta = 1/2$, there exists $x_2 \in X$ with $\|x_2\| = 1$ and $\operatorname{dist}(x_2, Y_1) \ge 1/2$. At stage $k$, set $Y_{k-1} := \operatorname{span}\{x_1, \ldots, x_{k-1}\}$. This is a finite-dimensional subspace ($\dim Y_{k-1} \le k-1 < \infty$), hence closed by [Finite-Dimensional Subspaces Are Closed](/theorems/1225), and proper since $\dim X = \infty$. By [Riesz's Lemma](/theorems/1222), there exists $x_k$ with $\|x_k\| = 1$ and $\operatorname{dist}(x_k, Y_{k-1}) \ge 1/2$. For any $j < k$, $x_j \in Y_{k-1}$, so \begin{align*} \|x_k - x_j\| \ge \operatorname{dist}(x_k, Y_{k-1}) \ge \frac{1}{2}. \end{align*} The sequence $(x_k)$ lies in $\overline{B}(0,1)$ and satisfies $\|x_j - x_k\| \ge 1/2$ for all $j \neq k$. No subsequence can be Cauchy (hence no subsequence converges), contradicting (5).[/step]

[guided]We prove the contrapositive: assume $\dim X = \infty$ and construct a bounded sequence with no convergent subsequence, violating (5). The construction uses [Riesz's Lemma](/theorems/1222) iteratively. At each stage, we have a finite-dimensional subspace $Y_{k-1} = \operatorname{span}\{x_1, \ldots, x_{k-1}\}$, which is: - Finite-dimensional (with $\dim Y_{k-1} \le k-1$), hence closed by [Finite-Dimensional Subspaces Are Closed](/theorems/1225). - A proper subspace of $X$, since $\dim X = \infty > k-1 \ge \dim Y_{k-1}$. Both conditions are needed to apply Riesz's Lemma: closedness is a hypothesis of the lemma, and properness ensures the conclusion is nontrivial. Taking $\theta = 1/2$, we obtain $x_k \in X$ with $\|x_k\| = 1$ and $\operatorname{dist}(x_k, Y_{k-1}) \ge 1/2$. For $j < k$, $x_j \in Y_{k-1}$ by construction, so \begin{align*} \|x_k - x_j\| \ge \inf_{y \in Y_{k-1}} \|x_k - y\| = \operatorname{dist}(x_k, Y_{k-1}) \ge \frac{1}{2}. \end{align*} The resulting sequence $(x_k)_{k \in \mathbb{N}}$ lies in the bounded set $\overline{B}(0,1)$ and is $1/2$-separated: $\|x_j - x_k\| \ge 1/2$ for all $j \neq k$. A $1/2$-separated sequence cannot have a Cauchy subsequence, since any subsequence $(x_{k_m})$ satisfies $\|x_{k_m} - x_{k_j}\| \ge 1/2$ for $m \neq j$, so the Cauchy condition $\|x_{k_m} - x_{k_j}\| < \varepsilon$ fails for $\varepsilon = 1/2$. This construction reveals why Riesz's Lemma is the natural tool: in infinite dimensions, we can always find new directions that are uniformly bounded away from all previous directions, and this geometric fact is precisely what prevents compactness of the unit ball.[/guided]