[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. For $(1) \Rightarrow (2)$, we use the isomorphism $X \cong \mathbb{R}^n$ to reduce to total boundedness of the closed unit ball in $\mathbb{R}^n$, which follows from a grid covering argument. For $(2) \Rightarrow (3)$, we combine total boundedness with the completeness of a closed ball in a normed space, then invoke the [Equivalent Characterisations of Compactness in Metric Spaces](/theorems/316). For $(3) \Rightarrow (1)$, we prove the contrapositive using [Riesz's Lemma](/theorems/???) to construct a sequence in the unit ball with no convergent subsequence when $X$ is infinite-dimensional. [/proofplan] [step:$(1) \Rightarrow (2)$: Reduce to $\mathbb{R}^n$ via the isomorphism of finite-dimensional normed spaces] Assume $X$ is finite-dimensional with $\dim(X) = n$. Let $\{e_1, \ldots, e_n\}$ be a basis for $X$. The coordinate map \begin{align*} \Phi: \mathbb{R}^n &\to X \\ (\alpha_1, \ldots, \alpha_n) &\mapsto \sum_{i=1}^n \alpha_i e_i \end{align*} is a linear isomorphism. Since all norms on a finite-dimensional space are equivalent, there exist constants $c, C > 0$ such that $c\|\alpha\|_{\ell^\infty} \le \|\Phi(\alpha)\|_X \le C\|\alpha\|_{\ell^\infty}$ for all $\alpha \in \mathbb{R}^n$. The closed unit ball $\overline{B}(0, 1) \subset X$ satisfies $\Phi^{-1}(\overline{B}(0, 1)) \subset \overline{B}_{\ell^\infty}(0, 1/c)$, the closed $\ell^\infty$-ball of radius $1/c$ in $\mathbb{R}^n$, which equals $[-1/c, \, 1/c]^n$. Fix $\varepsilon > 0$. We cover $[-1/c, \, 1/c]^n$. Partition each interval $[-1/c, \, 1/c]$ into subintervals of length at most $\varepsilon c / C$, requiring $N := \lceil 2C/(c^2\varepsilon) \rceil$ subintervals per coordinate. This produces $N^n$ sub-cubes, each of $\ell^\infty$-diameter at most $\varepsilon c / C$. Let $\{\alpha^{(1)}, \ldots, \alpha^{(N^n)}\}$ be the centres of these sub-cubes. For any $\alpha \in [-1/c, \, 1/c]^n$, there exists $j$ with $\|\alpha - \alpha^{(j)}\|_{\ell^\infty} \le \varepsilon c/(2C)$, so \begin{align*} \|\Phi(\alpha) - \Phi(\alpha^{(j)})\|_X = \|\Phi(\alpha - \alpha^{(j)})\|_X \le C\|\alpha - \alpha^{(j)}\|_{\ell^\infty} \le C \cdot \frac{\varepsilon c}{2C} = \frac{\varepsilon c}{2} \le \frac{\varepsilon}{2} < \varepsilon. \end{align*} Therefore $\{\Phi(\alpha^{(j)})\}_{j=1}^{N^n}$ is a finite $\varepsilon$-net for $\Phi([-1/c, \, 1/c]^n) \supset \overline{B}(0, 1)$, and $\overline{B}(0, 1)$ is totally bounded. [guided] Assume $\dim(X) = n < \infty$. The idea is that, up to a change of coordinates, the unit ball in $X$ is "essentially" a bounded subset of $\mathbb{R}^n$, which can be covered by a finite grid. Fix a basis $\{e_1, \ldots, e_n\}$ for $X$ and define the coordinate isomorphism $\Phi: \mathbb{R}^n \to X$ by $\Phi(\alpha) = \sum_{i=1}^n \alpha_i e_i$. On the finite-dimensional space $\mathbb{R}^n$, all norms are equivalent. Applying this to the pullback of $\|\cdot\|_X$ via $\Phi$ and the $\ell^\infty$-norm, there exist constants $c, C > 0$ with \begin{align*} c\|\alpha\|_{\ell^\infty} \le \|\Phi(\alpha)\|_X \le C\|\alpha\|_{\ell^\infty} \quad \text{for all } \alpha \in \mathbb{R}^n. \end{align*} The lower bound gives $\Phi^{-1}(\overline{B}(0, 1)) \subset [-1/c, \, 1/c]^n$: if $\|\Phi(\alpha)\|_X \le 1$, then $\|\alpha\|_{\ell^\infty} \le 1/c$. Now we cover the cube $[-1/c, \, 1/c]^n$ by small sub-cubes. Fix $\varepsilon > 0$. Partition each coordinate interval $[-1/c, \, 1/c]$ into $N = \lceil 2C/(c^2\varepsilon) \rceil$ subintervals of length at most $\varepsilon c / C$. This produces $N^n$ sub-cubes in $\mathbb{R}^n$, each of $\ell^\infty$-diameter at most $\varepsilon c / C$. For any point $\alpha$ in the cube, the centre $\alpha^{(j)}$ of the containing sub-cube satisfies $\|\alpha - \alpha^{(j)}\|_{\ell^\infty} \le \varepsilon c/(2C)$. The upper bound in the norm equivalence gives \begin{align*} \|\Phi(\alpha) - \Phi(\alpha^{(j)})\|_X \le C\|\alpha - \alpha^{(j)}\|_{\ell^\infty} \le C \cdot \frac{\varepsilon c}{2C} = \frac{c\varepsilon}{2} < \varepsilon. \end{align*} The images $\{\Phi(\alpha^{(j)})\}_{j=1}^{N^n}$ form a finite $\varepsilon$-net for $\Phi([-1/c, \, 1/c]^n)$, which contains $\overline{B}(0, 1)$. Hence $\overline{B}(0, 1)$ is totally bounded. The finiteness of the grid size $N^n$ depends crucially on $n$ being finite. In an infinite-dimensional space, the analogous grid would require infinitely many points, and this argument breaks down --- which is exactly the content of the reverse implication. [/guided] [/step] [step:$(2) \Rightarrow (3)$: Combine total boundedness with completeness of the closed ball] Assume $\overline{B}(0, 1)$ is totally bounded. The closed ball $\overline{B}(0, 1)$ is a closed subset of the normed space $X$. Every normed space (whether or not it is complete) has the property that a Cauchy sequence in a closed subset whose limit exists in the ambient space has its limit in the closed subset. However, we need $X$ to be complete for Cauchy sequences to converge. We proceed differently. By the [Sequential Characterisation of Total Boundedness](/theorems/1087), every sequence in $\overline{B}(0, 1)$ has a Cauchy subsequence. We claim $\overline{B}(0, 1)$ is complete. Since $\overline{B}(0, 1)$ is totally bounded, the space $X$ is finite-dimensional: we prove this as an intermediate claim. Total boundedness of $\overline{B}(0, 1)$ with $\varepsilon = 1/2$ gives a finite set $\{x_1, \ldots, x_N\}$ with $\overline{B}(0, 1) \subset \bigcup_{i=1}^N B(x_i, 1/2)$. Let $Y = \operatorname{span}\{x_1, \ldots, x_N\}$, which is finite-dimensional. We claim $Y = X$. Suppose for contradiction that $Y \subsetneq X$. Since $Y$ is finite-dimensional, it is closed in $X$. By [Riesz's Lemma](/theorems/???) (applied with $\theta = 1/2$), there exists $z \in X$ with $\|z\| = 1$ and $\operatorname{dist}(z, Y) > 1/2$. But $z \in \overline{B}(0, 1)$, so $z \in B(x_i, 1/2)$ for some $i$, giving $\|z - x_i\| < 1/2$. Since $x_i \in Y$, this contradicts $\operatorname{dist}(z, Y) > 1/2$. Therefore $X = Y$ is finite-dimensional, hence complete (since all norms on a finite-dimensional space are equivalent to the Euclidean norm, and $\mathbb{R}^n$ is complete). The closed ball $\overline{B}(0, 1)$ is a closed subset of the complete space $X$, hence complete. By the [Equivalent Characterisations of Compactness in Metric Spaces](/theorems/316), a metric space is compact if and only if it is complete and totally bounded. Since $\overline{B}(0, 1)$ is both, it is compact. [guided] Assume $\overline{B}(0, 1)$ is totally bounded. We want to show it is compact. The [Equivalent Characterisations of Compactness in Metric Spaces](/theorems/316) states that a metric space is compact if and only if it is complete and totally bounded. We have total boundedness; we need completeness. The difficulty is that we have not assumed $X$ is a Banach space, so $X$ itself might not be complete. However, we will show that total boundedness of the unit ball forces $X$ to be finite-dimensional, and finite-dimensional normed spaces are always complete. **Claim: $X$ is finite-dimensional.** Apply total boundedness with $\varepsilon = 1/2$: there exists a finite set $\{x_1, \ldots, x_N\} \subset X$ with $\overline{B}(0, 1) \subset \bigcup_{i=1}^N B(x_i, 1/2)$. Let $Y = \operatorname{span}\{x_1, \ldots, x_N\}$. Since $Y$ is the span of finitely many vectors, $\dim(Y) \le N < \infty$, and $Y$ is a closed subspace of $X$ (finite-dimensional subspaces of normed spaces are always closed). Suppose $Y \neq X$. By Riesz's Lemma (applied to the closed proper subspace $Y$ with parameter $\theta = 1/2$), there exists $z \in X$ with $\|z\| = 1$ and $\inf_{y \in Y} \|z - y\| > 1/2$. Since $z \in \overline{B}(0, 1)$ and $\overline{B}(0, 1) \subset \bigcup_{i=1}^N B(x_i, 1/2)$, there exists $i$ with $\|z - x_i\| < 1/2$. But $x_i \in Y$, so $\operatorname{dist}(z, Y) \le \|z - x_i\| < 1/2$, contradicting $\operatorname{dist}(z, Y) > 1/2$. This contradiction shows $Y = X$, so $\dim(X) = \dim(Y) \le N < \infty$. **Completing the argument.** Since $X$ is finite-dimensional, all norms on $X$ are equivalent, and $X$ is isomorphic (as a normed space) to $(\mathbb{R}^n, \|\cdot\|_2)$ for $n = \dim(X)$. Since $\mathbb{R}^n$ is complete under the Euclidean norm, $X$ is complete. The closed ball $\overline{B}(0, 1)$ is a closed subset of the complete space $X$, hence complete. Combined with total boundedness, the [Equivalent Characterisations of Compactness in Metric Spaces](/theorems/316) gives compactness. [/guided] [/step] [step:$(3) \Rightarrow (1)$: A compact unit ball forces finite dimensionality] Assume $\overline{B}(0, 1)$ is compact. We show $X$ is finite-dimensional by contradiction. Suppose $\dim(X) = \infty$. We construct a sequence in $\overline{B}(0, 1)$ with no convergent subsequence, contradicting the compactness of $\overline{B}(0, 1)$ (which, by the [Equivalent Characterisations of Compactness in Metric Spaces](/theorems/316), implies sequential compactness). **Construction by induction.** Choose $x_1 \in X$ with $\|x_1\| = 1$. Set $Y_1 = \operatorname{span}\{x_1\}$. Since $\dim(X) = \infty$, we have $Y_1 \subsetneq X$, and $Y_1$ is closed (finite-dimensional subspaces are closed). By Riesz's Lemma with $\theta = 1/2$, there exists $x_2 \in X$ with $\|x_2\| = 1$ and $\|x_2 - x_1\| = \operatorname{dist}(x_2, Y_1) > 1/2$. At stage $k$, set $Y_k = \operatorname{span}\{x_1, \ldots, x_k\}$. Since $\dim(X) = \infty$, $Y_k \subsetneq X$, and $Y_k$ is closed. Riesz's Lemma yields $x_{k+1} \in X$ with $\|x_{k+1}\| = 1$ and $\operatorname{dist}(x_{k+1}, Y_k) > 1/2$. Since $x_1, \ldots, x_k \in Y_k$, we have $\|x_{k+1} - x_j\| > 1/2$ for all $j \le k$. The resulting sequence $\{x_k\}_{k=1}^\infty$ lies in $\overline{B}(0, 1)$ (since $\|x_k\| = 1$ for all $k$) and satisfies $\|x_j - x_k\| > 1/2$ for all $j \neq k$. No subsequence can be Cauchy (and hence no subsequence converges), contradicting the compactness of $\overline{B}(0, 1)$. [guided] Assume $\overline{B}(0, 1)$ is compact, and suppose for contradiction that $\dim(X) = \infty$. We will construct a sequence of unit vectors that are "spread apart" from each other, contradicting sequential compactness. The tool is Riesz's Lemma, which asserts: if $Y$ is a closed proper subspace of a normed space $X$ and $0 < \theta < 1$, then there exists $x \in X$ with $\|x\| = 1$ and $\operatorname{dist}(x, Y) \ge \theta$. This is the mechanism by which infinite-dimensionality produces separated sequences. **Inductive construction.** Start with any $x_1 \in X$ with $\|x_1\| = 1$. At stage $k$, let $Y_k = \operatorname{span}\{x_1, \ldots, x_k\}$. This is a finite-dimensional subspace, hence closed. Since $\dim(X) = \infty$, we have $Y_k \neq X$, so Riesz's Lemma (with $\theta = 1/2$) provides $x_{k+1} \in X$ with $\|x_{k+1}\| = 1$ and $\operatorname{dist}(x_{k+1}, Y_k) > 1/2$. For any $j \le k$, we have $x_j \in Y_k$, so $\|x_{k+1} - x_j\| \ge \operatorname{dist}(x_{k+1}, Y_k) > 1/2$. Proceeding by induction, we obtain a sequence $\{x_k\}_{k=1}^\infty \subset \overline{B}(0, 1)$ with $\|x_j - x_k\| > 1/2$ for all $j \neq k$. **Contradiction.** Since $\overline{B}(0, 1)$ is compact, it is sequentially compact: every sequence in $\overline{B}(0, 1)$ has a convergent subsequence. But a convergent subsequence is Cauchy, so its terms eventually lie within distance $1/4$ of each other. The pairwise lower bound $\|x_j - x_k\| > 1/2$ prevents this. Therefore $\dim(X) < \infty$. The constant $1/2$ in Riesz's Lemma is not optimal --- one can take any $\theta < 1$. However, one cannot always achieve $\theta = 1$ (the case $\theta = 1$ would assert the existence of a unit vector at distance exactly $1$ from $Y$, which fails in general non-reflexive Banach spaces). The choice $\theta = 1/2$ suffices for the separation argument. [/guided] [/step]