[proofplan] We prove equivalence of any norm $\|\cdot\|$ on a finite-dimensional $K$-vector space $V$ with the max norm $\|\cdot\|_{\max}$ relative to a fixed basis $\{x_1, \ldots, x_n\}$. The upper bound $\|\cdot\| \leq D \|\cdot\|_{\max}$ follows directly from the strong triangle inequality. The lower bound $\|\cdot\| \geq C \|\cdot\|_{\max}$ is proved by induction on $\dim V$: the base case is immediate, and the inductive step uses the completeness of $K$ to deduce that each codimension-1 coordinate hyperplane $V_i$ is complete, hence closed. The positive distance from $x_i$ to the closed set $V_i$ provides the constant $C$. Completeness under any norm follows from equivalence with the max norm on a product of copies of the complete field $K$. [/proofplan] [step:Establish the upper bound $\|v\| \leq D \|v\|_{\max}$] Let $\{x_1, \ldots, x_n\}$ be a basis of $V$ over $K$. For any $v \in V$, write $v = \sum_{i=1}^n a_i x_i$ with $a_i \in K$. By definition, $\|v\|_{\max} = \max_{1 \leq i \leq n} |a_i|$. Define $D := \max_{1 \leq i \leq n} \|x_i\|$. Applying the strong triangle inequality (the norm $\|\cdot\|$ is non-archimedean since $K$ is non-archimedean) and multiplicativity: \begin{align*} \|v\| = \left\|\sum_{i=1}^n a_i x_i\right\| \leq \max_{1 \leq i \leq n} \|a_i x_i\| = \max_{1 \leq i \leq n} |a_i| \cdot \|x_i\| \leq \max_{1 \leq i \leq n} |a_i| \cdot D = D \|v\|_{\max}. \end{align*} [guided] The upper bound uses only the strong triangle inequality for $\|\cdot\|$ (which holds because $K$ is non-archimedean and $\|\cdot\|$ is a $K$-norm). We write $v = \sum a_i x_i$ and estimate: \begin{align*} \|v\| = \left\|\sum_{i=1}^n a_i x_i\right\| \leq \max_{1 \leq i \leq n} \|a_i x_i\| = \max_{1 \leq i \leq n} |a_i| \cdot \|x_i\|. \end{align*} The first inequality is the strong triangle inequality for $\|\cdot\|$. The equality $\|a_i x_i\| = |a_i| \cdot \|x_i\|$ is the absolute homogeneity of the norm. Bounding $\|x_i\|$ by $D = \max_i \|x_i\|$ and pulling the constant out of the max: \begin{align*} \max_{1 \leq i \leq n} |a_i| \cdot \|x_i\| \leq \max_{1 \leq i \leq n} |a_i| \cdot D = D \cdot \|v\|_{\max}. \end{align*} Note that this direction does not use completeness of $K$ at all --- it is purely algebraic. Completeness enters only in the lower bound. [/guided] [/step] [step:Prove the lower bound $\|v\| \geq C \|v\|_{\max}$ by induction on $\dim V$] We prove: for every finite-dimensional $K$-vector space $V$ with basis $\{x_1, \ldots, x_n\}$ and every norm $\|\cdot\|$, there exists $C > 0$ with $\|v\| \geq C \|v\|_{\max}$ for all $v \in V$. **Base case: $n = 1$.** Every element of $V$ has the form $v = a_1 x_1$ with $a_1 \in K$, and $\|v\|_{\max} = |a_1|$. Then \begin{align*} \|v\| = \|a_1 x_1\| = |a_1| \cdot \|x_1\| = \|x_1\| \cdot \|v\|_{\max}, \end{align*} so $C = \|x_1\| > 0$ works (since $x_1 \neq 0$ implies $\|x_1\| > 0$). **Inductive step.** Assume $n \geq 2$ and that the result holds for all spaces of dimension at most $n - 1$. For each $i \in \{1, \ldots, n\}$, define the codimension-1 subspace \begin{align*} V_i := \operatorname{span}_K\{x_1, \ldots, \hat{x}_i, \ldots, x_n\} = \operatorname{span}_K\{x_j : j \neq i\}, \end{align*} where $\hat{x}_i$ denotes omission. By the inductive hypothesis, $\|\cdot\|$ restricted to $V_i$ is equivalent to the max norm on $V_i$. Since the max norm on $V_i$ (with respect to the basis $\{x_j : j \neq i\}$) is complete (as $K$ is complete and completeness of a product of complete metric spaces holds), the norm $\|\cdot\|$ on $V_i$ is also complete. In particular, $V_i$ is a complete subspace of $(V, \|\cdot\|)$, hence closed. [guided] The inductive step is the heart of the proof. The idea is: for each basis vector $x_i$, the hyperplane $V_i$ (spanned by all other basis vectors) is complete under $\|\cdot\|$, hence closed. The translate $x_i + V_i$ is therefore also closed, and $0 \notin x_i + V_i$ (since $x_i \notin V_i$ by linear independence). The infimum of $\|\cdot\|$ on the closed set $x_i + V_i$ is therefore positive, and this infimum provides the constant. Why does completeness matter? In the inductive step, we need each $V_i$ to be closed in $V$. A subspace is closed if and only if it is complete (in a metric space, a complete subspace is always closed). The inductive hypothesis gives equivalence of $\|\cdot\|$ with $\|\cdot\|_{\max}$ on $V_i$ (which has dimension $n - 1 < n$). Since the max norm on $V_i$ inherits completeness from $K$ (the max norm on $K^{n-1}$ is complete when $K$ is complete, as convergence in the max norm is equivalent to coordinatewise convergence in $K$), completeness of $\|\cdot\|$ on $V_i$ follows from equivalence of norms. What fails without completeness? If $K$ is not complete, the inductive hypothesis might still give equivalence of norms on $V_i$, but $V_i$ might not be complete under $\|\cdot\|$, hence might not be closed in $V$. Without closedness, the infimum of $\|\cdot\|$ on $x_i + V_i$ might be $0$, and the constant $C$ degenerates. [/guided] [/step] [step:Extract the constant $C$ from the positive distance of each $x_i$ to $V_i$] For each $i \in \{1, \ldots, n\}$, define \begin{align*} c_i := \inf_{w \in V_i} \|x_i + w\|. \end{align*} [claim:$c_i > 0$ for each $i$] Since $V_i$ is closed in $(V, \|\cdot\|)$ (established in the previous step), the translate $x_i + V_i = \{x_i + w : w \in V_i\}$ is also closed. The element $0 \notin x_i + V_i$: if $0 = x_i + w$ for some $w \in V_i$, then $x_i = -w \in V_i$, contradicting the linear independence of $\{x_1, \ldots, x_n\}$ (since $x_i \notin \operatorname{span}\{x_j : j \neq i\}$). Since $x_i + V_i$ is closed and $0 \notin x_i + V_i$, and since the norm function $\|\cdot\|: V \to \mathbb{R}_{\geq 0}$ is continuous, the infimum $c_i = \inf_{z \in x_i + V_i} \|z\| = \operatorname{dist}(0, x_i + V_i) > 0$. [/claim] [proof] Suppose for contradiction that $c_i = 0$. Then there exists a sequence $(w_m)_{m \geq 1}$ in $V_i$ with $\|x_i + w_m\| \to 0$. This means $x_i + w_m \to 0$ in $(V, \|\cdot\|)$, i.e., $w_m \to -x_i$. Since $V_i$ is closed, $-x_i \in V_i$, hence $x_i \in V_i$. But $x_i \notin V_i$ by linear independence, a contradiction. Therefore $c_i > 0$. [/proof] Set $C := \min_{1 \leq i \leq n} c_i > 0$. Let $v = \sum_{j=1}^n a_j x_j$ be any nonzero element of $V$, and let $r \in \{1, \ldots, n\}$ be an index achieving $\|v\|_{\max} = |a_r|$. Since $v \neq 0$, we have $a_r \neq 0$. Then \begin{align*} \|v\|_{\max}^{-1} \|v\| = |a_r|^{-1} \left\|\sum_{j=1}^n a_j x_j\right\| = \left\|\sum_{j=1}^n \frac{a_j}{a_r} x_j\right\| = \left\|x_r + \sum_{j \neq r} \frac{a_j}{a_r} x_j\right\|. \end{align*} The element $\sum_{j \neq r} \frac{a_j}{a_r} x_j$ lies in $V_r$, so \begin{align*} \left\|x_r + \sum_{j \neq r} \frac{a_j}{a_r} x_j\right\| \geq \inf_{w \in V_r} \|x_r + w\| = c_r \geq C. \end{align*} Therefore $\|v\| \geq C \|v\|_{\max}$. [/step] [step:Conclude equivalence and completeness] Combining the two bounds, for all $v \in V$: \begin{align*} C \|v\|_{\max} \leq \|v\| \leq D \|v\|_{\max}. \end{align*} This shows $\|\cdot\|$ and $\|\cdot\|_{\max}$ are equivalent norms. For completeness: the max norm $\|\cdot\|_{\max}$ on $V \cong K^n$ is complete, since a sequence $(v_m)$ is Cauchy in $\|\cdot\|_{\max}$ if and only if each coordinate sequence $(a_{i,m})_{m \geq 1}$ is Cauchy in $(K, |\cdot|)$, and $K$ is complete by hypothesis. Since $\|\cdot\|$ is equivalent to $\|\cdot\|_{\max}$, they define the same Cauchy sequences and convergent sequences, so $(V, \|\cdot\|)$ is also complete. [/step]