[proofplan] Choose a basis of the finite-dimensional space and compare the given norm with the maximum coordinate norm on the associated finite-dimensional coordinate space determined by that basis. The triangle inequality gives an upper comparison, while compactness of the coordinate unit sphere gives a positive lower comparison. These two inequalities turn Cauchy sequences in the space into Cauchy coordinate sequences in the scalar field, where completeness of the scalar field supplies coordinate limits. Reassembling the limiting coordinates yields a vector in the space, and the comparison inequalities show that the original sequence converges to it. [/proofplan]

[step:Handle the zero-dimensional case] If $\dim_{\mathbb{F}}V=0$, then $V=\{0\}$. Every sequence in $V$ is the constant sequence $0$, hence every [Cauchy sequence](/page/Cauchy%20Sequence) in $(V,\|\cdot\|_V)$ converges to $0\in V$. Thus $(V,\|\cdot\|_V)$ is complete in this case. For the rest of the proof assume $\dim_{\mathbb{F}}V>0$. [/step]

[step:Choose coordinates and define the comparison norm] Since $V$ is finite-dimensional and nonzero, choose a positive integer $n$ and an ordered basis $(e_1,\dots,e_n)$ of $V$ over $\mathbb{F}$. Define the coordinate reconstruction map $\Phi:\mathbb{F}^n\to V$ by \begin{align*} \Phi(a_1,\dots,a_n)=\sum_{i=1}^{n} a_i e_i. \end{align*} Because $(e_1,\dots,e_n)$ is a basis, $\Phi$ is a linear bijection. Define the maximum norm $\|\cdot\|_\infty:\mathbb{F}^n\to[0,\infty)$ by \begin{align*} \|(a_1,\dots,a_n)\|_\infty=\max_{1\le i\le n}|a_i|. \end{align*} [/step]

[step:Compare the given norm with the maximum coordinate norm]We prove that there are constants $m>0$ and $M>0$ such that every $a=(a_1,\dots,a_n)\in\mathbb{F}^n$ satisfies \begin{align*} m\|a\|_\infty\le \|\Phi(a)\|_V\le M\|a\|_\infty. \end{align*} For the upper bound, define \begin{align*} M=\sum_{i=1}^{n}\|e_i\|_V. \end{align*} Since each $e_i\ne 0$, one has $M>0$. By homogeneity and the triangle inequality in $V$, \begin{align*} \|\Phi(a)\|_V=\left\|\sum_{i=1}^{n}a_i e_i\right\|_V\le \sum_{i=1}^{n}|a_i|\|e_i\|_V\le M\|a\|_\infty. \end{align*} For the lower bound, let $S\subset\mathbb{F}^n$ be the maximum-norm unit sphere \begin{align*} S=\{a\in\mathbb{F}^n:\|a\|_\infty=1\}. \end{align*} The set $S$ is closed because it is the inverse image of the [closed set](/page/Closed%20Set) $\{1\}\subset\mathbb{R}$ under the continuous map $a\mapsto\|a\|_\infty$, and it is bounded because $\|a\|_\infty=1$ for all $a\in S$. Hence $S$ is compact by the [Heine-Borel theorem](/theorems/309) for finite-dimensional real coordinate spaces, applied to $\mathbb{F}^n$ after identifying $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ in the complex case. The function $f:S\to[0,\infty)$ defined by \begin{align*} f(a)=\|\Phi(a)\|_V \end{align*} is continuous because the upper estimate gives \begin{align*} |f(a)-f(b)|\le \|\Phi(a)-\Phi(b)\|_V\le M\|a-b\|_\infty \end{align*} for all $a,b\in S$. Since $\Phi$ is injective and $0\notin S$, one has $f(a)>0$ for every $a\in S$. Therefore $f$ attains a positive minimum on $S$; define \begin{align*} m=\min_{a\in S}f(a)>0. \end{align*} If $a=0$, then $m\|a\|_\infty\le \|\Phi(a)\|_V$ is immediate. If $a\ne 0$, then $a/\|a\|_\infty\in S$, so homogeneity gives \begin{align*} \|\Phi(a)\|_V=\|a\|_\infty\left\|\Phi\left(\frac{a}{\|a\|_\infty}\right)\right\|_V\ge m\|a\|_\infty. \end{align*}[/step]

[guided]The purpose of this step is to make the given norm on $V$ comparable to a coordinate norm. This comparison is what lets us pass back and forth between Cauchy behavior in $V$ and Cauchy behavior of scalar coordinate sequences. First define the constant \begin{align*} M=\sum_{i=1}^{n}\|e_i\|_V. \end{align*} Since every basis vector $e_i$ is nonzero and norms vanish only at the zero vector, each $\|e_i\|_V$ is positive, hence $M>0$. For an arbitrary coordinate vector $a=(a_1,\dots,a_n)\in\mathbb{F}^n$, the triangle inequality and homogeneity of the norm give \begin{align*} \|\Phi(a)\|_V=\left\|\sum_{i=1}^{n}a_i e_i\right\|_V\le \sum_{i=1}^{n}\|a_i e_i\|_V=\sum_{i=1}^{n}|a_i|\|e_i\|_V. \end{align*} Because $|a_i|\le \|a\|_\infty$ for each index $i$, this becomes \begin{align*} \|\Phi(a)\|_V\le \sum_{i=1}^{n}\|a\|_\infty\|e_i\|_V=M\|a\|_\infty. \end{align*} The lower bound is subtler because the triangle inequality does not directly give a reverse inequality. We use compactness of the coordinate unit sphere. Let \begin{align*} S=\{a\in\mathbb{F}^n:\|a\|_\infty=1\}. \end{align*} This set is closed and bounded in the finite-dimensional scalar space $\mathbb{F}^n$. By the [Heine-Borel theorem](/theorems/315) for finite-dimensional real coordinate spaces, applied directly when $\mathbb{F}=\mathbb{R}$ and after the identification $\mathbb{C}^n\cong\mathbb{R}^{2n}$ when $\mathbb{F}=\mathbb{C}$, the set $S$ is compact. Define $f:S\to[0,\infty)$ by \begin{align*} f(a)=\|\Phi(a)\|_V. \end{align*} The upper estimate just proved implies that $f$ is Lipschitz, since for $a,b\in S$, \begin{align*} |f(a)-f(b)|\le \|\Phi(a)-\Phi(b)\|_V=\|\Phi(a-b)\|_V\le M\|a-b\|_\infty. \end{align*} Thus $f$ is continuous on the compact set $S$. Now $f$ has no zero on $S$. Indeed, if $f(a)=0$, then $\|\Phi(a)\|_V=0$, so $\Phi(a)=0$. Since $\Phi$ is injective, this forces $a=0$, contradicting $\|a\|_\infty=1$. Therefore $f(a)>0$ for all $a\in S$. A continuous positive function on a compact set attains a positive minimum, so we may define \begin{align*} m=\min_{a\in S} f(a)>0. \end{align*} Finally take any $a\in\mathbb{F}^n$. If $a=0$, then both sides of the desired lower estimate are zero. If $a\ne 0$, then the normalized vector $a/\|a\|_\infty$ belongs to $S$. By homogeneity of $\Phi$ and homogeneity of the norm on $V$, \begin{align*} \|\Phi(a)\|_V=\|a\|_\infty\left\|\Phi\left(\frac{a}{\|a\|_\infty}\right)\right\|_V. \end{align*} The normalized vector lies in $S$, so the last norm is at least $m$. Hence \begin{align*} \|\Phi(a)\|_V\ge m\|a\|_\infty. \end{align*} Combining the two estimates gives \begin{align*} m\|a\|_\infty\le \|\Phi(a)\|_V\le M\|a\|_\infty \end{align*} for every $a\in\mathbb{F}^n$.[/guided]

[step:Turn a Cauchy sequence in $V$ into Cauchy coordinate sequences] Let $(x_k)_{k=1}^{\infty}$ be a Cauchy sequence in $(V,\|\cdot\|_V)$. For each integer $k\ge 1$, let $a_k=(a_{k,1},\dots,a_{k,n})\in\mathbb{F}^n$ be the unique coordinate vector satisfying \begin{align*}x_k=\Phi(a_k)=\sum_{i=1}^{n}a_{k,i}e_i.\end{align*} For all integers $k,\ell\ge 1$, the lower comparison estimate applied to $a_k-a_\ell$ gives \begin{align*}m\|a_k-a_\ell\|_\infty\le \|x_k-x_\ell\|_V.\end{align*} Since $(x_k)$ is Cauchy in $\|\cdot\|_V$, it follows that $(a_k)$ is Cauchy in $(\mathbb{F}^n,\|\cdot\|_\infty)$. Therefore, for each fixed index $i\in\{1,\dots,n\}$, the scalar sequence $(a_{k,i})_{k=1}^{\infty}$ is Cauchy in $\mathbb{F}$. [/step]

[step:Use scalar completeness to assemble the limit vector] Since $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, the standard completeness theorem for the real and complex scalar fields implies that $\mathbb{F}$ is complete. Hence for each $i\in\{1,\dots,n\}$ there exists $a_i\in\mathbb{F}$ such that \begin{align*}\lim_{k\to\infty}a_{k,i}=a_i.\end{align*} Define $a=(a_1,\dots,a_n)\in\mathbb{F}^n$ and define $x\in V$ by \begin{align*}x=\Phi(a)=\sum_{i=1}^{n}a_i e_i.\end{align*} Then $\|a_k-a\|_\infty\to 0$ as $k\to\infty$, because each of the finitely many coordinate sequences converges. [/step]

[step:Prove convergence in the original norm and conclude completeness] By the upper comparison estimate, \begin{align*}\|x_k-x\|_V=\|\Phi(a_k-a)\|_V\le M\|a_k-a\|_\infty.\end{align*} Since $\|a_k-a\|_\infty\to 0$, the squeeze property for nonnegative real sequences gives \begin{align*}\lim_{k\to\infty}\|x_k-x\|_V=0.\end{align*} Thus every Cauchy sequence in $(V,\|\cdot\|_V)$ converges in $\|\cdot\|_V$ to an element of $V$. Therefore $(V,\|\cdot\|_V)$ is complete, so $V$ is a [Banach space](/page/Banach%20Space). [/step]