Prove that any finite dimensional normed linear space over a complete field is complete.

After several comments and corrections, I present the correct proof in the answer section.

  • $\begingroup$ Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say $\{\alpha_{i,r}\}^{n}_{i=1,r\in \Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required. $\endgroup$ – Pratyush Sarkar Dec 28 '18 at 6:41
  • $\begingroup$ @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension? $\endgroup$ – Omojola Micheal Dec 28 '18 at 6:43
  • $\begingroup$ You are using the fact that $\alpha_{i,r}\to \alpha_{i}\in \Bbb{R}\;\text{or}\;\Bbb{C},\;\text{as}\;r\to \infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $\epsilon'$, you have a corresponding $N$ (depending on $\epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = \max\{N_i: i = 1, 2, \dotsc, n\}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions. $\endgroup$ – Pratyush Sarkar Dec 28 '18 at 6:57
  • $\begingroup$ @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be. $\endgroup$ – Omojola Micheal Dec 28 '18 at 7:41
  • 1
    $\begingroup$ I removed some unnecessary parts which you forgot to delete. Looks fine now. $\endgroup$ – Pratyush Sarkar Dec 29 '18 at 1:01


Let $E$ be any finite dimensional normed linear space over a complete field, $\Bbb{R}$ or $\Bbb{C},$ say. Suppose $\dim E=n\geq 1,$ and let $\{e_i\}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars $\{\alpha_i\}^{n}_{i=1}$ such that, for arbitrary $x\in E$, \begin{align} x= \sum^{n}_{i=1} \alpha_i e_i .\end{align} Suppose $\{x_r\}_{r\in \Bbb{N}}$ is Cauchy in $E$ w.r.t $\|\cdot\|$ norm and $\epsilon'>0.$ Then, there exists $N$ such that for all $s\geq r\geq N,$ \begin{align} \|x_r-x_s\|<\epsilon'.\end{align} Now, $\|\cdot\|_1$ defined by $\|x\|_1=\sum^{n}_{i=1} |\alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $\|\cdot\|_1\sim \|\cdot\|,$ i.e., there exists $\gamma,\beta>0,$ such that \begin{align} \gamma\|x\|_1\leq\|x\|\leq \beta\|x\|_1,\;\forall\,x\in E.\end{align} Then, \begin{align}\gamma|\alpha_{i,r}-\alpha_{i,s}|\leq\gamma\sum^{n}_{i=1}|\alpha_{i,r}-\alpha_{i,s}|=\gamma\|x_r-x_s\|_1\leq\|x_r-x_s\|<\epsilon',\;s\geq r\geq N\end{align} and so the sequence $\{\alpha_{i,r}\}^{n}_{i=1,r\in \Bbb{N}}$ is Cauchy in $\Bbb{R}$ or $\Bbb{C}$. Now, $\alpha_{i,r}\to \alpha_{i}\in \Bbb{R}\;\text{or}\;\Bbb{C},\;\text{as}\;r\to \infty$ by completeness. This implies that for each $i\in\{1,2,\cdots,n\},$ there exists $N_i:=N(i,\epsilon')$ such that \begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq N_i.\end{align} Taking $M=\max\{N_i:1\leq i\leq n\},$ we have that \begin{align}|\alpha_{i,r}-\alpha_{i}|<\epsilon',\;r\geq M.\end{align}

Let $\epsilon>0$ and $n\in\Bbb{N}$, then for $\epsilon'=\dfrac{\epsilon}{n\beta},$ there exists $M$ such that \begin{align}|\alpha_{i}-\alpha_{i,s}|\leq\dfrac{\epsilon}{n\beta},\;s\geq M.\end{align} Taking sums, we have
\begin{align}\|x-x_s\|\leq\beta\|x-x_s\|_1=\beta\sum^{n}_{i=1}|\alpha_{i}-\alpha_{i,s}|\leq\sum^{n}_{i=1}\dfrac{\epsilon}{n},\;s\geq M.\end{align} Hence, \begin{align}\|x_s-x\|\leq\epsilon,\;s\geq M,\end{align} and so, we have that $x\in E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!


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