# Any finite dimensional normed linear space over a complete field is complete.

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.

• 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. – Pratyush Sarkar Dec 28 '18 at 6:41
• @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension? – Omojola Micheal Dec 28 '18 at 6:43
• 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. – Pratyush Sarkar Dec 28 '18 at 6:57
• @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be. – Omojola Micheal Dec 28 '18 at 7:41
• I removed some unnecessary parts which you forgot to delete. Looks fine now. – 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!