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I need verificate a proof of the following fact:

If $Y$ is a finite dimensional subspace of a normed space $X$, then $Y$ is a closed.

The most common proof of this fact uses the completes of all finite dimensional subspace, I understand it correctly, but I have curious to know why the next proof is not so popular, is there any error in the following argument?

Alternative proof: Let $Y$ be a finite dimensiona subespace of a normed vector space $X$. Suppose that $\dim(Y)= m$, then there exist $\left\{v_{1},\ldots,v_{m}\right\}\subset X$ independent vectors such that $\mathrm{span}\left\{v_{1},\ldots,v_{n}\right\}=Y$. Let $x\in X$ be a limit point of $Y$, then there are $\left\{x_{n}\right\}_{n\in\mathbb{N}}\subseteq Y$ such that $\lim_{n\rightarrow\infty}x_{n}=x$. We must show that $x\in Y$. In fact, note that $$x_{n}=\alpha_{1}^{(n)}v_{1}+\cdots+\alpha_{m}^{(n)}v_{m}$$ where $\alpha_{i}^{(n)}\in \mathbb{K}$ for $i=1,\ldots,m$ and $n\in \mathbb{N}$ ($\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$). Therefore, we have $$x=\lim_{n\rightarrow\infty}x_{n}=\lim_{n\rightarrow\infty}\left(\alpha_{1}^{(n)}v_{1}+\cdots+\alpha_{m}^{(n)}v_{m}\right)=\lim_{n\rightarrow\infty}\left(\alpha_{1}^{(n)}\right)v_{1}+\cdots+\lim_{n\rightarrow\infty}\left(\alpha_{m}^{(n)}\right)v_{m}.$$ Note that $\lim_{n\rightarrow\infty}\left(\alpha_{i}^{(n)}\right)$ exists since if this is not fulfilled then $\lim_{n\rightarrow\infty}x_{n}$ does not exist. Therefore, $x\in Y$.

Diego Fonseca
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  • How is the alternative proof not the same as the completeness argument? The proof is correct, by the way. – Olivier Mar 26 '17 at 04:06
  • Please give a few more details on the completeness argument, because as far as what has been given is concerned, I don't see where the proofs differ at all (at least in idea). Having said that, the second proof is certainly correct, at least. – Sarvesh Ravichandran Iyer Mar 26 '17 at 04:08
  • @Olivier You are right, it is not really an alternative proof, I name it that way to differentiate this argument from the traditional proof. – Diego Fonseca Mar 26 '17 at 15:34
  • @астонвіллаолофмэллбэрг You can find the completeness argument in http://math.stackexchange.com/questions/168275/proof-that-every-finite-dimensional-normed-vector-space-is-complete or https://www.quora.com/How-can-I-prove-that-every-finite-dimensional-vector-subspace-of-a-Banach-space-is-closed – Diego Fonseca Mar 26 '17 at 15:36
  • I consult this because the argument that I propose generates doubts to me, for example, it is not becoming evident that $ X $ is a normed vector space. I think it could be influencing when it is done: $$\lim_{n\rightarrow\infty}\left(\alpha_{1}^{(n)}v_{1}+\cdots+\alpha_{m}^{(n)}v_{m}\right)=\lim_{n\rightarrow\infty}\left(\alpha_{1}^{(n)}\right)v_{1}+\cdots+\lim_{n\rightarrow\infty}\left(\alpha_{m}^{(n)}\right)v_{m}.$$ – Diego Fonseca Mar 26 '17 at 15:42
  • The properties of limits you are using give a 'high level' argument. Make sure you can show why they hold. – Olivier Mar 26 '17 at 15:53

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