Proof of the lemma used in proving that a finite-dimensional normed space is complete I'm trying to understand the proof for the lemma:
$$\|\alpha _1 e_1 + \alpha _2 e_2 + \cdots + \alpha_n e_n\| \geq c (|\alpha_1|+|\alpha_2|+\cdots+|\alpha_n|)$$
where $c>0$ and the $e_i$s are finite and linearly independent.
The proof I'm referring to is found in many places, like Kreyszig’s book and on page 16 of this PDF file; and in other locations.
The proof is based on proving that no sequence of the form $(\alpha_1 e_1 + \alpha _2 e_2 + \cdots + \alpha _n e_n)$ could converge to a sequence with zero norm.
What I don't quite get is the following:


*

*Why the proof is not simply stated as: If the lemma is not true, then $\|\alpha _1 e_1 + \cdots + \alpha_n e_n\| = 0$ which is not possible since the $e_i$s are independent. I mean, why do they involve the sequence convergence business here?

*Even when they assume that a sequence has to converge to a zero normed sequence, why does it have to be with the constraint that $|\alpha_1^{(m)}|+|\alpha_2^{(m)}|+\cdots+|\alpha_n^{(m)}| = 1$? Shouldn’t we consider that any possible sequence converges to zero normed sequence without the restriction that the sum should always be one?
Thanks a lot!
 A: 1- The lemma they are trying to prove is that there exists some $c>0$ such that, for any $\alpha_1,\ldots,\alpha_n$, we have $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\geq c(|\alpha_1|+\cdots+|\alpha_n|)$. The key point here is that $c$ does not depend on $\alpha_1,\ldots,\alpha_n$. What you propose (showing that the norm is not $0$) only allows us to say we can pick some $c>0$ dependent on $\alpha_1,\ldots,\alpha_n$, which is a weaker statement.
2- It suffices to consider the case $|\alpha_1|+\cdots+|\alpha_n|=1$ to prove that if $|\alpha_1|+\cdots+|\alpha_n|=1$, then $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\geq c$. The full lemma can then be proved by dividing both sides by $|\alpha_1|+\cdots+|\alpha_n|$.
A: The lemma states that there exists $c\in\mathbb{R},c>0$ such that $\|\alpha_1e_1+\cdots+\alpha_ne_n\|\ge c|\alpha_1+\cdots+\alpha_n|$. The negation is: "for each $c\in\mathbb{R},c>0$, there exists $\alpha_1,\cdots,\alpha_n$ such that $\|\alpha_1e_1+\cdots+\alpha_n\|<c|\alpha_1+\cdots+\alpha_n|$". Maybe this example clarify a little: consider $l_2(\mathbb{N})$. Then $M=\left\{\frac{1}{n}e_n:n\in\mathbb{N}\right\}$ form a linearly independent set, but for every $c>0$, you can find $m\in\mathbb{N}$ such that $1/m<c$, so, $\left\|1\cdot \frac{1}{m}e_m\right\|_2=\frac{1}{m}<c|1|$. The problem here is that $M$ is infinite.
