Completeness of a normed vector space This is captured from a chapter talking about completeness of metric space in Real Analysis, Carothers, 1ed. 

I have been confused by two questions:


*

*What does absolutely summable mean in metric space? Does it mean the norm of xi(i=1,2,3,...) that belongs to norm vector space X is summable?

*2nd part of the proof should try to show that if ∑xn(n from 1 to infinity) converges in X whenever ||xn||(n from 1 to infinity) is summable, then X is complete. However, why does the author prove the subsequence of {xn} converges? 
Thanks^_^
 A: There is a fundamental distinction between the two series. Let $(x_i)$ be a sequence in $X$.
Then $\displaystyle \sum_{i=0}^{\infty}{\|x_i\|}$ is a series in $\mathbb{R}$. If   $\displaystyle \sum_{i=0}^{\infty}{\|x_i\|}$ is convergent, then the series $\displaystyle \sum_{i=0}^{\infty}{x_i}$ is said to be absolutely summable. 
On the other hand, $\displaystyle \sum_{i=0}^{\infty}{x_i}$ is said to be summable if it is convergent in the normed vector space. (i.e. $\displaystyle \sum_{i=0}^{\infty}{x_i}=x\in X$).

Now we address your second question. 
A standard theorem is that if a Cauchy sequence has a convergent subsequence then it is itself convergent and converges to the same limit as the subsequence.
Proof:
Take $(x_i)$ a Cauchy sequence. Let ($x_{i_k}$) be a convergent subsequence with limit $x$. Fix $\epsilon > 0$. For $N$ large enough, if $i_k,i>N$ 
$$\|x_i-x_{i_k}\|<\frac{\epsilon}{2} \textbf{ and } ||x_{i_k}-x||<\frac{\epsilon}{2}$$
The first inequality is due to the Cauchyness of the sequence. The second by the convergence of the subsequence.
By the triangle inequality, $\|x_i-x\|<\epsilon$ for $i>N$. Hence $(x_i)$ converges to $x$.

Comment on book proof:  As a consequence of this theorem, the author just needs to show that any Cauchy sequence in $X$ has a convergent subsequence in order to prove that it is convergent hence showing that $X$ is complete.
A: *

*$X$ is a normed space (not just a metric space). So a sequence $\{x_n\}$ is absolutely summable when $\sum\|x_n\| < \infty$.

*A common theorem in this game is that if $\{x_n\}$ is a Cauchy sequence, then given any subsequence $\{x_{n_k}\}$ converges then so does $\{x_n\}$. Moreover, if $x_{n_k} \to L$ then $x_n \to L$. As an idea of the proof of this:
$$
\|x_n - L\| \leq \|x_n - x_{n_k}\| + \|x_{n_k} - L\|
$$
You can control the size of $\|x_n - x_{n_k}\|$  by the Cauchy assumption, and the $\|x_{n_k}-L\|$ term by the assumption of convergence.
