Completeness of $\ell^2$ space I was reading up and it says that $(\ell^2,\|.\|_2)$ is complete. I know that a metric space $X$ in which every Cauchy sequence converges to an element of $X$ is called complete. And I know that a sequence is Cauchy if given $\epsilon$, there exists $N \in \mathbb{N}$ such that $|x_m-x_n|<\epsilon$ for all $n,m>N$.
I was reading the proof that  $(\ell^2,\|.\|_2)$ is complete, but I don't understand where does the $x_k^n$ comes from. What does it mean when it writes $x_k^n$? 

Let $(x_n)$ be Cauchy in $\ell^2$, i.e. $\forall \epsilon>0$ there exists $N \in \mathbb{N}$ such that $\sum_{k=1}^\infty|x_k^n-x_k^m|^2 <\epsilon^2$ for $n,m>N$.
For any fixed $k_0$, $|x_{k_0}^n-x_{k_0}^m|<\epsilon$ for $n,m >N$. So $(x_{k_0}^n)$ is Cauchy in $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C})$ and converges to say $y_{k_0}$.
Also, why did they square $\sum_{k=1}^\infty|x_k^n-x_k^m|^2 <\epsilon^2$?
 A: Each element in $l^2$ is again a sequence (of real numbers), i.e. if $x\in l^2$, then $x=(x_k)_k$. Now if $(x^n)$ is a sequence in $l^2$, then we might write $x^n=(x^n_k)_k=(x^n_1,x^n_2,\ldots)$ for each $n$. Then for a Cauchy sequence in $l^2$, we have 
$$
\epsilon^2 > \|x^n-x^m\|_{l^2}^2 = \sum_{k=1}^\infty |x_k^n-x_k^m|^2.
$$
Using this, you see that for any fixed $k$ (or $k_0$), the sequence $(x_k^n)$ is a Cauchy sequence of real numbers and then apply completeness of $\mathbb{R}$ to see that it converges. Then you must show that the resulting sequence $y=(y_k)_k$ is again in $l^2$, and that $x^n\to y$ in the $l^2$ norm.
As for the $\epsilon^2$, this is presumably because they are looking at at the square of the norm, rather than the norm itself.
A: To prove the completeness, we need to prove for any Cauchy sequence $z_1,z_2,\dotsm \in E$  satisfying that
$\forall \epsilon >0$, there exists a number $n_0$ such that for all $m,n\ge n_0$
$$
||z_m-z_n||^2=\sum_{k=1}^\infty |z_{m,k}-z_{n,k}|^2<\epsilon^2
$$
that implies $\forall k\in \mathbb N,\forall\epsilon>0,\exists n_0>0$
$$
|z_{m,k}-z_{n,k}|<\epsilon\quad \forall m,n>n_0
$$
that mean for every k, the sequence $(z_{n,k})$ is Cauchy sequence and convergent. Then we denote
$$
z_k=\lim_{n\rightarrow \infty} z_{n,k}\quad k=1,2,\dotsm\quad z=(z_k)
$$
We need to prove that there exists an element $z\in E$ such that $\forall \varepsilon>0,\exists M>0,\forall m>M$ such that 
$$
||z_m-z||=\sum_{k=1}^\infty |z_{m,k}-z_k|^2<\varepsilon^2
$$
By letting $n\rightarrow \infty $, we can prove the second equation by the first equation. 
Then we need to prove that $z\in E$
$$
\sqrt{\sum_{k=1}^\infty|z_k|^2}=\sqrt{\sum_{k=1}^\infty|z_{k}-z_{m,k}+z_{m,k}|^2}\\
\le \sqrt{\sum_{k=1}^\infty|z_{k}-z_{m,k}|^2}+\sqrt{\sum_{k=1}^\infty |z_{m,k}|^2}
< \infty
$$
That means $z_m \rightarrow z$ 
Q.E.D.
