Completeness of $\ell^2$ space

Solution 1:

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.

Solution 2:

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.