Check my proof please. The space $\ell_1$ is defined as

$$\ell_1:=\left\{(x_n)\in s(\Bbb K):\|(x_n)\|_1=\sum |x_n|<\infty\right\}$$

Here $\Bbb K$ represents $\Bbb R$ or $\Bbb C$. The operations of a normed vector space are easy to check, we can see that if $s,t\in\ell_1$ then $s+t\in\ell_1$, etc. Then I need to show that $\ell_1$ is complete.

This mean that for every Cauchy sequence in $\ell_1$ it converges to some element of $\ell_1$:

$$(\forall\epsilon>0,\exists N\in\Bbb N:\|s_n-s_m\|_1<\epsilon,\forall n,m\ge N)\implies (s_n)\to s\in\ell_1$$

Because $\|s_n\|_1=\sum |x_{k,n}|$ and $\|s_m\|_1=\sum |x_{k,m}|$ then we can write


Because $\Bbb K$ is a Banach space then $\lim_m x_{k,m}=x_k\in\Bbb K$. In other words because $(x_{k,n})_k$ is Cauchy, by the above expression, and every $x_{k,n}\in\Bbb K$ then $(x_{k,n})_k\to x_k$ and we can write

$$\lim_m\sum |x_{k,n}-x_{k,m}|=\sum |x_{k,n}-x_k|\le\epsilon$$


$$\left|\sum|x_{k,n}|-\sum|x_k|\right|=\sum ||x_{k,n}|-|x_k||\le\sum |x_{k,n}-x_k|\le\epsilon$$

From the last statement if $\sum |x_k|=\infty$, and because $\sum|x_{k,n}|=x\ge 0$ we would have the contradiction


then necessarily $\sum |x_k|$ converges. Then if we define $s=(x_k)$ we finally have that $(s_n)\to s$.$\Box$.


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