Lemma 12.2.2.label Let $E$ be a normed vector space, then the following are equivalent:
- (1)
$E$ is a Banach space.
- (2)
For any absolutely convergent series $\sum_{n = 1}^{\infty} x_{n}$ with $\seq{x_n}\subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^{\infty} x_{n}$.
Proof. (2) $\Rightarrow$ (1): Let $\seq{x_n}\subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}}\le 2^{-k}$ for all $k \in \natp$.
For each $k \in \natp$, let $y_{k} = x_{n_{k+1}}- x_{n_k}$, then $\sum_{n = 1}^{\infty} y_{k}$ is absolutely convergent, and there exists $y \in E$ such that $y = \sum_{n = 1}^{\infty} y_{n}$. Let $x = x_{n_1}+ y$, then $x_{n_k}\to x$ as $k \to \infty$. Since $\seq{x_n}$ is Cauchy, $x_{n} \to x$ as well by Lemma 6.6.5.$\square$