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Proof. Let $X$ be a normed space with norm $|\cdot |$ and $(x_n)$ be Cauchy. Then for all $\epsilon \gt 0, \ \exists N : m,n \gt N \implies |x_m - x_n| \lt \epsilon$ is the standard definition of Cauchy sequence. But it's easy to show that $||x_n| - |x_m|| \leq |x_m - x_n|$ and thus the sequence $|x_n|$ in $\mathbb{R}$ is Cauchy. By completeness of the reals under the absolute value norm, we have that $|x_n|$ approaches a limit and thus $(x_n)$ is absolutely convergent.

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Is your normed space complete? Or does your definition of absolute convergence not entail convergence as well? – Andrés E. Caicedo Oct 3 '13 at 4:55
No, not complete. This is a lemma to showing that a normed space is a Banach space iff absolutely convergent sequences converge. – Enjoys Math Oct 3 '13 at 4:56
Ah, I see. The argument you indicate is correct. – Andrés E. Caicedo Oct 3 '13 at 6:04
up vote 3 down vote accepted

The proof is correct. Applied more generally, it shows the following:

If $X$ and $Y$ are metric spaces, $(x_n)$ is Cauchy in $X$, and $f: X\to Y$ is a uniformly continuous map, then the sequence $f(x_n)$ has a limit.

The particular statement uses $Y=\mathbb R$ and $f(x)=\|x\|$ (which is a Lipschitz function).

That said, I don't understand the bigger picture. Apparently "absolutely convergent sequence" here means a sequence $(x_n)$ such that $\|x_n\|$ has a limit. This is the first time I see this term used anywhere (and I kind of hope it's the last one. It seems designed to confuse people.) More importantly, this notion of "absolutely convergent sequence" does not imply usual convergence, e.g., consider $x_n=(-1)^n$ in $\mathbb R$. And since $\mathbb R$ is a Banach space, this disproves the claim made in a comment, "a normed space is a Banach space iff absolutely convergent sequences converge".

Perhaps the intended claim was "a normed space is a Banach space iff absolutely convergent series converge". That is indeed correct, but then the argument given in the OP is not really relevant.

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