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A problem in my homework asks to show that the addition of two Cauchy sequences in $\mathbb R^n$ is also Cauchy. However, the metric is not specified. If we assume that we are dealing with the euclidean metric then the problem is relatively straightforward. However, when attempting to solve without this assumption, one needs to show for Cauchy sequences in $\mathbb R^n$ being $x_n$ and $y_n$ that $\forall \epsilon >0 \exists N \in \mathbb N$ such that $\forall l,k \ge N$ $$d(\mathbf x_l +\mathbf y_l, \mathbf x_k +\mathbf y_k) < \epsilon $$ Is it possible to show this? If so, hints on how to start would be appreciated.

Lucas Alanis
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    If they don't specify which metric, they surely mean the standard metric. – Seth Oct 04 '14 at 00:36
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    This is, in general, not true, as shown in http://math.stackexchange.com/questions/84854/sum-of-cauchy-sequences-cauchy – Desiato Oct 04 '14 at 01:39

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No. $\;\;\;\;$ $\langle -1,\hspace{-0.04 in}-2,\hspace{-0.04 in}-3,\hspace{-0.04 in}-4,\hspace{-0.04 in}-5,\hspace{-0.04 in}-6,\hspace{-0.04 in}-7,\hspace{-0.04 in}-8,...\rangle\:$ and $\:\langle 0,\hspace{-0.03 in}2\hspace{.02 in},\hspace{-0.03 in}2\hspace{.02 in},\hspace{-0.04 in}4,\hspace{-0.04 in}4,\hspace{-0.04 in}6\hspace{.02 in},\hspace{-0.03 in}6,\hspace{-0.03 in}8,...\rangle$
are both Cauchy with respect to the metric $\;\; \langle x\hspace{.02 in},\hspace{-0.03 in}y\rangle \: \mapsto \hspace{-0.05 in} \left|\hspace{.02 in}\operatorname{arctan}(x)\hspace{-0.03 in}-\hspace{-0.02 in}\operatorname{arctan}(\hspace{.03 in}y)\right| \;\;$,
but their sum is $\: \langle -1,\hspace{-0.04 in}0,\hspace{-0.04 in}-1,\hspace{-0.04 in}0,\hspace{-0.04 in}-1,\hspace{-0.04 in}0,\hspace{-0.04 in}-1,\hspace{-0.04 in}0,\hspace{-0.04 in}-1,\hspace{-0.04 in}0\rangle\:$, $\;$ which is not Cauchy with respect to any metric.