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Let $(X,+)$ be an abelian group and $d$ a metric on $X$. Suppose $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences. What conditions on the relation between the group operation and the metric are sufficient to show that $\{a_n+b_n\}$ is Cauchy? The proof in $\mathbb{R}$ simply relies on the fact that the Euclidean metric is induced by a norm:

$$d(a_m+b_m,a_n+b_n)=\|(a_m+b_m)-(a_n+b_n)\|\leq\|a_m-a_n\|+\|b_m-b_n\|$$

I believe this proof works so long as the triangle inequality relates $+$ and the norm which induces $d$.

I think a weaker assumption which still works is if the metric is translation invariant. Then,

$$\begin{align*}

d(a_m+b_m,a_n+b_n) &\leq d(a_m+b_m,a_n+b_m) + d(a_n+b_m,a_n+b_n) \\

&= d(a_m,a_n) + d(b_m,b_n)

\end{align*}$$

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Are there weaker assumptions under which the sum is Cauchy? What is an example in this case? Is there an example of a space and Cauchy sequences whose sum is not Cauchy?

This is probably not a standard way of phrasing the question. I’m just trying to strip down assumptions to clarify my understanding. I thought about this for a while but was having trouble coming up with examples.

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A sufficient condition weaker than translation invariance is that translations are uniformly equicontinuous, i.e that for every $\epsilon > 0$ there is $\delta > 0$ such that for all $x, y, z$, $d(x,y) < \delta$ implies $d(x+z, y+z) < \epsilon$. For then if $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences, take $N$ so for $n, m > N$, $d(a_n, a_m) < \delta$ and $d(b_n, b_m) < \delta$, and note that $d(a_n + b_n, a_m + b_m) \le d(a_n + b_n, a_m + b_n)

+ d(a_m + b_n, a_m + b_m) < 2 \epsilon$.

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