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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,

$$\begin{align*} d(a_m+b_m,a_n+b_n) &\leq d(a_m+b_m,a_n+b_m) + d(a_n+b_m,a_n+b_n) \\ &= d(a_m,a_n) + d(b_m,b_n) \end{align*}$$

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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The natural requirement is simply that $+$ be continuous with respect to $d$. – Brian M. Scott Nov 23 '11 at 6:14
No, continuity of $+$ is not enough. For example, consider $\mathbb R$ with the metric $d(x,y) = |\arctan(x) - \arctan y|$. Then $a_n = n$ and $b_n = (-1)^n - n$ are Cauchy but $a_n + b_n = (-1)^n$ is not Cauchy. – Robert Israel Nov 23 '11 at 7:05
Ok, so that's why I was having trouble proving that comment! Artan is even Lipschitz continuous, huh? Nice example! And that metric doesn't seem to be translation invariant, so it's not induced by a norm. This should be promoted to full answer. – dls Nov 23 '11 at 7:11
What about a space which isn't translation invariant, but preserves the Cauchy property under addition? – dls Nov 23 '11 at 7:26
This question is really a precursor to the question "when is the completion of a topological group also a topological group"?… – dls Nov 27 '11 at 18:03
up vote 6 down vote accepted

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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I ran into this while trying to prove the comment that regular continuity was sufficient; just define away the issue! I'm having trouble thinking of an example here which isn't translation invariant. But this brings up a question, related to the arctan example. If $d$ is induced by a norm, and $f : \mathbb{R} \rightarrow \mathbb{R}$, can we impose a growth-condition on $f$ so that $f \circ d$ is a metric? Can we refine this condition to say $f \circ d$ preserves Cauchy sequences under $+$? Maybe should make this own question unless answer is "no". – dls Nov 23 '11 at 16:25
Looking at the Wikipedia page for uniform space, they give the example $d(x,y)=|e^x-e^y|$ and state that this has a different uniform structure than the usual metric. Though I don't understand "uniform structure" yet, maybe I'm thinking, can you put a condition on the uniform structure which guarantees preservation of Cauchy sequences under addition? – dls Nov 23 '11 at 18:12
Instead of $f \circ d$, it's better to work with $d \circ (f \otimes f)$, i.e. $(x,y) \to d(f(x),f(y))$. For any one-to-one function from $X$ to a metric space $Y$ with metric $d$, $d \circ (f \otimes f)$ is a metric on $X$. This will typically not be translation-invariant. In the case $X=Y=\mathbb R$ with $d(x,y)=|x−y|$, it will satisfy my condition if there are constants $a$, $b$ with $0<a<f′(x)<b$ everywhere. – Robert Israel Nov 23 '11 at 19:58
Right, I see why $f \circ d$ is incorrect. So in the case $X=Y=\mathbb{R}$, we can think of level sets of the graph of $d$ over $\mathbb{R}^2$. If $d$ is translation invariant, all level sets are parallel to the line $y=x$. In the $\arctan$ example, this is clearly not the case... the level sets diverge away from this line as $x\rightarrow\pm\infty$. I think your condition says that the level sets are bounded between two lines parallel to the line $y=x$. Haven't actually tried computing this, but I think I've satisfied my curiosity for today! – dls Nov 23 '11 at 21:18

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