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I've seen the proof when you're talking about $\mathbb{R}$ and the metric is the absolute value function, but I'm trying to prove it for an arbitrary metric space. I figure the proof is pretty similar, for a given $\epsilon$, you choose $\delta$ to be the minimum of the $\delta$s required to make $d(f(x), f(y)) < \frac{\epsilon}{2}$ and $d(g(x), g(y) < \frac{\epsilon}{2}$ when $d(x, y) < \delta$. But then I don't know how to argue that $d(f+g(x), f+g(y)) \leq d(f(x), f(y)) + d(g(x), g(y))$. Any help?

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    $\begingroup$ You need some structure on the codomain of $f$ and $g$ in order to even define $f+g,$ so we cannot be talking about an arbitrary metric codomain. You need to settle that first. After that, the most common property of the metric you'd need is translation-invariance (or some slightly weaker form), under which the same (or similar) proof as in $\mathbb{R}$ works. $\endgroup$ Sep 19 '19 at 23:04
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EDITED:

After the comments of @Brian Moehring,

We assume that the distance is stable under translation and that both $f$ and $g$ target the same space,

$d(f(x)+a, f(y)+a)=d(f(x), f(y))$, So for every $x,y$ such that $d(x, y) < \delta$ we have,

$d(f(x)+g(x), f(y)+g(y)) \leq d(f(x)+g(x), f(y)+g(x))+d(f(y)+g(x), f(y)+g(y))\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

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  • $\begingroup$ To clarify: Not every distance is stable under translation (i.e. translation-invariant) so it would be better to state "Assuming the distance is stable under [...]" rather than "We know that the distance is stable under [...]" That said, absent any specific assumption relating $d$ and $+$ in the problem statement, we cannot prove anything, so this at least gives a good proof under that additional assumption. $\endgroup$ Sep 20 '19 at 0:17

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