# Show that the sum of two uniformly continuous functions is uniformly continuous in an arbitrary metric space

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?

• 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. Sep 19 '19 at 23:04

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,
$$d(f(x)+g(x), f(y)+g(y)) \leq d(f(x)+g(x), f(y)+g(x))+d(f(y)+g(x), f(y)+g(y))\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
• 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. Sep 20 '19 at 0:17