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Definition: A function $f: X \to Y$ between metric spaces $X$ and $Y$ is continuous in $x \in X$ iff for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x' \in X$ with $d(x,x') < \delta$: $$ d(f(x), f(x')) < \varepsilon. $$ Now I want to proof that the sum of continuous functions at a point $x$ is continuous at that point $x$ by just using this definition (i.e. not using the limit criterion).

If the metric is induced by a norm, $d(x,y) := ||x-y||$ then this is easy.

Let $x \in X$ be fixed, $f = f_1 + f_2$ with $f_1, f_2$ being continuous at $x$. Then for $\varepsilon > 0$ let $\delta_1$ and $\delta_2$ be the Delta's for the functions $f_1, f_2$. Set $\delta := \operatorname{max}(\delta_1, \delta_2)$, then \begin{align*} d(f(x), f(x')) & = d(f_1(x) + f_2(x), f_1(x') + f_2(x')) \\ & = ||f_1(x) + f_2(x) - (f_1(x') + f_2(x')) \\ & = ||f_1(x) - f_1(x') + f_2(x) - f_2(x')) \\ & \le ||f_1(x) - f_1(x')|| + ||f_2(x) - f_2(x')|| \\ & = d(f_1(x)),f_1(x')) + d(f_2(x), f_2(x')) \\ & \le 2\varepsilon. \end{align*} I want to generalize this proof for a metric, but therefore I needed the inequality $$ d(x+u, y+v) \le d(x,y) + d(u,v) $$ which I is false, what I learned here:

Does $d(x+u, y + v) \le d(x, y) + d(u,v)$ holds for every metric?

But how does I proof it then for a metric space?

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How do you define the sum of two functions between metric spaces? –  JSchlather Jan 22 '13 at 20:25
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you don't even have an addition operation; a metric space is not necessarily equipped with an addition –  user55315 Jan 22 '13 at 20:26
    
Are your functions defined from $X$ to $\mathbb{R}$ perhaps? –  Thomas E. Jan 22 '13 at 20:29
    
You should examine the metric spaces you work with more closely – what is addition in them? Then you could ask yourself whether $+ \colon Y × Y → Y$ is continous or not. If your metric spaces $X$, $Y$ happen to be complete, you can check this by finding out if the sum of convergent sequences in $Y$ happens to converge to the sum of their limits. Then this would imply the sum of continuous functions is continuous. Maybe you don't need $X$ to be complete for this argument. –  k.stm Jan 22 '13 at 20:30
    
ok, I undertand then there exists no addition in general. And what if i consider functions from $X$ to $\mathbb{R}$? –  Stefan Jan 22 '13 at 20:35

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