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Im getting trouble trying to formulate and making this proof. The exercise is stated like this:

Let $f:A\to F$ and $g:B\to F$ with $A,B\subseteq X$ where $F$ is a normed space and $X$ is a topological space. Prove that if $f$ and $g$ are continuous at $x_0$ then $f+g$ is continuous at $x_0$.

The proof is easy if $X$ would be a metric space instead of an arbitrary topological space.

The problem here is that for arbitrary topological spaces I cant use a sequential characterization of continuity.

Then I tried to play around with the topological definition of continuity and the algebra of open balls in normed spaces but I dont get any clue.

Some hint will be appreciated, thank you.

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  • $\begingroup$ The exercise is oddly stated. The domain of $f+g$ is $A\cap B$, so it's strange to give $f$ and $g$ different domains, and provide an ambient space. That doesn't make the exercise wrong, it's just odd. $\endgroup$ Nov 11 '16 at 23:26
  • $\begingroup$ Recall, what is the topological definition of continuity at a point? $\endgroup$ Nov 11 '16 at 23:27
  • $\begingroup$ @DanielFischer it is something like: if $f$ is continuous at $x_0$ then for any neighborhood $V$ of $f(x_0)$ exists some open set $U$ that contain $x_0$ such that $f(U)\subseteq V$. Im having problem to see the relation with the addition of an vector normed space ($f+g$) and this topological definition of continuity. $\endgroup$
    – Masacroso
    Nov 11 '16 at 23:41
  • $\begingroup$ You just use open sets on $X$ instead of the usual $\delta$. See my answer. $\endgroup$
    – Momo
    Nov 11 '16 at 23:54
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For $\epsilon>0$

There is an open set $U_1$ with $x_0\in U_1$ such that $||f(x)-f(x_0)||<\epsilon/2$

There is an open set $U_2$ with $x_0\in U_2$ such that $||g(x)-g(x_0)||<\epsilon/2$

So for the open set $U=U_1\cap U_2$ we have $x_0\in U$ and

$||(f+g)(x)-(f+g)f(x_0)||\leq ||f(x)-f(x_0)||+||g(x)-g(x_0)||<\epsilon/2+\epsilon/2=\epsilon$

Which was to be proved.

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  • $\begingroup$ Oh, very fine, thank you. The formulation is perfect. Trying to play around with open balls I lost the point. $\endgroup$
    – Masacroso
    Nov 11 '16 at 23:57

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