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Let $(f_{\alpha})_{\alpha \in A}$ be a family of real valued continous functions on a topological space $X$ such that for each $x\in X$ $f_\alpha (x)\neq 0$ for only finitely many $\alpha\in A$. Then we can define $f=\sum_{\alpha \in A}f_\alpha$ .

In general we can can not expect $f$ to be continous, for example if $X$ consists only of discrete points together with one limit point. Are there any conditions on $X$ that will guarantee the continouity of $f$?

Any help appreciated!

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I do not have the complete answer as of yet. But maybe a partial answer will already help you.

I noticed, that this is not even true if X = R together with the standard topology, since you can take any strictly monotonous convergent sequence $x_n$. Then construct a sequence of neighborhoods $U_n$ around each $x_n$ with $U_n\cap U_m$ is the empty set whenever $m\neq n$ (Construct this for example by taking the open balls around $x_n$ with diameter smaller then $\min(x_n-x_{n-1},x_{n+1}-x_n)$) Now one can construct a sequence of continuous $f_n$ such that $f_n(x_n) = n $ and the support of $f_n$ is restricted to $U_n$. Then for every point of the real line, there is at most one n such that $f_n\neq 0$ Therefore I can write the sum. Furthermore $f(x)=0$ but there exists a convergent sequence under which the images diverge. Therefore the limit of the sum of $f_n$ can not be continuous.

I think this construction can be modified to include at least all benach spaces.

PS: You could change your Assumptions to: For each x in X there exists a neighborhood $U_x$ such that $f_\alpha(U_x))={0}$ only for finitely many $\alpha$. Then my as well as your example fail to comply, and this might be true for at least any Banach space, since then any sequence converging to x would be at some point in $U_x$ and in $U_x$ we can restrict the sum to a finite sum.

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The standard condition (used in paracompactness considerations, e.g.) is that the family $(f_\alpha)^{-1}[(0,\rightarrow)]$, $\alpha \in A$ is locally finite. Every point of $X$ then has a neighbourhood $U_x$ on which all but finitely many $f_\alpha$ vanish. This clearly implies continuity.

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If $X$ is Hausdorff, the counterexample you give is essentially the only one: the implication holds iff the set of isolated points of $X$ has no limit point. Indeed, such continuous $f_\alpha$ have to be supported on a finite set of isolated points, and a discontinuity of $\sum f_\alpha$ can only happen at a limit point of isolated points.

For general $X$, the condition you need is that there is no disjoint union of finite open sets that has a limit point.

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