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Suppose $f$ is the a real valued function. If there exists an $M>0$, such that any finite number of distinct real numbers $x_{1},x_{2},...,x_{n}$ satisfy $|\sum_{k=1}^{n}f(x_{k})|\leq M$, how to prove that $\{x:f(x)\neq 0\}$ is at most countable?

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up vote 6 down vote accepted

Hint: The set $S_n=\{x: |f(x)|>1/n\}$ is finite for each $n\in\mathbb N$ (why?). The set $\{x:f(x)\neq 0\}$ is the union of all these $S_n$. The union of a countable collection of finite sets is...

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Hint: if there are uncountably many points where $f(x) \ne 0$, there are uncountably many of one sign, WOLOG positive. Then there can be only finitely many in $[1,\infty)$, only finitely many in $[\frac 12, 1)$, etc. The points where $f(x) \gt 0$ are now a countable union of finite sets.

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