Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 5 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...

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.