Is There a Better Way to Prove This?

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Discussion: This is a proposed solution to problem 2-13 in Apostol's "Mathematical Analysis". The method came to me after a lot of thought but it seems kind of bizarre and I'm wondering if there's a better way to prove this. I especially think the last part could be made more rigorous/explicit.

Also, I'm not even sure my proof is valid! Tell me what you guys think.

Also feel free to critique writing style, minor errors, choice of variable names, etc...

THEOREM: Let f be a real-valued function defined on the interval [0,1] with the following property: There exists a positive real number M such that for any finite collection \{x_1,\ldots,x_n\} of elements of [0,1], |f(x_1)+\cdots +f(x_n)|\leq M. Let S denote the set of all real numbers 0\leq x \leq 1 such that f(x)\not= 0. Then S is countable.

NOTATION: [x] denotes the greatest integer less than x. S_T denotes the set of all real numbers x in [0,1] such that f(x) \epsilon T.

PROOF: We prove the statement by contradiction. Assume S is uncountable. Then either S_{(-\infty,0)} or S_{(0,+\infty)} is uncountable (or both).
 
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Ack. Delete this thread please.
 
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