Is There a Better Way to Prove This?

[uman]

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).

 

[uman]

Ack. Delete this thread please.