# Evaluate this proof

1. Jun 23, 2008

### 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). We will prove the theorem for the case that $$S_{(0,+\infty)}$$ is uncountable. The proof for the other case is entirely analogous.

The fact that $$S_{(0,+\infty)}$$ is uncountable implies that either $$S_{(0,1)}$$ is uncountable or $$S_{[1,+\infty)}$$ is uncountable. In the latter case, we may simply choose $$[M]+1$$ members of $$S_{[1,+\infty)} \{s_1,\ldots,s_{[M]+1}\}$$. Then $$f(s_1)+\cdots+f(s_{[M]+1})>M$$, contradicting the hypothesis of the theorem.

If on the other hand $$S_{(0,1)}$$ is uncountable, then there must be some $$r$$ such that $$0<r<1$$ and $$S_{(r,1)}$$ is an infinite set. After we have proven this statement, we may choose $$[\frac{M}{r}]+1$$ elements of some such set. The sum of the values of $$f$$ given these arguments will be greater than $$M$$, again contradicting our original assumption.

Assume that no such $$r$$ exists. Then we may assign a positive integer to any $$x$$ in $$S_{(0,1)}$$ by simply choosing $$r$$ such that $$0<r<x$$ and enumerating the elements in the finite set $$S_{(r,1)}$$. This contradicts the fact that $$S_{(0,1)}$$ is uncountable. As noted earlier, this completes the proof for the case that $$S_{(0,+\infty)}$$ is uncountable. The other case is proved in exactly the same way.

2. Jun 24, 2008

### uman

:-( nobody?

3. Jun 24, 2008

### ice109

what the hell is T

4. Jun 24, 2008

### uman

That should read "by choosing $$r$$ such that $$0<r<f(x)$$. Here's a more explicit version of the last bit (I hope):

Assume that no such $$r$$ exists. Now consider the function $$n$$ defined on $$S_{(0,1)}$$ as follows: To evaluate $$n(x)$$, choose some $$r$$ such that $$0<r<f(x)$$ and such that there is no $$y$$ such that $$r<f(y)<f(x)$$ or such that $$f(y)=f(x)$$ and $$y<x$$. Then let $$n(x)$$ be the number of elements of $$S_{(r,1)}$$. If $$n(x_0)=n(x_1)$$, then clearly $$x_0=x_1$$, so $$n$$ is an injective function from $$S_{(0,1)}$$ to the set of positive integers, contradicting the assumption that it is uncountable.

Last edited: Jun 24, 2008