Evaluate "Proof of Theorem on Real-Valued Function Defined on Interval [0,1]

[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.

 

[uman]

:-( nobody?

 

[ice109]

what the hell is T

 

[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.