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Evaluate this proof

  1. Jun 23, 2008 #1
    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 [tex]f[/tex] be a real-valued function defined on the interval [tex][0,1][/tex] with the following property: There exists a positive real number [tex]M[/tex] such that for any finite collection [tex]{x_1,\ldots,x_n}[/tex] of elements of [tex][0,1][/tex], [tex]|f(x_1)+\cdots +f(x_n)|\leq M[/tex]. Let [tex]S[/tex] denote the set of all real numbers [tex]0\leq x \leq 1[/tex] such that [tex]f(x)\not=0[/tex]. Then S is countable.

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

    PROOF: We prove the statement by contradiction. Assume [tex]S[/tex] is uncountable. Then either [tex]S_{(-\infty,0)}[/tex] or [tex]S_{(0,+\infty)}[/tex] is uncountable (or both). We will prove the theorem for the case that [tex]S_{(0,+\infty)}[/tex] is uncountable. The proof for the other case is entirely analogous.

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

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

    Assume that no such [tex]r[/tex] exists. Then we may assign a positive integer to any [tex]x[/tex] in [tex]S_{(0,1)}[/tex] by simply choosing [tex]r[/tex] such that [tex]0<r<x[/tex] and enumerating the elements in the finite set [tex]S_{(r,1)}[/tex]. This contradicts the fact that [tex]S_{(0,1)}[/tex] is uncountable. As noted earlier, this completes the proof for the case that [tex]S_{(0,+\infty)}[/tex] is uncountable. The other case is proved in exactly the same way.
  2. jcsd
  3. Jun 24, 2008 #2
    :-( nobody?
  4. Jun 24, 2008 #3
    what the hell is T
  5. Jun 24, 2008 #4
    That should read "by choosing [tex]r[/tex] such that [tex]0<r<f(x)[/tex]. Here's a more explicit version of the last bit (I hope):

    Assume that no such [tex]r[/tex] exists. Now consider the function [tex]n[/tex] defined on [tex]S_{(0,1)}[/tex] as follows: To evaluate [tex]n(x)[/tex], choose some [tex]r[/tex] such that [tex]0<r<f(x)[/tex] and such that there is no [tex]y[/tex] such that [tex]r<f(y)<f(x)[/tex] or such that [tex]f(y)=f(x)[/tex] and [tex]y<x[/tex]. Then let [tex]n(x)[/tex] be the number of elements of [tex]S_{(r,1)}[/tex]. If [tex]n(x_0)=n(x_1)[/tex], then clearly [tex]x_0=x_1[/tex], so [tex]n[/tex] is an injective function from [tex]S_{(0,1)}[/tex] to the set of positive integers, contradicting the assumption that it is uncountable.
    Last edited: Jun 24, 2008
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