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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 [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.
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.