Results 1 to 6 of 6

Math Help - countable set

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    countable set

    Let f:[0,1]\rightarrow[0,1]be such that for every distinct x_1,x_2,...,x_n \in [0,1], f(x_1)+f(x_2)+..+f(x_n)\leq 1.
    Show that \{x:f(x)>0\} is countable.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Kat-M View Post
    Let f:[0,1]\rightarrow[0,1]be such that for every distinct x_1,x_2,...,x_n \in [0,1], f(x_1)+f(x_2)+..+f(x_n)\leq 1.
    Show that \{x:f(x)>0\} is countable.
    Assume the contrary, i.e. E = \{x:f(x)>0\} is uncountable.

    Consider the sets E_n = \{x: f(x) > 1/n\} for n = 1, 2, 3, \dots. The union of these sets is E, so one of them, say E_m, must contain infinitely many points; otherwise E could be written as the union of countably many finite sets and would be countable, contrary to our assumption.

    Now what does that say about the sum f(x_1)+f(x_2)+..+f(x_m), where x_i \in E_m?
    Last edited by awkward; August 7th 2009 at 05:25 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    154
    i understand the union of E_m make up the whole set E but i m still not getting why each E_m is countable. how does f(x_1)+f(x_2)+...+f(x_n)\leq1 help me figure it out?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Kat-M View Post
    i understand the union of E_m make up the whole set E but i m still not getting why each E_m is countable. how does f(x_1)+f(x_2)+...+f(x_n)\leq1 help me figure it out?
    The point was to show that there exists an m\in\mathbb{N} such that E_m := \left\{x\in [0,1]\mid f(x)>\frac{1}{m}\right\} is infinite (no matter whether countably infinite or uncountably so). Therefore, by secting finitely many x_1,\ldots,x_n from E_m, you can get arbitrarly large sums f(x_1)+\cdots+f(x_n), which contradicts f(x_1)+\cdots+f(x_n)\leq 1. Thus, the assumption that f(x)>0 for uncountably many x\in [0,1] must be rejected.
    Last edited by Failure; August 7th 2009 at 11:16 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    154
    even if each E_m is finite, i don know how f(x_1)+f(x_2)+..f(x_M)\leq 1could be always true. say m=3. If E_3=\{x:f(x)>1/3\} has finitely many elements in it and if you pick 3 elements from this set, then the sum f(x_1)+f(x_2)+f(x_3) is already greater than 1. or i m not supposed to pick points like that? i guess i m not understanding this funtion.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Kat-M View Post
    even if each E_m is finite, i don know how f(x_1)+f(x_2)+..f(x_M)\leq 1could be always true.
    Well, that f(x_1)+f(x_2)+..f(x_M)\leq 1 is a given fact about the particular function f in question. There is no need to prove that it holds. You just assume it holds and show, what you are asked to show.

    say m=3. If E_3=\{x:f(x)>1/3\} has finitely many elements in it and if you pick 3 elements from this set, then the sum f(x_1)+f(x_2)+f(x_3) is already greater than 1. or i m not supposed to pick points like that?
    Well, yes, if it is true that E_3=\{x\in [0,1]:f(x)>1/3\} has at least 3 members, then you get the contradiction 3\cdot \tfrac{1}{3}=1<f(x_1)+f(x_2)+f(x_3) to the assumed property of f that f(x_1)+f(x_2)+..f(x_n)\leq 1 for any n different elements x_1,\ldots,x_n of [0,1].

    Similarly, if we know that E_m=\{x\in [0,1]:f(x)>1/m\}, has at least m members (infinitely many, say), then you get the contradiction 1\leq n\cdot \tfrac{1}{m}<f(x_1)+\cdots f(x_n)\leq 1, for n\geq m different elements from E_m.

    i guess i m not understanding this function.
    You need not understand this function: you only have to assume that it has the properties that you are told it has and move on to show what you are told to show.

    But if you want an example of a function that has this property you may consider

    f(x) :=\begin{cases}x,& \text{if }x=\frac{1}{2^n}\text{ for some } n\in\mathbb{N}\backslash\{0\}\\0&\text{otherwise}\  end{cases}
    Last edited by Failure; August 8th 2009 at 02:33 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: January 9th 2013, 06:14 AM
  2. Replies: 2
    Last Post: November 11th 2010, 04:56 AM
  3. [SOLVED] Countable union of closed sets/countable interesection of open sets
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 8th 2010, 01:59 PM
  4. Replies: 1
    Last Post: February 9th 2010, 01:51 PM
  5. Replies: 11
    Last Post: October 11th 2008, 06:49 PM

Search Tags


/mathhelpforum @mathhelpforum