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Set theory problem (uncountable set)

  1. Oct 3, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I would like to show that if we have a non-negative real valued function f defined on f a set X, and that the set of points where f is non-vanishing is uncountable, then for any M > 0, I can find a sequence {x_n} of points in X such that

    [tex]\sum_n f(x_n)>M[/tex]


    2. Relevant equations



    3. The attempt at a solution
    This is not for a set theory class, its in the context of a measure and integration problem where I want to demonstrate a necessary and sufficient condition for a function to be integrable on a set X with respect to the cardinal measure mu(E) = |E|.

    Ok, so I would appreciate to get a confirmation that the following assertion is true, because if it is, then I will have solved the problem I think:

    "If for any natural number n,

    [tex]f^{-1}(]\frac{1}{n}, +\infty[)[/tex]

    is countable, then

    [tex]f^{-1}(]0, +\infty[)[/tex]

    is countable too."

    It seems so to me since it seems natural that a countable union of countable sets in countable and

    [tex]\bigcup_{n\geq 1} f^{-1}\left(]\frac{1}{n}, +\infty[\right) = f^{-1}\left(\bigcup_{n\geq 1}]\frac{1}{n}, +\infty[\right)=f^{-1}(]0, +\infty[)[/tex]


    Thx.
     
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  3. Oct 3, 2007 #2

    Hurkyl

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    A countable union of countables is countable: #(N x N) = #N.
     
  4. Oct 3, 2007 #3

    quasar987

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    sweet. By the end of the thread I didn't really had a doubt left but thx for confirming, and so fast!
     
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