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Closure of a countable subset of the reals

  1. Aug 20, 2010 #1


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    1. The problem statement, all variables and given/known data

    Let A = {ln(1 + q^2) : q is rational}. One needs to find Cl(A) in R with its euclidean topology.

    3. The attempt at a solution

    So, the set A is a countable subset of [0, +∞>. The closure is, by definition, the intersection of all closed sets containing A. So, Cl(A) would be [0, +∞> itself , right?

    I'm just curious about my solution, thanks in advance.

    By the way, another way to look at it would be the fact that A is dense in [0, +∞> (since every open interval in [0, +∞> intersects A), so Cl(A) = [0, +∞>. I'm not really sure about this, although it seems quite obvious. Can we, for every <a, b> in [0, +∞>, find some rational q such that ln(1 + q^2) is contained in <a, b>?
  2. jcsd
  3. Aug 21, 2010 #2
    As the rationals are dense in R, the closure will be the range of f(x) =ln(1 + x^2) i.e.,
    [0, +∞).
  4. Aug 21, 2010 #3


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    OK, thanks. But I was not sure about that, since the values of f are not rational in general.
  5. Aug 21, 2010 #4


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    Yes, that's why the range is all real numbers larger than or equal to 0 and not just rational numbers.
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