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Rudin 2.14

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. (Rudin: Principles of Mathematical Analysis, 2nd ed.)

    2. Relevant equations

    Every separable metric space has a countable base.

    3. The attempt at a solution

    Let F be closed. Using the above fact, I've shown that the isolated points of F are at most countable, likewise their closure. I'm trying to construct a perfect set by removing non-limit points of F' points from F', the set of limit points of F, but it's not quite falling into place yet. Is this a good direction to go?
     
  2. jcsd
  3. Jun 29, 2015 #2

    WWGD

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    Careful, that is a countable _local_ base, not a countable global base, i.e., every metric space is 1st-countable, but not necessarily 2nd-countable.
    EDIT: Sorry, I did not read carefully: the hypothesis of separable implies 2nd countable.
     
  4. Jun 29, 2015 #3

    micromass

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    It's a countable global base since the space is separable.
     
  5. Jun 29, 2015 #4

    WWGD

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    Yes, thanks, I just corrected it as you were writing; I did not read carefully-enough.
     
  6. Jun 29, 2015 #5

    micromass

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    Hint: consider the points ##x\in F## for which every neighborhood of ##x## meets ##F## in uncountably many points.
     
  7. Jun 30, 2015 #6
    Got it, thanks!
     
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