Rudin 2.14

  • Thread starter Rasalhague
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  • #1
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Homework Statement



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.)

Homework Equations



Every separable metric space has a countable base.

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?
 

Answers and Replies

  • #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.
 
  • #3
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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.
It's a countable global base since the space is separable.
 
  • #4
WWGD
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Yes, thanks, I just corrected it as you were writing; I did not read carefully-enough.
 
  • #5
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Hint: consider the points ##x\in F## for which every neighborhood of ##x## meets ##F## in uncountably many points.
 
  • #6
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Got it, thanks!
 

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