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Application of Baire category theorem

  1. Oct 30, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##(X,d)## be a complete metric space without isolated points and let ##D## be an enumerable dense subset of ##X##. Prove that ##D## is not a ##G_δ## (countable intersection of open sets).

    The attempt at a solution.

    Suppose that ##D## is a ##G_δ##. So ##D=\bigcap_{n \in \mathbb N} A_n## , ##A_i## open for every ##i \in \mathbb N##. Then ##D^c=\bigcup_{n \in \mathbb N} X \setminus A_n## is a countable union of closed sets. Here comes my question: How could I prove that this sets are nowhere dense? I know that ##\overline {D}=X## by hypothesis; then ##\overline {D}^c=\emptyset##, but I want to prove that ##\overline {D^c}^\circ=\emptyset##.
    If I could prove that the complement is nowhere dense, then ##X=A \cup B## where ##A=\bigcup_{n \in \mathbb N, d_n \in D} \{d_n\}## and ##B=D^c##. Since X has no isolated points, each ##\{d_n\}^\circ=\emptyset##. Then ##X## is the union of nowhere-dense closed sets, which is absurd by the Baire category theorem.
     
    Last edited: Oct 30, 2013
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  3. Oct 31, 2013 #2

    Dick

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    ##X \setminus A_n## is already closed. So to prove it's nowhere dense you just have to show it has empty interior. I.e. it doesn't contain any open subsets.
     
  4. Oct 31, 2013 #3
    You're right. ##\overline D^c=D^c##. So, I have to prove ##{D^c}^\circ=\emptyset##. I have to show that for every ##x \in D^c## and for every ##δ>0##, the ball ##B(x,δ) \not\subset D^c##. So, take ##x \in D^c## and ##δ>0## arbitrarily. Suppose ##B(x,δ) \subset D^c##. ##D## is dense in ##X##, so ##B(x,δ) \cap D≠\emptyset##. But then, there exists ##y## such that ##y \in D## and ##y \in D^c##, which is absurd. It follows that ##{D^c}^\circ=∅##.
     
  5. Oct 31, 2013 #4

    Dick

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    That's a little round about. I would just say that an open subset of ##X \setminus A_n## wouldn't contain any elements of D because they are all in ##A_n##. But that's impossible because D is dense. But I think it's the same idea.
     
  6. Oct 31, 2013 #5
    I have a doubt: with your idea, I would have proved that ##(X \setminus A_n)^\circ=\emptyset##. Then ##\bigcup_{n \in \mathbb N} (X \setminus A_n)^\circ=\emptyset##. But I am not sure if ##\bigcup_{n \in \mathbb N} (X \setminus A_n)^\circ=(\bigcup_{n \in \mathbb N} X \setminus A_n)^\circ={D^c}^\circ##. It's easy to see that given two sets ##A## and ##B##, it's always true that ##A^\circ \cup B^\circ \subset (A \cup B)^\circ##, but I don't know if the other inclusion always holds.
     
  7. Oct 31, 2013 #6

    Dick

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    The other inclusion doesn't hold in general. But you don't need that. You are going off on some kind of tangent here. If you prove that ##(X \setminus A_n)## has no interior, doesn't that show ##(X \setminus A_n)## is nowhere dense? Doesn't that in turn mean X is the countable union of nowhere dense sets? It's the union of all of those sets plus the union of the countable singletons in D.
     
  8. Oct 31, 2013 #7
    Yes, sorry. I forgot the original problem.
     
  9. Nov 1, 2013 #8

    Dick

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    No problem. Actually, I finally see what you are trying to do. I was skipping over some parts since they were so obviously wrong, I though that they were just bad notation. If you were actually trying to prove that ##X \setminus D## is nowhere dense think of the example where D is the rationals (Q) and X is the reals (R). The ##R \setminus Q## is not nowhere dense, in fact, it's dense. A countable union of nowhere dense sets doesn't have to be nowhere dense. It often helps to pick a concrete example like Q in R to hold in your mind when you are doing proofs like this.
     
  10. Nov 1, 2013 #9
    Thanks for the remark.
     
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