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Statement about topology of subsets of a metric space.

  1. Dec 10, 2013 #1
    The problem statement, all variables and given/known data.

    Prove that a closed subset in a metric space ##(X,d)## is the boundary of an open subset if and only if it has empty interior.

    The attempt at a solution.
    I got stuck in both implications:
    ##\implies## Suppose ##F## is a closed subspace with ##F=\partial S## for some open subset ##S## of ##(X,d)##. I want to show that ##F^°=\emptyset##. So, let ##x \in F^°##. Then, there is ##\delta## : ##B(x,\delta) \subset F##. But then, ##B(x,\delta) \subset \partial S##, which means there is ##y \in B(x,\delta) \cap S^c## and there is ##z \in B(x,\delta) \cap S##. I couldn't go farther than this.

    For the other implication, I need some help too. If ##F^°## is empty, then, for every ##x \in F## and every ##\delta>0##, ##B(x,\delta) \cap F^c \neq \emptyset##. As ##F## is closed, ##F^c## is open. I want to show that ##F=\partial F^c##, we've just seen that ##F \subset \partial F^c##. Now, let ##x \in \partial F^c##, then, for every ##n \in \mathbb N##, there is ##y_n \in B(x, \dfrac{1}{n}) \cap F##. But this means ##x## is a limit point of ##F##, as ##F## is closed ## \implies## ##x \in F##. I've proved ##\partial F^c \subset F##. This proves
    ##F=\partial F^c##, which means ##F## is the boundary of an open subset.

    Could anyone suggest me how to prove the forward implication? Is the other one correct?
     
  2. jcsd
  3. Dec 11, 2013 #2

    pasmith

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    If [itex]S[/itex] is open then it contains no boundary points, so [itex]F \subset X \setminus S[/itex]. Thus [itex]F^{\circ} \subset (X \setminus S)^{\circ}[/itex]. Is an interior point of [itex]X \setminus S[/itex] a limit point of [itex]S[/itex]?
     
  4. Dec 11, 2013 #3
    I don't see why an interior point of [itex]X \setminus S[/itex] would have to be a limit point of [itex]S[/itex]. For example, if I consider ##S=(0,1)##, then ##10 \in {S^c}^{\circ}## but ##10## is not a limit point of ##S##.
     
    Last edited: Dec 11, 2013
  5. Dec 11, 2013 #4

    pasmith

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    So is the answer to the question "Is an interior point of [itex]X \setminus S[/itex] a limit point of [itex]S[/itex]?" going to be "sometimes" or "never"?
     
  6. Dec 12, 2013 #5
    Sorry, I was getting it all wrong. I think I've understood what you were trying to say:

    As you've said, ##F \subset X \setminus S##. Then, ##F^{\circ} \subset {(X \setminus S)}^{\circ}##. Suppose there exists ##x \in F^{\circ}##. By hypothesis, ##F=\partial S##, thus ##x## is a limit point of ##S##. Now, ##x \in {(X \setminus S)}^{\circ}##, so there is ##\delta>0##: ##B(x,\delta) \subset X \setminus S##. But, as ##x## is a limit point of ##S##, there is ##y \in S \cap B(x,\delta)##, so ##y \in S \cap (X \setminus S)##, which is clearly absurd. From here it follows ##F^{\circ}=\emptyset##.

    The answer is going to be "never".

    Thanks!
     
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