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Math Help - Separated

  1. #1
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    Separated

    If  A,B\subset X are both open and disjoint, then why are they separated? (that is A\cap\overline{B}=\emptyset=\overline{A}\cap B)
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bram kierkels View Post
    If  A,B\subset X are both open and disjoint, then why are they separated? (that is A\cap\overline{B}=\emptyset=\overline{A}\cap B)
    Think about it this way. \text{cl }B=B\cup D\left(B\right) ( D is the derived set, the set of limit points) and so A\cap\text{cl }B=\left(A\cap B\right)\cup\left(A\cap D(B)\right)=A\cap D(B). Thus, if A\cap\text{cl }B\ne \varnothing then A must contain a limit point of B, but why is that stupid?
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    Since a sequence \{x_j\}_{j=1}^\infty in A does not enter B, the sequence cannot converge to any point of B. Thanks
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bram kierkels View Post
    Since a sequence \{x_j\}_{j=1}^\infty in A does not enter B, the sequence cannot converge to any point of B. Thanks
    Well, the idea's right. No neighborhood of a limit point of B can be contained entirely within A and thus can't be a point of A.

    Note though, I've seen you work in general topological spaces, and the definition you are implying of the closure in terms of sequences is only valid in first countable spaces.
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