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Math Help - Prove that the set of limit points is closed

  1. #1
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    Prove that the set of limit points is closed

    I have a few questions - they all look very straightforward but I haven't actually got answers to them and it's greatly annoying me...

    Let (X,T) be a Hausdorff topological space and let A be a non-empty subset of X.

    1) Prove that if an open set U in T has non-empty intersection with \overline A, where \overline A is the closure of A, then U has a non-empty intersection with A.

    2) Prove that A' is closed in A, where A' is the set of limit points of A. (Note, if we can prove that A ' contains all its limit points then we're done, but just because A ' contains all of the limit points of A it doesn't necessarily mean it contains all the limit points of A'...)

    3) Prove (\overline A)' = A'
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  2. #2
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    1. This can be proven by rewriting the statement of the theorem as, "If U is an open set of T that does not intersect A, then U does not intersect \overline{A}," using the definition of \overline{A} in terms of open sets.

    3. First, A'\subseteq(\overline{A})', for any open neighborhoods of a point minus that point intersecting A must also intersect \overline{A}. Now, all we need to prove is that any element of (\overline{A})' belongs to A'. For this, we can use the results of (1).
    Last edited by Scott H; February 12th 2009 at 04:47 PM.
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  3. #3
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    2.

    did you get an answer for number 2?
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