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Math Help - Real Analysis - Open/Closed Sets

  1. #1
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    Real Analysis - Open/Closed Sets

    Given A is a subset of R, let L be the set of all limit points of A.

    a. Show that the set L is closed

    b. Argue that if x is a limit point of A U L, then x is a limit point of A.
    Use this observation to furnish a proof of the following theorem:

    For every A is a subset of R, the closure of A bar is a closed set and is the smallest closed set containing A.
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  2. #2
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    If I could figure out how to do notation on here it would be easier, but basically if you assume an x not in the set of all limit points then there is an epsilon neighborhood around x such that the complement of L is open. Thus if L complement is open then L is closed. Where L is the set of all limit points.
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  3. #3
    Senior Member vincisonfire's Avatar
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    For (a) you can use the fact that a set is closed if and only if it contains all its boundary points. The complement of a boundary is open since it does not contains its boundary. Since the complement of the boundary is open then the boundary is a closed set.
    For (b) you need to show that the boundary of a boundary is the boundary itself.
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