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Math Help - Prove a set is open iff it does not contain its boundary points

  1. #1
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    Prove a set is open iff it does not contain its boundary points

    Hi! im new to this site, and id like to start off by asking for help with this problem ive been struggling with for the past hour or so.

    Problem: Let A be such that it is a subset of R (real numbers). Prove that A is open if and only if it does not contain its boundary points.
    [Also let A(c) be the complement of A]

    I started off by letting A be open. I said then that for each x in A, there exists an e-neighborhood V of x such that V is a subset of A.
    So, for every x in A, there exists a V(x) such that
    p in V(x) --> p is in A for some p in R -from definition of a subset
    ==>
    p not in A --> p is not in V(x)
    ==>
    p in A(c) --> p is not in V(x)
    ==>
    p in V(x) --> p not in A(c)

    So as you see, im not really sure where im going with this or even if its right.
    It seems to me that A being open implies that there is a neighborhood of x that yes, contains a point of A but not a point of A(c) [from the "p in A(c) --> p is not in V(x)" line]. However, i dont know what this means (like, does this just tell me p is not a boundary pt?) and i think im getting confused and i need to start over.
    I think that I need to get to a statement such that an arbitrary boundary point p is not in G, but i dont know how

    Any hints or clarification would be VERY appreciated. Thank you
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  2. #2
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    You're trying to do way too much! Here's a quick argument:

    Let x be any point of A, and V a neighborhood of x contained in A. Then V is a neighborhood of x that does not contain any point of A(c). So x is not a boundary point.


    Note: to show that A does not contain its boundary points, I start with an arbitrary point of A, and then argue that this specific point is not a boundary point.
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  3. #3
    Senior Member Tinyboss's Avatar
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    Define B to be the complement of A. Then A is open iff B is closed iff B is its own closure. Since the boundary is the intersection of the closure with the closure of the complement, the last condition means A contains none of its boundary.
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