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Math Help - proof on neighborhoods and accumulation points

  1. #1
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    proof on neighborhoods and accumulation points

    prove that: S is a subset of R. if every neighborhood of x contains at least one point of S other than x itself, then every neighborhood of x contains infinitely many points of S.

    i want to make sure my reasoning is valid. for the => direction, i assume that there exists some neighborhood of x that contains only finitely many points of S. then i say that there exists some neighborhood of x inside this neighborhood such that it contains no points of S other than x by picking it small enough, which is a contradiction.

    the <= direction seems clear

    also, neighborhoods as defined in this context are the open sets of points "a" such that |x-a| < e for any e.

    am i missing any subtleties in my argument? thanks.
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  2. #2
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    Re: proof on neighborhoods and accumulation points

    Quote Originally Posted by oblixps View Post
    prove that: S is a subset of R. if every neighborhood of x contains at least one point of S other than x itself, then every neighborhood of x contains infinitely many points of S.
    This may fly right over your head. But this is true if the space is a Hausdorff space, a T_2-space. Every two points are separated by disjoint open sets. Any metric space is a Hausdorff space.
    If you understand that definition then the proof is trivial.
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