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Math Help - Topology Closed Neighborhood Proof

  1. #1
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    Topology Closed Neighborhood Proof

    (8) N is a closed neighborhood of p in R_U iff N=R
    Where R represents the real numbers and
    R_U is the upper topological space for the real numbers (R, T_U)
    T_U = {A is a subset of R | a is an element of A implies (a - e, +inf) is a subset of A for some e > 0} (upper topology)

    I proved "=>"
    Need help showing "<="
    ie: If N=R, then N is a closed neighborhood of p in R_U
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  2. #2
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    Re: Topology Closed Neighborhood Proof

    The topology of \mathbb{R}_U is the set \{ \emptyset \} \cup \{ \mathbb{R} \} \cup \{ (r, \infty) | r \in \mathbb{R} \}.

    The direction you did was the "hard" direction. I think you forgot that the whole space is always a closed set, since it's the complement of the empty set, which is always open.

    Let p \in \mathbb{R}. Then p \in (p-1, \infty), which is open. Let N = \mathbb{R}. Will show N is a closed neighborhood of p.

    Have that N = \{\emptyset\}^c is closed. Since p \in (p-1, \infty) \subset N, it follows that N is a closed neighborhood of p.
    Last edited by johnsomeone; September 13th 2012 at 05:46 PM.
    Thanks from Kiefer
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