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Math Help - Regular space

  1. #1
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    Regular space

    I am working on this problem, but I am stuck in the middle. Could someone please give me a hand?
    I am trying to prove that the set of real numbers R with topology T on R consisting of all sets of the form (x,\infty), where  x\in R, is not a T_3 space.
    I have (0,\infty) is open in this space. So, A=R-(0,\infty)=(-\infty,0] is closed in this space and 1\notin (-\infty,0]. Suppose there exist U,V open in R such that A\subset U and 1 \in V. I claimed that U \cap V=\emptyset. I don't see how an open set in R can contain A but does not contain 1 because any open set in R must be of the form (x,\infty). However, I don't have a rigorous argument here.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by jackie View Post
    I am working on this problem, but I am stuck in the middle. Could someone please give me a hand?
    I am trying to prove that the set of real numbers R with topology T on R consisting of all sets of the form (x,\infty), where  x\in R, is not a T_3 space.
    I have (0,\infty) is open in this space. So, A=R-(0,\infty)=(-\infty,0] is closed in this space and 1\notin (-\infty,0]. Suppose there exist U,V open in R such that A\subset U and 1 \in V. I claimed that U \cap V=\emptyset. I don't see how an open set in R can contain A but does not contain 1 because any open set in R must be of the form (x,\infty). However, I don't have a rigorous argument here.
    I think you are making this harder than it need be. This space isn't even Haudorff since any neighborhood U of 0 contains 1 since there must exist a set of the form (a,\infty)\subseteq U with a<0 but then 1\in (a,\infty)\subseteq U
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