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Hausdorff Space and finite complement topology

  1. Apr 28, 2012 #1
    I want to come up with examples that finite complement topology of the reals R is not Hausdorff, because by definition, for each pair x1, x2 in R, x1 and x2 have some disjoint neighborhoods.

    My thinking is as follows: finite complement topology of the reals R is a set that contains open sets of the form, (- inf, a1) U (a1, a2) U ... U (an, inf). The complement of an open set is the finite elements {a1, a2, ... an}. However, any two points I pick out of any open set have disjoint neighborhoods. How is it possible?
     
  2. jcsd
  3. Apr 28, 2012 #2
    OK, pick two elements, how do they have disjoint neigborhoods???
     
  4. Apr 28, 2012 #3
    Ok, I see where my reasoning was wrong. The neighborhoods have to be open sets in this topology. If I pick two points, {0, 2}, in the open set (-inf, 1) U (1 inf), their neighborhoods can't be any open intervals and must be rays.

    Cheers!
     
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