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Regular spaces

  1. Nov 21, 2009 #1
    how do I show a topological space X with an order topology is regular. Ive shown it is hausdorff already.
     
  2. jcsd
  3. Nov 23, 2009 #2
    So, let S be a closed set, and x be a point not in S. Since S is closed, X\S is open, so write it as a union of open intervals. Since x is in X\S, fix an interval (a,b) in X\S containing x.
    Case 1: there exists an element c in (a,x) and an element d in (x,b). In that case, (-infinity,c) U (d, infinity) is an open set containing S disjoint from (c,d), an open set containing x.
    Case 2: There exists an element c in (a,x) but (x,b) is empty. Then (c,x] = (c,b) is open. In that case, (-infinity, c) U (x,infinity) is an open set containing S disjoint from (c,x], an open set containing x.
    Case 3: both (a,x) and (x,b) are empty. Then {x}=(a,b) is open. Then (-infinity, x) U (x, infinity) is an open set containing S, disjoint from {x}, an open set containing x.

    EDIT: I realize this isn't exactly correct, since you need to consider the case x is in an interval of the form (a, infinity) or (-infinity, b), but I think those can be handled the same way.
     
    Last edited: Nov 23, 2009
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