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Proving a Set in the Order Topology is Closed

  1. Jul 1, 2012 #1
    Proving a Set is Closed (Topology)

    1. The problem statement, all variables and given/known data

    Let [itex]Y[/itex] be an ordered set in the order topology with [itex]f,g:X\rightarrow Y[/itex] continuous. Show that the set [itex]A = \{x:f(x)\leq g(x)\}[/itex] is closed in [itex]X[/itex].


    2. Relevant equations



    3. The attempt at a solution

    I cannot for the life of me figure this out. As far as I can tell one either needs to show the set A gathers its limit points or that it is the pre-image (under f or g) of some closed set in Y. We know the order topology is Hausdorff, so that's something. I just don't even know how to get started on this one. Hopefully someone can help, thanks.
     
    Last edited: Jul 1, 2012
  2. jcsd
  3. Jul 2, 2012 #2
    This one is tricky.

    Instead of trying to prove directly that A is closed, try to show that the complement, X-A, is open. Show the set of all x such that f(x) > g(x) to be open. You can show this by showing that every point x in X is an interior point. Just find an open set around an arbitrary x.

    So let x be an arbitrary element of X such that f(x) > g(x). If there is no such x, then the set is empty, then and it is open trivially. So assume such an x exists and consider f(x) and g(x) in the order topology.

    Now draw the number line to represent Y and put f(x) and g(x) in their respective places.

    Then find two open sets in Y.

    Does that help get you started?
     
  4. Jul 2, 2012 #3
    So going from there we use the fact that the order topology is Hausdorff and find two open disjoint sets one containing f(x) and the other containing g(x), from here I would like to map back under f^-1 and g^-1 to two open sets in X and from here take the intersection of them which would contain x and would be open, but I can't see how we know anything about these two disjoint open sets in Y...
     
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