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Math Help - Need to show {x | f(x) ≤ g(x)} closed if f,g continuous.

  1. #1
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    Need to show {x | f(x) ≤ g(x)} closed if f,g continuous.

    I'm sure this is pretty elementary but I am intellectually needy.

    I need to show that with f,g :X \rightarrow Y both continuous and with Y in the order topology, the set \{x \ | \ f(x) \leq g(x) \} is closed.

    My work so far: I am trying to show that the complement is open. If we denote the given set by A, then the complement is A^c = \{x \ | \ f(x) > g(x) \}. For a particular x satisfying f(x) > g(x) we have one of two cases:

    (1) f(x) is the immediate successor of g(x). In this case, let U \subset Y be defined by U = \{y \ | \ y > g(x)\}, and let V \subset Y be defined by V = \{y \ | \ y < f(x) \}.

    (2) There exists some y_0 satisfying f(x) > y_0 > g(x) ; in this case, define U \subset Y as U = \{y \ | \ y > y_0\}, and define V \subset Y as V = \{y \ | \ y < y_0\} for some such y_0.

    In both of these cases, U,V are open sets (since Y has the order topology), and since f,g are continuous, we have that f^{-1}(U) and g^{-1}(V) are both open in X.

    That's the work that I've done so far, with some coaching from a professor. Now what I would like to do is create a collection of sets \{U_x\}_{x \in X}, and then claim that since f^{-1}(U_x) is open for each x ( U_x is open and f is continuous), we can take the union \bigcup_{x \in X} U_x and claim it to be open in X. This is where my reasoning gets a little blurry, though. Is this union really the same as \{x \ | \ f(x) > g(x)\}? It seems in a certain sense that it is, since each f^{-1}(U) produces the set of inputs such that the outputs of f(x) are greater than the outputs of g(x) for that particular x. But in another sense it seems obvious that this does NOT compose the set in question, since if we allow x to range over all possible values then the union will include a great many values for which in general it is not true that f(x) > g(x). ALSO, and perhaps more vexingly, the current argument does not make use of the continuity of g.

    Can anyone help with this? Thanks!

    -Steve
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  2. #2
    ynj
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    if f(x0)>g(x0) for certain x0,let u=(f(x0)-g(x0))/3
    then there exist v such that for any |x-x0|<v, |f(x)-f(x0)|<u,|g(x)-g(x0)|<u because of the continuuity.
    Thus f(x)>g(x) for any |x-x0|<v
    So it will be an open set.
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  3. #3
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    Quote Originally Posted by ynj View Post
    if f(x0)>g(x0) for certain x0,let u=(f(x0)-g(x0))/3
    then there exist v such that for any |x-x0|<v, |f(x)-f(x0)|<u,|g(x)-g(x0)|<u because of the continuuity.
    Thus f(x)>g(x) for any |x-x0|<v
    So it will be an open set.

    The problem does not state that the functions in question are real-valued, and so we cannot use things like "division by 3" in the answer. If they were both real-valued then the problem would be very easy. But we have no measures, no algebraic structure, no binary operations or any other structure on Y other than that it is a topological space with the order topology and nothing on X other than that it is an arbitrary topological space.
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  4. #4
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    Quote Originally Posted by numberstrong View Post
    A^c = \{x \ | \ f(x) > g(x) \}. For a particular x satisfying f(x) > g(x) we have one of two cases:

    (1) f(x) is the immediate successor of g(x). In this case, let U \subset Y be defined by U = \{y \ | \ y > g(x)\}, and let V \subset Y be defined by V = \{y \ | \ y < f(x) \}.
    (2) There exists some y_0 satisfying f(x_0) > y_0 > g(x_0) for an arbitrary x_0 in X; in this case, define U \subset Y as U = \{y \ | \ y > y_0\}, and define V \subset Y as V = \{y \ | \ y < y_0\} for some such y_0.
    For 2, let x_0 \in A^c. Then, f^{-1}(U) \cap g^{-1}(V) is an open set containing x_0 (f and g are continous and an intersection of open sets is an open set) and is contained in A^c. Since x_0 is an arbitrary point A^c = \{x \in X \ | \ f(x) > g(x) \} and the choice of y_0 is followed by the choice of x_0, we conclude that A^c is open.

    For 1, let x_0 \in A^c ; let U = \{y \in Y \ | \ y > g(x_0)\} and V = \{y \in Y \ | \ y < f(x_0) \}. Again, f^{-1}(U) \cap g^{-1}(V) is an open set containing x_0 and is contained in A^c (A choice of x_0 is arbitrary in A^c = \{x \in X \ | \ f(x) > g(x) \} as well ). Thus, A^c is open.

    Since A^c is open, we conclude that A is closed.
    Last edited by algtop; August 15th 2009 at 05:14 AM. Reason: Changed f(x) > y_0 > g(x) into f(x_0) > y_0 >g(x_0) for an arbitrary x_0 in X.
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