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Relatively open sets

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f be a continuous real-valued function of R^n. Show that [(x,y): x \in R^n, y > f(x)] is an open subset of R^{n+1}

    2. Relevant equations

    3. The attempt at a solution

    If I am thinking about this right... Since f(x) goes from (-oo, y), this is an open subset, and as f is continuous, the domain of f(x) must be open, so x\in U is open.
    Last edited: Nov 16, 2008
  2. jcsd
  3. Nov 16, 2008 #2
    I don't know if it is just me but I am having a hard time understanding what is going on in your set.

    As for functions on open sets, perhaps you may want to consult the different definitions of a continuous function (besides the traditional episilon delta defn).
  4. Nov 16, 2008 #3
  5. Nov 16, 2008 #4
    What is the complement of this set?
  6. Nov 16, 2008 #5
    Would the complement be as all the x are open, then the complement would be closed. Then the complement of the y would be y <= f(x), and as both of these are closed, then in R^{n+1} they would be closed. Thus as the complement is closed, it must be open?

    I guess one problem I am having is the y term. Are the f(x) chosen so that it is less than y? Thus y is one point. Or is it given an f(x), the y's go from (f(x), oo).
  7. Nov 16, 2008 #6
    It looks like y is chosen so it is less than f(x) for all f(x) in your range.
    If you can show that the complement is closed then you are done.
  8. Nov 16, 2008 #7
    I know if I show the complement is closed then it is open, but is there any reasonable way to show that the complement for this set is closed?
  9. Nov 16, 2008 #8
  10. Nov 17, 2008 #9
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