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Pre-image open or not?

  1. Jan 31, 2012 #1
    Hi all,

    I am struggling with the following:

    If X and Y are topological spaces. and f: X x Y → ℝ is a continuous function (product topology on X x Y, Euclidean topology on ℝ)

    Let g: X → ℝ defined by g(x) = sup { f(x,y) | y in Y }

    Then: If A=(r, ∞) for r in ℝ, g-1(A) is open. And If A=(-∞, t) for t in ℝ, g-1(A) is not always open.

    Why is that? How can I know if g-1(A) is open or not if I dont know anything about X??

    Does anyone have an idea?

    kind regards,
     
  2. jcsd
  3. Jan 31, 2012 #2
    This is one of those "apply the definitions" problems. Let [tex]x \in g^{-1}(t,\infty) [/tex]. By definition, there exists [tex]\epsilon >0[/tex] such that [tex]g(x)>t+2\epsilon[/tex]. By definition of supremum, there exists [tex]y \in Y [/tex] such that [tex]f(x,y)>t+\epsilon[/tex]. By definition of continuity, there exists an open set A containing (x,y) such that [tex]f(x,y)>t+\epsilon/2[/tex] for all [tex](x,y) \in A[/tex]. By definition of the product topology, there exists an open set B in X containing x such that [tex]\forall z \in B, \exists y_z \in Y, (z,y_z) \in A [/tex]. Hence [tex]g(z) > t [/tex] for all z in B.

    For the reverse, just construct a counterexample. Take the plane and consider the function [tex]f(x,y)=arctan(x^2 y^2)[/tex]
     
  4. Feb 1, 2012 #3
    Hi zhentil,

    Thank you very much for helping!

    Sorry, I don't see immediately why this holds:

    "By definition of continuity, there exists an open set A containing (x,y) such that
    f(x,y)>t+ϵ/2 for all (x,y)∈A"

    For this we need that some subset of { f(x,y) } is open?
     
  5. Feb 1, 2012 #4
    or is it the set were t + ε/2 < f(x,y) < t + ε ?
     
  6. Feb 1, 2012 #5
    [tex](t+\epsilon/2,\infty)[/tex] is an open set in R. Hence its pre-image is open by definition of continuity. (x,y) lies in that set.
     
  7. Feb 1, 2012 #6
    thank you.

    "For the reverse, just construct a counterexample. Take the plane and consider the function
    f(x,y)=arctan(x2y2)"

    Would g-1((-∞, b)) be open if Y is a compact space?
     
  8. Feb 7, 2012 #7

    Bacle2

    User Avatar
    Science Advisor


    Well, let's see. Let me review for next time I teach point set topology, so you can
    combine it with Zhentil's answer:

    What is g-1(a,b)? it is the collection of all x such that there is a y in Y
    with a<f(x,y)<b.

    We have g=Sof(x,y) , where f:XxY→f(XxY), and S:f(XxY)→ℝ , and g-1:=

    f-1os-1.

    Like Zhentil said, if there is y0 with f(x,y0)>a , then , by

    (assumed) continuity of f, there is a ball B(x,y0) where f(x,y)>a . This

    gives you openness in the subspace f(XxY). Now, compose with f-1, to

    get an open set in XxY, by assumed continuity of f(x,y).
     
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