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Showing that a given function is continous over a certain topology

  1. Oct 8, 2014 #1
    Consider the maps h: R^w (omega) ---> R^w (omega) , h (x1, x2, x3,......) = (x1,4x2, 9x3,.......)
    g: (same dimension mapping) , g (t) = (t, t, t, t, t,....)

    Is h continuous whn given the product topology, box topology, uniform topology?

    For the life of me i am having trouble trying to understand what i need to do to accomplish this. I know that the definition of continuous in topology is that the pre image f^-1 (v) of an open set v has to be open, but how do i apply that to this sort of situation?

    Take the function h, am i assuming that (x1, 4x2, 9x3,.....) is my open set or am i going to take an arbitrary basis element of my open set that looks like (x1, 4x2, 9x3,.....) apply the preimage to that, and then if it is open then i can conclude the function is continuous?

    I saw a solution for the function g, and they said it is not open in the box topology. They took th open set [-1/(n^2) , 1/(n^2)] and said taking the inverse image of g the set is {0}. How did they do this? I fail to see the connection. Please help i've been stuck here all day tryimg to make sense of this.

     
  2. jcsd
  3. Oct 8, 2014 #2

    RUber

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    I imagine that for g, they were taking the limit as ##n \to \infty##. This would give you an open set for the image, but in the limit, the pre-image is the closed singleton {0}.
     
  4. Oct 8, 2014 #3

    Yes, but how do they get that singleton is my question. Is it that you look at each inverse image of each coordinate function individually and take the intersection of those , which produces the common point 0?
     
  5. Oct 8, 2014 #4

    RUber

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    Assume that ##U## is an open set ##U=(-1/n^2, 1/n^2)##, for ##V=g^{-1}(U) ## to be open, there must be ##\epsilon \in V, \, \epsilon \neq 0 ##, such that ##g(\epsilon) \in U##. But for any epsilon, there is an ##n ## that makes ##g(\epsilon) \notin U##.
    You could also think of ##U = (-\epsilon, \epsilon)## if this helps.
    Wikipedia uses exactly this example to describe how box topology fails at continuity tests: http://en.wikipedia.org/wiki/Box_topology
     
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