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Inverse Image of an Open Set

  1. Aug 22, 2004 #1
    Why is the inverse image of the open set {t|t<1/2},
    {t|t>=1/2 or t<1/4}?

    the t>+1/2 sort of makes sense to me, but can't seem to grasp how t<1/4 is an inverse image
     
  2. jcsd
  3. Aug 22, 2004 #2

    HallsofIvy

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    What do you mean by inverse image?? A set has an image or inverse image under some function. What function are you talking about?

    (My first week in grad school, I was called upon to do a proof in class about "inverse image" of sets. I completely embarrased myself by assuming that, since the word "inverse" was used, f must have and inverse function!)
     
  4. Aug 22, 2004 #3

    mathwonk

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    one thing seems clear, the function will not likely be continuous, since an open set has a non open inverse image (unless the domain is a little special).
     
  5. Aug 22, 2004 #4
    http://www.csh.rit.edu/~pat/math/papers/topology/topology.pdf

    At the end of section 3.2 on Multipilication of Paths, they speak of the inverse image of the open set I mentioned earlier.

    I suppose the function is the step function y=2t when 0<=t<1/2
    and y=0 when t>=1/2

    Thanks for the replies
     
  6. Aug 22, 2004 #5

    HallsofIvy

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    Yes, and it specifically refers to the inverse image under a given function. In particular, the function of the problem you are referring to (on page 5 of your reference) is :
    &tau;&sigma;(t)= 2t if 0<= x< 1/2
    0 if x>= 1/2
    and notes that this is not a "path" because it is not continuous (as mathwonk pointed out from the given solution). (&tau;(t) and &sigma;(t) were defined separately.)

    We really need to know that before we can answer your question!

    Now, what is the inverse image of {t|t< 1/2}?
    That is, what are the values of t such that &tau;&sigma(x)< 1/2?

    Well, certainly 0< 1/2 so all t> =1/2 qualifies.
    In order that 2t< 1/2, we must have t<1/4. That is certainly less than 1/2 so it fits the formula.
    f(t)< 1/2 as long as t is either < 1/4 or >= 1/2 as claimed.
     
  7. Aug 22, 2004 #6
    Ah, I see. Thanks for clearing up.
     
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