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What is a Direct image and Inverse Image in Real Analysis?

  1. Aug 28, 2010 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand the definition of Direct and Inverse Images in Real Analysis I from my book, see attachment please.
     

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  3. Aug 28, 2010 #2
    Please explain how the set of the inverse image of G was found.
     

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  4. Aug 28, 2010 #3
    We're looking for all elements in the domain of f that can be mapped into elements of G, i.e., find all x such that f(x) is an element of G. For what values of x in the domain is 0 <= f(x) <= 4?
     
  5. Aug 28, 2010 #4
    Right, but how did they find 0 and 4 so that

    0 <= f(x) <= 4

    ?
     
  6. Aug 28, 2010 #5
    Read the text step-by-step.

    We're given E = {0 <= x <= 2}. The image of E is G = {0 <= f(x) <= 4}. Now we are finding the inverse image of G by the steps I described above. The purpose of this exercise is to just give you an example of where [tex] f(f^{-1}(A)) = A [/tex] and an example where [tex] f^{-1}(f(A)) \neq A. [/tex]
     
  7. Aug 28, 2010 #6

    HallsofIvy

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    The given function was [itex]f(x)= x^2[/itex] and [itex]E= \{0\le x\le 2\}[/itex].

    They got [itex]0\le f(x)\le 4[/itex] by squaring 0 and 2.

    But be careful, if it had been [itex]E= \{-2\le x\le 2\}[/itex] We would NOT have the image [itex]\{4\le f(x)\le 4}[/itex]. [itex]-2\le x\le 2[/itex] includes all numbers from -2 to 2 and the squares of those are all between 0 and 2, also the squares of the negatives are also positive, not negative: If [itex]E= \{-2\le x\le 2\}[/itex] we would still have the image as [itex]\{0\le f(x)\le 4\}[/itex]
     
    Last edited: Aug 29, 2010
  8. Aug 29, 2010 #7

    Mark44

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    Correction: all numbers between -2 and 2 (inclusive).
     
  9. Aug 29, 2010 #8

    HallsofIvy

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    Thanks for the correction, Mark44. I will now edit so I can claim I never made that mistake!
     
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