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Continuity of an inverse function

  1. May 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the a continuous function with compact domain has a continuous inverse. Also prove that the result does not hold if the domain is not compact.


    2. Relevant equations



    3. The attempt at a solution
    I tried using the epsilon delta definition of continuity but didn't get anywhere.
    This is supposed to be a pretty standard proof in any real analysis book but unfortunately the one I'm using has it listed as an exercise...
     
  2. jcsd
  3. May 7, 2008 #2

    Dick

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    First, you need to add that the function is 1-1 to your premises. I'll give you a hint as to why you need compactness by giving you the usual counterexample f:[0,2*pi)->R^2 defined by f(x)=(cos(x),sin(x)). Why does that show the result doesn't hold without compactness?
     
  4. May 7, 2008 #3
    Hmm, how do I prove the first part (that it is continuous)
     
  5. May 7, 2008 #4

    Dick

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    Think about it. Suppose f:X->Y. A function is continuous if for every convergent sequence in X, x_n->x that f(x_n)->f(x). f^(-1) is continuous if for every convergent sequence y_n->y in Y that f^(-1)(y_n)->f^(-1)(y). You want to show f^(-1) is continuous given f is continuous, 1-1, and X is compact. Hint: define x_n such that f(x_n)=y_n. Does the sequence x_n have a limit point? Why?
     
  6. May 7, 2008 #5
    But isn't f(x) = (cos(x), sin(x)) not a 1-1 function (because it's a circle)?
     
  7. May 7, 2008 #6

    Dick

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    No. Why? Show me two values of x that map to the same value of (cos(x),sin(x)) in the domain [0,2*pi).
     
  8. May 7, 2008 #7
    oh sorry, i also forgot to add that
    f:[a,b] => R
    I got the continuity part down (your hint really helped!), but I'm having trouble with the compact part.
     
  9. May 7, 2008 #8

    Dick

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    What kind of 'trouble with the compact part'? You are given that the domain is compact, you don't have to prove it.
     
  10. May 7, 2008 #9
    I'm having trouble with giving a counter example of a continuous 1-1 function f:[a,b] => R whose inverse is not continuous (does it even exist)?.
     
  11. May 7, 2008 #10

    Dick

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    You won't find one for f:[a,b]->R. [a,b] is compact. You are going to have to be more creative and pick a non-compact domain.
     
  12. May 7, 2008 #11
    If the domain is non-compact, does such a function exist?
     
  13. May 7, 2008 #12

    Dick

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    Yes. Pick the domain X to be [0,1) union [2,3]. Notice the open end on the first interval. Can you define a SIMPLE function f:X->R whose inverse isn't continuous? Concentrate on using the open endpoint.
     
  14. May 7, 2008 #13
    sorry im stumped... i tried playing with functions such as f(x)=1/(x-1) but im not sure what to do
     
  15. May 7, 2008 #14

    Dick

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    Try f(x)=x on [0,1) and f(x)=x-1 on [2,3]. Since I just gave that to you, for free, you have to explain to me why it works. Is it continuous? What's the range? Is it 1-1? Finally give me a formula for the inverse (like the one for f) and tell me why it's discontinuous.
     
  16. May 7, 2008 #15
    ooo that makes so much more sense:
    range of f is [0,2] but the inverse f^-1: [0,2] => [0,1) U [2,3] is clearly discontinuous at x=1.

    thanks so much
     
  17. May 7, 2008 #16

    Dick

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    Sure, but can you help a little more next time, and show some more of approaches you have tried?
     
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