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Continuous inverse funtion real analysis

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Let I be an interval in the real line, and let f map I --> R be a one-to-one, continuous function.
    Then prove that f^(-1) maps f(I) --> R is also continuous



    3. The attempt at a solution

    I've started with the definition of continuity but I don't see where to go next.
     
  2. jcsd
  3. Nov 1, 2008 #2

    morphism

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    Hint: by the intermediate value theorem, f is either increasing or decreasing on I. (Be careful though - the intermediate value only applies to closed and bounded intervals.)
     
  4. Nov 1, 2008 #3
    To CarmineCortez,
    Rudin gives a proof in his Principle of mathematics - Continuity but with f being a continuous 1-1 mapping of a compact metric space X onto a metric space Y. ( interval is of course compact) and you may check it out.

    I'm asking that, in OP's thread, what if the "interval" is replaced by "segment", (a,b) for example? I guess the conclusion should not hold true in this case and the proof should take some properties of interval. However, at least it seems okay to me that f is still strictly monotonic... puzzled.
     
  5. Nov 2, 2008 #4

    morphism

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    The conclusion will still follow even if I isn't closed, bounded, or both. Once you prove that f is increasing/decreasing everywhere, then an easy epsilon-delta argument will give you continuity.

    Of course if I is closed and bounded, then the entire proof is a 2-linear, provided you know a bit about compactness.
     
  6. Nov 2, 2008 #5
    yea, thanks! so my proof didn't fail me :) Hope the OP has solved it, too
     
  7. Nov 2, 2008 #6
    thnx
     
    Last edited: Nov 2, 2008
  8. Nov 2, 2008 #7
    I'm still having trouble, can the IVT be used for a interval that is not bounded, if not then I am lost with this one.
     
  9. Nov 3, 2008 #8
    well, I'm not sure if there's any other way to use IVT, this is how I did:
    suppose f is not monotonic, there should exist three distinct points, called x, y, z such that x<y<z and without loss of generality we assume that f(x)<f(y),f(y)>f(z). (note that being 1-1 function, it exclude the possiblity of '='). Using IVT on [x,y] or [y,z] we can conclude that f is monotonic.
    The rest of proof is similar. you can choose some interval you want and use IVT just on it. you do not need to use IVT on an unbounded interval (actually, I do not see how to use it for an unbounded interval)
    Hope that helps. I believe that someone else has better proof.
     
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