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Prove f continuous given IVP and 1-1

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f : (-1, 1) → ℝ. f satisfies the intermediate value property and is one-to-one on (-1, 1). Prove f is continuous on (-1, 1)

    3. The attempt at a solution
    I was thinking that the IVP and one-to-one implies that f should be strictly monotonic and that a strictly monotonic one-to-one function is continuous. Both of these seem very intuitive to me and yet I have no idea how to do the rigorous proof. For example if f is not strictly monotonic then it seems there would be a contradiction in f one-to-one but does that follow directly? Since you don't know anything else about the function..
     
  2. jcsd
  3. Apr 18, 2012 #2
    You know it's funny, on wikipedia both versions of the Intermediate Value Theorem require f to be continuous as a hypothesis, are you sure your question isn't asking you to prove [itex]f^{-1}[/itex] is continuous?
     
  4. Apr 18, 2012 #3
    The question as stated is correct. It states that f satisfies the intermediate value property: the result of the theorem rather than the theorem itself.
     
  5. Apr 18, 2012 #4
    Ah yes, I apologize. This problem is tough. You won't be able to prove it's uniformly continuous, thus recall that proving point-wise continuity allows delta to depend not only on epsilon but also on the point at which you are trying to prove continuity.

    Thus choose a point [itex]x \in (-1,1)[/itex] and use the IVP to find another suitable point [itex]y[/itex] (this choice will depend on epsilon). Think about what to do with |f(x) - f(y)| and with |x-y|. Then consider an arbitrary point w with a certain bound on |x-w| and its implications for |f(x) - f(w)|; I believe proving these implications will require applying the IVP and one-to-one in coordination, but I haven't formally worked out the details.
     
    Last edited: Apr 18, 2012
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