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Homework Help: Inverse proof

  1. Feb 21, 2009 #1

    jgens

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

    Prove that if f:X -> Y is a discontinuous bijection then f-1:Y -> X is also discontinuous.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    The contrapositive of this statement is that if f-1:Y -> X is continuous then f:X -> Y is continuous. Since f is bijective its invertable; hence, all I need to prove is, if f:X -> Y is continuous then f-1:Y -> X is continuous.

    If f is continuous there exits a value delta for every epsilon such that delta = g*epsilon, where g is a function of epsilon.

    We use the epislon delta method to prove f-1 is continuous. Hence, for all epsilon greater than zero there exists a delta greater than zero such that 0 < abs(y - k) < delta and 0 < abs(f-1(y) - f-1(k)) < epsilon. Hence, we may rewrite this as 0 < abs(f(x) - f(c)) < delta and 0 < abs(x - c) < epsilon. Since, this is merely the situation for f with epsilon and delta switched it follows that delta = epsilon/g. Therefore f-1:Y -> X is continuous. Q.E.D.

    I think some of my reasoning isn't particularly great, so any suggestions on how to correct this proof are welcome. Thanks.
     
  2. jcsd
  3. Feb 21, 2009 #2
    You can't prove an implication by proving its converse. While that statement is also true, it does not show that "if f-1:Y -> X is continuous then f:X -> Y is continuous". If you're trying to prove the contrapositive statement the straightforward constructive way, you have to start by assuming that f-1 is continuous.
     
    Last edited: Feb 21, 2009
  4. Feb 21, 2009 #3

    jgens

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    Sorry. I figured that since if a function is completely invertable then it's inverse is completely invertable as well, and since any function f that has a complete inverse g can also be called g-1, it followed that if f is a continuous bijection then f-1 is also continuous or equivalently stated if g-1 is a continuous bijection, then g is also continuous. I should have caught that and proved it probably. It should be simple to fix that part though.

    If f-1 is continuous then for every epsilon (e) there exists a delta (d) given by d = g*e where g is a function of e.

    We use the epsilon delta method to prove f is continuous. For all e > 0 there exists a d > 0 such that 0 < abs(x - c) < d and 0 < abs(f(x) - f(c)) < e. Which is equivalent to 0 < abs(f-1(y) - f-1(k)) < d and 0 < abs(y - k) < e. Since this is merely the situation for f-1 with e and d switched it follows that d = e/g. Q.E.D.

    I'm certain there still has to be something wrong with that though.
     
  5. Feb 21, 2009 #4
    Without certain conditions on X the statement to prove is false, anyway. A common counterexample is with X = [0,1) U (1,2], f(x)=x on [0,1] and f(x)=x+1 on (1,2].
    EDIT: X should be [0,2].
     
    Last edited: Feb 21, 2009
  6. Feb 21, 2009 #5

    jgens

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    Wouldn't f be discontinuous then? f has to be continuous.

    I'll go along with your other criticisms though. I'll try to explain my reasoning more clearly. First, what I mean by d = g*e: Suppose d = 2e +1, this can be expressed in the form d = g*e where g = 2 + 1/e, I'm simply trying to generalize something. Given 0 < abs(x - c) < e and 0 < abs(f(x) - f(c)) < d. If we go about the same process used to find e we would get e = g*d where g is a function of delta. Now I suppose my "proof" only works if you can explicitly express that in terms of d which is what I presume you were pointing out.
     
  7. Feb 21, 2009 #6

    jgens

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    As an aside, since my proof is invalid, would anyone mind trying to point me in the right direction?
     
  8. Feb 21, 2009 #7

    jgens

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    Nevermind, no need. I found a proof of it that I understand in my calculus book. Thanks for the repsonses!
     
  9. Feb 21, 2009 #8
    We can only go by what you wrote:
    Correcting the example I gave (I originally had it switched around, sorry): make that X=[0,2], then f is a discontinuous bijection with a continuous inverse.
     
    Last edited: Feb 21, 2009
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