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Math Help - Image of an Increasing Function

  1. #1
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    Image of an Increasing Function

    Hi,
    Could someone help me in proving this statement:

    Let I be an interval and assume that f: I--->R is an increasing function. Prove that if the image of f(I) is connected then f must be continuous.

    Thanks a lot
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AKTilted View Post
    Hi,
    Could someone help me in proving this statement:

    Let I be an interval and assume that f: I--->R is an increasing function. Prove that if the image of f(I) is connected then f must be continuous.

    Thanks a lot
    What have you tried?
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  3. #3
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    So far, I've written down the definition of an interval as follows:
    There exists a,b in S and c in S such that a < c < b. S

    Let S1 = Intersection (-inf,c)
    Let S2 = Intersection (c,inf).

    Then I've assumed f is not confinuous at a point a and started an epsilon-delta arguement.

    I think it has something to do with f(c) + epsilon/2, but I can't complete the proof...
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  4. #4
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    Use the notation f(c+) for the limit on the right at x=c.
    Likewise for the limit on the left, f(c-).
    Monotonic functions are quasi-continuous, so both exist.
    If f is not continuous at x=c then either f(c-)<f(c) or f(c)<f(c+).
    If f(c-)<f(c) consider \left( { - \infty ,f(c - )} \right] \cup \left[ {f(c),\infty } \right).
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AKTilted View Post
    So far, I've written down the definition of an interval as follows:
    There exists a,b in S and c in S such that a < c < b. S

    Let S1 = Intersection (-inf,c)
    Let S2 = Intersection (c,inf).

    Then I've assumed f is not confinuous at a point a and started an epsilon-delta arguement.

    I think it has something to do with f(c) + epsilon/2, but I can't complete the proof...
    Try thinking of it this way. Suppose that f was not continuous then you can find some x_0\in S, and some \varepsilon>0 such that for all ]math]\delta>0[/tex] there exists some y_\delta such that |x-y_\delta|<\delta and |f(x)-f(y_\delta)|\geqslant \varepsilon. Try creating a contradiction with this.

    Hint: Think about the IV property.
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