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Math Help - Topology - Darboux Theorem

  1. #1
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    Topology - Darboux Theorem

    Let I be an open interval in \Re and let f:I \rightarrow \Re be a differentiable function.

    Let T be the set T = \{(x,y) \in I \times I : x<y\} . I can show that this is a connected subset of \Re^2 with the standard topology.

    Let g: T \rightarrow \Re be a function defined by g(x,y) = (f(x) - f(y)) / (x-y). Prove that g(T) \subset f'(I) \subset \overline{g(T)}.

    Show that f'(I) is an interval.

    This last deduction would be trivial if g(T) was a connected set, but I don't see why this should necessarily be true. (It would be true if g was a continuous function but we are not given this. Nevertheless, I think there may be a way of showing it).
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  2. #2
    tah
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    Hi, if you want g(T) to be connect, isn't it sufficient to have g continuous on T?
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  3. #3
    tah
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    Hi, if you want g(T) to be connected, isn't it sufficient to have g continuous on T?
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  4. #4
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    Sure, but can you actually show this?!
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  5. #5
    tah
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    The derivative of f may be discontinuous then!! hummm
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  6. #6
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    I think this is the whole point of the Darboux Theorem. The derivative of f satisfies the intermediate value theorem even if f' is not continuous. In other words, there is nothing to stop f' being discontinuous.
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  7. #7
    tah
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    So instead of the continuity, you may prove certain "convexity" of g !
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    Unless you can actually prove this and why it helps, then I'm not totally convinced by this idea!
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  9. #9
    tah
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    I wanted to show something like the image by g of an interval in T is an interval in R.

    But h(x,y) = f(x) - f(y) is continuous on I\times I and \frac{1}{x-y} continuous on T. So why g shouldn't be continuous on T ?
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  10. #10
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    Yes, to be honest this was my concern exactly, even though it seems to contradict the "point" of the question! (in that the derivative doesn't have to be continuous). I think it's because of the form of T...
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  11. #11
    tah
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    But the derivative is obtained only at the limit where g could be not defined or discontinuous!
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