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Math Help - continues prove question..

  1. #1
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    continues prove question..

    prove that all the points of discontinuity
    of a monotonic and bounded function
    are points of discontinuity of the first type?
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  2. #2
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    Recall that a point x=c is a discontinuity of the first kind if \lim_{x\to{c-}}f(x) and \lim_{x\to{c+}}f(x) exist. Suppose that f(x) is monotonically increasing (an analogous argument will exist for monotonically decreasing f(x)). Then for an increasing infinite sequence of x_n<c with \lim_{n\to\infty}x_n=c, we see that f(x_n) is an increasing sequence. Consider this in combination with the boundedness of f(x), and what that says about the existence of \lim_{x\to{c-}}f(x). Similar work for a decreasing sequence x_n>c, \lim_{n\to\infty}x_n=c will give a similar result for \lim_{x\to{c+}}f(x).

    --Kevin C.
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  3. #3
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    a monotonic function is a function which is systematically increasing or decreasing.

    a bounded function means that there is at least one subsequence
    which converges to limit.

    so our function converges if its monotonic and bound.
    and every subsequence converge to the same limit.

    so you say that this limit is a point of discontinuity called C.

    and we have two cases
    the first is for monotonically increasing f(x)

    and that it converges to point C from the left (minus side)

    but C cannot be the limit of the function
    after a discontinuity point

    the function must go on and converge to real limit
    it cannot stop on point C

    ??
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    To OP: Possible alternate approach...wait for someone to confirm.


    You could also approach this from the contradiction approach. Let f be a montonically increasing real function f. Next suppose that c is a discontinuity of f but it is of the second kind. This implies that \lim_{x\to{c^+}}f(x) or \lim_{x\to{c^-}}f(x) does not exist. I will show case of \lim_{x\to c^-}f(x) and you can extrapolate the other case. So suppose that \lim_{x\to c^-}f(x) does not exist. This implies that there does not exist a p>0( the case is similar for negative p) such that for every \varepsilon>0 there exist a \delta>0 such that c-x<\delta\implies |p-f(x)| <\varepsilon. So for any \delta we can find a x_1<c such that c-x_1<c-x<\delta but |p-f(x_1)|>|p-f(x)|\implies f(x_1)<f(x), but this violates that f is monotonically increasing.
    Last edited by Mathstud28; December 27th 2008 at 12:06 PM.
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  5. #5
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    \varepsilon>0 mean that on certain surrounding

    so  \delta>0

    what is the meaning of  \delta>0 ??



    c is our point of discontinuity
    what is x??

    why
     c-x<\delta\implies f(c) - f(x) <\varepsilon

    ??
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  6. #6
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    by this
    <br />
c-x<\delta\implies |p-f(x)| <\varepsilon<br />
    this is a definition of a limit

    can you explain in simple words
    why are doing this
    what is the meaning of this part

    <br /> <br />
c-x_1<\delta<br />
    but
    <br /> <br />
|p-f(x_1)|>|p-f(x)|\implies f(x_1)<br />
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