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Math Help - The UnBounded Function

  1. #1
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    The UnBounded Function

    Here is my own problem.

    Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

    *I have been able to show that f is discontinous everywhere.

    Now I want to find an example of such a function.

    * shows how difficult it might be.

    Or show it does not exist.
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is my own problem.

    Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

    *I have been able to show that f is discontinous everywhere.

    Now I want to find an example of such a function.

    * shows how difficult it might be.

    Or show it does not exist.
    If it is unbounded on any interval of the real line, how can this really be considered a function?

    -Dan
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is my own problem.

    Let f be a function on (-oo,+oo) such that for any interval I, f is unbounded on I.

    *I have been able to show that f is discontinous everywhere.

    Now I want to find an example of such a function.

    * shows how difficult it might be.

    Or show it does not exist.
    As we can construct a sequence of nested closed intervals I_n, and our
    function is unbounded on all of them it must be infinite (undefined on
    (-infty, infnty)) at the common point of the I_n's. This implies that it is undefined
    everywhere so does not exist!

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    As we can construct a sequence of nested closed intervals I_n, and our
    function is unbounded on all of them it must be infinite (undefined on
    (-infty, infnty)) at the common point of the I_n's. This implies that it is undefined
    everywhere so does not exist!

    RonL
    Did you use the "Nested Interval Theorem"?
    If so can you please be very formal, because I never used that theorem except once. Thus, I do not know it well.
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    Quote Originally Posted by ThePerfectHacker View Post
    Did you use the "Nested Interval Theorem"?
    If so can you please be very formal, because I never used that theorem except once. Thus, I do not know it well.

    I've changed my mind the idea that I had in mind does not work.

    RonL
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    How about this for an idea:

    1. let f(x)=0 for x irrational

    2. Arrange the rationals [0,1) in the order 0, 1/2, 1/3, 2/3, 1/4, 2/4, 3,4,...

    3. Then if r_n is the n-th rational in this sequence then put f(r_n)=n

    4. Put f(x) = f(frac(x)), frac(x) the fractional part of x.

    Conjecture: F as defined above has the property you are looking for
    (and if it does not it can be modified slightly to have)

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    How about this for an idea:

    1. let f(x)=0 for x irrational

    2. Arrange the rationals [0,1) in the order 0, 1/2, 1/3, 2/3, 1/4, 2/4, 3,4,...

    3. Then if r_n is the n-th rational in this sequence then put f(r_n)=n

    4. Put f(x) = f(frac(x)), frac(x) the fractional part of x.

    Conjecture: F as defined above has the property you are looking for
    (and if it does not it can be modified slightly to have)

    RonL
    I did not check it yet. Because I am about to leave for class.

    But, remember the important point I mentioned. IT MUST be discontinous everywhere. It provides an easy check.

    Note, if in your function it turns out that f(0)=0 then we have a problem. It is continous at 0.
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    Quote Originally Posted by ThePerfectHacker View Post
    I did not check it yet. Because I am about to leave for class.

    But, remember the important point I mentioned. IT MUST be discontinous everywhere. It provides an easy check.

    Note, if in your function it turns out that f(0)=0 then we have a problem. It is continous at 0.
    f(0)=1

    but it would not be continuous at 0 even if f(0) did equal 0, as for any n, f(1/n)>n, so there would be points
    x arbitarily close to 0 for which f(x) is arbitarily large.

    RonL
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    I was discussing this with somebody else today he gave the following construction.

    f(x)=0 for irrationals.

    f(a/b)=b for rationals where gcd(|a|,b)=1 and a!=0

    f(0)=1.

    I think that works.

    Because small intervals contain rationals with large denominators.

    I think we should not forget this function, because it might be useful for disproving certain conjectures about functions.
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    Quote Originally Posted by ThePerfectHacker View Post
    I was discussing this with somebody else today he gave the following construction.

    f(x)=0 for irrationals.

    f(a/b)=b for rationals where gcd(|a|,b)=1 and a!=0

    f(0)=1.

    I think that works.

    Because small intervals contain rationals with large denominators.

    I think we should not forget this function, because it might be useful for disproving certain conjectures about functions.
    Look at this this is like the sort of thing that I had in mind.

    Looking t what I suggested I now note that it needs the refinment (to make
    my f single valued):

    f(x) = 0 for x irrational.

    for x rational f(x) = n, where x=r_n and n is the least n for which
    this is true.


    RonL
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