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Why can monotone functions only have jump discontinuities?

  1. Jun 23, 2011 #1
    I'm trying to understand why monotone functions can only have jump discontinuities. I've seen formal proofs, but I don't find them intuitively convincing; my imagination can still conceive of an increasing function which is just a bunch of scattered points. To take a simple case, why can't you have a monotone function whose range only contains rational numbers, or only contains irrational numbers?

    Or to take a more complicated case, Conway's base 13 function satisfies the conclusion of the intermediate value theorem even though it is nowhere continuous. Is there some kind of operation which "pushes up points" on the graph of this function which would make it increasing?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Jun 24, 2011 #2

    HallsofIvy

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    You can. And all discontinuities are jump discontinuities.

     
  4. Jun 24, 2011 #3
    I thought a jump discontinuity occurs when both one-sided limits occur, but are unequal. It seems like for the function I'm describing, one-sided limits would not exist anywhere. Isn't there some theorem to the effect that, if f is discontinuous everywhere on some interval containing a, then the one-sided limits of f(x) as x goes to a cannot exist?

    Also, I thought a function can only have countably many jump discontinuities, while the function I'm describing would be discontinuous over the entire real line.
     
  5. Jun 24, 2011 #4
    OK, I just realized that my description of the function wasn't what I intended. So let me restate my question in this way: why can't you have a monotone function whose range contains only irrational numbers and contains all the irrational numbers in some interval (a,b)?
     
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