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Real Analysis: Continuity and discontinuity

  1. Nov 21, 2013 #1
    1) For the following choice of A, construct a function f: R → R that has discontinuities at every point x in A and is continuous on the complement of A.

    A = { x : 0 < x < 1} My function is f(x) = 10 if x in (0,1) and Q and f(x) = 20, if x in (0,1) and irrational number, f(x) = 30, elsewhere. By the way, this is piece wise function.

    Does this example work?

    Also for A = {1/n: n is natural number) Does f(n) = 10 if 1/n is natural number and f(n) = 0, elsewhere work?
     
  2. jcsd
  3. Nov 21, 2013 #2
    That will also be discontinuous in ##0## and ##1##.

    That will be discontinuous in ##0##.
     
  4. Nov 21, 2013 #3
    For the first one does make 10 elsewhere work?

    For second one, does f(n) = 1/n where n is natural number and 0 elsewhere work?
     
    Last edited: Nov 21, 2013
  5. Nov 22, 2013 #4

    pasmith

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    That fails to be continuous at 0 and 1.

    You need [itex]\displaystyle\lim_{x \to 0} f(x) = f(0)[/itex] and [itex]\displaystyle\lim_{x \to 1} f(x) = f(1)[/itex]. If you just assign different constants to rational and irrational numbers then those limits will not exist and your function will not be continuous at 0 and 1.

    For f to be continuous at 0, you must have [itex]\displaystyle\lim_{n \to \infty} f(1/n) = f(0)[/itex].
     
    Last edited: Nov 22, 2013
  6. Nov 22, 2013 #5
    I think you can use an idea similar to this:
    http://en.wikipedia.org/wiki/Thomae's_function
    But irrationals have to have other value, then you may have a function that is continous only in the extremes.
     
  7. Nov 22, 2013 #6

    PeroK

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    For the first one, you need f to be continuous at 0 and 1. This is a problem for x in (0,1), where you need both rationals and irrationals to be heading towards the same limit.

    So, how can you do that?

    Suppose, for example, f(0) = 0 and f(1) = 1. And f(x) = x on the irrationals in (0, 1) and everything outside (0, 1).

    Then, how can you define f on the rationals within (0,1)?

    You need lim q -> 0 f(q) = 0 and lim q -> 1 f(q) = 1.
     
  8. Nov 22, 2013 #7
    Here's an idea for the first part:

    Split (1/2, 1) into a countable number of subintervals. Specifically, (1/2, 3/4), (3/4, 7/8), (7/8, 15/16), etc. Let the functional value for all the irrationals in (1/2, 1) be zero. Let the functional value for the rationals in each subinterval be (1/n), i.e. the functional value for the rationals in (1/2, 3/4) is 1/1, the functional value for the rationals in (3/4, 7/8) is 1/2, etc.
     
  9. Nov 23, 2013 #8

    PeroK

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    That would work, but you don't need to do anything so complicated. 1-q on the rationals in (1/2, 1) would do in this case!
     
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