Find a function f:R->R (R the set of real numbers), such that
1. f has a discontinuity in every rational number, but is continous everywhere else, and
2. f is monotonic: x<y → f(x)<f(y)
Note: Textbooks frequently present examples of functions that meet only the first condition; requiring monotonicity makes for a slightly more challenging problem.
(In reply to re(2): Oops... Maybe another hint?
I'm not getting any closer. Here are my answers to your hint questions (and I'm adding the constraint that f_r be monotonic).
1. Given a rational number r, what is the simplest function f_r you can think of, that is discontinuous at r and continuous elsewhere?
f_r(x) = x for x < r
f_r(x) = x+1 for x = r
f_r(x) = x+2 for x > r
2. Now imagine that you have a whole bunch of rational numbers, how could you use the corresponding f_r's to construct a function that is discontinuous at all your rational numbers?
Assume that the "bunch" of rationals are arranged in ascending order and numbered r0, r1... rN. The function would be:
f_r(x) = x for x<r0
f_r(x) = x+1 for x=r0
f_r(x) = x+2 for r0<x<r1
f_r(x) = x+3 for x=r1
3. Can that be extended if the "whole bunch" consists of ALL rational numbers?
Um...no. First of all, you can't put ALL the rational numbers in ascending order--you can't even start. Second, it's impossible to declare which two unique rational numbers an irrational number is between for purposes of deciding how much to add to x to obtain the function's value.
I must not be thinking of the same "simplest" function that you have in mind, JLo. Maybe Steve (or someone else) can make more of this hint. I do suspect that the function will have an infinite definition (finite for any particular x in the domain), but I don't see how it will be defined for the irrationals.
(hmm.. I just thought of something...I'll be back..)
--- (later that day...) ---
Dead end. I thought by considering the m/n representation of every rational, I could devise a function of m and n that would show promise. No luck.
Edited on August 20, 2006, 11:50 pm