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? by JLo)
JLo:
Maybe I'm being dense, but this hint doesn't get me any closer.
1) The simplest monotonic function which is discontinuous at a point
and continuous elsewhere is a step function, because of the jump at the
point.
2) If we have a finite number of rational numbers, we can clearly construct a function that jumps at each rational number.
3) But I'm still not figuring out how to construct a function that
jumps at every rational number. We already knew that the weird
function would have a jump at every rational number. And this
means, incidentally, that if it is monotonic increasing then it is
strictly monotonic increasing, because there are rational numbers
between every pair of irrational numbers. But this hint still
leaves me where I was at my second post.
Except now I am hoping VERY MUCH that this function really works as advertised. Because I am prepared to be impressed.
Steve
re(3): Oops... Maybe another hint?
