Let I be a non-empty Interval and $f:I\rightarrow \mathbb{R}$ a monotone, non decreasing function. Show that $f$ has countable many Classification of discontinuities.
My prove is nearly complete. But I'm missing one step. I thought I showed that there is a injection between the set of points where $f$ is not continuous and the rationals, but according to my assistant professor, I'm only allowed to do that if I show that there are rationals in the interval $(f_-(x),f_+(x))$, for$$ f_−(x) = \sup \{f(x′) | x′∈ [a,b], x′< x\}, $$
$$ f_+(x) = \inf \{f(x′)) | x′∈ [a,b], x′> x\}. $$
I know I can do that by applying the Archimedean property, but for that I need to show ( formally correct ) that $f_-(x)<f_+(x)$, where x is my point of discontinuity.
For me it's clear that $f_-(x)= f_+(x)$ can't be true, otherwise it wouldn't be a point of discontinuous so it has to be $f_-(x) < f_+(x)$ but how can this be proved?
thanks in advance