I found this exercise:>Let $f$ be a monotone function, then it can have only a countable number of discontinuities.
I've seen around here a lot of posts about the same but I never get to any conclusion. What happens with Identity function restricted to the set of irrationals? It is monotone, but, isn't it discontinuous at every point of its domain? Which, by the way is uncountable.
I know that if $f$ were defined on an interval, it would be easy to prove, but this is not the case...
Could anyone clarify me please?
Actually, the identity function, restricted to the irrationals, is continuous at every point of its domain. If you want a $\varepsilon-\delta$ proof, then just pick, for any $\varepsilon>0$, $\delta=\varepsilon$. So, it is not a counter-example.