Let $f$ be a real-valued monotone function defined on an interval $I$. Then we know that the set $D \subset I$ of discontinuities of the first kind is at most countable. Then can I say that the complement $C=I\backslash D$ of $D$ is dense, please?
I suppose this is true since we can imagine getting rid of all rationals (may be bigger than $D$) in $I$ and still end up with a dense set. If so, can I claim that the complement of a countable set in an uncountable set is dense, please? Thank you!
Assume the complement is not dense, then there is an open set $(a,b)\subseteq I$ such that $I\setminus D\cap (a,b)=\varnothing$ But since we have a bijection of $(a,b)$ with $(0,1)$--namely $f(x) = {1\over b-a}(x-a)$--which is known to be uncountable by Cantor's diagonalization argument to be uncountable, it is impossible that $(a,b)\subseteq D$, hence no open subset of $I$ lacks points of $I\setminus D$, and so it is dense.