Let $g(x) :\mathbb{R}\to\mathbb{R}$ be a non-decreasing function in the sense that $g(y)\geq g(x) \; \forall y \geq x$. Can $g(x)$ be nowhere continuous? My guess is that any such function that is non-decreasing must have at most countably infinite number of discontinuities. Otherwise, it will cease to be discontinuous due to non-decreasing nature. However, I couldn't come up with a rigorous proof of this. Am I right in my guess?
My attempt was to assume that for each $x_0 \in \mathbb{R}$ there exists a fixed $\varepsilon_{x_0}>0$ such that for all $x \in (x_0,\infty)$ we have the relation that $f(x)>f(x_0)+\varepsilon_{x_0}$ by non-decreasing nature because for each arbitrary $\delta$ neighborhood of $x_0$ there is at least one $x'$ in it that satisfies the relation (hence all $x>x'$[non-decreasing] works but $x'$ can be made arbitrarily close to $x_0$). My intuition is that such a strictly increasing relation can only be forced for countable number of points. Is there a nice way to show it? Or I am on the wrong track?
Nvm: This proof strategy is wrong. Take the Floor function.
And yes, that works! Uncountable sums of strictly positive but small real numbers will become unbounded very easily and the relation won't make sense anymore. Thank you.
– Someone Apr 25 '20 at 10:36