Let $f$ be a strictly increasing function (that is, $f(b) > f(a)$ if $b > a$). Show that $f$ is continuous at at some point.
Hint: Use the fact that uncountably many positive real numbers can not have a finite sum.
This leads me to the following idea. Let $J = \mathbb{R} \setminus f(\mathbb{R})$.
If $J = \varnothing$, then $f$ is surjective. Consider any $x$, and fix $\varepsilon > 0.$
Take $a$ and $b$ such that $f(a) = f(x) – \varepsilon$ and $f(b) = f(x) + \varepsilon.$ Since the function is strictly increasing, we must have $x \in (a,b)$ and for any other $x'$ in the interval we have:
$a < x' < b \implies f(x) < f(x') < f(b) \implies |f(x') – f(x)| < \varepsilon.$
Then choose $\delta = \min \{|x-a|, |x-b| \}$ to ensure $a \leq x – \delta < u < x+\delta \leq b.$ Thus if $|x-u| < \delta,$ we have $|f(x) – f(u)| < \varepsilon$ as desired.
The case that $J$ is countable can be dealt with quite similarly. This leaves the case where $J$ is uncountable, and without loss of generality we can consider $J \cap (0,\infty)$ uncountable as well by considering the function $g$ where $g(x) = f(x) + a$.
This seems to be the point where the hint might come in to play? Any further hints would be greatly appreciated.
Best Answer
Notice that by monotonicity, all the intervals of the form $(\lim\limits_{x\to x_0^-}f(x),f(x_0))$ and $(f(x_0),\lim\limits_{x\to x_0^+}f(x))$ are disjoint for all $x\in \mathbb R$.
Suppose that $f$ is discontinuous at $x_0$, this means that $\lim\limits_{ x_\to x_o^+}f(x)$ or $\lim\limits_{x\to x_0^-}f(x)$ is different from $x_0$. and so one of the two open open intervals is non-empty.
This implies that $f$ must have a numerable number of discontinuities, since each non-empty open interval contains rational points.