A strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ is continuous at least one point. 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. 
 A: 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.
A: Suppose $f$ is such a function. Let $a,b\in \Bbb{R}$ with $a < b$. Given $x\in (a,b)$, define
$$d_x = \inf_{y > x} (f(y)-f(x))$$
The fact that $f$ is increasing means that the set of all $f(y)-f(x)$ with $y>x$ is bounded below by $0$, guaranteeing that this $\inf$ exists. Now choose points $x_0, x_1, ..., x_n$ such that 
$$a=x_0 < x_1 < ... < x_{n-1} < x_n = b$$
then
$$f(b) - f(a) \geq (f(x_n) - f(x_{n-1}))+...+(f(x_1)-f(x_0)) \geq d_{x_{n-1}}+...+d_{x_0}$$
Since the values $n$ and $x_1, ..., x_{n-1}$ are arbitrary, it follows that
$$f(b)-f(a) \geq \sum_{x\in [a,b)}d_x$$
Can you finish the proof? 
