Does there exist a function $f : \mathbb{R} \rightarrow \mathbb{R}$ that is strictly increasing and discontinuous everywhere?
My line of thought (possibly incorrect): I know there are increasing functions such as $f(x) = x$, and there are everywhere-discontinuous functions such as the Dirichlet function. I also know that when there is a discontinuity at a point $c$, there is a finite gap $\epsilon$ such that there are points $d$ arbitrarily close to $c$ such that $|f(d) - f(c)| > \epsilon$. This is where my thinking gets unclear - does it make sense to have a "gap" at every real number?
Answer
There is no such function. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is strictly increasing. For each $a\in\mathbb{R}$ let $f^-(a) =$ $\lim\limits_{x\to a^-}f(x)$ and $f^+(a) = \lim\limits_{x\to a^+}f(x)$. Then $f$ is discontinuous at $a$ if and only if $f^-(a) < f^+(a)$. Let $D = \{a\in\mathbb{R}:f\text{ is not continuous at }a\}$, and for each $a\in D$ let $q_a$ be a rational number in the non-empty open interval $I_a = (f^-(a),f^+(a))$.
It’s not hard to check that if $a,b \in D$ with $a
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