Answer: Monotone Function with Countably Many Discontinuities

Key Idea / Intuition

A monotone function can only have jump discontinuities — at each point of discontinuity, there is a positive "gap" between the left and right limits. The beautiful insight is that these gaps are disjoint intervals sitting inside $\mathbb{R}$, and the rationals thread through each one, giving a canonical injection from discontinuities into $\mathbb{Q}$.

---

Formal Proof / Solution

Step 1: Monotone functions have only jump discontinuities.

Since $f$ is increasing, at every point $x \in (0,1)$ the one-sided limits exist: $f(x^-) = \lim_{t \nearrow x} f(t), \quad f(x^+) = \lim_{t \searrow x} f(t),$ and satisfy $f(x^-) \leq f(x) \leq f(x^+)$.

A point $x$ is a discontinuity if and only if $f(x^-) < f(x^+),$ i.e., there is a positive jump of size $\delta_x = f(x^+) - f(x^-) > 0$.

(Handle endpoints $0$ and $1$ similarly with one-sided limits.)

Step 2: The jump intervals are pairwise disjoint.

For each discontinuity $x$, associate the open interval $I_x = \bigl(f(x^-),\, f(x^+)\bigr) \subset \mathbb{R}.$

Claim: if $x < y$ are both discontinuities, then $I_x \cap I_y = \emptyset$.

Indeed, since $f$ is increasing, $f(x^+) \leq f(y^-)$, so the interval $I_x$ lies entirely to the left of $I_y$: $f(x^-) < f(x^+) \leq f(y^-) < f(y^+).$

Step 3: Inject discontinuities into $\mathbb{Q}$.

Since the intervals $\{I_x\}$ are pairwise disjoint, non-empty, and open, by density of $\mathbb{Q}$ in $\mathbb{R}$, each $I_x$ contains a rational number $q_x \in \mathbb{Q}$.

The map $x \mapsto q_x$ is injective (distinct discontinuities have disjoint intervals, hence different chosen rationals).

This gives an injection $\{\text{discontinuities of } f\} \hookrightarrow \mathbb{Q},$ and since $\mathbb{Q}$ is countable, the set of discontinuities is at most countable. $\blacksquare$

---

Remark (Sharpness). The result is sharp: every countable set $S \subset [0,1]$ can be realized as the exact set of discontinuities of some increasing function. For instance, enumerate $S = \{x_1, x_2, \ldots\}$ and set $f(x) = \sum_{n:\, x_n \leq x} 2^{-n},$ which is increasing with a jump of $2^{-n}$ at each $x_n$.