Froda's countable bound is attained: a bounded nondecreasing function on $\mathbb{R}$ discontinuous exactly at the points $1-1\mathrm{/}(k+1)$ for $k\in \mathbb{N}$, an infinite discontinuity set inside a bounded interval

Put

$E:=\{1-\frac{1}{\iota (k+1)}:k\in \mathbb{N}\}=\{0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots \}\subseteq [0,1)$

(The canonical natural $\iota (n)=n\cdot 1_F$ of a field, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. $E$ is countably infinite (Finite, countably infinite, countable, uncountable);
2. there is a nondecreasing $f:\mathbb{R}\to \mathbb{R}$ with $0\le f\le 1$ whose set of discontinuities is exactly $E$, every one of them a jump (Converse to Froda: for every at most countable $E\subseteq \mathbb{R}$ there is a bounded nondecreasing $f:\mathbb{R}\to \mathbb{R}$ whose set of discontinuities is exactly $E$, every one of them a jump, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences, Discontinuity of $f$ at a point of its domain, and its classification: removable discontinuity, jump discontinuity and essential discontinuity, equivalently Rudin's discontinuities of the first and of the second kind);
3. $E$ is contained in the bounded interval $[0,1)$, so a monotone function may have infinitely many discontinuities inside a bounded interval.

Indexing. $\mathbb{N}$ contains $0$, so the points are $1-1\mathrm{/}\iota (k+1)$ for $k\in \mathbb{N}$ and never $1-1\mathrm{/}\iota (k)$, which is undefined at $k=0$; the first point of $E$ is $1-1\mathrm{/}\iota (1)=0$.

The point $1$ is not in $E$ and $f$ is continuous there. $E$ has $1$ as a limit point but does not contain it, and claim 2 asserts continuity at every point outside $E$, so in particular at $1$: a monotone function may be continuous at a limit point of its own discontinuity set.

• What the example is for. Froda's theorem bounds the discontinuity set of a monotone function by countability and by nothing else; in particular it does not bound it by finiteness, even inside a bounded interval. The set $E$ above is the simplest witness: infinitely many jumps accumulating at a single point, all within $[0,1)$.

• The accumulation point is a point of continuity. The real $1$ is not a member of $E$, so claim 2 gives continuity of $f$ at $1$, even though every neighbourhood of $1$ contains infinitely many discontinuities of $f$. Being a limit of discontinuities is not itself an obstruction to continuity.

• A denser example is available. Taking $E=\mathbb{Q}$ instead gives a monotone function discontinuous on a dense set (A bounded nondecreasing $f:\mathbb{R}\to \mathbb{R}$ whose set of discontinuities is exactly $\mathbb{Q}$, obtained from the prescribed-jump construction applied to one fixed enumeration of the rationals); the present example is the smaller and more concrete one, and it is the one where the points can be listed.