TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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Converse to Froda: for every at most countable $E \subseteq R$ there is a bounded nondecreasing $f : R \to R$ whose set of discontinuities is exactly $E$ , every one of them a jump

Statement

Let $E \subseteq R$ be at most countable (Finite, countably infinite, countable, uncountable). Then there is a function $f : R \to R$ such that

$f$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences) and $0 \leq f (x) \leq 1$ for every real $x$ , so $f$ is bounded (Lower bound, bounded below, bounded set);
$f$ is continuous at every $x \notin E$ and discontinuous at every $x \in E$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), so the discontinuity set of $f$ is exactly $E$ ;
every discontinuity of $f$ is a jump (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), with $\lim_{x \to c^{-}} f (x) = f (c) < \lim_{x \to c^{+}} f (x)$ at every $c \in E$ .

Together with Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $N$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used this settles the question completely: the sets that occur as discontinuity sets of monotone functions on $R$ are exactly the at most countable ones.

The construction. For $E = \emptyset$ take $f : = 0$ . Otherwise fix a surjection $s : N \to E$ (A nonempty set is at most countable iff it is a surjective image of $N$ ) and set

$f (x) : = \sum_{k = 0}^{\infty} a_{k} (x), a_{k} (x) : = {\begin{cases} 1 / 2^{k + 1} & if s (k) < x, \\ 0 & otherwise, \end{cases}$

(Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ ): the mass $1 / 2^{k + 1}$ is placed at the point $s (k)$ and is collected by $f$ strictly to the right of it. Repetitions in the enumeration are harmless; they only make the jump at a point larger.

Facts & Assumptions

Given: An at most countable $E \subseteq R$ .

[L1]

A nonempty at most countable set is the image of a surjection $s : N \to E$ (A nonempty set is at most countable iff it is a surjective image of $N$ , Finite, countably infinite, countable, uncountable).

[L2]

A series of nonnegative terms converges if and only if its partial sums are bounded above, and its sum is then the supremum of its partial sums; in particular every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series, Lower bound, bounded below, bounded set).

[L3]

Finite sums: $\sum_{k < n}$ is monotone in the terms, splits as $\sum_{k < n} = \sum_{k < m} + \sum_{k = m}^{n - 1}$ for $m \leq n$ , scales, and telescopes as $\sum_{k < n} (c_{k + 1} - c_{k}) = c_{n} - c_{0}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L4]

$\sum_{k \geq 0} r^{k}$ converges to $1 / (1 - r)$ for $∣ r ∣ < 1$ , the first term being $r^{0} = 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Integer powers $a^{m}$ ); a series converges if and only if each of its tails does, and $\sum_{k \geq 0} u_{k} = \sum_{k < N} u_{k} + \sum_{k \geq N} u_{k}$ (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail); a convergent sequence of reals comes within every positive $ε$ of its limit from some index on (Limits and Cauchy sequences of reals).

[L5]

A nonempty finite set of reals, presented as ${c_{0}, \dots, c_{m}}$ , has a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set); and strictly between any two distinct reals there lies a real (The rationals embed densely in the reals).

[L6]

A nondecreasing function on an order-convex set has both one-sided limits at every interior point, and is discontinuous there exactly when they differ, in which case the discontinuity is a jump (One-sided limits of a monotone function always exist: for $f$ nondecreasing on an interval $I$ and $c \in I$ , $\lim_{x \to c^{-}} f (x) = \sup {f (x) : x \in I, x < c}$ whenever $I$ has points below $c$ , $\lim_{x \to c^{+}} f (x) = \inf {f (x) : x \in I, x > c}$ whenever it has points above $c$ , and these satisfy $\lim_{x \to c^{-}} f (x) \leq f (c) \leq \lim_{x \to c^{+}} f (x)$ , A monotone function on an interval has no discontinuity of the second kind: at every point both relevant one-sided limits exist, and an interior point $c$ is a discontinuity exactly when $\lim_{x \to c^{-}} f (x) < \lim_{x \to c^{+}} f (x)$ , 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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · constructive

1.1

If $E = \emptyset$ , the constant function $0$ is nondecreasing, takes values in $[0, 1]$ , is continuous at every real, and has empty discontinuity set; all three claims hold vacuously for claim 3. Assume from here on that $E \neq \emptyset$ and fix a surjection $s : N \to E$ .

L1construct

1.2

For every $n \in N$ , $\sum_{k < n} 1 / 2^{k + 1} = 1 - 1 / 2^{n}$ : each term is $1 / 2^{k + 1} = 1 / 2^{k} - 1 / 2^{k + 1}$ , so the sum telescopes to $1 / 2^{0} - 1 / 2^{n} = 1 - 1 / 2^{n}$ .

2.1

Define $a_{k} (x) : = 1 / 2^{k + 1}$ when $s (k) < x$ and $a_{k} (x) : = 0$ otherwise, and note $0 \leq a_{k} (x) \leq 1 / 2^{k + 1}$ for every $k$ and every real $x$ .

step 1.1construct

2.2

For every real $ε > 0$ there is $n \in N$ with $1 / 2^{n} < ε$ : the partial sums $t_{n} : = \sum_{k < n} 1 / 2^{k}$ converge to $2$ , and $t_{n} = 2 - 2 / 2^{n}$ by the same telescoping as in step 1.2, so $∣ t_{n} - 2 ∣ = 2 / 2^{n} < ε$ for all large $n$ , whence $1 / 2^{n} < ε / 2 < ε$ for those $n$ . Consequently the partial sums $1 - 1 / 2^{n}$ of $\sum_{k} 1 / 2^{k + 1}$ have supremum $1$ , so that series converges with sum $1$ .

step 1.2L2L3L4

3.1

For every real $x$ the series $\sum_{k} a_{k} (x)$ converges and $0 \leq f (x) \leq 1$ : its terms are nonnegative and its partial sums satisfy $\sum_{k < n} a_{k} (x) \leq \sum_{k < n} 1 / 2^{k + 1} = 1 - 1 / 2^{n} \leq 1$ , so they are bounded above by $1$ and the sum, being their supremum, lies in $[0, 1]$ .

step 2.1step 1.2L2L3

3.2

Left continuity holds at every real $c$ : given real $ε > 0$ take $n$ with $1 / 2^{n} < ε$ ; let $F : = {k < n : s (k) < c}$ ; if $F = \emptyset$ put $x_{0} : = c - 1$ , and otherwise put $x_{0}$ to be a real with $\max {s (k) : k \in F} < x_{0} < c$ , which exists because the maximum of the nonempty finite set ${s (k) : k \in F}$ is a real strictly below $c$ .

step 2.2L5

3.3

Right continuity holds at every $c \notin E$ : given real $ε > 0$ take $n$ with $1 / 2^{n} < ε$ ; since $c \notin E$ and $s$ has image $E$ , no $k$ has $s (k) = c$ , so every $k < n$ has $s (k) < c$ or $s (k) > c$ . Let $G : = {k < n : s (k) > c}$ ; if $G = \emptyset$ put $y_{0} : = c + 1$ , and otherwise put $y_{0}$ to be a real with $c < y_{0} < \min {s (k) : k \in G}$ .

step 1.1step 2.2L5

4.1

$f$ is nondecreasing: if $x \leq y$ then $s (k) < x$ implies $s (k) < y$ , so $a_{k} (x) \leq a_{k} (y)$ for every $k$ , hence $\sum_{k < n} a_{k} (x) \leq \sum_{k < n} a_{k} (y)$ for every $n$ , and taking suprema gives $f (x) \leq f (y)$ .

step 2.1step 3.1L2L3

4.2

For all reals $x \leq y$ and every $n \in N$ with $a_{k} (x) = a_{k} (y)$ for every $k < n$ , one has $f (y) - f (x) \leq 1 / 2^{n}$ : for $N \geq n$ the splitting $\sum_{k < N} a_{k} (y) = \sum_{k < n} a_{k} (y) + \sum_{k = n}^{N - 1} a_{k} (y) \leq \sum_{k < n} a_{k} (x) + \sum_{k = n}^{N - 1} 1 / 2^{k + 1}$ holds, the last sum being at most $\sum_{k \geq n} 1 / 2^{k + 1} = 1 - (1 - 1 / 2^{n}) = 1 / 2^{n}$ ; so every partial sum of $\sum_{k} a_{k} (y)$ is at most $f (x) + 1 / 2^{n}$ , and so is their supremum $f (y)$ .

step 2.1step 1.2step 3.1L2L3L4

4.3

Let $c \in E$ and fix $k_{0}$ with $s (k_{0}) = c$ . For every $y > c$ and every $N > k_{0}$ the finite sum $\sum_{k < N} a_{k} (y)$ exceeds $\sum_{k < N} a_{k} (c)$ by at least $1 / 2^{k_{0} + 1}$ , because the list $k \mapsto a_{k} (y) - a_{k} (c)$ has nonnegative entries, so the finite sum of its first $N$ entries is at least its entry at the index $k_{0}$ , which is $a_{k_{0}} (y) - a_{k_{0}} (c) = 1 / 2^{k_{0} + 1} - 0$ . Hence $f (y) - 1 / 2^{k_{0} + 1} \geq \sum_{k < N} a_{k} (c)$ for every $N$ , the case $N \leq k_{0}$ holding because the partial sums of a nonnegative series are nondecreasing; so $f (y) - 1 / 2^{k_{0} + 1}$ is an upper bound of those partial sums and therefore at least their supremum $f (c)$ .

step 1.1step 2.1step 3.1L2L3

5.1

With $x_{0}$ as in step 3.2 and any $x$ with $x_{0} < x \leq c$ : for $k < n$ with $s (k) < c$ we have $s (k) \leq \max {s (j) : j \in F} < x_{0} < x$ , so $a_{k} (x) = 1 / 2^{k + 1} = a_{k} (c)$ ; and for $k < n$ with $s (k) \geq c \geq x$ we have $a_{k} (x) = 0 = a_{k} (c)$ . So $a_{k} (x) = a_{k} (c)$ for every $k < n$ , and step 4.2 applied to the pair $x \leq c$ gives $0 \leq f (c) - f (x) \leq 1 / 2^{n} < ε$ .

step 2.1step 4.1step 4.2step 3.2

5.2

With $y_{0}$ as in step 3.3 and any $y$ with $c \leq y < y_{0}$ : for $k < n$ with $s (k) < c \leq y$ we get $a_{k} (y) = 1 / 2^{k + 1} = a_{k} (c)$ , and for $k < n$ with $s (k) > c$ we have $s (k) \geq \min {s (j) : j \in G} > y_{0} > y$ , so $a_{k} (y) = 0 = a_{k} (c)$ . So $a_{k} (y) = a_{k} (c)$ for every $k < n$ , and step 4.2 applied to the pair $c \leq y$ gives $0 \leq f (y) - f (c) \leq 1 / 2^{n} < ε$ .

step 2.1step 4.1step 4.2step 3.3

5.3

So $f$ is discontinuous at $c$ : for $ε : = 1 / 2^{k_{0} + 1} > 0$ and any real $δ > 0$ the point $y : = c + δ / 2$ satisfies $∣ y - c ∣ < δ$ and $∣ f (y) - f (c) ∣ \geq ε$ , so no $δ$ witnesses the continuity condition at $c$ .

step 4.3

6.1

Hence $f$ is continuous at every $c \notin E$ : fix a real $ε > 0$ , take $x_{0}$ as in step 3.2 and $y_{0}$ as in step 3.3 for that same $ε$ , and put $δ : = \min {c - x_{0}, y_{0} - c} > 0$ ; then every real $x$ with $∣ x - c ∣ < δ$ satisfies $x_{0} < x < y_{0}$ and therefore $∣ f (x) - f (c) ∣ < ε$ , by step 5.1 when $x \leq c$ and by step 5.2 when $x \geq c$ .

step 5.1step 5.2L5

6.2

Every point of $E$ is an interior point of the order-convex set $R$ , so both one-sided limits of $f$ exist there; step 5.1 gives $\lim_{x \to c^{-}} f (x) = f (c)$ and step 4.3 gives $\lim_{x \to c^{+}} f (x) \geq f (c) + 1 / 2^{k_{0} + 1} > f (c)$ . The two one-sided limits therefore differ, and the discontinuity at $c$ is a jump.

step 5.1step 4.3step 5.3L6

7.1

Claims 1, 2 and 3 hold for the function $f$ constructed in steps 1.1 and 2.1: claim 1 by steps 3.1 and 4.1, claim 2 by steps 6.1 and 5.3, and claim 3 by step 6.2.

step 3.1step 4.1step 6.1step 5.3step 6.2discharge-construct∎

Remarks

Why the mass is collected strictly to the right. The definition uses $s (k) < x$ rather than $s (k) \leq x$ , and that is what makes $f$ left continuous everywhere, as steps 3.2 and 5.1 show without any hypothesis on $c$ . The value $f (c)$ at a point of $E$ is therefore the left limit, and the whole jump sits on the right. Using $s (k) \leq x$ would produce a right continuous function with the same discontinuity set; nothing else would change.
Repetitions in the enumeration are harmless. If $s$ takes the value $c$ at several indices, the jump at $c$ is the total mass $\sum {1 / 2^{k + 1} : s (k) = c}$ rather than a single term. Step 4.3 uses only one index $k_{0}$ and so needs no such sum; it establishes a lower bound for the jump, which is all that discontinuity requires.
Boundedness is free, and it is worth recording. The total mass available is $\sum_{k \geq 0} 1 / 2^{k + 1} = 1$ , so $f$ maps $R$ into $[0, 1]$ however large $E$ is. A bounded nondecreasing function on $R$ can therefore have a dense set of discontinuities; the companion page takes $E = Q$ and gets exactly that.

Depends on

Used by

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80 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Classification of discontinuities (Wikipedia) (standard reference, not scraped)
Froda's theorem (Wikipedia) (standard reference, not scraped)
Discontinuities of monotone functions (Wikipedia) (standard reference, not scraped)

Converse to Froda: for every at most countable E⊆RE⊆R there is a bounded nondecreasing f:R→Rf:R→R whose set of discontinuities is exactly EE, every one of them a jump