CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A continuous injection on $[0, 1] \cup [2, 3]$ that is not monotone, so the interval hypothesis cannot be dropped from the strict-monotonicity theorem

Statement refuted

Refuted claim: every continuous injective function $f : A \to R$ on a subset $A \subseteq R$ is strictly monotone (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, Injection, surjection, bijection, Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

A continuous injective function on an interval is strictly monotone proves this under the hypothesis that $A$ is order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). The refuted claim drops that hypothesis, and it is false.

Counterexample

Let $A : = [0, 1] \cup [2, 3]$ and define $f : A \to R$ by

$f (x) : = x for x \in [0, 1], f (x) : = 5 - x for x \in [2, 3] .$

Then $f$ is continuous on $A$ and injective, and it is not monotone: $f (0) = 0 < 1 = f (1)$ while $f (2) = 3 > 2 = f (3)$ . The set $A$ is not order-convex, since $0, 3 \in A$ and $3 / 2 \notin A$ .

Facts & Assumptions

Given: The set $A = [0, 1] \cup [2, 3]$ and the function $f$ above.

[L1]

$f$ is continuous at $c \in A$ when for every real $ε > 0$ there is a real $δ > 0$ with $∣ f (x) - f (c) ∣ < ε$ for every $x \in A \cap N_{δ} (c)$ (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, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L2]

Constants, the identity and their sums and scalar multiples are continuous on every subset of $R$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claims 1 and 5).

[L3]

$f$ is increasing when $f (x) < f (y)$ for all $x < y$ in $A$ , decreasing when $f (x) > f (y)$ for all $x < y$ in $A$ , and monotone when nondecreasing or nonincreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L4]

$A$ is order-convex when $x, y \in A$ and $x \leq z \leq y$ imply $z \in A$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

A continuous injective function on an order-convex subset of $R$ is strictly monotone, and its inverse on the image is continuous (A continuous injective function on an interval is strictly monotone, Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

Verification

technique · direct

1.1

$f [[0, 1]] = [0, 1]$ and $f [[2, 3]] = [2, 3]$ : on $[0, 1]$ the map is the identity, and on $[2, 3]$ the map $x \mapsto 5 - x$ sends $2$ to $3$ and $3$ to $2$ and is order-reversing, so its image is $[2, 3]$ .

1.2

$f$ is continuous on $A$ . Let $c \in [0, 1]$ and let $ε > 0$ be real; take $δ : = \min {1, ε}$ . Every $x \in A$ with $∣ x - c ∣ < δ$ satisfies $x < c + 1 \leq 2$ , so $x \in [0, 1]$ and $∣ f (x) - f (c) ∣ = ∣ x - c ∣ < ε$ .

L1L2

1.3

Let $c \in [2, 3]$ and let $ε > 0$ be real; take $δ : = \min {1, ε}$ . Every $x \in A$ with $∣ x - c ∣ < δ$ satisfies $x > c - 1 \geq 1$ , so $x \in [2, 3]$ and $∣ f (x) - f (c) ∣ = ∣ (5 - x) - (5 - c) ∣ = ∣ x - c ∣ < ε$ .

L1L2

1.4

$f$ is not monotone: $0 < 1$ with $f (0) = 0 < 1 = f (1)$ rules out nonincreasing, and $2 < 3$ with $f (2) = 3 > 2 = f (3)$ rules out nondecreasing.

1.5

$A$ is not order-convex: $0 \in A$ , $3 \in A$ and $0 \leq 3 / 2 \leq 3$ , but $3 / 2 \notin [0, 1] \cup [2, 3]$ .

2.1

$f$ is injective: it is injective on $[0, 1]$ , being the identity there; it is injective on $[2, 3]$ , since $5 - x = 5 - y$ gives $x = y$ ; and the two images $[0, 1]$ and $[2, 3]$ are disjoint, so no point of one piece has the same value as a point of the other.

step 1.1

3.1

So $f$ is a continuous injection on $A$ that is not monotone, refuting the claim; and the hypothesis that fails is exactly order-convexity of the domain, which is what the theorem assumes.

step 2.1step 1.2step 1.3step 1.4step 1.5L5∎

Remarks

The inverse is still continuous here, and that is a coincidence of this example. $f$ is a bijection onto $[0, 1] \cup [2, 3]$ and its inverse is the same kind of piecewise map, continuous by the same argument. So this example does not refute the continuity of the inverse; what it refutes is monotonicity, and Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ derives continuity of the inverse from monotonicity, so on a domain that is not order-convex that route is unavailable even when the conclusion happens to hold.
Two pieces are enough, and the gap does the work. The values on $[0, 1]$ and on $[2, 3]$ never interfere, because the two images are disjoint; injectivity is therefore free and the two pieces may be oriented oppositely. On an order-convex domain the intermediate value theorem forbids exactly that, which is the content of steps 1.2 to 4.1 of A continuous injective function on an interval is strictly monotone.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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

Monotonic function (Wikipedia) (standard reference, not scraped)
Chapter 4: Continuous Functions (Trinity College Dublin) (standard reference, not scraped)

A continuous injection on [0,1]∪[2,3][0,1]∪[2,3] that is not monotone, so the interval hypothesis cannot be dropped from the strict-monotonicity theorem