CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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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 $f : A \to \mathbb{R}$ on a subset $A \subseteq \mathbb{R}$ is strictly monotone (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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 $\mathbb{R}$ , 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 $\mathbb{R}$ : 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 $f : A \to \mathbb{R}$ by

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

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 $\varepsilon > 0$ there is a real $\delta > 0$ with $|f(x) - f(c)| < \varepsilon$ for every $x \in A \cap N_\delta(c)$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

Constants, the identity and their sums and scalar multiples are continuous on every subset of $\mathbb{R}$ (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 $\mathbb{R}$ , with the dictionary to monotone sequences).

[L4]

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

[L5]

A continuous injective function on an order-convex subset of $\mathbb{R}$ 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 $\varepsilon > 0$ be real; take $\delta := \min\{1, \varepsilon\}$ . Every $x \in A$ with $|x - c| < \delta$ satisfies $x < c + 1 \le 2$ , so $x \in [0,1]$ and $|f(x) - f(c)| = |x - c| < \varepsilon$ .

L1L2

1.3

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

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 $0 \le 3/2 \le 3$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 85 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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] \cup [2,3][0,1]∪[2,3] that is not monotone, so the interval hypothesis cannot be dropped from the strict-monotonicity theorem