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

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A continuous injective function on an interval is strictly monotone

Statement

Let $I \subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) and let $f : I \to \mathbb{R}$ be continuous on $I$ (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) and injective (Injection, surjection, bijection). Then $f$ is strictly monotone (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$ , with the dictionary to monotone sequences): either $f(x) < f(y)$ whenever $x < y$ in $I$ , or $f(x) > f(y)$ whenever $x < y$ in $I$ .

Both hypotheses are needed and neither can be weakened to the other. Continuity alone does not give injectivity, and injectivity alone does not give monotonicity: the companion page exhibits a continuous injection on $[0,1] \cup [2,3]$ , a set that is not order-convex, that is not monotone. So it is order-convexity of the domain, and not merely continuity, that forces the conclusion.

Facts & Assumptions

Given: An order-convex $I \subseteq \mathbb{R}$ and a continuous injective $f : I \to \mathbb{R}$ .

[A1]

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

[A2]

$f$ is injective: $f(u) = f(v)$ implies $u = v$ (Injection, surjection, bijection).

[A3]

$f$ is continuous at every point of $I$ ; the restriction of $f$ to a subset $S \subseteq I$ is continuous at every point of $S$ , since the $\varepsilon$ - $\delta$ condition of 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 quantifies over fewer points when the domain shrinks.

[L1]

Intermediate value theorem: if $u \le v$ , $g : [u,v] \to \mathbb{R}$ is continuous and $y$ lies between $g(u)$ and $g(v)$ in either order, then $g(p) = y$ for some $p \in [u,v]$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a,b]$ takes every value between $f(a)$ and $f(b)$ ).

[L2]

Strictly between any two distinct reals there lies a real (The rationals embed densely in the reals).

[L3]

$f$ is increasing when $f(x) < f(y)$ for all $x < y$ in $I$ , decreasing when $f(x) > f(y)$ for all $x < y$ in $I$ , and strictly monotone when it is one or the other (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$ , with the dictionary to monotone sequences).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $f$ is not strictly monotone: $f$ is not increasing and $f$ is not decreasing.

assume-contra

1.2

Three-point claim. For all $u < v < w$ in $I$ , either $f(u) < f(v) < f(w)$ or $f(u) > f(v) > f(w)$ . Suppose not. By injectivity the three values are pairwise distinct, so the failure means that $f(v)$ is not between $f(u)$ and $f(w)$ ; hence either $f(v) > f(u)$ and $f(v) > f(w)$ , or $f(v) < f(u)$ and $f(v) < f(w)$ .

2.1

Being decreasing means $f(x) > f(y)$ for all $x < y$ in $I$ , so its failure gives $a, b \in I$ with $a < b$ and not $f(a) > f(b)$ , that is $f(a) \le f(b)$ ; injectivity together with $a \ne b$ gives $f(a) \ne f(b)$ , so $f(a) < f(b)$ .

step 1.1A2L3

2.2

In the first case of step 1.2 pick a real $y$ with $\max\{f(u), f(w)\} < y < f(v)$ ; in the second pick $y$ with $f(v) < y < \min\{f(u), f(w)\}$ . Such a $y$ exists because the two bounds are distinct reals.

step 1.2L2

3.1

With $y$ as in step 2.2, $[u,v] \subseteq I$ and $[v,w] \subseteq I$ by order-convexity, and $f$ restricted to each is continuous; $y$ lies strictly between $f(u)$ and $f(v)$ , and strictly between $f(v)$ and $f(w)$ . So there are $p \in [u,v]$ and $q \in [v,w]$ with $f(p) = y = f(q)$ .

step 2.2A1A3L1

4.1

Since $f(p) = y \ne f(v)$ and $f(q) = y \ne f(v)$ we have $p \ne v \ne q$ , so $u \le p < v < q \le w$ and in particular $p \ne q$ ; but $f(p) = f(q)$ contradicts injectivity. The three-point claim of step 1.2 is therefore established.

step 1.2step 3.1A2

5.1

Let $x \in I$ with $x < a$ . Applying the three-point claim to $x < a < b$ gives $f(x) < f(a) < f(b)$ or $f(x) > f(a) > f(b)$ ; the second is impossible because $f(a) < f(b)$ . So $f(x) < f(a)$ .

step 2.1step 4.1

6.1

Let $x \in I$ with $x > a$ . If $x < b$ , the three-point claim applied to $a < x < b$ gives $f(a) < f(x) < f(b)$ , the alternative being impossible as in step 5.1; if $x = b$ then $f(a) < f(x)$ by step 2.1; and if $x > b$ , the claim applied to $a < b < x$ gives $f(a) < f(b) < f(x)$ . In every case $f(x) > f(a)$ .

step 2.1step 4.1

6.2

Let $c, d \in I$ with $c < d \le a$ ; we show $f(c) < f(d)$ . If $d = a$ then $c < a$ and step 5.1 gives $f(c) < f(a) = f(d)$ . If $d < a$ then the three-point claim applied to $c < d < a$ gives $f(c) < f(d) < f(a)$ or $f(c) > f(d) > f(a)$ , and the second contradicts $f(d) < f(a)$ from step 5.1; so $f(c) < f(d)$ .

step 4.1step 5.1

7.1

Let $c, d \in I$ with $a \le c < d$ ; we show $f(c) < f(d)$ . If $c = a$ then $d > a$ and step 6.1 gives $f(d) > f(a) = f(c)$ . If $a < c$ then the three-point claim applied to $a < c < d$ gives $f(a) < f(c) < f(d)$ or $f(a) > f(c) > f(d)$ , and the second contradicts $f(c) > f(a)$ from step 6.1; so $f(c) < f(d)$ .

step 4.1step 6.1

7.2

The only remaining case is $c < a < d$ , where steps 5.1 and 6.1 give $f(c) < f(a) < f(d)$ directly.

step 5.1step 6.1

8.1

Steps 7.1, 6.2 and 7.2 cover every pair $c < d$ in $I$ : either $a \le c$ , which is step 7.1, or $c < a$ , and then $d \le a$ , which is step 6.2, or $a < d$ , which is step 7.2. So $f(x) < f(y)$ whenever $x < y$ in $I$ , that is, $f$ is increasing. This contradicts step 1.1, which assumed that $f$ is not increasing; the assumption of step 1.1 is therefore false and $f$ is strictly monotone.

step 1.1step 7.1step 6.2step 7.2L3discharge-contradiction∎

Remarks

The three-point claim is the whole content. Steps 1.2, 2.2, 3.1 and 4.1 say that a continuous injection on an interval cannot fold: the middle of three points always has the middle value. Everything after that is bookkeeping, comparing an arbitrary pair with one fixed pair $a < b$ on which the direction is known.
Where the intermediate value theorem enters. Once only, in step 3.1, and it is what makes order-convexity of $I$ indispensable: the segments $[u,v]$ and $[v,w]$ must lie inside the domain for the theorem to apply. That is exactly the hypothesis the companion page's counterexample removes.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 31 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)
Intermediate value theorem (Wikipedia) (standard reference, not scraped)
Continuous Injection of Interval is Strictly Monotone (ProofWiki) (standard reference, not scraped)