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A continuous injective function on an interval is strictly monotone

Statement

Let $I \subseteq R$ be order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and let $f : I \to R$ be continuous on $I$ (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) 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 $R$ , 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 $I \subseteq R$ and a continuous injective $f : I \to R$ .

[A1]

$I$ is order-convex: $x, y \in I$ and $x \leq z \leq y$ imply $z \in I$ (Intervals of $R$ : 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 $ε$ - $δ$ condition of 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 quantifies over fewer points when the domain shrinks.

[L1]

Intermediate value theorem: if $u \leq v$ , $g : [u, v] \to R$ 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 $R$ , with the dictionary to monotone sequences).

Proof

technique · contradiction

1.1assume-contra

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

1.2A2

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 $f (v) < f (w)$ .

2.1step 1.1A2L3

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 $f (a) \leq f (b)$ ; injectivity together with $a \neq b$ gives $f (a) \neq f (b)$ , so $f (a) < f (b)$ .

2.2step 1.2L2

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.

3.1step 2.2A1A3L1

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 $f (p) = y = f (q)$ .

4.1step 1.2step 3.1A2

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

5.1step 2.1step 4.1

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 $f (x) < f (a)$ .

6.1step 2.1step 4.1

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 $f (x) > f (a)$ .

6.2step 4.1step 5.1

Let $c, d \in I$ with $c < d \leq a$ ; 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 $f (c) < f (d)$ .

7.1step 4.1step 6.1

Let $c, d \in I$ with $a \leq c < d$ ; 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 $f (c) < f (d)$ .

7.2step 5.1step 6.1

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

8.1step 1.1step 7.1step 6.2step 7.2L3discharge-contradiction∎

Steps 7.1, 6.2 and 7.2 cover every pair $c < d$ in $I$ : either $a \leq c$ , which is step 7.1, or $c < a$ , and then $d \leq a$ , 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.

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.

A continuous injective function on an interval is strictly monotone

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources