TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Every continuous map of a closed bounded interval into itself has a fixed point

Statement

Let $a, b \in R$ with $a \leq b$ and let $φ : [a, b] \to R$ be continuous on $[a, b]$ (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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with

$φ (x) \in [a, b] for every x \in [a, b] .$

Then $φ$ has a fixed point: there is $c \in [a, b]$ with $φ (c) = c$ .

This is the one-dimensional case of Brouwer's theorem, and here it is elementary. The whole content is that $φ (x) - x$ is $\geq 0$ at the left endpoint and $\leq 0$ at the right, so the intermediate value theorem produces a zero. Nothing about contraction, and no metric hypothesis, is needed: the map is not assumed to shrink distances, and the fixed point need not be unique.

Both hypotheses on the interval are used. The interval must be closed, or the fixed point can escape through an endpoint; and it must be bounded, or there need be no fixed point at all, as $x \mapsto x + 1$ on $[0, \infty)$ shows.

Facts & Assumptions

Given: Reals $a \leq b$ and a continuous $φ : [a, b] \to R$ with $φ (x) \in [a, b]$ for every $x \in [a, b]$ .

[L1]

Sums, scalar multiples and the identity: the identity $x \mapsto x$ is continuous on $[a, b]$ , and a sum of two functions continuous on $[a, b]$ is continuous on $[a, b]$ , as is a scalar multiple (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, 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).

[L2]

Intermediate value theorem: if $g$ is continuous on $[a, b]$ with $a \leq b$ and $y$ lies between $g (a)$ and $g (b)$ in either order, then $g (c) = y$ for some $c \in [a, b]$ (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[L3]

Membership in $[a, b]$ means $a \leq x \leq b$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Ordered-field arithmetic in $R$ : adding and subtracting preserves order, and $u - v \geq 0$ exactly when $u \geq v$ (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

Define $g : [a, b] \to R$ by $g (x) : = φ (x) - x$ . By [L1] the function $g$ is continuous on $[a, b]$ , being the sum of $φ$ and $(- 1)$ times the identity.

1.2

By hypothesis $φ (a) \in [a, b]$ , so $φ (a) \geq a$ and hence $g (a) = φ (a) - a \geq 0$ by [L4]. Likewise $φ (b) \in [a, b]$ gives $φ (b) \leq b$ and hence $g (b) = φ (b) - b \leq 0$ .

L3L4

2.1

So $g (b) \leq 0 \leq g (a)$ : the value $0$ lies between $g (b)$ and $g (a)$ . By [L2], applied to $g$ on $[a, b]$ with $y : = 0$ , there is $c \in [a, b]$ with $g (c) = 0$ .

step 1.1step 1.2L2choose

3.1

Then $φ (c) - c = 0$ , that is $φ (c) = c$ , with $c \in [a, b]$ : the map $φ$ has a fixed point.

step 2.1L4∎

Remarks

Uniqueness is not claimed and is false in general. The identity map of $[a, b]$ into itself is continuous and fixes every point. What forces uniqueness is a contraction hypothesis, which is the setting of the Banach fixed point theorem in a complete metric space; that theorem also produces the fixed point as a limit of iterates, whereas the argument above only asserts that one exists.
The interval may not be replaced by an arbitrary compact set. The map $x \mapsto - x$ carries the compact set ${- 1, 1}$ into itself, is continuous, and fixes nothing. Order-convexity, not compactness alone, is what the intermediate value theorem needs.
A worked instance is A worked fixed point on $[1, 2]$ for the map $x \mapsto (x + 2 / x) / 2$ , from the one-dimensional fixed point theorem ↗ on the companion page, where $φ (x) = (x + 2 / x) / 2$ maps $[1, 2]$ into itself and its unique fixed point is $\sqrt{2}$ .

Depends on

Used by

A worked fixed point on [1,2] for the map x ↦ (x + 2/x)/2, from the one-dimensional fixed point theorem Example

Dependency tree · two levels

29 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

Brouwer fixed-point theorem (Wikipedia) (standard reference, not scraped)
Intermediate value theorem (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3 (standard reference, not scraped)
K. Conrad, The Contraction Mapping Theorem (standard reference, not scraped)