CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31verified 2026-08-08 (gpt-5.6-terra-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every nonempty convex subset of $R^{n}$ is contractible

Statement

Let $n \geq 1$ . Every nonempty convex subset $C \subseteq R^{n}$ , with its Euclidean subspace topology, is contractible. More precisely, for each $c \in C$ the formula

$H (x, t) = (1 - t) x + t c$

is a homotopy from ${id}_{C}$ to the constant map at $c$ .

Facts & Assumptions

Given: A nonempty convex subset $C \subseteq R^{n}$ and a point $c \in C$ .

[L1]

Any two continuous maps into a nonempty convex subset of $R^{n}$ are homotopic by the straight-line formula (Any two continuous maps into a nonempty convex subset of $R^{n}$ are homotopic by straight lines).

[L2]

A nonempty space is contractible exactly when its identity map is nullhomotopic (A nonempty space is contractible if and only if its identity map is nullhomotopic).

Proof

technique · direct

1.1L1

Apply [L1] to ${id}_{C} : C \to C$ and the constant map $c_{c} : C \to C$ . The resulting homotopy is $H (x, t) = (1 - t) x + t c$ .

2.1step 1.1L2∎

Thus ${id}_{C}$ is nullhomotopic, so $C$ is contractible by [L2].

Depends on

Dependency tree · 2 levels

Sources

Algebraic Topology lecture notes (UC Riverside)

Every nonempty convex subset of RnRn is contractible