False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: every proper embedding of an n-manifold lands in R^n

Statement

False claim: every smooth $n$ -manifold admits a proper embedding into $R^{n}$ .

Facts & Assumptions

Given: The circle $S^{1}$ .

Refutation

technique · direct

1.1givenalgebra

Suppose $S^{1}$ embedded in $R$ . Because $S^{1}$ is compact and connected, its image would be a compact connected subset of $R$ , hence a closed interval $[a, b]$ .

2.1step 1.1algebra

Removing any point from $S^{1}$ leaves a connected space, but removing an interior point from $[a, b]$ disconnects it, while removing an endpoint leaves a noncompact interval. Therefore $S^{1}$ is not homeomorphic to any closed interval. This contradicts step 1.1.

3.1step 2.1∎

So the claim already fails in dimension $1$ .

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Theorem 6.15 (standard reference, not scraped)