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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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R\mathbb{R} is not homeomorphic to Rn\mathbb{R}^n for any n≥2n\ge2

Statement

For every n≥2n\ge2, there is no homeomorphism R→Rn\mathbb R\to\mathbb R^n.

Facts & Assumptions

Given: n≥2n\ge2.

[L1]

The punctured space Rn∖{0}\mathbb R^n\setminus\{0\} is polygonally connected, hence connected (For n≥2n\ge2, the punctured space Rn∖{0}\mathbb{R}^n\setminus\{0\} is polygonally connected, Every path-connected space is connected, and every path component lies inside a component).

[L2]

A continuous image of a connected space is connected (A continuous image of a connected space is connected, and connectedness is a topological property).

[L3]

A connected subset of R\mathbb R is order-convex (The connected subspaces of R\mathbb{R} with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in R\mathbb{R}").

[L4]

A homeomorphism is a continuous bijection with continuous inverse (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Proof

technique · contradiction
1.1

Suppose h:R→Rnh:\mathbb R\to\mathbb R^n is a homeomorphism, and put a:=h−1(0)a:=h^{-1}(0).

assume-contraL4choose
1.2

Restricting h−1h^{-1} to Rn∖{0}\mathbb R^n\setminus\{0\} gives a continuous surjection onto R∖{a}\mathbb R\setminus\{a\}. The source is connected by [L1], so the target is connected by [L2].

L1L2L4
1.3

Choose a−1<a<a+1a-1<a<a+1. Both endpoints lie in R∖{a}\mathbb R\setminus\{a\}, but aa does not, so this subset is not order-convex and therefore not connected by [L3].

L3
2.1

Steps 1.2 and 1.3 contradict one another. Thus no such homeomorphism exists.

step 1.2step 1.3discharge-contradiction∎

Depends on

Used by

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