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## Euclidean spaces of different dimensions are not homeomorphic

I would like to give a proof for the fact that if $m\neq n,\ m,n \in \Bbb{N}^*$ then $\Bbb{R}^n$ and $\Bbb{R}^m$ are not homeomorphic.

1) for the case where one of the numbers $m,n$ is $1$, we can give a simple proof

2) for the general proof, we can use the Generalized Jordan Curve Theorem.

(cf. Spivak, Intro to Differential Geometry, Chapter 1, Ex. 8,9)

1. December 5, 2010 at 2:59 am

Poti sa vezi foarte simplu asta uitandu-te la omologia relativa fata de complementul unui punct:
e usor de demonstrat ca $H_i(\mathbb{R}^n,\mathbb{R}^n-\{x\}) = 0$ daca $i\neq 0, n-1$ si $\mathbb{Z}$ altfel. Presupunand prin absurd ca ai un homeomorfism iti rezulta imediat ca $m=n$.

2. December 5, 2010 at 4:24 pm

Sau chiar mai simplu, daca ai un homeomorfism il poti prelungi la compactificarile Alexandrov ale celor doua spatii. Iti rezulta un homeomorfism intre sfera n-dim si sfera m-dim. Grupurile de omologie ale sferei n-dim sunt $H_i = \mathbb{Z}, i=0,n$ si $H_i=0, i\neq 0,n$ . De aici rezulta imediat ca m=n.

• December 6, 2010 at 5:08 pm

Multumesc pentru comentariile facute. O sa incerc sa adaptez o solutie cu informatiile postate.