Home > Analysis, Problem Solving, Topology > Euclidean spaces of different dimensions are not homeomorphic

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)

Advertisements
  1. Dragos
    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. Dragos
    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.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: