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http://people.math.sc.edu/howard/Notes/brouwer.pdf https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/

These papers give fairly elementary proofs of Brouwer's fixed point and invariance of domain theorems. Having established these tools, is it possible to prove that an open subset of $ \mathbb{R}^n $ and an open subset of $ \mathbb{R}^m $ can't be homeomorphic unless $ n = m $ without refering to all those more advanced things which are normally used here, such as homology theory and algebraic topology?

Ormi
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If $V$, open subset of $\Bbb R^m$, were homeomorphic to an open subset of $\Bbb R^n$, $U$, let $f: U \to V$ be a homeomorphism. (Suppose WLOG that $m \leq n$.) Compose with a linear inclusion map $\Bbb R^m \hookrightarrow \Bbb R^n$ to get a continuous injective map $U \to \Bbb R^n$ with image contained in the subspace $\Bbb R^m$, as it factors as $U \to V \subset \Bbb R^m \hookrightarrow \Bbb R^n$.

If $m < n$, the image cannot be open - any neighborhood of a point in the hyperplane contains points not in the hyperplane. Therefore $m=n$. So invariance of domain implies invariance of dimension.

Of course, this says even more: there's not even a continuous injection $U \to V$ between open sets of $\Bbb R^n, \Bbb R^m$ respectively, $n>m$.

user5826
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    And of course you don't need the fix point theorem. This is one of many applications which show what a ridiculously strong result the invariance of domain actually is. – Daniel Valenzuela Mar 19 '15 at 23:36
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Sperner showed in his doctoral thesis [Sperner 1928] that invariance of open sets, invariance of domain and invariance of dimension can be proved already with elementary combinatorial methods alone.

They follow simply from the following theorem of Lebesgue:

A bounded point set $G$ in the $n$-dimensional number space is given, which contains inner points. Given a bounded point set $G$ in the $n$-dimensional number space containing inner points. The points of $G$ be distributed to a finite number of closed sets $M_{i}$ ($i=1,2,3,...,s$), so that every point of $G$ occurs at least in one of the sets $M_{i}$. Then there is at least one point in $G$ which lies in at least $n+1$ sets, if only the $M_{i}$ were chosen sufficiently small.

Sperner proved this theorem of Lebesgue by simple combinatorial methods.

Sperner's lemma can also be used in a proof of Brouwer's fixed-point theorem.

[Sperner 1928] Sperner, Emanuel: Neuer Beweis für die Invarianz der Dimensionszahl und des Gebietes. In: Abh. Math. Sem. Univ. Hamburg. Band 6, 1928, 265–272

Sperner, Emanuel: Gesammelte Werke. Benz, W.; Karzel, H.; Kreuzer , A. (Hrsg.). Heldermann Verlag, Lemgo, Germany, 2005

Sehie Park: Ninety years of the Brouwer Fixed Point Theorem. Vietnam Journal of Mathematics 27 (1999) (3) 187-222

Huang, J.: On the Sperner lemma and ist applications. 2004

Pak, K.: Sperner’s Lemma. Formalized Mathematics 18 (2010) (4) 189-196

Fox, J.: MAT 307: Combinatorics Lecture 3: Sperner's lemma and Brouwer's theorem. 2009

Sperner's Lemma and Brouwer fixed point theorem. 2014

Maliwal, A.: Sperner's Lemma, The Brouwer Fixed Point Theorem, the Kakutani Fixed Point Theorem, and their applications in social sciences. Electronic Theses and Dissertations. 2574. 2016

Anderson, A.: Sperner's Lemma and Brouwer's Fixed Point Theorem. 2021

Encyclopedia of Mathematics: Sperner lemma

Encyclopedia of Mathematics: Brouwer theorem

Wikipdia: Sperner's lemma

IV_
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  • Sperner's paper is really beautiful, and it does prove both invariance of dimension and domain, but I believe that it does not prove the Brouwer fixed-point theorem. It was later that people realized that the BFPT was an easy consequence of Sperner's lemma. – Nat Kuhn Nov 02 '20 at 02:13
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The following 3 page note proves that open sets in Euclidean spaces are homeomorphic only if their dimensions agree. http://personal.colby.edu/~sataylor/teaching/S09/MA331/InvarianceOfDomain.pdf

It uses Sperner's lemma and the notion of topological dimension (or Lebesgue covering dimension: https://en.wikipedia.org/wiki/Lebesgue_covering_dimension), which is defined using open covers, and is a topological invariant, by definition.

Hexhist
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