2
$\begingroup$

There is a proof of the Brouwer Fixed Point Theorem via the Jordan Curve Theorem ?

The Brouwer Fixed Point Theorem. Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow B$ has a fixed point.

The Jordan curve theorem. The image of a continuous injective mapping $J:S^1\rightarrow \mathbb R^2$ divides the plane into exactly two components, one of which is unbounded and the other bounded. Moreover, both of these components have the image of the mapping $J$ as their boundary.

The Jordan Curve Theorem via the Brouwer Fixed Point Theorem

$\partial B=S^{1}$

Any hints would be appreciated.

$\endgroup$
2
$\begingroup$

You might enjoy David Gale's paper The Game of Hex and the Brouwer Fixed-Point Theorem, The Monthly, Dec 1979, pp. 818-827. He gives direct proofs in both directions between the Brouwer Theorem and the:

Hex Theorem: The game of Hex cannot end in a tie.

As Gale explains, the Hex Theorem is completely elementary (Gale gives an algorithm that takes in a completed Hex board and finds a win for one of the two players). But the Hex Theorem can be thought of as a sort of "finite" or "discrete" version of the Jordan theorem (the weak form, needed for Brouwer, only says that at least one player wins; the strong form, that both cannot win, is closer to Jordan), and the former does follow from the latter (Beck, Bleicher, and Crow, Excursion into Mathematics, 1969, pp. 327-339). For further discussion relationing the Hex and Jordan Theorems, see the three or so paragraphs after "Hex Theorem" in op. cit., p. 820.

$\endgroup$
  • $\begingroup$ As a side-note: The Jan 2014 AMM has an article with the following abstract: "A short, inductive proof is presented of the fact that a Hex board cannot be colored such that winning conditions are satisfied for both players." Appropriately titled, the article has as its citation: Huneke, S. C. An Inductive Proof of Hex Uniqueness. *The American Mathematical Monthly, 121*(1), 78-80. $\endgroup$ – Benjamin Dickman Dec 17 '13 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.