International Journal of Mathematics and Mathematical Sciences

Volume 2011, Article ID 793848, 9 pages

http://dx.doi.org/10.1155/2011/793848

## The Bolzano-Poincaré Type Theorems

College of Science, Cardinal Stefan Wyszyński, University in Warsaw, ul. Dewajtis 5, 01-815 Warszawa, Poland

Received 24 February 2011; Revised 7 May 2011; Accepted 30 May 2011

Academic Editor: Giuseppe Marino

Copyright © 2011 Przemysław Tkacz and Marian Turzański. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In 1883–1884, Henri Poincaré announced the result about the structure of the set of zeros of function , or alternatively the existence of solutions of the equation . In the case the Poincaré Theorem is well known Bolzano Theorem. In 1940 Miranda rediscovered the Poincaré Theorem. Except for few isolated results it is essentially a non-algorithmic theory. The aim of this article is to introduce an algorithmical proof of the Theorem “On the existence of a chain” and for an algorithmical proof of the Bolzano-Poincaré Theorem and to show the equivalence of Poincaré, Brouwer and “On the existence of a chain” theorems.

#### 1. Introduction

It is well known how influential topology was for the development of many other branches of mathematics and economics. Among many others, let us recall significant place of fixed point theorems of Brouwer and Banach which served as a main tool in solving problems in differential equations, theory of fractals and problems of market equilibrium. Some of these applications raised a question of computability of the fixed points. In [1, 2] Steinhaus presented following conjecture: *Let some segments of the chessboard be mined. Assume that the king cannot go across the chessboard from the left edge to the right one without meeting a mined square. Then the rook can go from upper edge to the lower one moving exclusively on mined segments*.

According to Surówka [3] several proofs of the Steinhaus Chessboard Theorem seem to be incomplete or use induction on the size of the chessboard [4].

The simple proof of the Steinhaus Chessboard Theorem was presented in [5]. In [6] the following generalization of the Steinhaus Chessboard Theorem was published: Theorem [On the existence of a chain] *For an arbitrary decomposition of n-dimensional cube ** onto ** cubes and an arbitrary coloring function ** for some natural number ** there exists an **th colored chain ** such that ** and **. *

This theorem was the main tool in the proof (see [6]) of the Bolzano-Poincaré theorem (see [7, 8]). In the first part of our paper an algorithm of finding the chain will be presented and will be shown that the theorem “on the existence of a chain”, the Bolzano-Poincaré theorem, and the Brouwer fixed point theorem are equivalent (for more informations see [9, 10]).

#### 2. Theorems

Let be the *n-*dimensional cube in .

Its *th opposite faces* are defined as follows:
Let
be the *boundary* of the cube .

Let be an arbitrary natural number.

We call the family the decomposition of into cubes.

The map is said to be a *coloring function* of the decomposition .

The sequence where for is said to be an *th colored chain*, if for all and for .

The set is said to be the *n-dimensional combinatorial cube. *

Its *th opposite faces* are defined as follows:
Let
be the *boundary* of the *n-dimensional combinatorial cube. *

Let be the *th basic vector. *

An ordered set is said to be an *n-simplex* if there exists permutation of set such that .

Any subset is said to be an *(n−1)-face of the n-simplex **. *

Every map is said to be a *coloring map* of .

The set we call *n'-colored* if .

*Observation 1. *Let be an *n-*simplex. Then for each if for each , then there exists exactly one *n-*simplex such that else there does not exist such .

*Observation 2. *Any (*n−*1)-face of an *n-*simplex is an (*n−*1)-face of exactly one or of two *n-*simplexes from depending on whether or not it lies on for some , .

*Observation 3. *Each *n-*colored *n-*simplex has exactly two *n-*colored (*n−*1)-faces.

Theorem 2.1 (on the existence of a chain). *For an arbitrary decomposition of n- dimensional cube onto cubes and an arbitrary coloring function for some natural number there exists an th colored chain such that and . **
The algorithm (based on the proof from [6]) is as follows.*

*Step 1. *Let us define the coloring map :

*Step 2. *Let us take *n-*colored *n-*simplex where ,

We say that the *n-*colored (*n−1*)-face is “used”.

Let .

*Step 3. *Take “unused” *n-*colored (*n−1*)-face of the *n-*simplex .

If this face is contained in for some , then go to Step 5. Else this is (*n−1*)-face of exactly one *n-*simplex different to .

Since that moment this (*n−1*)-face is said to be “used”. Go to the Step 4.

*Step 4. *Let us create the sequence of *n-*simplexes .

Let .

Go to Step 3.

*Step 5. *After finitely many iterations we obtain the sequence such that for . And the *n-*simplex has the *n-*colored (*n−1*)-face which is a subset of for some , . Hence where , .

Let us take the smallest index such that for some , then let us find the biggest index such that .

*Step 6. *Then from the chain choose successively points in the way that for and for , and , .

*Step 7. *For the sequence we have the chain where and for .

END

Theorem 2.2 (Bolzano-Poincar). *Let , be a continuous map such that and for then there exists such that .*

Theorem 2.3 (Brouwer fixed point theorem). *Let , be a continuous map then there exists such that .*

Theorem 2.4. *The following theorems are equivalent:*

(1)*Theorem on the existence of a chain*(2)*Bolzano-Poincaré theorem*(3)*Brouwer fixed point theorem.*

*Proof. *“(1)(2)” let us assume that for each . Let us define sets:

for , each set is open.

We have .

Let us consider the space with the metric . From the Lebesgue lemma of covering it follows that there exists such that for every and we have for every there exist such that .

Let us define coloring function :
From theorem “on the existing of a chain” there exists th colored sequence connecting th opposite faces of the cube .

The set is closed and connected.

The intersections , Hence there exists such that and . Since is the continuous map, hence is connected in . Hence the set is an interval containing . From the Bolzano theorem there exists such that .

Contradiction.

“(2)(3)” let . The function fulfills the assumptions of the Bolzano-Poincaré theorem. hence there exist such that .

So .

“(3)(1)” let us assume that there exists decomposition of -dimensional cube onto cubes and a coloring function such that for each there is no th colored chain connecting and .

Let .

Let be the family of components of .
The subsets of :
are closed and disjoint.

with the Euclidean metric is a normal space, hence there exists a continuous map such that and .

For each let us define the map where .

Observe that is continuous map. Take an arbitrary .

There exists such that . The cube is a subset of or for some . We have or .

Hence the function has no fixed point. Contradiction.

#### 3. Poincaré Theorem for

##### 3.1. The Basic Algorithm

Let be an arbitrary natural number.

We have the decomposition of into cubes.

Assume w.l.o.g. that and for . Let be the Euclidean metric.

Observe that there exist such that for each and for each we have , , .

###### 3.1.1. Surface

Let be a natural number, such that .

The center of each is defined as follows: Let us define coloring map Algorithm for surface is as follows.

*Step 1. *Let

*Step 2. *If then END.

Otherwise do Step 3.

*Step 3. *Add elements of the set to .

Next
and go to Step 2.

Since is finite, hence after finitely many steps set is empty (the procedure ends).

Let us consider the family . We may assume that is closed under finite intersections.

The elements , such that are called squares, edges, and vertices.

*Observation 4. *The separates cube between and .

*Observation 5. *Each edge if it is an edge of exactly 1 square, else it is an edge of 2 or 4 squares.

###### 3.1.2. Modification of

Let us divide each element of onto 27 cubes (in the natural way).

Denote the set consisting of all this cubes by .

Create coloring map as follows: Now.

*Observation 6. *Any edge of is an edge of exactly one or of two squares from depending on whether or not it lies on .

Let us define coloring :
where is the center of square .

The edge is said to be 2-coloured if there exists squares such that and .

*Observation 7. *The vertex of 2-coloured edge is a subset of exactly one or even number of 2-coloured edges depending on whether or not it lies on .

*Observation 8. *The components of are broken lines without self-cutting.

*Observation 9. *The number of broken lines lying on and connecting and is odd.

Lemma 3.1. *The number of 2-coloured edges from , which one of vertices lies on is odd.*

*Proof. *Let us consider components of the set .

We have odd number of broken lines connecting and and the number of the rest components is arbitrary.

Let us see that .

So, the number of 2-coloured edges from , which one of vertices lies on is odd if it lies on broken line connecting and else it is even (using the definition of ).

According to Observation 9 this ends the proof.

###### 3.1.3. Broken Line Connecting and

*Step 1. *Let ,

*Step 2. *Take .

Add to .

The vertex is said to be used.

Go to Step 3.

*Step 3. *Take unused vertex of the last added edge to the set .

If END.

Otherwise,

If go to Step 2.

Else go to Step 4.

*Step 4. *Take unused vertex of the last added edge to the set .

Next take 2-coloured edge such that .

Now vertice is said to be used.

Add to the set .

Go to Step 3.

First of all the number of 2-coloured edges from , which one of vertices lies on is odd (Lemma 3.1).

The second each vertex of 2-coloured edge is a subset of exactly one or even number of 2-coloured edges depending on whether or not it lies on (Observation 7).

This arguments allows one to say that procedure is well defined.

Now our broken line connecting and is created as follows:

let be the last added element to .

If then is previous added element to

else Stop.

We obtained the sequence of edges . Let us define coloring where :

It is easy to see that and .

So starting from we search with order the first edge such that .

##### 3.2. Topological Part

For each , we have(i) such that and ,(ii) such that and where are squares from and ,(iii) such that and where is a cube from , is a cube from and .

Define the sets .

For each there exist such that Without loss of generality we can assume that .

Moreover, . So for each the fact yields So ends proof.

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