# The Bolzano-Poincaré Type Theorems

Authors: Przemysław Tkacz; Marian Turzański
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

In 1883–1884, Henri Poincaré announced the result about the structure of the set of zeros of functionf:In→Rn, or alternatively the existence of solutions of the equation f(x)=0. In the case n=1 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 n=3 an algorithmical proof of the Bolzano-Poincaré Theorem and to show the equivalence of Poincaré, Brouwer and “On the existence of a chain” theorems.


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## Body

## 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 cubeInontokncubes and an arbitrary coloring functionF:T(k)→{1,…,n}for some natural numberi∈{1,…,n}there exists anith colored chainP1,…,Prsuch thatP1∩Ii+≠∅andPr∩Ii-≠∅.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

LetIn:=[0,1]n be the n-dimensional cube in Rn.Itsith opposite faces are defined as follows: (2.1)Ii-:={x∈In:x(i)=0},Ii+:={x∈In:x(i)=1}.
Let (2.2)∂In:=⋃i=1n(Ii-∪Ii+)
be the boundary of the cube In.Letk be an arbitrary natural number.We call the family(2.3)T(k):={[i1k,i1+1k]×⋯×[ink,in+1k]:ij∈{0,…,k-1}}
the decomposition of In into kn cubes.The mapF:T(k)→{1,…,n} is said to be a coloring function of the decomposition T(k).The sequenceP1,…,Pr where Pl∈T(k) for l=1,…,r is said to be an ith colored chain, if for all l∈{1,…,r}F(Pl)=i and Pj∩Pj+1≠∅ for j=1,…,r-1.The setC={-1/2k,1/2k,…,1+1/2k}n is said to be the n-dimensional combinatorial cube.Itsith opposite faces are defined as follows: (2.4)Ci-={z∈C:z(i)=-12k},Ci+={z∈C:z(i)=1+12k}.
Let(2.5)∂C=⋃i=1nCi-∪Ci+
be the boundary of the n-dimensional combinatorial cube.Letei=(0,…,0,1/k,0,…,0),ei(i)=1/k be the ith basic vector.An ordered setS=[z0,…,zn]⊂C is said to be an n-simplex if there exists permutation α of set {1,…,n} such that z1=z0+eα(1)⋯zn=zn-1+eα(n).Any subset[z0,…,zi-1,zi+1,…,zn]⊂S,i=0,…,n is said to be an (n−1)-face of the n-simplexS.Every mapΦ:C→{1,…,n} is said to be a coloring map of C.The setA⊂C we call n'-colored if Φ(A)={1,…,n′}.Observation 1.
LetS=[z0,…,zn]⊂C be an n-simplex. Then for each zi∈S if  [z0,…,zi-1,zi+1,…,zn]⊄Cpϵ for each p∈{1,…,n}, ϵ∈{+,-} then there exists exactly one n-simplex  S[i]⊂C such that   S∩S[i]=[z0,…,zi-1,zi+1,…,zn] else there does not exist such S[i]⊂C.Observation 2.
Any (n−1)-face of an n-simplex S⊂C is an (n−1)-face of exactly one or of two n-simplexes from C depending on whether or not it lies on Cpϵ for some p∈{1,…,n}, ϵ∈{+,-}.Observation 3.
Eachn-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 cubeIn onto  kn cubes and an arbitrary coloring function F:T(k)→{1,…,n} for some natural number i∈{1,…,n} there exists an ith colored chain P1,…,Pr such that P1∩Ii+≠∅ and Pr∩Ii-≠∅.
The algorithm (based on the proof from [6]) is as follows.Step 1.
Let us define the coloring mapΦ:C→{1,…,n}:
(2.6)Φ(z)={F(t)forz∈C∖∂Candz∈t1forz∈C1-∪C2+jforz∈(Cj-∪Cj+1+)∖(⋃l=1j-1(Cl-∪Cl+1+)),j=2,…,n-1nforz∈(Cn-∪C1+)∖(⋃l=1n-1(Cl-∪Cl+1+)).Step 2.
Let us taken-colored n-simplex S1=[z01,…,zn-11,zn1] where z01=(-1/2k,-1/2k,…,-1/2k),
(2.7)z11=z01+e1,…,zn-11=zn-21+en-1,zn1=(12k,12k,…,12k)=zn-11+en.
We say that then-colored (n−1)-face [z01,…,zn-11] is “used”.
LetS=S1.Step 3.
Take “unused”n-colored (n−1)-face of the n-simplex S.
If this face is contained inCpϵ for some p∈{1,…,n}, ϵ∈{+,-} then go to Step 5. Else this is (n−1)-face of exactly one n-simplex S′ different to S.
Since that moment this (n−1)-face is said to be “used”. Go to the Step 4.Step 4.
Let us create the sequence ofn-simplexes S1,…,Sl,S′.
LetS=S′.
Go to Step3.Step 5.
After finitely many iterations we obtain the sequenceS1,…,Sm⊂C such that Φ(Sl∩Sl+1)={1,…,n} for l=1,…,m-1. And the n-simplex Sm has the n-colored (n−1)-face which is a subset of Cpϵ for some p∈{1,…,n}, ϵ∈{+,-}. Hence Sm=[z0m,z1m,…,znm] where z0m=(1-1/2k,1-1/2k,…,1-1/2k), z1m=z0m+e1,z2m=z1m+e2,…,znm=zn-1m+en.
Let us take the smallest indexl1∈{1,2,…,m} such that Sl1∩Ci+≠∅ for some i∈{1,…,n}, then let us find the biggest index l2∈{1,2,…,l1} such that Sl2∩Ci-≠∅.Step 6.
Then from the chainSl2+1,…,Sl1 choose successively points z1,z2,…,zr in the way that Φ(zj)=i for j=1,2,…,r and zj≠zj+1 for j=1,2,…,r-1, z1∈C∖∂C and z1-ei∈Ci-, zr∈Sl1.Step 7.
For the sequencez1,…,zr we have the chain P1,…,Pr where Pj∈T(k) and zj∈Pj for j=1,…,r.
ENDTheorem 2.2 (Bolzano-Poincare´).
Letf:In→Rn, f(x)=(f1(x),…,fn(x)) be a continuous map such that fi(Ii-)⊂(-∞,0] and fi(Ii+)⊂[0,∞) for i=1,…,n then there exists x0∈In such that f(x0)=(0,…,0).Theorem 2.3 (Brouwer fixed point theorem).
Letg:In→In, g(x)=(g1(x),…,gn(x)) be a continuous map then there exists x0∈In such that g(x0)=x0.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 x∈Inf(x)≠(0,…,0). Let us define sets:
Ui={x∈In:fi(x)≠0} for i=1,…,n, each set  Ui is open.
We haveIn=U1∪⋯∪Un.
Let us consider the spaceRn with the metric δ(x,y)=max⁡{|xi-yi|:i=1,…,n}. From the Lebesgue lemma of covering it follows that there exists λ>0 such that for every k∈N and 1/k<λ we have for every t∈T(k) there exist j∈{1,…,n} such that t⊂Uj.
Let us define coloring functionF:T(k)→{1,…,n}:
(2.8)F(t):=min⁡{j:t⊂Uj}.
From theorem “on the existing of a chain” there exists ith colored sequence P1(k),…,Pr(k)(k) connecting ith opposite faces of the cube In.
The setW:=⋃l=1r(k)Pl(k) is closed and connected.
The intersectionsW∩Ii-≠∅≠W∩Ii+, Hence there exists x,y∈W such that fi(x)<0 and fi(y)>0. Since f(x) is the continuous map, hence fi(W) is connected in R. Hence the set fi(W) is an interval containing [fi(x),fi(y)]. From the Bolzano theorem there exists c∈W such that fi(c)=0.
Contradiction.
“(2)⇒(3)” let f(x)=x-g(x). The function f(x) fulfills the assumptions of the Bolzano-Poincaré theorem. hence there exist c∈In such that f(c)=0.
Sog(c)=c.
“(3)⇒(1)” let us assume that there exists decomposition of n-dimensional cube In onto kn cubes and a coloring function F:T(k)→{1,…,n} such that for each i∈{1,…,n} there is no ith colored chain connecting Ii- and Ii+.
LetCi={t∈T(k):F(t)=i}.
Letℒi be the family of components of ⋃Ci⊂In. 							(2.9)Ci-={l∈Li:l∩Ii-≠∅},Ci+={l∈Li:l∩Ii+≠∅},Ci0={l∈Li:l∩(Ii-∪Ii+)=∅}.
The subsets of In:
(2.10)Ai=⋃Ci-∪⋃Ci0∪{x∈In:x(i)∈[0,12k]},Bi=⋃Ci+∪{x∈In:x(i)∈[1-12k,1]}
are closed and disjoint.
In with the Euclidean metric is a normal space, hence there exists a continuous map fi:In→[-1/2k,1/2k] such that fi(Ai)=1/2k and fi(Bi)=-1/2k.
For eachx∈In let us define the map g(x):=x+f(x) where f(x)=(f1(x),…,fn(x)).
Observe thatg:In→In is continuous map. Take an arbitrary x∈In.
There existst∈T(k) such that x∈t. The cube t is a subset of Ai or Bi for some i∈{1,…,n}. We have gi(x)=x(i)+1/2k or gi(x)=x(i)-1/2k.
Hence the functiong(x) has no fixed point. Contradiction.


## 3. Poincaré Theorem forn=3

### 3.1. The Basic Algorithm

Letk be an arbitrary natural number.We have the decomposition ofI3 into k3 cubes.Assume w.l.o.g. thatfi(Ii-)⊂(-∞,0) and fi(Ii+)⊂(0,∞) for i=1,2,3. Let d:I3×I3→R be the Euclidean metric.Observe that there existϵ*>0 such that for each x∈I3,d(x,Ii-)<ϵ* and for each y∈I3,d(y,Ii+)<ϵ* we have fi(x)<0, fi(y)>0, i=1,2,3.


#### 3.1.1. Surface

Letk be a natural number, such that 1/k<ϵ*.The center of eacht∈T(k),t=[i1/k,(i1+1)/k]×⋯×[i3/k,(i3+1)/k] is defined as follows: (3.1)tc=(i1k+12k,i2k+12k,i3k+12k).
Let us define coloring map ϕ1:T(k)→{0,1}(3.2)ϕ1(t)={0f1(tc)≤01f1(tc)>0.
Algorithm for surface is as follows.Step 1.
Let(3.3)A0={t∈T(k):t∩I1-≠∅},A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1}.Step 2.
IfC={t∈T(k)∖A0:dim⁡[t∩A0]=2,ϕ1(t)=0}=∅ then END.
Otherwise do Step3.Step 3.
Add elements of the setC to A0.
Next(3.4)A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1,t∩t′≠∅}
and go to Step 2.
SinceT(k) is finite, hence after finitely many steps set C is empty (the procedure ends).
Let us consider the familyB. We may assume that B is closed under finite intersections.
The elementsb∈B, such that dim⁡[b]=2,dim⁡[b]=1,dim⁡[b]=0 are called squares, edges, and vertices.Observation 4.
The⋃B separates cube I3 between I1- and I1+.Observation 5.
Each edgeb∈B if b⊂∂I3 it is an edge of exactly 1 square, else it is an edge of 2 or 4 squares.


#### 3.1.2. Modification ofB

Let us divide each element of{a∈A0:a∩⋃B≠∅} onto 27 cubes (in the natural way).Denote the set consisting of all this cubes byT′.Create coloring mapϕ1′:T′→{0,1} as follows:
(3.5)ϕ1′(t′)={0t′∩A1=∅1t′∩A1≠∅.
NowB′={t∩t′:t,t′∈T′,ϕ1′(t)=0,ϕ1′(t′)=1,t∩t′≠∅}.Observation 6.
Any edge ofB′ is an edge of exactly one or of two squares from B′ depending on whether or not it lies on ∂I3.
Let us define coloringϕ2:{t∈B′:tisasquare}→{0,1}:
(3.6)ϕ2(t)={0f2(tc)≤01f2(tc)>0,
where tc is the center of square t.
The edget∈B′ is said to be 2-coloured if there exists squares s,s′∈B′ such that s∩s′=t and ϕ2({s,s′})={0,1}.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∂I3.Observation 8.
The components of⋃B′∩∂I3 are broken lines without self-cutting.Observation 9.
The number of broken lines lying onI3- and connecting I2- and I2+ is odd.Lemma 3.1.
The number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd.Proof.
Let us consider components of the set⋃B′∩I3-.
We have odd number of broken lines connectingI2- and I2+ and the number of the rest components is arbitrary.
Let us see that⋃B′⊂I3∖I1-∪I1+.
So, the number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd if it lies on broken line connecting I2- and I2+ else it is even (using the definition of ϕ2).
According to Observation9 this ends the proof.


#### 3.1.3. Broken Line ConnectingI3- and I3+

Step 1.
LetE0={t∈B′:tisa2-colourededge,t∩I3-≠∅},
(3.7)E1=∅.Step 2.
Takee∈E0∖E1.
Adde to E1.
The vertexv∈e∩I3- is said to be used.
Go to Step3.Step 3.
Take unused vertexu of the last added edge to the set E1.
Ifu∈I3+ END.
Otherwise,
Ifu∈I3- go to Step 2.
Else go to Step4.Step 4.
Take unused vertexu of the last added edge to the set E1.
Next take 2-coloured edgee∈B′∖E1 such that v∈e.
Now verticev is said to be used.
Adde to the set E1.
Go to Step3.First of all the number of 2-coloured edges fromB′, which one of vertices lies on I3- 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∂I3 (Observation 7).This arguments allows one to say that procedure is well defined.Now our broken line connectingI3+ and I3- is created as follows:lete1 be the last added element to E1.Ifei∩I3-=∅ then ei+1 is previous added element to E1else Stop.We obtained the sequence of edges{e1,e2,…,em}⊂B′. Let us define coloring ϕ3:{t∈B′:tisanedgeofei}→{0,1} where i∈{1,…,m}: (3.8)ϕ3(t)={0f3(t)≤01f3(t)>0.It is easy to see thatϕ3(e1)={1} and ϕ3(em)={0}.So starting frome1 we search with order the first edge ek∈{e1,e2,…,em} such that ϕ3(ek)={0,1}.


### 3.2. Topological Part

For eachk∈N, 1/k<ϵ* we have(i)
vk,vk′∈ek such that f3(vk)≤0 and f3(vk′)>0,(ii)
uk,uk′∈tu∪tu′ such that f2(uk)≤0 and f2(uk′)>0 where tu,tu′ are squares from B and ek∩tu≠∅≠ek∩tu′,(iii)
wk,wk′∈tw∪tw′ such that f1(wk)≤0 and f1(wk′)>0 where tw is a cube from A0, tw′  is a cube from A1 and ek∩tw≠∅≠ek∩tw′.Define the setsWk:=conv{vk,vk′,uk,uk′,wk,wk′}.For eachWk there exist ck1,ck2,ck3∈Wk such that (3.9)f1(ck1)=f2(ck2)=f3(ck3)=0.
Without loss of generality we can assume that lim⁡k→∞ck1=c.Moreover,lim⁡k→∞diam⁡[Wk]=0. So for each ck′∈Wk the fact d(c,ck′)≤d(c,ck1)+d(ck1,ck′) yields(3.10)lim⁡k→∞⁡ck1=lim⁡k→∞⁡ck2=lim⁡k→∞⁡ck3=c.
So f(c)=0 ends proof.


## 3.1. The Basic Algorithm

Letk be an arbitrary natural number.We have the decomposition ofI3 into k3 cubes.Assume w.l.o.g. thatfi(Ii-)⊂(-∞,0) and fi(Ii+)⊂(0,∞) for i=1,2,3. Let d:I3×I3→R be the Euclidean metric.Observe that there existϵ*>0 such that for each x∈I3,d(x,Ii-)<ϵ* and for each y∈I3,d(y,Ii+)<ϵ* we have fi(x)<0, fi(y)>0, i=1,2,3.


### 3.1.1. Surface

Letk be a natural number, such that 1/k<ϵ*.The center of eacht∈T(k),t=[i1/k,(i1+1)/k]×⋯×[i3/k,(i3+1)/k] is defined as follows: (3.1)tc=(i1k+12k,i2k+12k,i3k+12k).
Let us define coloring map ϕ1:T(k)→{0,1}(3.2)ϕ1(t)={0f1(tc)≤01f1(tc)>0.
Algorithm for surface is as follows.Step 1.
Let(3.3)A0={t∈T(k):t∩I1-≠∅},A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1}.Step 2.
IfC={t∈T(k)∖A0:dim⁡[t∩A0]=2,ϕ1(t)=0}=∅ then END.
Otherwise do Step3.Step 3.
Add elements of the setC to A0.
Next(3.4)A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1,t∩t′≠∅}
and go to Step 2.
SinceT(k) is finite, hence after finitely many steps set C is empty (the procedure ends).
Let us consider the familyB. We may assume that B is closed under finite intersections.
The elementsb∈B, such that dim⁡[b]=2,dim⁡[b]=1,dim⁡[b]=0 are called squares, edges, and vertices.Observation 4.
The⋃B separates cube I3 between I1- and I1+.Observation 5.
Each edgeb∈B if b⊂∂I3 it is an edge of exactly 1 square, else it is an edge of 2 or 4 squares.


### 3.1.2. Modification ofB

Let us divide each element of{a∈A0:a∩⋃B≠∅} onto 27 cubes (in the natural way).Denote the set consisting of all this cubes byT′.Create coloring mapϕ1′:T′→{0,1} as follows:
(3.5)ϕ1′(t′)={0t′∩A1=∅1t′∩A1≠∅.
NowB′={t∩t′:t,t′∈T′,ϕ1′(t)=0,ϕ1′(t′)=1,t∩t′≠∅}.Observation 6.
Any edge ofB′ is an edge of exactly one or of two squares from B′ depending on whether or not it lies on ∂I3.
Let us define coloringϕ2:{t∈B′:tisasquare}→{0,1}:
(3.6)ϕ2(t)={0f2(tc)≤01f2(tc)>0,
where tc is the center of square t.
The edget∈B′ is said to be 2-coloured if there exists squares s,s′∈B′ such that s∩s′=t and ϕ2({s,s′})={0,1}.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∂I3.Observation 8.
The components of⋃B′∩∂I3 are broken lines without self-cutting.Observation 9.
The number of broken lines lying onI3- and connecting I2- and I2+ is odd.Lemma 3.1.
The number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd.Proof.
Let us consider components of the set⋃B′∩I3-.
We have odd number of broken lines connectingI2- and I2+ and the number of the rest components is arbitrary.
Let us see that⋃B′⊂I3∖I1-∪I1+.
So, the number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd if it lies on broken line connecting I2- and I2+ else it is even (using the definition of ϕ2).
According to Observation9 this ends the proof.


### 3.1.3. Broken Line ConnectingI3- and I3+

Step 1.
LetE0={t∈B′:tisa2-colourededge,t∩I3-≠∅},
(3.7)E1=∅.Step 2.
Takee∈E0∖E1.
Adde to E1.
The vertexv∈e∩I3- is said to be used.
Go to Step3.Step 3.
Take unused vertexu of the last added edge to the set E1.
Ifu∈I3+ END.
Otherwise,
Ifu∈I3- go to Step 2.
Else go to Step4.Step 4.
Take unused vertexu of the last added edge to the set E1.
Next take 2-coloured edgee∈B′∖E1 such that v∈e.
Now verticev is said to be used.
Adde to the set E1.
Go to Step3.First of all the number of 2-coloured edges fromB′, which one of vertices lies on I3- 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∂I3 (Observation 7).This arguments allows one to say that procedure is well defined.Now our broken line connectingI3+ and I3- is created as follows:lete1 be the last added element to E1.Ifei∩I3-=∅ then ei+1 is previous added element to E1else Stop.We obtained the sequence of edges{e1,e2,…,em}⊂B′. Let us define coloring ϕ3:{t∈B′:tisanedgeofei}→{0,1} where i∈{1,…,m}: (3.8)ϕ3(t)={0f3(t)≤01f3(t)>0.It is easy to see thatϕ3(e1)={1} and ϕ3(em)={0}.So starting frome1 we search with order the first edge ek∈{e1,e2,…,em} such that ϕ3(ek)={0,1}.


## 3.1.1. Surface

Letk be a natural number, such that 1/k<ϵ*.The center of eacht∈T(k),t=[i1/k,(i1+1)/k]×⋯×[i3/k,(i3+1)/k] is defined as follows: (3.1)tc=(i1k+12k,i2k+12k,i3k+12k).
Let us define coloring map ϕ1:T(k)→{0,1}(3.2)ϕ1(t)={0f1(tc)≤01f1(tc)>0.
Algorithm for surface is as follows.Step 1.
Let(3.3)A0={t∈T(k):t∩I1-≠∅},A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1}.Step 2.
IfC={t∈T(k)∖A0:dim⁡[t∩A0]=2,ϕ1(t)=0}=∅ then END.
Otherwise do Step3.Step 3.
Add elements of the setC to A0.
Next(3.4)A1={t∈T(k):t∩A0≠∅,ϕ1(t)=1},B={t∩t′:t∈A0,t′∈A1,t∩t′≠∅}
and go to Step 2.
SinceT(k) is finite, hence after finitely many steps set C is empty (the procedure ends).
Let us consider the familyB. We may assume that B is closed under finite intersections.
The elementsb∈B, such that dim⁡[b]=2,dim⁡[b]=1,dim⁡[b]=0 are called squares, edges, and vertices.Observation 4.
The⋃B separates cube I3 between I1- and I1+.Observation 5.
Each edgeb∈B if b⊂∂I3 it is an edge of exactly 1 square, else it is an edge of 2 or 4 squares.


## 3.1.2. Modification ofB

Let us divide each element of{a∈A0:a∩⋃B≠∅} onto 27 cubes (in the natural way).Denote the set consisting of all this cubes byT′.Create coloring mapϕ1′:T′→{0,1} as follows:
(3.5)ϕ1′(t′)={0t′∩A1=∅1t′∩A1≠∅.
NowB′={t∩t′:t,t′∈T′,ϕ1′(t)=0,ϕ1′(t′)=1,t∩t′≠∅}.Observation 6.
Any edge ofB′ is an edge of exactly one or of two squares from B′ depending on whether or not it lies on ∂I3.
Let us define coloringϕ2:{t∈B′:tisasquare}→{0,1}:
(3.6)ϕ2(t)={0f2(tc)≤01f2(tc)>0,
where tc is the center of square t.
The edget∈B′ is said to be 2-coloured if there exists squares s,s′∈B′ such that s∩s′=t and ϕ2({s,s′})={0,1}.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∂I3.Observation 8.
The components of⋃B′∩∂I3 are broken lines without self-cutting.Observation 9.
The number of broken lines lying onI3- and connecting I2- and I2+ is odd.Lemma 3.1.
The number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd.Proof.
Let us consider components of the set⋃B′∩I3-.
We have odd number of broken lines connectingI2- and I2+ and the number of the rest components is arbitrary.
Let us see that⋃B′⊂I3∖I1-∪I1+.
So, the number of 2-coloured edges fromB′, which one of vertices lies on I3- is odd if it lies on broken line connecting I2- and I2+ else it is even (using the definition of ϕ2).
According to Observation9 this ends the proof.


## 3.1.3. Broken Line ConnectingI3- and I3+

Step 1.
LetE0={t∈B′:tisa2-colourededge,t∩I3-≠∅},
(3.7)E1=∅.Step 2.
Takee∈E0∖E1.
Adde to E1.
The vertexv∈e∩I3- is said to be used.
Go to Step3.Step 3.
Take unused vertexu of the last added edge to the set E1.
Ifu∈I3+ END.
Otherwise,
Ifu∈I3- go to Step 2.
Else go to Step4.Step 4.
Take unused vertexu of the last added edge to the set E1.
Next take 2-coloured edgee∈B′∖E1 such that v∈e.
Now verticev is said to be used.
Adde to the set E1.
Go to Step3.First of all the number of 2-coloured edges fromB′, which one of vertices lies on I3- 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∂I3 (Observation 7).This arguments allows one to say that procedure is well defined.Now our broken line connectingI3+ and I3- is created as follows:lete1 be the last added element to E1.Ifei∩I3-=∅ then ei+1 is previous added element to E1else Stop.We obtained the sequence of edges{e1,e2,…,em}⊂B′. Let us define coloring ϕ3:{t∈B′:tisanedgeofei}→{0,1} where i∈{1,…,m}: (3.8)ϕ3(t)={0f3(t)≤01f3(t)>0.It is easy to see thatϕ3(e1)={1} and ϕ3(em)={0}.So starting frome1 we search with order the first edge ek∈{e1,e2,…,em} such that ϕ3(ek)={0,1}.


## 3.2. Topological Part

For eachk∈N, 1/k<ϵ* we have(i)
vk,vk′∈ek such that f3(vk)≤0 and f3(vk′)>0,(ii)
uk,uk′∈tu∪tu′ such that f2(uk)≤0 and f2(uk′)>0 where tu,tu′ are squares from B and ek∩tu≠∅≠ek∩tu′,(iii)
wk,wk′∈tw∪tw′ such that f1(wk)≤0 and f1(wk′)>0 where tw is a cube from A0, tw′  is a cube from A1 and ek∩tw≠∅≠ek∩tw′.Define the setsWk:=conv{vk,vk′,uk,uk′,wk,wk′}.For eachWk there exist ck1,ck2,ck3∈Wk such that (3.9)f1(ck1)=f2(ck2)=f3(ck3)=0.
Without loss of generality we can assume that lim⁡k→∞ck1=c.Moreover,lim⁡k→∞diam⁡[Wk]=0. So for each ck′∈Wk the fact d(c,ck′)≤d(c,ck1)+d(ck1,ck′) yields(3.10)lim⁡k→∞⁡ck1=lim⁡k→∞⁡ck2=lim⁡k→∞⁡ck3=c.
So f(c)=0 ends proof.



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*Source: 793848-2011-10-12.xml*