Sufficent Conditions for the preservation of Polygonal-Connectedness in an arbitrary normed space
Savvas Andronicou Emmanouil Milakis

	
	



			


	Sufficent Conditions for the preservation of Polygonal-Connectedness in an arbitrary normed space
    Savvas Andronicou and Emmanouil Milakis
    
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		In this article we prove that if (X,·_X) is a normed space and U is a polygonally-connected subset of X with M:={S_i: i∈ I}⊂𝒫( U), a non-empty arbitrary family of discrete, non-empty  subsets of U, then the property of polygonal-connectedness is also preserved in the resulting set U∖( ⋃_i∈ I S_i) , under appropriate conditions.
	
	
		Mathematics Subject Classification: 54D05, 30L99
	
	Keywords: Polygonal-connectedness, arbitrary normed spaces	
	
	

§ INTRODUCTION

	
	The property of polygonal-connectedness appears in many settings from pure applications to abstract theories. For instance in a series of papers by Klee (see <cit.>, <cit.>) it played a critical role when one tries to characterise convex sets in different topologies. More recently, in a very interesting result <cit.>, it was proven that the set of all framelets is polygonally connected  both in L^2() and in multiplicative L^2(,dx/|x|) norm, with a generalisation in higher dimensions, answering a fundamental question in the theory of wavelets. Finally one could mention recent results in image processing (see <cit.> and references therein) where  digitization can be defined as a process of transforming a given set in Euclidean space ^n  into a discrete set by considering the Gauss digitization i.e its intersection with ^n. As it is apparent in <cit.> , due to the approximations induced by the digitization process, the polygonal connectedness of the initial set is not always preserved. 
		
	The main objective of the present article is to investigate certain properties of polygonally connected sets in arbitrary normed spaces. It can be considered as a continuation of paper <cit.> where the authors examined the question of preservation of path-connectedness, in particular, the finding of natural conditions on sets with holes in an arbitrary metric spaces in order to preserve the property of path-connectedness. A similar question stands in an arbitrary normed space for polygonally connected sets. 
	
		 Question: Assume that (X,·_X) is an arbitrary normed space and U is a polygonally-connected subset of X. If one removes a family M of discrete, non-empty  subsets from U, is it true that resulting set U∖⋃ M is also polygonally-connected? 
	

	
	
	The purpose of the present paper is provide the structural condition on sets with holes that allows the property of polygonal-connectedness to remain valid, providing in that way a positive answer to the above mentioned question. Our approach is pure analytic and deals with a delicate construction of polygonal lines which preserve the connectedness of the structure.  To the best of our knowledge, these constructions are absent from the literature. Note that although our intention is to provide an answer to a pure theoretical question, it seems that one will end up constructing similar polygonal lines when treating problems from numerical analysis and computer science, for instance when constructing Algorithms for finding convex hulls of simple polygons in general settings (see for instance <cit.>, <cit.>, <cit.>). To this perspective, and in addition to the applications in image processing <cit.> mentioned above, we anticipate that our results will also gain attention among scientists working in these areas as well. For other applications in Analysis of PDEs in nodal sets the reader is referred to the introduction of <cit.>.     
	
	The structure of the paper is as follows. Firstly we give the list of notations to be used and in Section 2 we provide the appropriate definitions and present the main results. Section 3 is devoted to the proof of the main theorem.
	
	
		

§ NOTATIONS

	Next we will provide a list of symbols and notions to be used in the sequel. Let us consider (X,d) to be a metric space with  A,B⊂ X and x_0∈ X. We denote 

	B_d(x_0,δ):={z∈ X : d(z,x_0)<δ} the usual open ball with center x∈ X and radius δ>0 

	int(A):={x∈ X: ∃δ>0, B_d(x,δ)⊂ A}, the internal of the set A

	ext(A):=int(X∖ A), the external of the set A

	Cl(A)≡A̅:={x∈ X: x limit point of the set A}, the closure of the set A

	∂ A:=A̅∩ Cl( X∖ A), the boundary of the set A

	iso(A):={x∈ X: ∃δ_x>0, B_d(x,δ_x)∩ A={x}}, the set of isolated points of the set A

	
    dist(x_0,A):=inf{ d(x_0,y): y∈ A },     A≠∅
    
    			0,     A=∅, 
    	   dist(A,B):=inf{ d(x,y): x∈ A, x∈ B },     A≠∅
    
    		0,     A=∅, or B=∅

where the quantity dist(x_0,A) is the distance of point x_0 from the set  A and dist(A,B)  is the distance from the set A to the set B. 

	For a continuous function γ:I→ X, where I is an interval in ,  we use the following symbolism:
	
		
  * γ(I):={γ(t)|  t∈ I} the trace of the curve γ on the interval I.
		
  * For x,y∈γ(I), let x=γ(t_x), y=γ(t_y) and without loss of generality, let t_x<t_y, then we set: 
		γ_xy:={γ(t)| t∈[t_x,t_y]}, i.e the part of curve γ that connects the points x and y.
	
	

§ DEFINITIONS AND MAIN RESULT

	
	In this section we give some necessary definitions for connected metric spaces and polygonally-connected subsets of normed spaces.
	
		A metric space ( X,d) is called connected if there is no open partition i.e there is no pair of sets {A,B} such that the following hold:
		
    (i) X=A∪ B, (ii)  A,B≠∅, (iii)  A∩ B=∅, (iv) A, B  are both open.

	
	
		Let (X,·_X) be a normed space and U⊂ X.
		
			
  * If x,y∈ X then the set [x,y]:={(1-t)x+ty: t∈[0,1]} is called line segment with endpoints x,y.
			
  * If (x_j)_j=1^n is a finite sequence of points of X, (not necessary distinct), then the set ⋃_j=2^n[x_j-1,x_j] is called polygonal line with initial point x_1 and final point x_n. The line seqments [x_1,x_2], [x_2,x_3], …, [x_n-1,x_n] are called edges of the polygonal line.
			
  * For any points x,y of the space X, with x≠ y, we say that are connected through a polygonal line, if there exists a polygonal line with initial point x and final point y.
			
  * A set U⊂ X is called polygonally connected, if for every x,y∈ U, there exists a polygonal line ⋃_j=2^n[x_j-1,x_j] that connects the points x,y and ⋃_j=2^n[x_j-1,x_j]⊂ U.
			
		
	
	
	
	
	
		Let (X,·_X) be a normed space and U a polygonally-connected subset of X with M:={S_i: i∈ I}⊂𝒫( U), an arbitraty family of discrete, non-empty subsets of U. We call that the set U satisfies the (U-M) property if and only if, for every polygonal line, ⋃_j=2^n[x_j-1,x_j] the following statement holds:
		
    For all j∈{2,…,n}, the set [x_j-1,x_j]∩(⋃_i∈ I∂ S_i ) is finite.

	


Next we are ready to present our main result.

	
		Let (X,·_X) be a normed space and U a polygonally-connected subset of X with M:={S_i: i∈ I}⊂𝒫( U), an non-empty, arbitrary family of discrete[By discrete family of sets M:={S_i: i∈ I}, we mean that, the map P:I→ M is injective i.e for every i,j∈ I, if i≠ j then S_i=P(i)≠ P(j)=S_j.], non-empty subsets of U. We assume that:
		
			
  (i) U satisfies the (U-M) property.
			
  (ii) For all i∈ I, S_i is open, ∂ S_i≠∅ and S̅_i int(U), with S̅_i∩S̅_j=∅, ∀ i,j∈ I, i≠ j.
			
  (iii) For all i∈ I, there exists δ_i>0 such that 
			
    K_δ_i:={ x∈ U∖( ⋃_i∈ I S_i) : dist(x, ∂ S_i)<δ_i }

		is polygonally-connected. 
		
	Then, the resulting set U∖(⋃_i=1^n S_i) is polygonally-connected.
	
























   
    
 	
	
	
	
	
	
	
	

	
	
	
	

§ PROOF OF MAIN RESULT

	The proof of the Theorem <ref> is implemented through the following fundamental to our analysis proposition:
	
	
		Let (X,d) be an arbitrary metric space, K⊂ X, with ∂ K≠∅ and x_0∈ int(K), y_0∈ ext(K). If γ:I→ X a continuous function defined on the interval I⊂ such that, x_0=γ(t_x_0),y_0=γ(t_y_0), where t_x_0,t_y_0∈ I, then
		
    γ(Ĩ) ∩∂ K≠∅

		where Ĩ:=
			[t_x_0,t_y_0],    t_x_0<t_y_0

			[t_y_0,t_x_0],    t_y_0<t_x_0.
	


	
		Without loss of generality, let t_x_0<t_z, i.e Ĩ:=[t_x_0,t_z].
		Let suppose the opposite of what we seek to prove i.e γ(Ĩ) ∩∂ K=∅. We use the fact that the metric space X receives a partition  in terms of the set K, i.e
		
    X=int(K) ∪∂ K∪ ext(K) .

		As a result from the above, we obtain
		
    γ(Ĩ)   =γ(Ĩ)∩ X 
       =(γ(Ĩ)∩ int(K) ) ∪(γ(Ĩ)∩∂ K ) ∪( γ(Ĩ)∩ ext(K) )
       =^(  γ(Ĩ) ∩∂ K=∅)  (γ(Ĩ)∩ int(K) )∪( γ(Ĩ)∩ ext(K) )
       =V_1∪ V_2

		where V_1:=γ( Ĩ) ∩ int(K) and V_2:=γ( Ĩ) ∩ ext(K). First we see that from the Principle of Inheritance, the sets V_1, V_2 are open in the metric subspace (γ(Ĩ),d_γ(Ĩ)). Moreover, V_1≠∅, since x_0∈γ(Ĩ)∩ int(K) and V_2≠∅, since from the previous claim, we have y_0∈ ext(K) and y_0∈γ( Ĩ) . Clearly, we see that V_1∩ V_2=∅. 
		Consequently, the pair of sets {V_1, V_2} is an open (with respect to the relative metric d_γ(Ĩ)) partition of γ(Ĩ). From the last we conclude that the set γ(Ĩ) is not connected. The latter is contradicted with the fact that, since Ĩ is an interval in , therefore a connected set and the function  γ:Ĩ→ X, is continuous, it follows from known theorem [Let (X,d ), (Y,ρ) metric spaces, and f:X→ Y be a continuous function. If (X,d ) is connected space, then the image f(X) is connected subset of Y.] that γ(Ĩ) will be connected. As a result from the above analysis, we receive that:  γ(Ĩ)∩∂ K≠∅. 
		
Next section is devoted solely to the proof of the main theorem.  As it will be apparent, the proof will be achieved  by distinguishing several appropriate cases.





 §.§ Proof of Theorem <ref>

	
		Let x,y∈ U∖( ⋃_i∈ IS_i), with x≠ y. Since by the assumption of the theorem, U is polygonally-connected, there exists a polygonal line inside U that connects the points x and y. In specific, let ⋃_j=2^n[x_j-1,x_j]⊂ U where x_1:=x and x_n:=y. Now we have the following cases:

		Case: ( ⋃_j=2^n[x_j-1,x_j])⋂(⋃_i∈ IS_i )=∅. 
		
		In this case we obtain that ⋃_j=2^n[x_j-1,x_j]⊂ U∖(⋃_i∈ IS_i ). Moreover, observe that it is possible to have the situation ( ⋃_j=2^n[x_j-1,x_j]) ∩( ⋃_i∈ I∂ S_i)≠∅. Indeed, from the assumption of the case, we receive that ⋃_j=2^n[x_j-1,x_j]⊂⋃_i∈ I S^c_i. Moreover, since S_i is open, it folows that S_i∩∂ S_i=∅, ∀ i∈ I. Equivalently, ∂ S_ĩ⊂ S^c_ĩ⊂⋃_i∈ I S^c_i, ∀ĩ∈ I. Therefore, ⋃_i∈ I∂ S_i⊂⋃_i∈ I S^c_i.

		Case: ( ⋃_j=2^n[x_j-1,x_j])⋂(⋃_i∈ IS_i )≠∅.

		 We define 
		 
    J:={ j∈{2,3,…,n}: ∃ i∈ I,  [x_j-1,x_j]∩∂ S_i ≠∅ and  [x_j-1,x_j]∩ S_i≠∅}.

	 
	 	The set J is not empty.
	 
 
 	By our consideration above, there exists z∈⋃_j=2^n[x_j-1,x_j] and z∈⋃_i∈ IS_i. Equivalently, there exists j^*∈{2,3,…,n} and i_0∈ I such that, z∈[x_j^*-1,x_j^*] and z∈ S_i_0. Then, we distinguish the following cases:
 	
 		
  * [x_j^*-1,x_j^*]∩∂ S_i_0≠∅.

 		If this case holds, then we obviously have that j^*∈ J
 		
  * [x_j^*-1,x_j^*]∩∂ S_i_0=∅.

 		If this case holds then we obtain that, [x_j^*-1,x_j^*]⊂ int(S_i_0)=^(S_i_0 open)S_i_0. Indeed, we observe that,
 		
    [x_j^*-1,x_j^*]   =( [x_j^*-1,x_j^*]∩ int(S_i_0)) ∪( [x_j^*-1,x_j^*]∩∂ S_i_0) ∪([x_j^*-1,x_j^*]∩ ext( S_i_0) ) 
       =( [x_j^*-1,x_j^*]∩ int(S_i_0)) ∪([x_j^*-1,x_j^*]∩ ext( S_i_0) )=V_1∪ V_2
 
 	where V_1, V_2 are open in the metric subspace [x_j^*-1,x_j^*] with V_1∩ V_2=∅. Because V_1≠∅, (since z∈ V_1), and [x_j^*-1,x_j^*] is connected as a continuous image of the interval [0,1], we conclude that V_2=∅. Therefore [x_j^*-1,x_j^*]=V_1⊂ int(S_i_0)=S_i_0.

 	Next, we define
 	
    L:={j∈{j^*,j^*+1,…,n-1}: [x_j,x_j+1]∩∂ S_i_0≠∅}.

 	We claim that, L≠∅. Indeed, by the assumption of the theorem, y∈ U∖( ⋃_i∈ IS_i), and as a result from the last, we obtain that y≡ x_n∉ S_i_0. Therefore, since  S_i_0 is open, it follows that y∈∂ S_i_0∪ int(S^c_i_0). Then, we distinguish the following cases:
 	
 		
  * y∈∂ S_i_0. 
In this case, since y≡ x_n∈[x_n-1,x_n], by the definition of L, we receive n-1∈ L.
 		
  * y∈ int(S^c_i_0)≡ ext(S_i_0). 
In this case we define the path, γ_z,y:[a,b]→ X where γ_z,y:=γ_[z,x_j^*]⊕( ⊕_j=j^*^n-1γ_[x_j,x_j+1]). Obviously the path γ_z,y is continuous and connects the points z∈ S_i_0=int (S_i_0) and y≡ x_n∈ int(S^c_i_0).  Finally, from the Proposition <ref>, it holds that γ_z,y([a,b]) ∩∂ S_i_0≠∅. Therefore, there exists t_0∈[a,b] such that γ_z,y(t_0)∈∂ S_i_0. Equivalently, there exists j̃∈{j^*,…,n-1} such that γ_z,y(t_0)∈[x_j̃,x_j̃+1]∩∂ S_i_0. From the last, we conclude that j̃∈ L.
 	
 From both cases, we get that L≠∅ and we are in the position to define j̅:=min L. By the definition of j̅, it follows that 
 
    [x_j̅,x_j̅+1]∩∂ S_i_0≠∅ and∀ j∈{j^*,j̅-1}, [x_j,x_j+1]∩∂ S_i_0=∅.
 
We claim now that, [x_j̅,x_j̅+1]∩ S_i_0≠∅. Indeed, let us suppose that [x_j̅,x_j̅+1]∩ S_i_0=∅. We define

    t̃   :=inf{t∈[0,1]: (1-t)x_j̅+t x_j̅+1∈∂ S_i_0}≡inf L_x_j̅, x_j̅+1
    z̃   :=(1-t̃)x_j̅+t̃x_j̅+1.

We notice that, L_x_j̅, x_j̅+1:={t∈[0,1]: (1-t)x_j̅+t x_j̅+1∈∂ S_i_0} is a non-empty set since [x_j̅,x_j̅+1]∩∂ S_i_0≠∅. Moreover, L_x_j̅, x_j̅+1 is lower bounded from 0, therefore t̃ is well defined. Now we claim that t̃∈[0,1]. Indeed, since [x_j̅,x_j̅+1]∩∂ S_i_0≠∅, there exists t^*∈ L_x_j̅, x_j̅+1 and by the definition of t̃, we receive that 0≤t̃≤ t^*≤ 1.

 Furthermore, z̃∈∂ S_i_0. Indeed, since t̃ is a limit point of L_x_j̅, x_j̅+1, there exists a sequence (t_k)_k∈⊂ L_x_j̅, x_j̅+1 such that, t_kt̃. Now we define  the map s:[0,1]→ X, with s(t):=(1-t)x_j̅+t x_j̅+1. From the continouity of s it follows that, s(t_k)s(t̃)≡z̃. Since, t_k∈ L_x_j̅, x_j̅+1, ∀ k∈, it follows that s(t_k)∈∂ S_i_0. Therefore, z̃ is a limit point of ∂ S_i_0 and due to the fact that ∂ S_i_0 is closed, we obtain finally that z̃∈∂ S_i_0.

We distinguish the following cases for t̃. In specific, if t̃=0, then z̃=x_j̅. Therefore, x_j̅∈∂ S_i_0 and as a result from the last, we receive that [x_j̅-1,x_j̅]∩∂ S_i_0≠∅, which is a contradiction from the second part of (<ref>). Now, if t̃∈ (0,1], then from the definition of t̃, we obtain that for all  t∈[0,t̃), z_t:=(1-t)x_j̅+ t x_j̅+1∉∂ S_i_0. Therefore, z_t∈ ext(S_i_0), ∀ t∈[0,t̃). (since [x_j̅,x_j̅+1]∩ S_i_0=∅). For t=0, we receive that, z_0=x_j̅ and from the last we conclude that x_j̅∈ ext(S_i_0). At the same time, we know from the initial part of the proof that z∈[x_j^*-1,x_j^*]⊂ S_i_0. Next, we define the path, γ_z,x_j̅:[ã,b̃]→ X where γ_z,x_j̅:=γ_[z,x_j^*]⊕(⊕_j=j^*^j̅-1γ_[x_j,x_j+1]). Obviously, the path γ_z,x_j̅ is continuous and connects the points z∈ S_i_0 and x_j̅∈ ext(S_i_0). Finally from Proposition <ref>, it holds that γ_z,x_j̅([ã,b̃]) ∩∂ S_i_0≠∅. Therefore, there exists t̃_̃0̃∈[ã,b̃] such that γ_z,x_j̅(t̃_̃0̃)∈∂ S_i_0. Equivalently, there exists j̃∈{j^*,…,j̅-1}, ([x_j^*-1,x_j^*]⊂ S_i_0 and S_i_0∩∂ S_i_0≠∅, since S_i_0 is open ) such that  γ_z,x_j̅(t̃_̃0̃)∈[x_j̃,x_j̃+1]∩∂ S_i_0. I.e j̃∈ L and j̃<j̅. The last contradicts with the definition of j̅:=min L.

From the above analysis we conclude that [x_j̅,x_j̅+1]∩ S_i_0≠∅. Then by the definition of j̅ it follows that [x_j̅,x_j̅+1]∩∂ S_i_0≠∅, therefore by the definition of the set J, we obtain j̅+1∈ J.



Next, we define the following indexes

    k_1:=min J, k_m:=min(J∖{k_m-1}), ∀ m∈{2,3,…,j_0} (for j_0≥ 2)

where j_0:=max J. I.e for j_0≥ 2, J={k_1<k_2<…<k_j_0}. 

	If k_1≥ 3, then ∀ i∈ I, ∀ l∈{2,3,…,k_1-1},  [x_l-1,x_l]⊂∂ S_i ∪ ext(S_i).


	For every l∈{2,3,…,k_1-1} it is obvious that l∉ J. Let l_0∈{2,3,…,k_1-1} be fixed, randomly selected. Therefore, by the definition of the set J, we conclude that, for all i∈ I, the conditions [x_l_0-1,x_l_0] ∩∂ S_i≠∅ and [x_l_0-1,x_l_0] ∩ S_i≠∅ cannot hold at the same time. The last is equivalent to the following: ∀ i∈ I,
	
    [x_l_0-1,x_l_0] ∩∂ S_i=∅ and [x_l_0-1,x_l_0] ∩  S_i=∅
    or [x_l_0-1,x_l_0] ∩∂ S_i≠∅ and [x_l_0-1,x_l_0] ∩  S_i=∅
    or [x_l_0-1,x_l_0] ∩∂ S_i=∅ and [x_l_0-1,x_l_0] ∩  S_i≠∅.

	For a random i∈ I, if the condition (<ref>) holds, we receive that, [x_l_0-1,x_l_0] ⊂ ext(S_i)⊂∂ S_i∪ ext(S_i). Moreover, for a random i∈ I if the condition (<ref>) holds, we obtain that  [x_l_0-1,x_l_0] ⊂∂ S_i∪ ext(S_i). Now we will show that the condition (<ref>) cannot be true. Indeed,  let that condition to be true i.e we allow the case where for a random ĩ∈ I the condition (<ref>) holds. Then using the fact that the line interval [x_l_0-1,x_l_0] is connected, we obtain [x_l_0-1,x_l_0] ⊂ S_ĩ. As a result there exists z∈[x_l_0-1,x_l_0] , z∈ int(S_ĩ). Also we notice that x_1=x∉ S_ĩ, i.e x_1∈∂ S_ĩ∪ ext(S_ĩ). We consider the polygonal line γ^*_x_1,z:=∪_q=2^l_0-1γ_[x_q-1,x_q]∪γ_[x_l_0-1,z]. Since the map γ^*_x_1,z is continuous, using the Proposition <ref>, we receive that γ^*(I) ∩∂ S_ĩ≠∅. Therefore, there exists q_0∈{2,3,…,l_0-1} such that [x_q_0-1,x_q_0]∩∂ S_ĩ≠∅. We show that the last statement cannot be true. In specific, we prove that ∀ q∈{2,…,l_0-1}, [x_q-1,x_q]∩∂ S_ĩ=∅. Let us suppose that q_0=l_0-1 i.e [x_l_0-2,x_l_0-1]∩∂ S_ĩ≠∅. We know already that x_l_0-1∈ S_ĩ (since [x_l_0-1,x_l_0]⊂ S_ĩ). From the last we conclude by the definition of the set J, that l_0-1∈ J, therefore k_1=min J≤ l_0-1. This is a contradiction, since l_0≤ k_1-1<k_1 and as a result  l_0-1<k_1 . Furthermore, we obtain that [x_l_0-2,x_l_0-1]∩∂ S_ĩ =∅. Thus, [x_l_0-2,x_l_0-1]⊂ S_ĩ∪ ext(S_ĩ). Again using the connectedness argument of the line interval [x_l_0-2,x_l_0-1] and the fact that [x_l_0-2,x_l_0-1]∩ S_ĩ≠∅, we receive that [x_l_0-2,x_l_0-1]⊂ S_ĩ hence, q_0∈{2,…,l_0-2}. If we suppose now that q_0=l_0-2, repeating the previous procedure, we conclude that [x_l_0-3,x_l_0-2]⊂ S_ĩ. Continuing this procedure, until q_0=2, we conclude that [x_1,x_2]⊂ S_ĩ which is a contradiction since x≡ x_1∈[x_1,x_2]⊂ S_ĩ and x∉∪_i∈ IS_i. Thus, we proved that ∀ q∈{2,…,l_0-1}, [x_q-1,x_q]∩∂ S_ĩ=∅. In consequence, the condition (<ref>) does not hold. Hence, ∀ i∈ I, [x_l_0-1,x_l_0]⊂∂ S_i∪ ext(S_i).


For each m∈{1,2,…,j_0} we define:

    Q_m:={ i∈ I: [x_k_m-1,x_k_m]∩∂ S_i≠∅ and [x_k_m-1,x_k_m]∩ S_i≠∅}.


	For all m∈{1,2,…,j_0} the set Q_m is finite, non-empty.


	Let m∈{1,2,…,j_0} be randomly selected. Initially we prove that Q_m≠∅. By the definition of k_m, there exists i^*∈ I such that [x_k_m-1,x_k_m]∩ S_i^*≠∅. Therefore, i^*∈ Q_m. Next, we prove that, Q_m is finite. Let us suppose the opposite of what we seek to prove i.e Q_m is infinite. Consequetly there exists countably infinite  P⊂ Q_m. Let P:={i_λ: λ∈}. Since, P⊂ Q_m, for each λ∈ there exists y_λ∈[x_k_m-1,x_k_m]∩∂ S_i_λ. We claim that the sequence (y_λ)_λ∈ consists of distinct terms. Indeed, let λ_1,λ_2∈ with λ_1≠λ_2 such that y_λ_1=y_λ_2. As a result from the last, we receive ∂ S_i_λ_1∩∂ S_i_λ_2≠∅, which contradicts with the assumption of the theorem, that S̅_i_λ_1∩S̅_i_λ_2=∅. We conclude then, that {y_i_λ: λ∈} is countably infinite. From the fact that
	
    {y_i_λ: λ∈}⊂[x_k_m-1,x_k_m]∩( ∪_λ∈∂ S_i_λ)⊂ [x_k_m-1,x_k_m]∩( ∪_i∈ I∂ S_i)

we obtain that the set [x_k_m-1,x_k_m]∩( ∪_i∈ I∂ S_i) is infinite, where it contradicts with the (U-M) assumption of the theorem.
  
For all m∈{1,2,…,j_0}, i∈ Q_m we define:

    L^(m)_i:={t∈[0,1]: z^(m)(t):=(1-t)x_k_m-1+t x_k_m∈∂ S_i}
    
    	t^(m)_i,min:=inf L^(m)_i and t^(m)_i,max:=sup L^(m)_i.



For every m∈{1,2,…,j_0} the following hold:
	
		
  (i) For all i∈ Q_m, L^(m)_i≠∅, finite. Let L^(m)_i:={t^(m)_i,1<t^(m)_i,2<…<t^(m)_i,r_m,i} where r_m,i:=card(L^(m)_i)
		
  (ii) For all i,j∈ Q_m, with  i≠ j,  L^(m)_i∩ L^(m)_j=∅. 
	


	Let  m∈{1,2,…,j_0} be randomly selected.
	
		
  (i) Let i∈ Q_m, be randomly selected. Therefore, [x_k_m-1,x_k_m]∩ S_i≠∅ and [x_k_m-1,x_k_m]∩∂ S_i≠∅. From the last, it follows that there exists y∈[x_k_m-1,x_k_m] and y∈∂ S_i. Equivalently, there exists t̃∈[0,1], y=z^(m)(t̃)∈∂ S_i. Hence, t̃∈ L^(m)_i.

		We proceed now to prove that for all i∈ Q_m the set L^(m)_i is finite. Let i^*∈ Q_m, be randomly selected. From the definition of L^(m)_i^* it holds that ∀ t∈  L^(m)_i^*, z^(m)(t)∈∂ S_i^*. Therefore, z^(m)(L^(m)_i^*)⊂ [x_k_m-1,x_k_m]∩∂ S_i^*⊂ [x_k_m-1,x_k_m]∩(∪_i∈ I∂ S_i). From the (U-M) assumption, we have that [x_k_m-1,x_k_m]∩(∪_i∈ I∂ S_i) is finite, therefore z^(m)(L^(m)_i^*) is finite. Since, z^(m) is injective, we conclude that L^(m)_i^* is finite.
		
  (ii)  Let us suppose the opposite of what we seek to prove i.e there exist i,j∈ Q_m with i≠ j such that L^(m)_i∩ L^(m)_j≠∅. Therefore, there exists t_0∈  L^(m)_i and t_0∈  L^(m)_j. Thus, z^(m)(t_0)∈∂ S_i and z^(m)(t_0)∈∂ S_j. From the last, we obtain, ∂ S_i∩∂ S_j≠∅ which contradicts with the assumption of theorem, S̅_i ∩S̅_j=∅. 
	

Next, we define for every m∈{1,2,…,j_0} a special type of  sequences: ( R_i_(v,m))_v∈⊂ [0,1], (i_(v,m))_v∈⊂ Q_m, ( t^(m)_i_(v,m))_v∈⊂⋃_v∈ R_i_(v,m). In specific, for every m∈{1,2,…,j_0} we define by induction the following sequences:

    i_(1,m)∈ Q_m such that t^(m)_i_(1,m):=min{t^(m)_i,min: i∈ Q_m}≡ t^(m)_i_(1,m),min

and
	 
    R_i_(1,m):={ t∈⋃_q∈ Q_m∖{i_(1,m)} L^(m)_q : ∃δ̃_t>0, z^(m)(t+δ̃)∈⋃_q∈ Q_m∖{i_(1,m)}S_q and t>t^(m)_i_(1,m),max}
    
    	 	i_(2,m):={ q∈ Q_m∖{i_(1,m)}:∃ s∈{1,2,…,r_m,q} t^(m)_q,s=min R_i_(1,m)} ,	    R_i_(1,m)≠∅
    
    	 		i_(1,m),	    R_i_(1,m)=∅
    
    	 		t^(m)_i_(2,m):=min R_i_(1,m),     R_i_(1,m)≠∅
    
    	 			t^(m)_i_(1,m),max,     R_i_(1,m)=∅

 For all  v≥ 3,
	 	
    R_i_(v-1,m):={ t∈⋃_q∈ Q_m∖{i_(1,m),…,i_(v-1,m)} L^(m)_q : ∃δ̃_t>0, z^(m)(t+δ̃)∈⋃_q∈ Q_m∖{i_(1,m),…,i_(v-1,m)}S_q and t>t^(m)_i_(v-1,m),max}
    
    	 		i_(v,m):={ q∈ Q_m∖{i_(1,m),…,i_(v-1,m)}:∃ s∈{1,2,…,r_m,q} t^(m)_q,s=min R_i_(v-1,m)} ,	    R_i_(v-1,m)≠∅
    
    	 			i_(v-1,m),	    R_i_(v-1,m)=∅
    
    	 			t^(m)_i_(v,m):=min R_i_(v-1,m),     R_i_(v-1,m)≠∅
    
    	 				t^(m)_i_(v-1,m),max,     R_i_(v-1,m)=∅



The next Claim guarantees that the above recursive definition of sequences is well-defined.

	For every m∈{1,2,…,j_0} the following hold:
	 
		
  (i) For every v∈ the set R_i_(v,m) is finite
		
  (ii) For every v∈ with v≥ 2, the sequence of indexes ( i_(v,m)) _v≥2 is well defined i.e For all v∈ with v≥ 2, if  R_i_(v-1,m)≠∅, then there exists unique q∈ Q_m∖{i_(1,m),…,i_(v-1,m)} with the property: ∃ s∈{1,2,…,r_m,q} t^(m)_q,s=min R_i_(v-1,m)
		
  (iii) For every k,λ∈, with k,λ≥ 2, if k≠λ and R_i_(k-1,m),R_i_(λ-1,m)≠∅ then i_(k,m)≠ i_(λ,m)
		
  (iv) If there exists v_0∈ such that R_i_(v_0,m)=∅ then for all v≥ v_0, R_i_(v,m)=∅
		
  (vi) For every k,λ∈ with k,λ≥ 2, if k<λ and R_i_(k-1,m),R_i_(λ-1,m)≠∅ then t^(m)_i_(k,m) < t^(m)_i_(λ,m)
		
  (vii) There exists v∈ with v≥ 2 such that R_i_(v-1,m)=∅
	

	
		Let m∈{1,2,…,j_0} be randomly selected.
		
			
  (i) For all v∈, by the definition of R_i_(v,m) we have
			
    R_i_(v,m)⊂⋃_q∈ Q_m∖{i_(1,m),…,i_(v,m)}L^(m)_q⊂⋃_q∈ Q_mL^(m)_q

		where from the Claim <ref> (i), it holds that L^(m)_q is finite for every q∈ Q_m. Hence, ⋃_q∈ Q_mL^(m)_q is finite, which follows that R_i_(v,m) is also finite.
		
			
  (ii) Let v∈ with v≥ 2 such that R_i_(v-1,m)≠∅ and let q_1,q_2∈ Q_m∖{i_(1,m),…,i_(v-1,m)} with q_1≠ q_2 which satisfy the following:
			
    ∃ s_1∈{1,2,…,r_m,q_1}, t^(m)_λ_1,s_1:=min R_i_(v-1,m)
    ∃ s_2∈{1,2,…,r_m,q_2}, t^(m)_λ_2,s_2:=min R_i_(v-1,m)

			By the definition of numbers t^(m)_q_1,s_1,t^(m)_q_2,s_2, we obtain that t^(m)_q_1,s_1∈ L^(m)_q_1 and t^(m)_q_2,s_2∈ L^(m)_q_2. Therefore, L^(m)_q_1∩ L^(m)_q_2≠∅, which leads to a contradiction with the fact that since q_1≠ q_2, from Claim <ref>(ii) it holds that, L^(m)_q_1∩ L^(m)_q_2=∅. 
			
			
  (iii) Let k,λ∈, with k,λ≥ 2 and k<λ such that R_i_(k-1,m),R_i_(λ-1,m)≠∅. Therefore, the indexes i_(k,m), i_(λ,m) are well defined. Thus by the definition of index i_(λ,m), it follows that
 i_(λ,m)∈ Q_m∖{i_(1,m),i_(2,m),…,i_(k,m),…,i_(λ-1,m)}. Consequently, i_(λ,m)≠ i_(k,m).
			
			
  (iv) We prove this by induction. From the assumption we have that R_i_(v_0,m)=∅. Let r ≥ v_0 such that R_i_(r,m)=∅. We claim that R_i_(r+1,m)=∅. Indeed, let suppose that R_i_(r+1,m)≠∅. Therefore, there exists t_0∈⋃_q∈ Q_m∖{i_(1,m),…,i_(r,m),i_(r+1,m)} L^(m)_q such that the following hold:
			
    t_0>t^(m)_i_(r+1,m),max and z^(m)(t_0+δ_t_0)∈⋃_q∈ Q_m∖{i_(1,m),…,i_(r,m),i_(r+1,m)} S_q

			 We observe that ⋃_q∈ Q_m∖{i_(1,m),…,i_(r,m)i_(r+1,m)} S_q⊂⋃_q∈ Q_m∖{i_(1,m),…,i_(r,m)} S_q. Moreover, since R_i_(r,m)=∅, from the definition of t^(m)_i_(r+1,m), it holds that t^(m)_i_(r+1,m)=t^(m)_i_(r,m),max. Hence, t_0>t^(m)_i_(r+1,m),max≥ t^(m)_i_(r+1,m)= t^(m)_i_(r,m),max. I.e t_0∈ R_i_(r,m) which contradicts with the assumption of the induction step, R_i_(r,m)=∅. Consequently, we proved that R_i_(r+1,m)=∅ and by the Induction we get the desired result.
			
			
  (vi) Since R_i_(λ-1,m)≠∅, by the definition of t^(m)_i_(λ,m) we receive that t^(m)_i_(λ,m)=min R_i_(λ-1,m). Moreover, from the definition of R_i_(λ-1,m), we obtain that for all t∈ R_i_(λ-1,m), t>t^(m)_i_(λ-1,m),max. Therefore,
			
    R_i_(λ-1,m)∋ t^(m)_i_(λ,m)> t^(m)_i_(λ-1,m),max≥  t^(m)_i_(λ-1,m)
 
			 Since, R_i_(λ-1,m)≠∅ then ∀ v∈{k,k+1,…,λ-2}, R_i_(v,m)≠∅. Indeed, if we suppose that there exists v^*∈{k,k+1,…,λ-2} such that R_i_(v^*,m)=∅, then from Claim <ref>(iv) it follows that R_i_(λ-1,m)=∅ where the last contradicts with the assumption of the (vi).

			 
			As we proved before, since R_i_(k,m)≠∅ it follows that: t^(m)_i_(k+1,m)>t^(m)_i_(k,m).
			Similarly, since R_i_(k+1,m)≠∅ it follows that: t^(m)_i_(k+2,m)>t^(m)_i_(k+1,m).
		    Continuing the above procedure for finite steps, we obtain finally
		    
    t^(m)_i_(k,m)<t^(m)_i_(k+1,m)< t^(m)_i_(k+2,m)<…<t^(m)_i_(λ-1,m)

		    Combining, (<ref>) and (<ref>) we conclude that t^(m)_i_(k,m)<t^(m)_i_(λ,m).
		    
			
  (vii) Let us suppose the opposite of what we seek to prove i.e ∀ v≥ 2, R_i_(v-1,m)≠∅. From the Claim <ref>(iii), we obtain that the sequnece of indexes ( i_(v,m))_v≥ 2 consists of distinct terms. Therefore, the set {i_(v,m): v∈, v≥ 2}⊂ Q_m is infinite. Consequently, Q_m is infinite. The last contradicts with the Claim <ref>.
		
	
From the above Claim <ref>(vii), we are in the position to define, for every m∈{1,2,…,j_0}:

    v^(m)_0:=min{v∈: R_i_(v,m)=∅}.

Therefore, we have the following strictly increasing[Look the Claim <ref>(vi)] sequence ( t^(m)_i_(v,m))^v^(m)_0_v=1 of points in [0,1] and the corresponding sequence ( z^(m)( t^(m)_i_(v,m)) )^v^(m)_0_v=1 of points  in [x_k_m-1,x_k_m]. From now on, we symbolize z^(m)_v:=z^(m)(t^(m)_i_(v,m)) and z^(m)_v, max:=z^(m)(t^(m)_i_(v,m), max).

			 

Next, we proceed to the final part of the paper, the construction of the desired polygonal line in U∖(⋃_i∈ IS_i ) which connects the points x≡ x_1 and  y≡ x_n.



For every m∈{1,2,… j_0} there are two possible cases for x_k_m. In specific, x_k_m∈⋃_i∈ IS_i or x_k_m∉⋃_i∈ IS_i. Thus, for each case for x_k_m, we construct a special type of polygonal line, namely Γ^(1)_x_k_m, if x_k_m∉⋃_i∈ IS_i and Γ^(2)_x_k_m, if x_k_m∈⋃_i∈ IS_i. Before proceeding to case separation, we construct a special type of polygonal line, Γ̃_x_k_m-1, x_k_m, ∀ m∈{1,2,…,j_0}, which as we will see later, will be a main part of the most complex ones polygonal lines Γ^(1)_x_k_m, Γ^(2)_x_k_m.  



	
  * Construction of Polygonal line Γ̃_x_k_m-1, x_k_m,  m∈{1,2,…,j_0}

	In the following, for any x̃,ỹ∈ K_δ_i  we denote by Γ̂_x̃,ỹ any polygonal line in K_δ_i, i∈ I (for a definition of K_δ_i, see theorem formulation) that connects the points  x̃,ỹ. This polygonal line exists by the assumption of the theorem, the set K_δ_i⊂ U∖(⋃_i∈ IS_i ) is polygonally-connected.
	
		For every m∈{1,2,…,j_0}, the following hold:
		
			
  (i) If x_k_m-1∉⋃_i∈ IS_i, then  [x_k_m-1, z^(m)_1] ⊂ U∖( ⋃_i∈ IS_i)
			
  (ii) If x_k_m∉⋃_i∈ IS_i, then  [z^(m)_v_0, max, x_k_m] ⊂ U∖( ⋃_i∈ IS_i)
			
  (iii) If v_0^(m)≥ 3 then for every v∈{1,2,…, v_0^(m)-1}, [z^(m)_v, max, z^(m)_v+1] ⊂ U∖( ⋃_i∈ IS_i)	
			
  (iv) Let x_k_m-1∈ S_q for some q∈ I. Then, q∈ Q_m and also q=i_(1,m). Respectively, if x_k_m∈ S_q for some q∈ I, then, q∈ Q_m
			
  (v) Let x_k_m-1,x_k_m∈∪_i∈ IS_i. If v_0^(m)=1 then, there exists (unique) q∈ I such that, x_k_m-1, x_k_m∈ S_q.
			
  (vi) If x_k_m-1, x_k_m∈ S_q, for some q∈ Q_m,	 then v_0^(m)=1
			
  (vii) Let x_k_m-1,x_k_m∈∪_i∈ IS_i. If v_0^(m)≥ 2 then, there exists (unique) q_1,q_2∈ I with q_1≠ q_2, such that x_k_m-1∈ S_q_1 and x_k_m∈ S_q_2
			
  (viii) If x_k_m-1∈ S_q_1 and x_k_m∈ S_q_2 for some q_1,q_2∈ Q_m with q_1≠ q_2, then v_0^(m)≥ 2
		
	

	Let m∈{1,2,…,j_0}. 
	
		
  (i) Let x_k_m-1∉⋃_i∈ IS_i.

		We prove that [x_k_m-1, z^(m)_1] ⊂ U∖( ⋃_i∈ IS_i) i.e, ∀ i∈ I, [x_k_m-1, z^(m)_1]∩ S_i=∅. Let us suppose the opposite of what se seek to prove: There exists i^*∈ I, [x_k_m-1, z^(m)_1] ∩ S_i^*≠∅. Since,
		
    [x_k_m-1, z^(m)_1]:={ (1-t)x_k_m-1+tz^(m)_1: t∈[0,1] }={ (1-t)x_k_m-1+t x_k_m: t∈[0,t^(m)_i_(1,m)]}

		there exists t^*∈[0,t^(m)_i_(1,m)] such that z^*≡ z^(m)(t^*):=(1-t^*)x_k_m-1+t^* x_k_m∈ S_i^*. Initially we prove that t^*<t^(m)_i_(1,m). For this task we need the following intermediate step:
 
		Intermediate Step (I): i^*≠ i_(1,m).
		
			Let i^*= i_(1,m). Thus, i^*∈ Q_m. We notice that x_k_m-1∈∂ S_i_(1,m)∪ ext(S_i_(1,m)). In case where, x_k_m-1∈ ext(S_i_(1,m)), since z^*∈ S_i_(1,m)  from the Proposition <ref>, it follows that z^(m)([0,t^*]) ∩∂ S_i_(1,m)≠∅. The same holds, if we make the assumption that x_k_m-1∈∂ S_i_(1,m). Therefore, there exists, t̃∈[0,t^*], such that z̃:=z^(m)(t̃) ∈∂ S_i_(1,m). We observe that, t̃<t^*, since, if t̃=t^* we extract that z̃=z^*, i.e ∂ S_i_(1,m)∩ S_i_(1,m)≠∅,  where the last is contradiction since S_i_(1,m) is open. Now, from the definition of L^(m)_i_(1,m), we receive that, t̃∈ L^(m)_i_(1,m). Moreover, from the definition of t^(m)_i_(1,m):=min{t^(m)_q, min: q∈ Q_m}, we receive that t^(m)_i_(1,m)≤min L^(m)_i_(1,m)≤t̃. We see that the last contradicts with the fact that, t̃<t^*≤ t^(m)_i_(1,m). Consequently, i^*≠ i_(1,m).
			
			We observe that, t^*<t^(m)_i_(1,m). Indeed, if t^*=t^(m)_i_(1,m), then z^*=z^(m)_1. Since, z^*∈ S_i^* and z^(m)_1∈∂ S_i_(1,m) we obtain S̅_i^*∩S̅_i_(1,m)≠∅, which is a contradiction, since from the Intermediate Step (I), i^*≠ i_(1,m) therefore by the assumption of theorem, S̅_i^*∩S̅_i_(1,m)=∅.

			Intermediate Step (II): i^*∈ Q_m.
			
				We work similarly as we did in the Intermediate Step (I). Initially, we know that, x_k_m-1∈∂ S_i^*∩ ext(S_i^*). If x_k_m-1∈∂ S_i^* then, since z^*∈ S_i^*, we conclude that i^*∈ Q_m. Let now, x_k_m-1∈ ext(S_i^*). Again, since z^*∈ S_i^*, from the Proposition <ref> we receive, z^(m)([0,t^*])∩∂ S_i^*≠∅. Therefore[It holds that [x_k_m-1, z^*]:={ (1-t)x_k_m-1+tz^*: t∈[0,1] }={ (1-t)x_k_m-1+t x_k_m: t∈[0,t^*]}], [x_k_m-1, z^*]∩∂ S_i^*≠∅.(and [x_k_m-1, z^*]∩ S_i^*≠∅, since z^*∈ S_i^* ). Hence, i^*∈ Q_m.
			
			Let now z̅∈ [x_k_m-1, z^*] ∩∂ S_i^*=z^(m)([0,t^*])∩∂ S_i^*, i.e there exists (unique) t̅∈[0, t^*] such that z̅=z^(m)(t̅). By the definition of L^(m)_i^* we obtain, t̅∈ L^(m)_i^*. Furthermore, t̅< t^*. Indeed, if we suppose the equality, then we have z̅= z^*. Consequently, ∂ S_i^*∩ S_i^*≠∅ which is a contradiction since the set S_i^* is open. In addition, we know from the definition of t^(m)_i_(1,m), that t^(m)_i_(1,m)≤min L^(m)_i^*≤t̅. Therefore,  t^(m)_i_(1,m)≤t̅<t^*. The last, contradicts with previous result, we proved, i.e t^*<t^(m)_i_(1,m).  From the above analysis, we finally receive the desired result, ∀ i∈ I, [x_k_m-1, z^(m)_1]∩ S_i=∅.
		
  (ii) Let x_k_m∉⋃_i∈ IS_i.

		We prove that [ z^(m)_v_0^(m), max, x_k_m]  ⊂ U∖( ⋃_i∈ IS_i), i.e, ∀ i∈ I, [ z^(m)_v_0^(m), max, x_k_m] ∩ S_i=∅. Let us suppose the opposite of what we seek to prove. Therefore, there exists i^*∈ I, such that, [ z^(m)_v_0^(m), max, x_k_m] ∩ S_i^*≠∅.
		From the last, we obtain that there exists z^*∈[ z^(m)_v_0^(m), max, x_k_m] ∩ S_i^*. Thus, there exists[It holds that, [ z^(m)_v_0^(m), max, x_k_m] :={(1-t) z^(m)_v_0^(m), max+t x_k_m: t∈[0, 1] }={ (1-t) x_k_m-1+t x_k_m: t∈[ t^(m)_i_(v_0^(m), m), max, 1] }] t^*∈[t^(m)_i_( v_0^(m), m ) , max, 1] such that z^*≡ z^(m)(t^*)∈ S_i^*. For convenience, we set t_v_0^(m):=t^(m)_i_( v_0^(m), m ) , max.

		In order to lead to a contradiction, we prove the following Intermediate Steps.

		Intermediate Step (I): t^*>t_v_0^(m)
		
			Let us suppose the opposite of what we seek to prove, i.e t^*=t_v_0^(m). Then, z^*≡ z^(m)(t^*)=z^(m)( t_v_0^(m)) ≡ z^(m)_v_0^(m), max. Since, z^*∈ S_i^* and z^(m)_v_0^(m), max∈∂ S_i_( v_0^(m), m) we conclude that S_i^*∩∂ S_i_( v_0^(m), m) ≠∅. If i^*=i_(v_0, m) then the expression S_i^*∩∂ S_i_( v_0^(m), m) ≠∅ leads to a contradiction, since the set S_i^*=S_i_( v_0^(m), m) is open. On the other hand, if i^*≠ i_( v_0^(m), m) then using the fact that S_i^*∩∂ S_i_( v_0^(m), m) ≠∅ we conclude S̅_i^*∩S̅_i_( v_0^(m), m) ≠∅, which is a contradiction from the assumption of the theorem. Consequently, we obtain t^*>t_v_0^(m). 
		
		Intermediate Step (II): i^*∈ Q_m and in specific, [z^*, x_k_m]∩∂ S_i^*≠∅.
		
			Initially, we observe that [z^*,x_k_m]∩ S_i^*≠∅, therefore [x_k_m-1,x_k_m]∩ S_i^*≠∅. By the assumption we have that x_k_m∉ S_i^*. Thus, x_k_m∈∂ S_i^*∪ ext(S_i^*). We distinguish the following cases:
			
				
  * Let x_k_m∈∂ S_i^*. Then [x_k_m-1,x_k_m]∩∂ S_i^*≠∅, therefore i^*∈ Q_m.
				
  * Let x_k_m∈ ext(S_i^*). Since z^*∈ S_i^*, from Proposition <ref>, it holds that 
				
    z^(m)([t^*,1])∩∂ S_i^*≠∅

			Furthermore, since [z^*, x_k_m]=z^(m)([t^*, 1])⊂ z^(m)([0, 1])=[x_k_m-1, x_k_m], we obtain [x_k_m-1, x_k_m] ∩∂ S_i^*≠∅, therefore i^*∈ Q_m.
			
		From both cases, we also obtain, [z^*, x_k_m]∩∂ S_i^*≠∅.
		
		Let z̃∈[z^*,x_k_m]∩∂ S_i^*. I.e. there exists t̃∈[t^*,1], z̃=z^(m)(t̃). We observe that, t̃>t^*. Indeed, if t̃=t^* then z̃=z^*. From the last, we receive that S_i^*∩∂ S_i^*≠∅ which contradicts with the fact that S_i^* is open set. Therefore, t̃>t^*. Using the Intermediate Step (I), we have t̃>t^*>t_v_0^(m).

		Intermediate Step (III): i^*∈ Q_m ∖{ i_(1,m), i_(2,m),…, i_( v_0^(m), m) }.
		
			From the Intermediate Step (II), we already know that i^*∈ Q_m. We now prove  that, i^*≠ i_(λ,m), ∀λ∈{ 1,2,…,v_0^(m)}. First we see that, 
			
    t̃>t^*>t_v_0^(m)≡ t^(m)_i_( v_0^(m), m) , max≥ t^(m)_i_(v_0^(m), m )>t^(m)_i_( v_0^(m)-1, m) , max≥ t^(m)_i_(v_0^(m)-1, m) >…>t^(m)_i_(1, m), max

		Let us suppose that there exists λ̃∈{ 1,2,…,v_0^(m)} such that i^*=i_(λ̃, m). Thus, L^(m)_i^*=L^(m)_i_(λ̃, m). At the same time, we know that  t̃∈ L^(m)_i^*, since z̃=z^(m)(t̃)∈∂ S_i^*. Hence, t̃∈ L^(m)_i_(λ̃, m). By the definition of t^(m)_i_(λ̃, m), max, we obtain that  t̃≤ t^(m)_i_(λ̃, m), max, which contradicts with the fact that  t̃>t^(m)_i_(λ̃, m), max. Consequently, i^*≠ i_(λ, m), ∀λ∈{ 1,2,…, v_0^(m)}.
		
	   Intermediate Step (IV): [ z^(m)_v_0^(m), max, z^*] ∩∂ S_i^*≠∅.
		
			We know that z^(m)_v_0^(m), max∈∂ S_i_(v_0^(m), m ) ⊂ ext (S_i^*) and z^*∈ S_i^*. Thus, from the Proposition <ref>,we obtain that z^(m)([t_v_0^(m), t^* ]  )∩∂ S_i^*≠∅.
		
	Let z^**∈[ z^(m)_v_0^(m), max, z^*] ∩∂ S_i^*, i.e. there exists[It holds that [ z^(m)_v_0^(m), max, z^*]:={ (1-t) z^(m)_v_0^(m), max+ t z^*: t∈[0, 1] }={ (1-t) x_k_m-1+ t x_k_m: t∈[ t_v_0^(m), t^*]  }] t^**∈[t_v_0^(m), t^* ] such that z^**=z^(m)(t^**).
	

	   Intermediate Step (V): t^*>t^**>t_v_0^(m).
		
		If t^**=t_v_0^(m), then z^**= z^(m)_v_0^(m), max  where z^**∈∂ S_i^* and z^(m)_v_0^(m), max∈∂ S_i_(v_0^(m), m ) . Hence, ∂ S_i^*∩∂ S_i_(v_0^(m), m ) ≠∅. Therefore, S̅_i^*∩S̅_i_(v_0^(m), m ) ≠∅. The last leads to a contradiction, since, from the Intermediate Step (III), we have i^*≠ i_(v_0^(m), m ) and as a result from the assumption of the theorem, we receive that S̅_i^*∩S̅_i_(v_0^(m), m ) =∅. Consequently, t^**>t_v_0^(m).

		On the other hand, if t^*=t^**, it follows that z^*=z^**. Therefore, S_i^*∩∂ S_i^*≠∅. The last contradicts with the fact that S_i^* is open set.
		
		Intermediate Step (VI): t^**∈ R_i_( v_0^(m), m).
		 	
		 	Initially, we observe that t^**∈⋃{ L^(m)_q: q∈ Q_m∖{ i_(1,m), i_(2,m),…,i_( v_0^(m), m) }}. Indeed, since z^**=z^(m)(t^**)∈∂ S_i^*, it follows from the definition of L^(m)_i^* that t^**∈ L^(m)_i^*. Moreover, from the Intermediate Step (III) we know that i^*∈ Q_m∖{ i_(1,m), i_(2,m),…,i_(v_0^(m), m ) }. Consequently, we receive that t^**∈⋃{ L^(m)_q: q∈ Q_m∖{ i_(1,m), i_(2,m),…,i_( v_0^(m), m)}}. Furthermore, from the Intermediate Step (V), we know that t^**>t_v_0^(m). It remains to prove the existence of δ_t^**>0 such that z^(m)(t^**+δ_t^**)∈⋃{ S_q: q∈ Q_m∖{ i_(1,m), i_(2,m),…,i_( v_0^(m), m) }}. Indeed, we set δ_t^**:=t^*-t^**>0, where δ_t^**>0, since from Intermediate Step (V) we know that t^*>t^**.
 Moreover, z^(m)(t^**+δ_t^**)=z^(m)(t^*)=z^*∈ S_i^*⊂⋃{ S_q: q∈ Q_m∖{ i_(1,m), i_(2,m),…,i_( v_0^(m), m) }}. Hence, t^**∈ R_i_( v_0^(m), m) .
		 
		From the last Intermediate Step (VI), we are in a contradiction, since R_i_( v_0^(m), m) =∅ (see above the definition of v_0^(m). ) Finally, we receive the desired result, [ z^(m)_v_0^(m), max, x_k_m]  ⊂ U∖( ⋃_i∈ IS_i). 
		
		
  (iii) Let v_0^(m)≥ 3 and a random v^*∈{ 1,2,…, v_0^(m)-1}. We prove that, [ z^(m)_v^*, max, z^(m)_v^*+1] ⊂ U∖( ⋃_i∈ IS_i). Let us suppose the opposite of what we seek to prove. Therefore, there exists i^*∈ I, such that [z^(m)_v,max, z^(m)_v+1]∩ S_i^*≠∅. From the last, we obtain that there exists z^*∈[z^(m)_v,max, z^(m)_v+1]∩ S_i^*. Thus, there exists[It holds that [z^(m)_v,max, z^(m)_v+1]:={ (1-t)z^(m)_v,max+tz^(m)_v+1: t∈[0,1]} ={ (1-t)x_k_m-1+t x_k_m: t∈[ t^(m)_i_(v,m),max, t^(m)_i_(v+1,m)] }] t^*∈[t^(m)_i_(v,m),max,t^(m)_i_(v+1,m)] such that z^*≡ z^(m)(t^*)∈ S_i^*. In order to lead to a contradiction, we make the following Intermediate Steps, where their proofs are omitted since is similar with the proofs we did in (ii).

		Intermediate Step (I): t^*< t^(m)_i_(v+1,m).

		Intermediate Step (II): i^*∈ Q_m and in specific [ z^*, z^(m)_v+1]∩∂ S_i^*≠∅.

		Let z̃∈[ z^*, z^(m)_v+1]∩∂ S_i^*, i. e. there exists[It holds [ z^*, z^(m)_v+1]:={ (1-t) z^* +t z^(m)_v+1: t∈[0,1] }={ (1-t)x_k_m-1+t x_k_m: t∈[t^*,t^(m)_i_(v+1,m)] }] t̃∈[t^*, t^(m)_i_(v+1,m)] such that z̃=z^(m)(t̃) We observe that, t̃>t^*. Indeed, if t̃=t^* then z̃=z^*. From the last, we receive that S_i^*∩∂ S_i^*≠∅ which contradicts with the fact that S_i^* is open set. Therefore, t̃>t^*. Using now the Intermediate Step (I), we receive t̃>t^*>t^(m)_i_(v,m), max. 

		Intermediate Step (III): i^*∈ Q_m∖{ i_(1,m), i_(2,m),…,i_(v,m)}.

		Intermediate Step (IV): [z^(m)_v, max,z^* ]∩∂ S_i^*≠∅.

		Let z^**∈[z^(m)_v, max,z^* ]∩∂ S_i^*, i. e. there exists t^**∈[t^(m)_i_(v,m), max,t^* ] such that z^**=z^(m)(t^**).

		Intermediate Step (V): t^*>t^**>t^(m)_i_(v,m), max

		Therefore, from the Intermediate Steps (I) and (V), we receive that t^(m)_i_(v,m), max<t^**<t^(m)_i_(v+1,m).

		Intermediate Step (VI): t^**∈ R_i_(v,m)

		At this stage we have a contradiction,  since by the definition of t^(m)_i_(v+1,m)=min R_i_(v,m) but t^**∈ R_i_(v,m) and t^**<t^(m)_i_(v+1,m).	Finally, we receive that [ z^(m)_v^*, max, z^(m)_v^*+1] ⊂ U∖(⋃_i∈ IS_i).
		
		
		
		
		
		
		
		
  (iv) Let x_k_m-1∈ S_q for some q∈ I. We prove first that q∈ Q_m and in the end, that q=i_(1,m).

		Since, x_k_m-1∈ S_q, it follows that, [x_k_m-1, x_k_m]∩ S_q≠∅. Let us suppose that q∉ Q_m. Thus, [x_k_m-1, x_k_m]∩∂ S_q=∅. From previous Claim, we know that Q_m≠∅. Let q̃∈ Q_m. Definitely, q≠q̃. Hence, S̅_q∩S̅_q̃=∅. Moreover, [x_k_m-1, x_k_m]∩∂ S_q̃≠∅. Let, z̃∈ [x_k_m-1, x_k_m]∩∂ S_q̃, i. e. there exists, t̃∈[0,1] such that, z̃=z^(m)(t̃)∈∂ S_q̃⊂ ext(S_q). Using Proposition <ref>, we receive that, z^(m)([0,t̃] )∩∂ S_q≠∅. Thus, ∅≠[x_k_m-1, z̃]∩∂ S_q⊂[x_k_m-1, x_k_m]∩∂ S_q. The last contradicts with the previous statement, [x_k_m-1, x_k_m]∩∂ S_q=∅. From the above, analysis, we conclude that, q∈ Q_m.

		Now, we proceed to the proof that, q=i_(1,m). Let us suppose the opposite of what we seek to prove, i. e q≠ i_(1,m). Therefore, by the assumption of the theorem, S̅_q∩S̅_i_(1,m)=∅. Since, x_k_m-1∈ S_q and z^(m)_1∈∂ S_i_(1,m), (i.e. z^(m)_1∈ ext(S_q)) it follows from the Proposition <ref>, that z^(m)( [0,t^(m)_i_(1,m)]) ∩∂ S_q≠∅. Let z^*=z^(m)(t^*)∈∂ S_q for t^*∈[ 0, t^(m)_i_(1,m)]. It holds that, t^*<t^(m)_i_(1,m). Indeed, if we suppose that, t^*=t^(m)_i_(1,m), then it follows that ∂ S_q∋ z^*=z^(m)∈∂ S_i_(1,m). Thus, S̅_q∩S̅_i_(1,m)≠∅ which is a contradiction. Consequently, t^*<t^(m)_i_(1,m). Furthermore, since t^*∈ L^(m)_q, by the definition of, t^(m)_q, min, we obtain that t^(m)_q, min≤ t^*<t^(m)_i_(1,m). On the other hand, by the defintion of t^(m)_i_(1,m)=min{ t^(m)_i_(s, min): s∈ Q_m }, we receive that t^(m)_i_(1,m)≤ t^(m)_q, min where we are in contradiction with the previous finding. 

		With a similar, way we prove the case, x_k_m∈ S_q.
		
  (v)	Let v_0^(m)=1 and let x_k_m-1, x_k_m∈⋃_i∈ IS_i. Therefore, there exists, (unique) q_1,q_2∈ I such that x_k_m-1∈ S_q_1 and x_k_m∈ S_q_2. From the Claim <ref> (iv), we receive that q_1,q_2∈ Q_m and also q_1=i_(1,m). Let us suppose that q_1≠ q_2. Since, z^(m)_1, max∈∂ S_i_(1,m)⊂ ext(S_q_2) and x_k_m∈ S_q_2, from the Proposition <ref> we receive that
		
    z^(m)([t^(m)_i_(1,m), max, 1 ]  ) ∩∂ S_q_2≠∅.

	  Let z^*=z^(m)(t^*)∈∂ S_q_2, where t^*∈[t^(m)_i_(1,m), max, 1 ]. We observe that t^(m)_i_(1,m)<t^*<1. Indeed, t^*<1, since if we suppose that t^*=1 we obtain that z^*=z^(m)(t^*)= z^(m)(1)=x_k_m where z^*∈∂ S_q_2 and z^(m) (1)=x_k_m∈ S_q_2. Therefore, S_q_2∩∂ S_q_2≠∅, which contradicts with the fact S_q_2 is an open set. Next, we claim that t^(m)_i_(1,m)<t^*. This is true from the fact that if we suppose that t^(m)_i_(1,m)=t^*, we obtain ∂ S_q_1∋ z^(m)_1, max=z^*∈∂ S_q_2, thus S̅_q_1∩S̅_q_2≠∅. On the other hand, from the condition q_1≠ q_2 it follows from assumption of theorem,that S̅_q_1∩S̅_q_2=∅. Consequently, t^(m)_i_(1,m)<t^*.
	  Furthermore, t^*∈ R_i_(1,m). Indeed, first we notice that q_2∈ Q_m∖{q_1=i_(1,m)}. Also, by the definition of t^*, it is obvious that t^*∈ L^(m)_q_2. In addition, if we set as δ_t^*:=1-t^*>0 (since t^*<1), then we obtain, that z^(m)(t^*+δ_t^*)=z^(m)(1)=x_k_m∈ S_q_2. Therefore, by the definition of R_i_(1,m), we receive that t^*∈ R_i_(1,m). From the last, we also conclude that, R_i_(1,m)≠∅. Therefore, v_0^(m)≥ 2, which contradicts with the assumption of Claim <ref> (v), that v_0^(m)=1.
	  
	  
  (vi) Let x_k_m-1,x_k_m∈ S_q, for some q∈ Q_m. From the Claim <ref> (iv), it holds that q=i_(1,m).

		Intermediate Step (I): ∀ t>t^(m)_i_(1,m), max, z^(m)(t)∈ S_i_(1,m).
		
			Let us suppose the opposite of what we seek to prove, i. e. there exists t̃>t^(m)_i_(1,m) such that z̃≡ z^(m)(t̃)≠ S_i_(1,m). Then, from <ref>, it holds that z^(m)([t̃, 1] )∩∂ S_i_(1,m)≠∅. Let z^*=z^(m)(t^*)∈∂ S_i_(1,m) where t^*∈[t̃, 1]. Thus, t^*≥t̃>t^(m)_i_(1,m), max. At the same time, by the definition of L^(m)_i_(1,m), it holds that t^*∈ L^(m)_i_(1,m). Therefore, t^*≤ t^(m)_i_(1,m), max=sup L^(m)_i_(1,m). The last contradicts with the previous finding, t^*>t^(m)_i_(1,m). Consequently, the proof is complete.
		
		Next, we claim that R_i_(1,m)=∅. Indeed, if R_i_(1,m)≠∅ then there exists t'∈ L^(m)_q^* where q^*∈ Q_m∖{i_(1,m)} such that t'>t^(m)_i_(1,m), max, i. e. z^(m)(t')∈∂ S_q^*⊂ S_i_(1,m). The last contradicts with the Intermediate Step (I). Consequently, R_i_(1,m)=∅, hence v_0^(m)=1.
		
		
		
		
		
		
		
		
		
  (vii) Let v_0^(m)≥ 2 and x_k_m-1, x_k_m∈⋃_i∈ I S_i. Thus, there exist q_1,q_2∈ I, such that x_k_m-1∈ S_q_1 and x_k_m∈ S_q_2. From the Claim <ref> (iv) it holds that q_1,q_2∈ Q_m. Let us suppose that q_1=q_2=q, i. e. x_k_m-1,x_k_m∈ S_q. Using now the Claim <ref>(vi), we obtain that v_0^(m)=1. The last contradicts with the assumption of the Claim <ref> (vii), v_0^(m)≥ 2. Hence, q_1≠ q_2.
		
  (viii) Let x_k_m-1∈ S_q_1 and x_k_m∈ S_q_2, where q_1,q_2∈ Q_m with q_1≠ q_2. Let us suppose the opposite of what we seek to prove, i. e v_0^(m)=1. Then, from the Claim <ref> (v) there exists q∈ I such that x_k_m-1,x_k_m∈ S_q. Therefore, S_q_1∩ S_q≠∅ and S_q_2∩ S_q≠∅. If q_1≠ q then by the assumption of the theorem, we receive that S_q_1∩ S_q=∅, which contradicts with the previous finding. Consequently, q_1=q. Similarly, we obtain that q_2=q. Thus, q_1=q_2, which leads to a contradiction with the assumpion of Claim <ref> (viii). Hence, v_0^(m)≥ 2.
		
		
		
		
	



	We distinguish the following subcases:

	Case: x_k_m-1, x_k_m∉⋃_i∈ IS_i
	

	
    Γ̃_x_k_m-1, x_k_m:=
    				[x_k_m-1,z^(m)_1]∪Γ̂_z^(m)_1, z^(m)_1, max∪[z^(m)_1, max, x_k_m],     v_0^(m)=1
    
    				[x_k_m-1, z^(m)_1]∪(⋃_v=1^v_0^(m)-1(  Γ̂_z^(m)_v, z^(m)_v, max∪[z^(m)_v, max, z^(m)_v+1]) ) ∪
    ∪(Γ̂_z^(m)_v_0^(m), z^(m)_v_0^(m), max∪[z^(m)_v_0^(m), max, x_k_m] ),     v_0^(m)≥ 2

	
	

	
	
		Case: x_k_m-1∈⋃_i∈ IS_i and  x_k_m∉⋃_i∈ IS_i
	

	
    Γ̃_x_k_m-1, x_k_m:=Γ̂_z^(m)_1, z^(m)_1, max∪[z^(m)_1, max,x_k_m],     v_0^(m)=1
    (⋃_v=1^v_0^(m)-1(  Γ̂_z^(m)_v, z^(m)_v, max∪[z^(m)_v, max, z^(m)_v+1]) )∪(Γ̂_z^(m)_v_0^(m), z^(m)_v_0^(m), max∪[z^(m)_v_0^(m), max, x_k_m] ),     v_0^(m)≥ 2

		


	
	
			Case: x_k_m-1∉⋃_i∈ IS_i and  x_k_m∈⋃_i∈ IS_i
	

	
    Γ̃_x_k_m-1, x_k_m:=
    				[x_k_m-1, z^(m)_1]∪Γ̂_z^(m)_1, z^(m)_1, max,     v_0^(m)=1
    
    				[x_k_m-1, z^(m)_1]∪(⋃_v=1^v_0^(m)-1(  Γ̂_z^(m)_v, z^(m)_v, max∪[z^(m)_v, max, z^(m)_v+1]) )∪Γ̂_z^(m)_v_0^m, z^(m)_v_0, max,     v_0^(m)≥ 2

		
		
	

	

	Case: x_k_m-1∈⋃_i∈ IS_i and  x_k_m∈⋃_i∈ IS_i
	
    Γ̃_x_k_m-1, x_k_m:=Γ̂_z^(m)_1, z^(m)_1, max,    v_0^(m)=1
    ⋃_v=1^v_0^(m)-1(  Γ̂_z^(m)_v, z^(m)_v, max∪[z^(m)_v, max, z^(m)_v+1]) ∪Γ̂_z^(m)_v_0^(m), z^(m)_v_0^(m), max,     v_0^(m)≥ 2

	


  * Construction of Polygonal line: Γ^(1)_x_k_m,   ( x_k_m∉⋃_i∈ IS_i )

	Let m∈Λ':={ m∈{1,2,…,j_0}: x_k_m∉⋃_i∈ I S_i}∖Λ̃ 
	where 
	
    Λ̃:={ m∈{1,2,…,j_0}: x_k_m∈⋃_i∈ I∂ S_i and {(1-t)x_k_m-1+ t x_k_m: t∈[0,1)}⊂⋃_i∈ IS_i}

	In this case we proceed to the construction of the polygonal line  Γ^(1)_x_k_m.
	
	 Let m∈{1,2,…,j_0} such that x_k_m∉⋃_i∈ IS_i. Furthermore, if [x_k_m, x_k_m+1]∩( ⋃_i∈ I S_i )≠∅, then k_m+1∈ J therefore, k_m+1=k_m+1.
	
   
	Let x_k_m∉⋃_i∈ IS_i and [x_k_m, x_k_m+1]∩( ⋃_i∈ I S_i )≠∅. Therefore, there exists (unique) i^*∈ I (since S̅_i∩S̅_j=∅, ∀ i,j∈ I, i≠ j) such that [x_k_m, x_k_m+1]∩ S_i^*≠∅. Consequently, there exists z̃∈ Int(S_i^*)≡ S_i^* such that z̃∈[x_k_m, x_k_m+1]. Since x_k_m∉ S_i^*, it holds that x_k_m∈∂ S_i^*∪ ext(S_i^*). We distinguish the following cases for x_k_m:
	
		
  * x_k_m∈∂ S_i^*

		Then [x_k_m,x_k_m+1]∩∂ S_i^*≠∅
		
  * x_k_m∈ ext(S_i^*)

		We consider the usual curve γ_[x_k_m, x_k_m+1]:[0, 1]→ X that describes the line seqment with extreme points x_k_m, x_k_m+1, (i. e x_k_m=γ_[x_k_m, x_k_m+1](0), x_k_m+1=γ_[x_k_m, x_k_m+1](1) and [x_k_m, x_k_m+1]=γ_[x_k_m, x_k_m+1]( [0, 1])). Now, we consider the usual curve γ_[x_k_m, z̃]:[0, t_z̃]→ X that describes the line seqment with extreme points x_k_m and z̃, (i. e  x_k_m=γ_[x_k_m, z̃](0) and z̃=γ_[x_k_m, z̃](t_z̃) and [x_k_m, z̃]=γ_[x_k_m, z̃]( [0,t_z̃]). ) Since, x_k_m∈ ext(S_i^*), z̃∈ Int(S_i^*) from the Proposition <ref> we receive that: 
		
    γ_[x_k_m, z̃]([0,t_z̃] ) ∩∂ S_i^*≠∅

	From the fact that γ_[x_k_m, z̃]([0, t_z̃] ) ⊂γ_[x_k_m, x_k_m+1]([0,1] ) we obtain that γ_[x_k_m, x_k_m+1]([0,1] )∩∂ S_i^*≠∅. Hence, [x_k_m, x_k_m+1]∩∂ S_i^*≠∅.
	
From both cases, we obtain that k_m+1∈ J, since [x_k_m, x_k_m+1]∩ S_i^*≠∅ and [x_k_m, x_k_m+1]∩∂ S_i^*≠∅.
   
Next, we define:
	
    Γ^(1)_x_k_m, (-):=Γ̃_x_k_m-1, x_k_m

	and
	
    Γ^(1)_x_k_m, (+):=⋃_λ=1^n-k_m[x_k_m+λ-1, x_k_m+λ],    
    			 		[x_k_m, x_k_m+1]⊂(⋃_i∈ IS_i )^c,
    
    			 		 m=j_0, n>k_m
    ( ⋃_λ=1^k_m+1-(k_m+1)[x_k_m+λ-1, x_k_m+λ])∪Γ̃_x_k_m+1-1, x_k_m+1,    
    				[x_k_m, x_k_m+1]⊂(⋃_i∈ IS_i )^c,
    
    				 m<j_0, n>k_m
    Γ̃_x_k_m+1-1, x_k_m+1,     [x_k_m, x_k_m+1] ∩( ⋃_i∈ IS_i) ≠∅

Since [x_k_m, x_k_m+1]⊂(⋃_i∈ IS_i )^c, it holds that k_m+1>k_m+1.
	Finally, we define the polygonal line Γ^(1)_x_k_m, as follows:
	
    Γ^(1)_x_k_m:=Γ^(1)_x_k_m, (-)∪Γ^(1)_x_k_m, (+)

	
	
  * Construction of Polygonal line: Γ^(2)_x_k_m,  ( x_k_m∈⋃_i∈ IS_i ) 

	Let m∈{1,2,…,j_0} such that x_k_m∈⋃_i∈ IS_i. Hence, there exists (unique) i^*∈ I such that x_k_m∈ S_i^*. We notice that k_m<n, since if it holds that k_m=n, we obtain S_i^*∋ x_k_m=x_n≡ y, where the last contradicts with the assumption of theorem, y∉⋃_i∈ IS_i. In this case we proceed to the construction of the polygonal line  Γ^(2)_x_k_m.

	
		Let m∈{1,2,…,j_0} such that x_k_m∈ S_i^*, for some i^*∈ I. Then, there exists λ^*∈ such that [x_k_m+λ^*-1, x_k_m+λ^*]∩∂ S_i^*≠∅.
	
	
			  We notice that x_k_m∈ S_i^* and y∈∂ S_i^*∪ ext(S_i^*). Next, we distinguish the following cases for y.
			
				
  - y=x_n∈∂ S_i^*

				Therefore, [x_n-1,x_n] ∩∂ S_i^* where x_n-1=x_k_m+λ^*-1 for some (unique) λ^*∈ with λ^*≥ 3.
				
  - y∈ ext(S_i^*)

				We consider the polygonal line γ_x_k_m,y:Ĩ→ X with γ_x_k_m,y:=⋃_r=1^n-k_m[x_k_m+r-1, x_k_m+r] that connects the points x_k_m and y. Since γ_x_k_m,y is a continuous curve and x_k_m∈ S_i^*, y∈ ext(S_i^*), from Proposition <ref>, it holds that γ_x_k_m,y(Ĩ)∩∂ S_i^*≠∅. Consequently, there exists t_0∈Ĩ, such that γ_x_k_m,y(t_0)∈[x_k_m+λ-1, x_k_m+λ]∩∂ S_i^* for some λ∈{1,…,n-k_m}.
			
			From both cases is concluded that there exists λ∈ such that [x_k_m+λ-1, x_k_m+λ]∩∂ S_i^*≠∅.
		

Next we define: 

    λ_0^(m):=max{λ∈{ 1,2,…,n-k_m}: [ x_k_m+λ-1, x_k_m+λ]∩∂ S_i^*≠∅}
    
    	t_λ_0^(m)^*:=min{ t∈[0,1]: (1-t) x_k_m+λ_0^(m)-1+t x_k_m+λ_0^(m)∈∂ S_i^*}
    
    	z_λ_0^(m)^*:=( 1-t_λ_0^(m)^*)  x_k_m+λ_0^(m)-1+t_λ_0^(m)^* x_k_m+λ_0^(m)


In case where λ_0^(m)<n-k_m, we define
	
    H:={λ∈{ 1,2,…,n-( k_m+λ_0^(m)) }: k_m+λ_0^(m)+λ∈ J}

	
Finally, we define the Polygonal line Γ^(2)_x_k_m, as follows:

    Γ^(2)_x_k_m:=Γ̂_z^(m)_v_0^(m), max, z^*_λ_0^(m),     k_m+λ_0^(m)∈ J
    Γ̂_z^(m)_v_0^(m), max, z^*_λ_0^(m)⋃[z^*_λ_0^(m), x_k_m+λ_0^(m)],    
    						k_m+λ_0^(m)∉ J, H=∅,
     n=k_m+λ_0^(m)
    Γ̂_z^(m)_v_0^(m), max, z^*_λ_0^(m)⋃[z^*_λ_0^(m), x_k_m+λ_0^(m)]  ⋃_λ=1^n-(k_m+λ_0^(m))[x_k_m+λ_0^(m)+(λ-1), x_k_m+λ_0^(m)+λ] ,    
    						k_m+λ_0^(m)∉ J, H=∅,
     n>k_m+λ_0^(m)
    Γ̂_z^(m)_v_0^(m), max, z^*_λ_0^(m)⋃[z^*_λ_0^(m), x_k_m+λ_0^(m)]  ,    
    					k_m+λ_0^(m)∉ J,
     
    					H≠∅, min H=1
    Γ̂_z^(m)_v_0^(m), max, z^*_λ_0^(m)⋃[z^*_λ_0^(m), x_k_m+λ_0^(m)]  ⋃_λ=1^min H-1[x_k_m+λ_0^(m)+(λ-1), x_k_m+λ_0^(m)+λ],    
    				k_m+λ_0^(m)∉ J,
    
    				 H≠∅, min H>1,






—————————————————————————–

Finally, we define the polygonal line Γ_x,y that connects the points x=x_1 and y=x_n.

    Γ_x,y:=Γ^*,     k_1=2
    ( ⋃_l=2^k_1-1[x_l-1,x_l]) ∪Γ^*,    k_1≥ 3

where Γ^*:=⋃_l∈ΛΓ_x_k_l with 

    Γ_x_k_l:=Γ^(1)_x_k_l,     x_k_l∉⋃_i∈ IS_i
    Γ^(2)_x_k_l,     x_k_l∈⋃_i∈ IS_i

and Λ:={1,2,…,j_0}∖( Λ̃∪Λ^*),
where 

    Λ^*:={ m∈{1,2,…, j_0}: x_k_m-1∈⋃_i∈ I∂ S_i and {(1-t)x_k_m-1+t x_k_m: t∈(0,1]}⊂⋃_i∈ I S_i }.

	

	
	
	
	
	
	
	
	
	
	99999
	
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		Savvas Andronicou
 University of Cyprus 
 Department of Mathematics & Statistics 
 P.O. Box 20537

		Nicosia, CY- 1678 CYPRUS
		
 andronikou.savvas@ucy.ac.cy
	
	
	
	
	
	
	
	
		Emmanouil Milakis
 University of Cyprus 
 Department of Mathematics & Statistics 
 P.O. Box 20537

		Nicosia, CY- 1678 CYPRUS
		
 emilakis@ucy.ac.cy
	
	
 