The arc is the only chainable continuum admitting a mean

The arc is the ony chainabe continuum admitting a mean Aejandro Ianes and Hugo Vianueva September 4, 26 Abstract Let X be a metric continuum. A mean on X is a continuous function : X X! X such that for each x; y 2 X, (x; y) = (y; x) and (x; x) = x. In this paper we prove that if X is chainabe and admits a mean, then X is an arc. This answers a question stated by Phiip Bacon in 197. Introduction Troughout this paper the etter X wi denote a metric continuum. A mean on X is a continuous function : X X! X such that for each x; y 2 X, (x; y) = (y; x) and (x; x) = x. A chain in X is a sequence U = fu 1 ; : : : ; U n g of open subsets of X such that X = U 1 [ : : : [ U n and U i \ U j 6= ; if and ony if ji jj 1. Given a positive number ", the chain U = fu 1 ; : : : ; U n g is said to be an "-chain provided that diameter(u i ) < " for each i 2 f1; : : : ; ng. We say that X is chainabe provided that, for each " > there exists an "-chain in X. The probem of determining which continua X admit means has been studied by a number of authors. In [2] P. Bacon proved that if a continuum X admits a mean then X is unicoherent, he aso showed ([1]), answering a question by A. D. Waace ([15]), that the sin( 1 x )-continuum admits no means. This was the rst exampe of an acycic continuum admitting no means. More information about means can be found in [6], [9, Section 76] and [1]. In [1] P. Bacon posed the foowing questions: (1) Is the arc the ony chainabe continuum that admits a mean? (2) Is the arc the ony continuum containing an open dense haf-ine that admits a mean? Question (2) has been recenty answered, in the positive, by the rst named author ([7]). With respect to question (1), answering a question by J. J. Charatoni, the rst named author, showed that the simpest indecomposabe continuum (aso caed the bucethande continuum or the Brouwer-Janiszewsi- Knaster continuum) does not admit means ([8]). Recenty, D. P. Beamy ([4]) 1

has shown that each Knaster-type continuum (i.e., the inverse imit of arcs with open bonding mappings) di erent from the arc admits no mean. Some partia answers to question (1) can be obtained by using the resuts contained in the papers [3], [5] and [11]. In this paper we give the na answer to question (1) by showing the foowing. Theorem 1. If X is chainabe and X admits a mean, then X is an arc. This paper is devoted to prove Theorem 1. A property in the pane We denote by R 2 the Eucidean pane. Given points p; q 2 R 2, where p 6= q, et pq be the convex segment in R 2 joining p and q. Theorem 2. Let p = (; ), q = (1; ) and r = ( 1 2 ; 1) in R2. Let be the convex triange in R 2 with vertices p; q and r: Suppose that H and K are cosed disjoint subsets of such that pr [ rq H and K \ pq 6= ;. Then there exists an arc in, with end points p and q such that \ K = ; and \ H pq. Proof. Let V = H. Then V is an open subset of and K V. Let K = K \ pq. Since the components of V are open in and they cover the nonempty compact set K, there exist n 2 N and components V 1 ; : : : ; V n of V such that K V 1 [ : : : [ V n and K intersects each V i. For each i 2 f1; : : : ; ng, et K i = K \ V i. Notice that K i is compact. Let m i = minfx 2 [; 1] : (x; ) 2 K i g and M i = maxfx 2 [; 1] : (x; ) 2 K i g. Since ; 6= K \ V i K \ pq \ V i, m i and M i are we de ned. Since V i is arcwise connected, there exists a continuous function i : [; 1]! V i such that i () = (m i ; ) and i (1) = (M i ; ). Since r =2 V i, we can appy Theorem 2 of [12, 57, III, p. 438], to the cosed sets V i and K i [Im i and the points r 2 V i and (m i ; ) 2 K i [Im i, then there exists a ocay connected subcontinuum C i of such that C i V i (K i [ Im i ) and C i separates r and (m i ; ) in. Thus C i intersects the connected set rp [ p(m i ; ). Since rp \ V i = ;, there exists a point p i = (u i ; ) 2 p(m i ; ) \ C i. Simiary, there exists a point q i = (v i ; ) 2 (m i ; )q \ C i. Since p, (m i ; ) and q do not beong to C i, we have that < u i < m i < v i < 1. Let i : [; 1]! C i be a continuous one-to-one function such that i () = (u i ; ) and i (1) = (v i ; ), we may assume that Im i \p(m i ; ) = f(u i ; )g and Im i \(m i ; )q = f(v i ; )g. Thus Im i intersects the boundary of ony at the points (u i ; ) and (v i ; ) and we can appy the emma of the -curve ([12, 61, II, Theorem 2, p. 511]) and concude that Im i has exacty two components D i and E i, where D i and E i are the respective component of Im i which contain the connected 2

sets F i = p i q i fp i ; q i g and G i = p i p [ pr [ rq [ qq i fp i ; q i g. Since F i [ Im i is a connected subset of Im i, F i [ Im i D i, So (M i ; ) 2 D i \ pq p i q i. This impies that M i < v i. Notice that Im i V i K i = V i (K \V i ) V i K, so Im i \ (K [ H) = ;. Let i; j 2 f1; : : : ; ng such that i 6= j. If u j 2 [u i ; v i ], then F i [ Im j is a connected subset of Im i, so q j 2 (F i [ Im j ) \ pq D i \ pq p i q i. Thus v j 2 [u i ; v i ]. Simiary, it can be shown that if v j 2 [u i ; v i ], then u j 2 [u i ; v i ]. We have shown that [u i ; v i ] and [u j ; v j ] are disjoint or one is contained in the other. Therefore, taing the maxima intervas of the form [u i ; v i ], there exist m 2 N and i 1 ; : : : ; i m 2 f1; : : : ; ng such that < u i1 < v i1 < u i2 < v i2 < : : : < u im < v im < 1 and [u i1 ; v i1 ][[u i2 ; v i2 ][: : :[[u im ; v im ] = [u 1 ; v 1 ][[u 2 ; v 2 ][: : :[[u m ; v m ]. Given a point w = (u; ) 2 K \pq = K, there exists i 2 f1; : : : ; ng such that w 2 K i. Thus u 2 [m i ; M i ] (u i ; v i ) (u i1 ; v i1 ) [ (u i2 ; v i2 ) [ : : : [ (u im ; v im ). This proves that K \ pq ((u i1 ; v i1 ) [ (u i2 ; v i2 ) [ : : : [ (u im ; v im )) fg. Let : [; 1]! be the continuous, one-to-one function such that () = p, 1 (1) = q and ([; 2m+1 ]) = pp 1 i 1, ([ 2m+1 ; 2 2m+1 ]) = Im 2 i 1, ([ 2m+1 ; 3 2m+1 ]) = 3 q i1 p i2, ([ 2m+1 ; 4 2m+1 ]) = Im 4 i 2, ([ 2m+1 ; 5 2m+1 ]) = q i 2 p i3,...,([ 2m 1 2m+1 ; 2m 2m+1 ]) = 2m Im im, ([ 2m+1 ; 1]) = q i m q. Finay, et = Im. It is easy to chec that has the required properties. PL mappings A continuous function f : [; 1]! [; 1] is caed a PL mapping (piecewise inear mapping), provided that there exists a partition P : = t < t 1 < : : : < t n = 1 of [; 1] such that, for each i 2 f1; : : : ; ng and each t 2 [t i 1 ; t i ], f(t) = t ti 1 t i t i 1 f(t i ) + ti t t i t i 1 f(t i 1 ) in this case we say that f is supported by P. It is easy to see that the cass of PL mappings is cosed under compositions. A PL mapping f is said to be a jump mapping provided that f() = and f(1) = 1. The foowing theorem can be proved with the techniques of the paper [14]. We incude its proof here for competeness. Theorem 3. If f and g are jump mappings, then there exist jump mappings and such that f = g. Proof. Let A = f(x; y) 2 [; 1] 2 : f(x) = g(y)g. The set A is a compact subset of [; 1] 2 such that (; ); (1; 1) 2 A. Let L be the component of A such that (; ) 2 L. 3

We are going to prove that (1; 1) 2 L. Suppose to the contrary that (1; 1) =2 L. Then (see [13, Theorem 5.2, p. 72]) there exist compact disjoint subsets H and K of [; 1] 2 such that A = H [K, (; ) 2 H and (1; 1) 2 K. By [12, 57, III, Theorem 2, p. 438], there exists a separator C between (; ) and (1; 1) which is a ocay connected continuum disjoint from A. Thus there exist points (x ; y ) 2 C \((fg[; 1])[([; 1]f1g)) and (u ; v ) 2 C \(([; 1]fg)[(f1g[; 1])). Notice that f(x ) g(y ) and f(u ) g(v ). Since C is connected, there exists a point (t; s) 2 C such that f(t) = g(s). This impies that (t; s) 2 C \ A, a contradiction. Hence (1; 1) 2 L. Since f and g are PL mappings, there exists a partition = t < t 1 < : : : < t n = 1 such that, for each i 2 f1; : : : ; ng and each t 2 [t i 1 ; t i ], f(t) = t t i 1 t i t i 1 f(t i ) + ti t t i t i 1 f(t i 1 ) and g(t) = t ti 1 t i t i 1 g(t i ) + ti t t i t i 1 g(t i 1 ). It is easy to prove that, if i; j 2 f1; : : : ; ng, x; u 2 [t i 1 ; t i ], y; v 2 [t j 1 ; t j ], f(x) = g(y) and f(u) = g(v), then the segment (x; y)(u; v) is contained in A. Let p = (x; y) 2 A, now we prove that there exists " p > such that if (u; v) beongs to the set D(" p ; p) = A \ ((x " p ; x + " p ) (y " p ; y + " p )), then the segment (x; y)(u; v) is contained in A. In order to prove this caim, by the paragraph above, it is enough to tae " p > such that if u 2 [x " p ; x+" p ]\[; 1] and v 2 [y " p ; y + " p ] \ [; 1], then both points x and u beong to a set of the form [t i 1 ; t i ] and both points v and y beong to a set of the form [t j 1 ; t j ]. By the connectedness of the segments of the form (x; y) and (u; v) the caim proved in the paragraph above impies that, if p = (x; y) 2 L, and " p > is as before, then for each (u; v) 2 D(" p ; p), the segment (x; y)(u; v) is contained in L. Since the famiy fd(" p ; p) : p 2 Lg is an open cover of L and L is connected, it foows that L is connected by poygonas. In particuar, there exist m 2 N and points p = (u ; v ), p 1 = (u 1 ; v 1 ); : : : ; p m = (u m ; v m ) in L such that p = (; ), p m = (1; 1) and p p 1 [ p 1 p 2 [ : : : [ p m 1 p m L. De ne ; : [; 1]! [; 1] by (t) = (mt i + 1)u i + (i mt)u i 1 and (t) = (mt i + 1)v i + (i mt)v i 1, if t 2 [ i 1 m ; i m ]. Ceary, and have the required properties. Theorem 4. If f is a PL mapping such that f(1) = 1 and g is a jump mapping, then there exist a jump mapping and a PL mapping such that (1) = 1 and f = g. Proof. In the case that f() =, both mappings f and g are jump mappings, so the existence of and foows from Theorem 3. Thus, suppose that f() >. Let h; : [; 1]! [; 1] be the mappings given by 4

2tf(); if t 2 [; 1 h(t) = 2 ]; f(2t 1); if t 2 [ 1 2 ; 1]; and (t) = ; if t 2 [; 1 2 ]; g(2t 1); if t 2 [ 1 2 ; 1]: Ceary, h and are jump mappings. By Theorem 3, there exist jump mappings and such that h =. Let s = max 1 ( 1 2 ). Since (1) = 1, < s < 1. De ne ; : [; 1]! [; 1] by (t) = 2(t + (1 t)s ) 1 and (t) = maxf2(t + (1 t)s ) 1; g. For each t 2 [; 1], since s t + (1 t)s 1 and (1) = 1, by the de nition of s, (t + (1 t)s ) 2 [ 1 2 ; 1]. Thus is a we de ned jump mapping, is a PL mapping and (1) = 1. In order to chec that f = g, et t 2 [; 1]. If (t + (1 t)s ) 1 2, then 2(t + (1 t)s ) 1, so g((t)) = g(2(t + (1 t)s ) 1) = ((t + (1 t)s )) = h((t + (1 t)s )) = f(2((t + (1 t)s ) 1) = f((t)). Thus g((t)) = f((t)). And in the case that (t+(1 t)s ) 1 2, g((t)) = g() =. Notice that h((t+(1 t)s )) = ((t+(1 t)s )) =. On the other hand, since (t+(1 t)s ) 2 [ 1 2 ; 1], = h((t+(1 t)s ) = f(2(t+(1 t)s ) 1) = f((t)). Hence g((t)) = = f((t)). In both cases g((t)) = f((t)). Therefore, f = g. Theorem 5. If f is a PL mapping such that f() = and g is a jump mapping, then there exist a jump mapping and a PL mapping such that () = and f = g. Proof. Let f 1 ; g 1 : [; 1]! [; 1] be given by f 1 (t) = 1 f(1 t) and g 1 (t) = 1 g(1 t). Then f 1 is a PL mapping such that f 1 (1) = 1 and g 1 is a jump mapping. By Theorem 4, there exist a jump mapping 1 and a PL mapping 1 such that 1 (1) = 1 and f 1 1 = g 1 1. De ne ; : [; 1]! [; 1] by (t) = 1 1 (1 t) and (t) = 1 1 (1 t). Then is a jump mapping and is a PL mapping such that () =. Moreover, for each t 2 [; 1], f((t)) = f(1 1 (1 t)) = 1 f 1 ( 1 (1 t)) = 1 g 1 ( 1 (1 t)) = g(1 1 (1 t)) = g((t)). Therefore, f = g. Theorem 6. Let P : = t < t 1 < : : : < t n = 1 and Q : = s < s 1 < : : : < s m = 1 be partitions of [; 1]. Let f; g : [; 1]! [; 1] be PL mappings such that f is supported by P and g is supported by Q. Suppose that there exists a function : f; 1; : : : ; mg! f; 1; : : : ; ng such that () =, (m) = n, f(t (i) ) = g(s i ) for each i 2 f; 1; : : : ; mg and j(i) (i 1)j 1 for each i 2 f1; : : : ; mg. Then there exists a jump mapping such that f = g. 5

Proof. Let : [; 1]! [; 1] be the PL mapping de ned by the conditions (s i ) = t (i) for each i 2 f; 1; : : : ; mg. Note that () = t = and (1) = t (m) = 1. Thus is a jump mapping. In order to chec that f = g, et i 2 f1; : : : ; mg and et s 2 [s i 1 ; s i ]. Then (s) = s s i 1 s i s i 1 (s i ) + si s s i s i 1 (s i 1 ) = s si 1 s i s i 1 t (i) + si s s i s i 1 t (i 1). Thus (s) is a convex combination of the numbers t (i) and t (i 1), so one of the foowing two expresions is a convex combination for (s) (depending on which inequaity: t (i 1) t (i) or t (i) t (i 1) hods) (s) = (s) t (i 1) t (i) t (i 1) t (i) + t (i) (s) t (i) t (i 1) t (i 1), (s) = (s) t (i) t (i 1) t (i) t (i 1) + t (i 1) (s) t (i 1) t (i) t (i). By hypothesis, j(i) (i 1)j 1. If (i) > (i 1), since f is supported by P, f((s)) = (s) t (i 1) t (i) t (i 1) f(t (i) ) + t (i) (s) t (i) t (i 1) f(t (i) 1 ) = (s) t (i 1) t (i) t (i 1) f(t (i) ) + t (i) (s) t (i) t (i 1) f(t (i 1) ) = (s) t (i 1) t (i) t (i 1) g(s i ) + t (i) (s) t (i) t (i 1) g(s i 1 ). The equaity s s i 1 s i s i 1 t (i) + si s s i s i 1 t (i 1) = (s) t (i 1) t (i) t (i 1) t (i) + t (i) (s) t (i) t (i 1) t (i 1) impies that s s i 1 s i s i 1 = (s) t (i 1) s i s t (i) t (i 1) and si s s i s i 1 = t (i) (s) t (i) t (i 1). Thus f((s)) = s si 1 s i s i 1 g(s i )+ s i s i 1 g(s i 1 ) = g(s). Hence f((s)) = g(s). The case (i) < (i 1) is simiar. Finay, if (i) = (i 1), then (s) = t (i) = t (i 1). Hence, f((s)) = f(t (i) ) = f(t (i 1) ) = g(s i ) = g(s i 1 ). Since g is supported by Q, g(s) = g(s i ). Therefore f((s)) = g(s). The proof of the theorem is compete. Chains For a chainabe continuum X, with metric d, and a positive number ", we say that a chain U = fu 1 ; : : : ; U n g is a separated chain in X provided that U 1 is not contained in c X (U 2 ), U n is not contained in c X (U n 1 ) and c X (U i )\ c X (U j ) 6= ; if and ony if ji jj 1. If, in addition, diameter(u i ) < " for each i 2 f1; : : : ; ng, then U is said to be a separated "-chain. It is easy to see that if V = fv 1 ; : : : ; V m g is a -chain in X, then the sequence fv 1 [ V 2 ; V 3 [ V 4 ; : : :g (the ast eement in this sequence is V m, if m is od and, it is V m 1 [ V m, if m is even) is a separated 2-chain in X. Thus for each " > there exists a separated "-chain in X. Given a chain U = fu 1 ; : : : ; U n g, there is a natura order in U (given by the order of the subindices) which wi be denoted with the usua symbos <, >, and. So, if U; V 2 U, we de ne UV = S fw 2 U : U W V g, if U V and UV = S fw 2 U : V W Ug, if V U. We aso say that an eement W 2 U is between U; V 2 U provided that U W V, if U V, and V W U, if V U. Given a separated chain U = fu 1 ; : : : ; U n g in X, with n 3, the tightness t(u) of U is de ned as 6

t(u) = minfd( S fc X (U) : U < U i g; S fc X (U) : U i < Ug) : i 2 f2; : : : ; n 1gg Given two separated chains V and U = fu 1 ; : : : ; U n g in X, with n 3, we say that V utrare nes U provided that: (a) V is a t(u) 3 -chain, (b) there exist V; W 2 V such that V \ (U 2 [ : : : [ U n ) = ;, W \ (U 1 [ : : : [ U n 1 ) = ; and, (c) for each V 2 V, there exists U 2 U such that V U. Ceary, for each separated chain U in X and for each " >, there exists a separated "-chain V in X such that V utrare nes U. Given two separated chains V and U in X such that V utrare nes U and given U; V 2 U, with U 6= V, we say that V fods from V to U provided that there exist P; Q; R 2 V such that P < Q < R or R < Q < P, RP UV, P [ R V and Q U. We aso say that V maes a zigzag between U and V with eements P; Q; R; S 2 V if P < Q < R < S, SP UV, P [ R U and Q [ S V. Given " >, metric spaces Y and Z, and an onto mapping f : Y! Z, f is said to be an "-mapping provided that diameter(f 1 (z)) < " for each z 2 Z. The foowing emma is easy to prove. Lemma 7. Let X be a chainabe continuum and et U and V be separated chains in X such that V utrare nes U. Then: (a) If U; V 2 U, where U < V and P; Q 2 V satisfy P \U 6= ; and Q\V 6= ;, then for each W 2 U such that U < W < V, there exists R 2 V such that R is between P and Q, R \ W 6= ; and the ony eement of U which intersects R is W. (b) If U; V 2 U, where U 6= V, then there exist P; Q 2 V such that P U, Q V, P Q UV and P Q intersects W for each W 2 U which is between U and V. (c) If A is a subcontinuum of X, U (resp., V ) is the rst (resp., ast) eement of U intersecting A, then A UV and A intersects each eement W 2 U which is between U and V. (d) If U; V 2 U and U < V, then there exists a subcontinuum A of X such that A c X (UV ), A\ c X (U) 6= ; and A\ c X (V ) 6= ;. (e) If U is an "-chain, U; V 2 U and c X (U)\ c X (V ) = ;, then there exists an onto "-mapping ' : c X (UV )! [; 1] such that c X (U) = ' 1 () and c X (V ) = ' 1 (1). Two basic resuts Throughout this paper the etter X wi denote a continuum, with metric d, we de ne the metric D on XX by D((u; v); (x; y)) = 1 2 (d(u; x)+d(v; y)). Given 7

two nonempty subsets K and L of X we de ne d(k; L) = inffd(x; y) : x 2 K and y 2 Lg. For subsets K; L X X, the symbo D(K; L) is de ned in a simiar way. Given a chainabe continuum X, a mean : X X! X, a separated chain U and eements U and V of U, we de ne D(c X (U)) = f(u; u) 2 X X : u 2 c X (U)g and D(U; V ) = S fe : E is a component of (c X (UV ) c X (UV )) \ 1 (c X (U)) and E \ D(c X (U)) 6= ;g: Theorem 8. Let X be a chainabe continuum, : X X! X a mean, U a separated chain in X and U; V 2 U. Suppose that c X (U)\ c X (V ) = ; and D(U; V ) \ ((c X (UV ) c X (V )) [ (c X (V ) c X (UV )) = ;. Then there exists > such that, if V is a separated -chain in X, then V utrare nes U and V does not fod from V to U. Proof. Let L = c X (UV ) c X (UV ), J = D(c X (U)) and M = (c X (UV ) c X (V )) [ (c X (V ) c X (UV )). Let N = (L \ 1 (c X (U))) [ M. Then N is a compact subset of L. Given a component E of N, we caim that either E \ J = ; or E \ M = ;. Suppose to the contrary that E \ J 6= ; and E \ M 6= ;. Fix a point p 2 E \ J. Since c X (U)\ c X (V ) = ;, M \ J = ;, so p =2 M. Let C be the component of E M containing p. Since E M is a proper nonempty subset of the continuum E, by [13, Theorem 5.6, p. 74], ; 6= c E (C)\ bd E (E M) c E (C) \ M. On the other hand, since C L \ 1 (c X (U)), there exists a component F of L \ 1 (c X (U)) such that C F. Then ; 6= c E (C) \ M F \ M. Since p 2 F \ J, F D(U; V ). Thus D(U; V ) \ M 6= ;, contrary to our assumption on D(U; V ). We have proved that E \ J = ; or E \ M = ;. Therefore, no component of N intersects both sets N \ J and M. By [13, Theorem 5.2, p. 72], there exist disjoint compact sets K and G such that N = K [ G, N \ J K and M G. For each point u 2 U, (u; u) 2 N \ J. Thus N \ J 6= ;. Fix < < D(K; L) such that, if V is a separated -chain in X, then V has at east three eements and V utrare nes U. Let V be a separated -chain in X. We are going to prove that V does not fod from V to U. Suppose to the contrary that that V fods from V to U. Then there exist P; Q; R 2 V such that P < Q < R or R < Q < P, P R UV, P [ R V and Q U. Let > be such that < and, if D((u; v); (x; y)) <, then d((u; v); (x; y)) < t(v). Let W be a separated -chain such that W utrare nes V. By Lemma 7 (b), there exist S; T 2 W such that S P, T R, ST P R and ST intersects W 8

for each W 2 V which is between P and R. By Lemma 7 (d), there exists an onto -mapping ' : c X (ST )! [; 1] such that c X (S) = ' 1 () and c X (T ) = ' 1 (1). Let : c X (ST ) c X (ST )! [; 1] 2 be given by (x; y) = ('(x); '(y)). Then is an onto -mapping. Suppose that there exist points (u; v); (x; y) 2 c X (ST ) c X (ST ) such that (u; v) 2 K, (x; y) 2 G and (u; v) = (x; y). Then D((u; x); (v; y)) < < D(K; G), a contradiction. We have shown that ((c X (ST ) c X (ST )) \ K) \ ((c X (ST ) c X (ST )) \ G) = ;. We show that the boundary B of [; 1] 2 in R 2 is contained in ((c X (ST ) c X (ST )) \ G). Given a point (; s) 2 B, since ' is onto, there exist points x 2 S P V and y 2 c X (ST ) such that (x; y) = (; s). Thus (x; y) 2 M G. Hence fg [; 1] ((c X (ST ) c X (ST )) \ G). The rest of points of B can be treated in a simiar way. Thus B ((c X (ST ) c X (ST )) \ G). Let denote the triange contained in [; 1] 2 with vertices (; ); (; 1) and (1; 1) and et denote the diagona of [; 1] 2. By the choice of S and T, there exists a point x 2 ST \ Q U. Then (x ; x ) 2 N \ J K. Thus (x ; x ) 2 \ ((c X (ST ) c X (ST )) \ K). Hence we can appy Theorem 2 to the triange, the cosed disjoint subsets H = ((c X (ST ) c X (ST ))\G)\ and K = ((c X (ST ) c X (ST ))\K)\, since (fg [; 1]) [ ([; 1] f1g) H and (x ; x ) 2 K \. Thus there exists a one-to-one continuous function : [; 1]! such that () = (; ), (1) = (1; 1), Im \ K = ; and Im \ H. Since is a -mapping, there exists " > such that if A [; 1] 2 and diameter(a) < ", then diameter( 1 (A)) <. Since is uniformy continuous, there exists > such that, if jt sj <, then (t) (s) < ". Let = t < t 1 < t 2 < : : : < t m = 1 be a partition of the interva [; 1] such that t i t i 1 < for each i 2 f1; 2; : : : ; mg. For each i 2 f; 1; 2; : : : ; mg, choose an eement (x i ; y i ) 2 c X (ST ) c X (ST ) such that (x i ; y i ) = (t i ), in the case that (t i ) 2, (t i ) = (t; t) for some t 2 [; 1], since ' is onto, we can choose x i 2 c X (ST ) such that '(x i ) = t, then (x i ; x i ) = (t i ), so in the case that (t i ) 2, we can assume that x i = y i. In particuar, since () = (; ) and (1) = (1; 1), x 2 c X (S) and x m 2 c X (T ). For each i 2 f; 1; 2; : : : ; mg, et p i = (x i ; y i ). Then p = x = (x ; x ) 2 c X (S) c X (P ) and p m = x m = (x m ; x m ) 2 c X (T ) c X (R): Given i 2 f1; 2; : : : ; mg, since t i t i 1 <, (t i ) (t i 1 ) < ". Since (x i ; y i ), (x i 1 ; y i 1 ) 2 1 (f(t i ); (t i 1 )g), by the choice of ", D((x i ; y i ); (x i 1 ; y i 1 )) <. By the choice of, d(p i ; p i 1 ) < t(v). Since p 2 c X (P ), p m 2 c X (R), P < Q < R and d(p i ; p i 1 ) < t(v) for each i 2 f1; 2; : : : ; mg, there exists j 2 f1; 2; : : : ; mg such that p j 2 Q. Thus p j 2 9

c X (U) and (x j ; y j ) 2 1 (c X (U))\(c X (ST ) c X (ST )) 1 (c X (U))\L N = K [ G. Hence (x j ; y j ) 2 K [ G. We now that (x j ; y j ) 2 Im and Im \ K = ;, this impies that (x j ; y j ) =2 K. Thus (x j ; y j ) 2 G. This impies that (x j ; y j ) 2 H \ Im. By the choice of (x j ; y j ), x j = y j. Then x j = (x j ; y j ) = p j 2 c X (U). Thus (x j ; x j ) 2 N \ J K. Hence (x j ; y j ) 2 G \ K, a contradiction. We have shown that V does not fod from V to U. This competes the proof of the theorem. Theorem 9. Let X be a chainabe continuum and : X X! X a mean. Then for each " >, there exists > such that, if U is a separated -chain in X and U; V 2 U are such that d(c X (U);c X (V )) ", then there exists > such that for each separated -chain V, V utrare nes U, and V does not fod from V to U or V does not fod from U to V. Proof. Let " > : Since is uniformy continuous, there exists > such that, if D((x; y); (u; v)) <, then d((x; y); (u; v)) < ". Let U be a separated -chain and et U; V 2 U be such d(c X (U);c X (V )) ". Caim. D(U; V ) \ ((c X (UV ) c X (V )) [ (c X (V ) c X (UV )) = ; or D(V; U) \ ((c X (UV ) c X (U)) [ (c X (U) c X (UV )) = ;. In order to prove this caim, suppose that it is not true. By Lemma 7 (e), there exists an onto -mapping ' : c X (UV )! [; 1] such that c X (U) = ' 1 () and c X (V ) = ' 1 (1). De ne : c X (UV ) c X (UV )! [; 1] 2 by (x; y) = ('(x); '(y)). Then is an onto -mapping. Given a component E of (c X (UV ) c X (UV )) \ 1 (c X (U)) such that E \ D(c X (U)) 6= ;, there exists x 2 c X (U) such that (x ; x ) 2 E, so (; ) 2 (E). Since is symmetric, E is a symmetric subset of c X (UV ) c X (UV ). This impies that (D(U; V )) is a symmetric subcontinuum of [; 1] 2 containing (; ), and by our assumption on D(U; V ), (D(U; V )) intersects the set [; 1]f1g. Simiary, (D(V; U)) is a symmetric subcontinuum of [; 1] 2 containing (1; 1) and intersecting the set fg[; 1]. This impies that (D(U; V ))\ (D(V; U)) 6= ;. Tae points (x; y) 2 D(U; V ) 1 (c X (U)) and (u; v) 2 D(V; U) 1 (c X (V )) such that (x; y) = (u; v). Then D((x; y); (u; v)) < and d((x; y); (u; v)) < ". This contradicts the choice of U and V and competes the proof of the caim. Suppose, without oss of generaity, that D(U; V ) \ ((c X (UV ) c X (V )) [ (c X (V ) c X (UV )) = ;. Let > be as in Theorem 8. Hence, if V is a separated -chain in X, then V utrare nes U and V does not fod from V to U. 1

The hereditariy decomposabe case A nondegenerate continuum X is decomposabe provided that there exist two proper subcontinua A and B of X such that X = A [ B. The continuum X is said to be hereditariy decomposabe if each nondegenerate subcontinuum of X is decomposabe. Given two points p; q 2 X, we say that X is irreducibe between p and q, provided that there is no proper subcontinuum of X containing both points p and q. Given a subcontinuum A of a chainabe continuum X and a chain U in X, we say that two eements U and V of U bound A provided that A UV, U V and fw 2 U : U W V g is a minima subchain of U containing A. Note that W \ A 6= ; for each W 2 U such that U W V. Note aso that, if A is not contained in the intersection of two eements of U then U and V are unique. Theorem 1. If X is a hereditariy decomposabe chainabe continuum and X admits a mean, then X is an arc. Proof. Let : X X! X be a mean. Suppose to the contrary that X is not an arc. By [13, Theorem 12.5, p. 233] there exist two points p and q of X such that X is irreducibe between p and q. By [12, Theorem 3, p. 216], there exists a monotone mapping ' : X! [; 1] such that '(p) =, '(q) = 1 and int X (' 1 (t)) = ; for each t 2 [; 1]: Since X is not an arc, ' is not one-to-one. Thus, there exists t 2 I such that W = ' 1 (t ) is nondegenerate. Note that W c X (' 1 ([; t )))[ c X (' 1 ((t ; 1])): So, W = (c X (' 1 ([; t )) \ W ) [ (c X (' 1 ((t ; 1])) \ W ). Without oss of generaity we can assume that Y = c X (' 1 ((t ; 1])) \ W is nondegenerate. Since X is monotone, each set of the form ' 1 ([t; 1]) is a subcontinuum of X, this impies that ' 1 ((t ; 1]) is connected. Since X is chainabe, X is hereditariy unicoherent ([13, Theorem 12.2, p. 23]). Thus Y is a subcontinuum of X. Since Y itsef is chainabe, there exist points p ; q 2 Y such that Y is irreducibe between p and q. Hence there exists a monotone mapping : Y! [; 1] such that (p ) =, (q ) = 1 and int Y ( 1 (t)) = ; for each t 2 [; 1]: Let " = 1 4 d( 1 ([; 1 3 ]); 1 ([ 2 3 ; 1])). By Theorem 9, there exists > such that, if U is a separated -chain and U; V 2 U are such that d(c X (U);c X (V )) ", then there exists > such that for each separated -chain V, V utrare nes U, and V does not fod from V to U or V does not fod from U to V. Let > be such that 4 < minf; d( 1 2 ([; 3 ]); 1 (1)); d( 1 (); 1 ([ 1 3 ; 1])); "g. 11

Let U be a separated -chain in X. Let U p ; U q 2 U be such that p 2 U p and q 2 U q. At this point we have two possibe orders for U. So, we choose the order that satis es U p < U q. Given eements W 1 ; W 2 2 U such that p 2 W 1 and q 2 W 2, we have that diameter(u p [ W 1 ), diameter(u q [ W 2 ) < " 2. If (U p [ W 1 ) \ (U q [ W 2 ) 6= ;, then d(p ; q ) < ", contradicting the choice of ". Therefore, (U p [ W 1 ) \ (U q [ W 2 ) = ;. Since U p < U q, we concude that W 1 < W 2. Let U; V 2 U be such that U and V bound Y. Caim. d(c X (U);c X (V )) ". In order to prove the caim, et U 1 ; U 2 ; U 3 ; U 4 ; U 5 ; U 6 ; U 7 ; U 8 2 U be such U 1 and U 2 bound 1 (); U 3 and U 4 bound 1 ([; 2 3 ]); U 5 and U 6 bound 1 ([ 1 3 ; 1]) and U 7 and U 8 bound 1 (1): Note that we may assume that U U 3 U 1 U 2 U 4 V and U U 5 U 7 U 8 U 6 V: We show that U 1 U 2 \ U 5 U 6 = ; and U 3 U 4 \ U 7 U 8 = ;: Suppose that there exists a point x 2 U 1 U 2 \ U 5 U 6. Then there exist P; Q 2 U such that x 2 P \ Q and U 1 P U 2 ; U 5 Q U 6 : We can tae points y 2 P \ 1 () and z 2 Q \ 1 ([ 1 3 ; 1]): Then d(y; z) d(y; x) + d(x; z) diameter(p )+ diameter(q) < 2 < d( 1 (); 1 ([ 1 3 ; 1])) d(y; z), a contradiction. We have shown that U 1 U 2 \ U 5 U 6 = ;. Simiary, U 3 U 4 \ U 7 U 8 = ;. Since p 2 U 1 U 2 and q 2 U 5 U 6, there exist W 1 ; W 2 2 U such that p 2 W 1, q 2 W 2, U 1 W 1 U 2 and U 5 W 2 U 6. Then W 1 < W 2. Since U 1 U 2 \ U 5 U 6 = ;, we concude that U 2 < U 5. Simiary, U 4 < U 7. If U \ 1 ([ 1 3 ; 1]) 6= ;, then U \ U 5U 6 6= ;. This is impossibe since U U 2 < U 5 U 6. Hence U \ 1 ([ 1 3 ; 1]) = ;. On the other hand, U \ Y 6= ;, so U \ 1 ([; 1 3 ]) 6= ;. Simiary, V \ 1 ([ 2 3 ; 1]) 6= ;. If d(c X (U);c X (V )) < ", then there exist points x 2 c X (U) and y 2 c X (V ) such that d(x; y) < ". Since diameter(c X (U)) and diameter(c X (V )) < ", we concude that d( 1 ([; 1 3 ]); 1 ([ 2 3 ; 1])) < 3", contradicting the de nition of ". Therefore, d(c X (U);c X (V )) " and the caim is proved. Since U is a separated -chain, there exists > such that for each separated -chain V, V utrare nes U, and V does not fod from V to U or V does not fod from U to V. Since T fc X (' 1 ((t ; t + 1 n ])) : n 2 Ng is contained in Y, there exists n 2 N such that ' 1 ((t ; t + 1 n ]) UV. Fix points u 1 2 Y \ U and v 1 2 Y \ V. Since u 1 2 ' 1 (t )\ c X (' 1 ((t ; 1])), we can choose a point u 2 2 (U (' 1 ([t + 1 n ; 1])) \ ' 1 ((t ; 1], so u 2 2 U \ ' 1 ((t ; t + 1 n )). Simiary, we can choose a point v 2 2 V \ ' 1 ((t ; t + 1 n )). Let A 2 = ' 1 ('(u 2 )'(v 2 )), where '(u 2 )'(v 2 ) is the subinterva of the rea ine joining the points '(u 2 ) and '(v 2 ). Since ' is monotone, A 2 is a subcontinuum of X such that A 2 ' 1 ((t ; t + 1 n ]) UV, A 2 \ U 6= ; and A 2 \ V 6= ;. Let m > n be such that ' 1 ((t ; t + 1 m ]) \ A 2 = ;. Proceeding as before, there exists a subcontinuum 12

A 3 of X such that A 3 ' 1 ((t ; t + 1 m ]) UV, A 3 \ U 6= ; and A 3 \ V 6= ;. Thus A 3 \ A 2 = ;. Simiary, there exists a subcontinuum A 4 of X such that A 4 ' 1 ((t ; t + 1 n ]) (A 2 [ A 3 ) such that A 4 \ U 6= ; and A 4 \ V 6= ;. For each i 2 f2; 3; 4g, x points r i 2 A i \ U and s i 2 A i \ V. Let K be the convex hu of the set '(A 2 ) [ '(A 3 ) [ '(A 4 ) (t ; t + 1 n ]. Then ' 1 (K) is a subcontinuum of UV. Let > be such that 2 < minf; d(fr 2 ; r 3 ; r 4 g; X V ); d(' 1 (K); X UV )g and 2 < d(a i ; A j ), if i 6= j. U); d(fs 2 ; s 3 ; s 4 g; X Tae a separated -chain V (and then V is a separated -chain). We wi obtain a contradiction by proving that V fods from V to U and V fods from U to V. Let V 1 ; W 1 ; V 2 ; V 3; V 4 ; W 2 ; W 3 ; W 4 2 V be such that: V 1 and W 1 bound ' 1 (K), V 2 and W 2 bound A 2, V 3 and W 3 bound A 3 and V 4 and W 4 bound A 4. By the choice of the sets V 2 W 2, V 3 W 3 and V 4 W 4 are pairwise disjoint. We may assume that V 1 V 2 W 2 < V 3 W 3 < V 4 W 4 W 1. Let R 2 ; R 3 ; R 4 ; S 2 ; S 3 ; S 4 2 V be such that, for each i 2 f2; 3; 4g, r i 2 R i, s i 2 S i and V i R i ; S i W i. By the choice of, R 2 [ R 3 [ R 4 U, S 2 [ S 3 [ S 4 V and V 1 W 1 UV. Since R 2 < S 3 < R 4 and S 2 < R 3 < S 4, we obtain that V fods from V to U and V fods from U to V. This contradiction competes the proof of the theorem. The other case Proof of Theorem 1. As usua, et D be the metric in X X given by D((u; v); (x; y)) = 1 2 (d(u; x) + d(v; y)), where d is a metric for X. Suppose that : X X! X is a mean. By Theorem 1, we may assume that there exists a nondegenerate subcontinuum Y of X such that Y is not the union of two of its proper subcontinua (Y is indecomposabe). We are going to nd a contradiction by constructing a function h : f1; 2; : : : ; 4Ng! X (where N is a positive integer) with the property that diameter(im h) 3 4diameter(Y ) and diameter(im h) 1 2diameter(Y ). Caim 1. If U is a separated ( 1 3diameter(Y ))-chain in X and U and V are the eements of U which bound Y, then there exists 1 > such that each separated 1 -chain V satis es that V utrare nes U and V maes a zigzag between U and V with eements P; Q; R; S 2 V such that U P < Q < R < S V, where U and V are the eements in V which bound Y. 13

In order to prove Caim 1, et U be a separated ( 1 3diameter(Y ))-chain in X and et U and V be the eements of U which bounds Y. Fix points p 2 Y \ U and q 2 Y \ V. Let K 1 ; K 2 ; K 3 and K 4 be four pairwise di erent composants of Y ([13, Theorem 11.15, p. 23]). Since each K i is dense in Y ([13, 5.2 (a), p. 83]), we can choose points p i 2 K i \ U and q i 2 K i \ V. Then there exist proper subcontinua A 1 K 1, A 2 K 2, A 3 K 3 and A 4 K 4 of Y such that p i ; q i 2 A i for each i 2 f1; 2; 3; 4g. Thus A 1 ; A 2 ; A 3 and A 4 are pairwise disjoint. Let 1 > be such that 1 < min(fd(a i ; A j ) : i; j 2 f1; 2; 3; 4g and i 6= jg [ fd(y; X UV ); d(fp 1 ; p 2 ; p 3 ; p 4 g; X U); d(fq 1 ; q 2 ; q 3 ; q 4 g; X V )g and each separated 1 -chain utrare nes U. Let V be a separated 1 -chain. Then V utrare nes U. We show that V maes a zigzag between V and U. Let U ; U 1 ; U 2 ; U 3 ; U 4 ; V ; V 1 ; V 2 ; V 3 ; V 4 2 V be such that U and V bounds Y and U i and V i bounds A i for each i 2 f1; 2; 3; 4g. By the choice of 1, it foows that U 1 V 1 [ U 2 V 2 [ U 3 V 3 [ U 4 V 4 U V UV and U 1 V 1 ; U 2 V 2 ; U 3 V 3 and U 4 V 4 are pairwise disjoint. Thus we may assume that U U 1 V 1 < U 2 V 2 < U 3 V 3 < U 4 V 4 V. For each i 2 f1; 2; 3; 4g, et S i ; T i 2 V be such that p i 2 S i, q i 2 T i, U i S i ; T i V i. By the choice of 1, S 1 [ S 2 [ S 3 [ S 4 U and T 1 [ T 2 [ T 3 [ T 4 V. Since U S 1 < T 2 < S 3 < T 4 V, we have that V maes a zigzag between U and V. This competes the proof of Caim 1. Let > be such that < 1 16 (diameter(y )) and, if D((u; v); (x; y)) < 4, then d((u; v); (x; y)) < 1 16 (diameter(y )). Fix a separated -chain U. Let U; V 2 U be such that U and V bound Y. If c X (U)\ c X (V ) 6= ;, then U \ V 6= ; (U is a separated chain). Thus diameter(uv ) < 2. This is a contradiction since Y UV. This proves that c X (U)\ c X (V ) = ;. By Lemma 7 (a), there exists an onto -mapping f : c X (UV )! [; 1] such that c X (U) = f 1 () and c X (V ) = f 1 (1). Let > be such that, if x; y 2 c X (UV ) and jf(x) d(x; y) <. f(y)j < 2, then Let 1 > be as in Caim 1 appied to U, U and V. We may assume that 1 < d(y; X UV ), 1 < and 1 has the property that, if x; y 2 c X (UV ) and d(x; y) < 2 1, then jf(x) f(y)j <. 14

Fix a separated 1 -chain V. By the choice of 1, V utrare nes U and V maes a zigzag between U and V with eements P; Q; R; S 2 V such that U P < Q < R < S V, where U and V are the eements in V which bound Y. Then P S UV, P [ R U and Q [ S V. Since each eement T 2 V such that U T V intersects Y, by the choice of 1, we concude that T UV. Thus U V UV. We can assume that P is the rst eement in the set ft 2 V : U T V g such that P U ; we aso assume that Q is the rst eement in the set ft 2 V : P T V g such that Q V ; R is the rst eement in the set ft 2 V : Q T V g such that R U and S is the rst eement in the set ft 2 V : R T V g such that S U. If U \ V 6= ;, then diameter(y ) diameter(u V ) < 2 1 < 1 2diameter(Y ), a contradiction. Hence U \ V = ;. Thus V has at east three eements, so t(v) is we de ned. Since V is separated, c X (U )\ c X (V ) = ;. Caim 2. D(U ; V ) \ ((c X (U V ) c X (V )) [ (c X (V ) c X (U V )) 6= ;. To prove Caim 2, suppose, to the contrary that D(U ; V ) \ ((c X (U V ) c X (V )) [ (c X (V ) c X (U V )) = ;. By Theorem 8, there exists 1 > such that if V is a separated 1 -chain in X, then V utrare nes V and V does not fod from V to U. Let 2 > be as in Caim 1 appied to V, U and V, we may aso as that 2 < 1. Let V be a separated 2 -chain in X. By the choice of 1, V utrare nes V and V does not fod from V to U. On the other hand, by the choice of 2, V maes a zigzag between U and V and then V fods from V to U, a contradiction. This competes the proof of Caim 2. By Caim 2, there is a component E of (c X (U V ) c X (U V ))\ 1 (c X (U )) such that E \ D(c X (U )) 6= ; and E \ ((c X (U V ) c X (V )) [ (c X (V ) c X (U V )) 6= ;. We ony consider the case E \ (c X (V ) c X (U V )) 6= ;, the other one is anaogous. Let 1 ; 2 : X X! X be the respective projections on the rst and second coordinates. Let C = 1 (E) [ 2 (E). Fix an eement (u ; u ) 2 E \ D(c X (U )). Then u 2 1 (E) \ 2 (E), so C is a subcontinuum of X such that C \ c X (U ) 6= ;, C \ c X (V ) 6= ; and C c X (U V ). Fix an eement (v ; z ) 2 E \ (c X (V ) c X (U V )). Then u 2 C \ c X (U ) and v 2 C \ c X (V ). Let 3 > be such that 3 < minf 1 ; t(v)g and 3 has the property that, 15

if D((u; v); (x; y)) < 3, then d((u; v); (x; y)) < t(v). Fix a separated 3 -chain W such that W utrare nes V. Let U 1 ; V 1 2 W be such that U 1 and V 1 bound C. For each W 2 W, x a point p W 2 W ( S fc X (S) : S 2 W fw gg). Given a point (x; y) 2 E, x; y 2 C, so there exist S; T 2 W such that (x; y) 2 S T and U 1 S; T V 1. We have shown that the famiy F = fs T : S; T 2 W and U 1 S; T V 1 g is an open cover of E. Since E is connected, there exists n 2 N and S 1 ; : : : ; S n ; T 1 ; : : : ; T n 2 W such that U 1 S 1 ; : : : ; S n ; T 1 ; : : : ; T n V 1, (u ; u ) 2 (S 1 T 1 ) \ E, (v ; z ) 2 (S n T n ) \ E and, for each i 2 1; : : : ; n 1, ((S i T i ) \ E) \ ((S i+1 T i+1 ) \ E) 6= ;. For each i 2 f1; : : : ; n 1g, x a pair ((i); (i)) 2 (S i T i )\(S i+1 T i+1 )\E. Hence ((i); (i)) 2 c X (U ). Put () = u = () 2 S 1 \ T 1. Caim 3. There exists j 2 f; 1; : : : ; n 1g and there exists an eement x 2 f(j); (j)g such that x 2 R and R is the ony eement of V containing x in its cosure. We prove Caim 3. Since S 1 \ T 1 6= ;, S i \ S i+1 6= ; and T i \ T i+1 6= ; for each i 2 f1; : : : ; n 1g, the set fs 1 ; : : : ; S n ; T 1 ; : : : ; T n g can be reordered as a subchain W of W. Since u 2 S 1 \ c X (U ), S 1 \ U 6= ;. Since v 2 S n \ c X (V ), S n \ V 6= ;. By Lemma 7 (a), since U < R < V, there exists an eement R 2 W such that R \ R 6= ;, and the ony eement of V which intersects R is R, so the ony eement of V whose cosure interesects R is R. Thus R R. Since R contains either one eement of the form (j) or one eement of the form (j), we concude that there exists j 2 f; 1; : : : ; n 1g and there exists an eement x 2 f(j); (j)g such that x 2 R and R is the ony eement of V containing x in its cosure. This ends the proof of Caim 3. Let j R = minfj 2 f; 1; : : : ; n 1g : there exists an eement x 2 f(j); (j)g such that x 2 R and R is the ony eement of V containing xg. By symmetry, we may assume that (j R ) 2 R and R is the ony eement of V containing (j R ) in its cosure. Since () = u 2 c X (U ) and U < R, < j R. Let W 1 = fw 2 V : U W Rg. Since W 1 is a subchain of V, we can put W 1 = fw ; : : : ; W m g, where U = W < : : : < W m = R. Note that P; Q 2 W 1. So, Q = W i for some i 2 f; 1; : : : ; mg. Since U P < Q < R and P \ Q = ; (P U and Q V ), 1 < i < m. Appying Lemma 7 (a), considering that the famiy ft 1 ; : : : ; T jr g can be put as a subchain of W, it can be shown that for each i 2 f1; : : : ; m 1g, there exists j i 2 f1; : : : ; j R g such that: (j i ) 2 W i and W i is the ony eement of V containing (j i ) in its cosure. 16

We may assume that j i is the ast eement in f1; : : : ; j R g with the described properties. De ne j m = j R. Since E c X (U V ) c X (U V ), C c X (U V ). Caim 4. (i) 2 c X (U ) [ W 1 [ : : : [ W m for each i 2 f; 1; : : : ; j R g. In order to prove Caim 4, suppose to the contrary that there exists i 2 f; 1; : : : ; j R g such that (i) =2 c X (U ) [ W 1 [ : : : [ W m. Since () = u c X (U ), < i. Let W 2 V, be such that (i) 2 W. If W m < W, since the famiy fs 1 ; : : : ; S i g can be put as a subchain of W, appying Lemma 7 (a), there exists 2 f1; : : : ; i 1g such that () 2 W m = R and R is the ony eement of V containing (). This contradicts the choice of j R and proves that W W m, thus W < W = U. Since (i) 2 C c X (U V ) = S fc X (T ) : U T V g. Since V is a separated chain, the ony eement in the famiy fc X (T ) : U T V g that can be intersected by W is c X (U ). Thus (i) 2 c X (U ), a contradiction. This competes the proof of Caim 4. In a simiar way it can be proved the foowing caim. Caim 5. (i) 2 c X (U ) [ W 1 [ : : : [ W m for each i 2 f; 1; : : : ; j R g. De ne : f; 1; : : : ; g! f; 1; : : : ; g by 8 < : (i) = ; if i j R and (i) 2 c X (U ), minf 2 f1; : : : ; mg : (i) 2 W g, if i j R and (i) =2 c X (U ), 2m (2j R i), if j R < i. Since () = u 2 c X (U ), () =. Given i 2 f1; : : : ; mg, j i 2 f; 1; : : : ; j R g, (j i ) 2 W i and W i is the ony eement of V containing (j i ) in its cosure, in particuar, (j i ) =2 c X (U ). Thus (j i ) = i. In particuar, (j i ) = i. Hence ( ) = 2m (j i ) =. Let i 2 fj i ; : : : ; j R g, we are going to show that i (i) m. Since (j i ) = i, we may assume that j i < i. Suppose to the contrary that (i) < i. Note that (i) 2 c X (U ) or (i) 2 W (i). In the rst case, et W 2 V be such that (i) 2 W. Since W \ U 6= ; and 1 < i, W < W i = Q. Thus, in both cases, there exists W 2 V such that (i) 2 W and W < W i = Q < R. Consider the famiy ft i ; : : : ; T jr g W. Since T i \ W 6= ; and (j R ) 2 T jr \ W m and ft i ; : : : ; T jr g can be rearrenged as a subchain of W, by Lemma 7 (a), there exists 2 fi; : : : ; j R g such that Q is the ony eement of V containing () in its cosure. This contradicts the maximaity of j i and competes the proof that i (i) m. Given i 2 fj R ; : : : ; g, j i 2j R i j R. Since we have shown that i (2j R i) m, we concude that m 2m (j R i). This proves that (i) 2 f; 1; : : : ; g for each i 2 f; 1; : : : ; g. 17

Caim 6. For each i 2 f1; 2; : : : ; g, j(i) (i 1)j 1. To prove Caim 6, et i; j 2 f; 1; : : : ; g be such that ji jj = 1. By the choice of (i) and (j), there exists T 2 W such that (i); (j) 2 T. By the choice of 3 and W, d((i); (j)) < t(v). Since W 1 ; W 2 2 V, if (i) 2 c X (W 1 ) and (j) 2 c X (W 2 ) for some 1 ; 2 2 f; 1; : : : ; mg, then j 1 2 j 1. We consider three cases. Case 1. i; j j R. If (i) 2 c X (U ), then (j) 2 c X (U ) or (j) 2 W 1 c X (U ), in both cases, j(i) (j)j 1. If (j) 2 c X (U ), simiary, j(i) (j)j 1. If (i) =2 c X (U ) and (j) =2 c X (U ), then (i) 2 W (i) and (j) 2 W (j). Thus j(i) (j)j 1. Case 2. j R < i; j : In this case, 1 j i 2j R i; 2j R j < j R and j2j R i (2j R j)j = 1. Appying the rst case, j(2j R i) (2j R j)j 1. Hence j(i) (j)j 1. Case 3. i = j R and j = j R + 1. In this case, (i) = m and (j) = 2m (j R 1). By the rst case, j(j R ) (j R 1)j 1. Thus j(i) (j)j 1. Therefore, Caim 6 is proved. Since (j i ) 2 c X (U ) [ W 1 [ : : : [ W m, we can de ne m 2 f; : : : ; mg as: m =, if (j i ) 2 c X (U ), and m = minfj 2 f1; : : : ; mg : (j i ) 2 W j g, if (j i ) =2 c X (U ). Let m 1 = + 1 + m. De ne : f; 1; : : : ; m 1 g! f; 1; : : : ; 2m i g by: 8 >< >: (i) =, if i j R and (i) 2 c X (U ), minf 2 f1; : : : ; mg : (i) 2 W g, if i j R and (i) =2 c X (U ), (2j R i), if j R < i, m + i ( + 1), if < i m 1. Since () = u 2 c X (U ), () =. Note that (m 1 ) = and (i) m for each i. Caim 7. For each i 2 f1; : : : m 1 g, j(i) (i 1)j 1. To prove Caim 7, et i; j 2 f; 1; : : : ; m 1 g be such that ji jj = 1. We consider ve cases. 18

Case 1. < i; j m 1. In this case, j(i) (j)j = ji jj = 1. Case 2. i = and j = + 1. In this case, (i) = ( ) = (j i ) = m = (j). Case 3. i; j j R. By the choice of (i) and (j), there exists T 2 W such that (i); (j) 2 T. By the choice of 3 and W, d((i); (j)) < t(v). Thus, if (i) 2 c X (W 1 ) and (j) 2 c X (W 2 ) for some 1 ; 2 2 f; 1; : : : ; mg, then j 1 2 j 1. Thus: If (i) 2 c X (U ), then (j) 2 c X (U ) or (j) 2 W 1 c X (U ), in both cases, j(i) (j)j 1. If (j) 2 c X (U ), simiary, j(i) (j)j 1. If (i) =2 c X (U ) and (j) =2 c X (U ), then (i) 2 W (i) and (j) 2 W (j). Thus j(i) (j)j 1. Case 4. j R < i; j : In this case, 1 j i 2j R i; 2j R j < j R. Appying Case 3, we obtain that j(2j R i) (2j R j)j 1. Hence j(i) (j)j 1. Case 5. i = j R and j = j R + 1. In this case, (j) = (j R 1). By Case 3.1, j(i) (j)j 1. Therefore, Caim 7 is proved. Notice that c X (U ) [ W 1 [ : : : [ W m c X (U V ) c X (UV ). Let J = f(i) : i 2 f; 1; : : : ; j R gg [ f(i) : i 2 f; 1; : : : ; j R gg. De ne g : J! [; 1] by f(u ), if (i) 2 c g((i)) = X (U ); and f((j (i) )), if (i) =2 c X (U ), f(u ), if (i) 2 c g((i)) = X (U ); f((j (i) )), if (i) =2 c X (U ): Note that, if i 2 f; 1; : : : ; j R g and (i) =2 c X (U ), then (i) 2 f1; : : : ; mg, so j (i) 2 f; 1; : : : ; j R g, then (j (i) ) 2 c X (U V ). Thus f((j (i) )) and g((i)) are we de ned. Simiary, g((i)) is we de ned for each i 2 f; 1; : : : ; j R g. Given i 2 f1; : : : ; mg, (j i ) 2 W i (c X (U )[: : :[W i 1 ), so (j i ) = i. Thus g((j i )) = f((j i )). Let : [; 1]! [; 1] be the PL mapping de ned by the foowing conditions: 1 m () = g(()), ( ) = g((j 1 )), : : :, ( ) = g((j m )), ( m+1 ) = g((j m 1 )), ( m+2 m+(m i) ) = g((j m 2 )), : : :, ( ) = g((j i )) 19

Let : [; 1]! [; 1] be the PL mapping de ned by the foowing conditions: () = g(()), ( 1 j ) = g((1)), : : :, ( R ) = g((j R )), ( j R+1 ) = g((j R 1)), ( j R+2 ) = g((j R 2)), : : :, ( j R+( ) ) = g((j i )). Let : [; 1]! [; 1] be the PL mapping de ned by the foowing conditions: () = g(()), ( 1 j ) = g((1)), : : :, ( R ) = g((j R )), ( j R+1 ) = g((j R 1)), ( j R+2 ) = g((j R 2)), : : :, ( j R+( ) ) = g((j i )). Let : [; 1]! [; 1] be the PL mapping de ned by the foowing conditions: 1 () = g((j R )), ( ) = g((j R 1)), : : :, ( ) = g((j i )). Since i > and (j i ) 2 Q V, so g((j i )) = f((j i )) = 1. Since (j R ) 2 R U, g((j R )) = g((j m )) = f((j m )) = f((j R )) =. Therefore, (1) = 1, (1) = 1, () = and (1) = 1. Hence is a jump mapping. We want to appy Theorem 6 to the mappings and. Caim 8. ( (i) ) = ( i 2j R i ) for each i 2 f; 1; : : : ; g. To prove Caim 8, we consider three cases. Case 1. i j R and (i) 2 c X (U ). In this case (i) =, ( (i) i ) = g(()) = f(u ), ( 2j R i ) = g((i)) = f(u ). Hence, ( (i) i ) = ( 2j R i ). Case 2. i j R and (i) =2 c X (U ). By de nition of (i), (i) 2 f1; : : : ; mg. By de nition of g((i)), g((i)) = i f((j (i) )). Thus ( 2j R i ) = f((j (i) )). Since f((j )) = g((j )) for each 2 f1; : : : ; mg and (i) 2 f1; : : : ; mg, we obtain that ( (i) ) = g((j (i) )) = f((j (i) )). Hence, ( (i) i ) = ( 2j R i ). Case 3. j R < i. In this case, 1 j i 2j R i < j R. As we proved after the de nition of this inequaities impy that 1 < i (2j R i) m. Thus g((j (2jR 1))) = f((j (2jR 1))) and (i) = 2m (2j R i) 2 fm; : : : ; g. By the de nition of, ((i)) = g((j 2m (i) )) = g((j (2jR i))) = f((j (2jR i))). On the i other hand, ( 2j R i ) = g((2j R i)). Since 1 < (2j R i), (2j R i) =2 i c X (U ), so ( 2j R i ) = g((2j R i)) = f((j (2jR i))) = ((i)). This competes the proof of Caim 8. 2

Let : [; 1]! [; 1] be the PL mapping which is the common extension of the foowing two mappings: (2t), if t 2 [; 1 2 ], and m (( )(4 4t) + 4t 3), if 3 4 t 1. Since i < m, < m. Notice that is supported by the 1 partition < 2( ) < : : : < 2( ) = 1 2 < 3 4 < 3 4 + 1 4 ( 1 m ) < : : : < 3 4 + 1 2m i m 4 ( m ) = 1, which divides the interva [; 1] into m 1 = + 1 + m subintervas. i Caim 9. ( 2( ) ) = ( (i) ) for each i 2 f; : : : ; g. To prove Caim 9, we consider three cases. Case 1. i j R and (i) 2 c X (U ). (i) In this case, (i) = and ( 2m i ) = g(()) = f(u ) ). On the other hand, i 2( ) 1 2, so i ( 2( ) ) = ( i ) = g((i)) = f(u ). Hence, i ( 2( ) ) = ( (i) ). Case 2. i j R and (i) =2 c X (U ), By de nition of (i), (i) 2 W (i) (c X (U ) [ W 1 [ : : : [ W (i) 1 ). By i de nition of g((i)), g((i)) = f((j (i) )). Thus ( 2(2j R i ) = ( i ) ) = g((i)) = f((j (i) )). Since f((j )) = g((j )) for each 2 f1; : : : ; mg and (i) 2 f1; : : : ; mg, we obtain that ( (i) ) = g((j (i) )) = f((j (i) )). Hence, i ( 2(2j R i ) = ( (i) ) ). Case 3. j R < i. i In this case, j i 2j R i < j R, so ( 2( ) ) = ( i ) = g((2j R i)) = ( 2j R i 2j ) = ( R i 2( ) ) = (by the rst two cases) ( (2j R i) ) = ( (i) ). This ends Case 3 and competes the proof of Caim 9. It is easy to chec that, if < i m 1, then ( 3 4 + 1 4 ( i ( +1) m )) = ( m+i ( +1) ). Therefore, we can appy Theorem 6 to obtain a jump mapping such that =. Let : [; 1]! [; 1] be the PL mapping given by (t) = ( t 2 ). = and () =. Then We aso can appy Theorem 6 to, such that =. and and obtain a jump mapping 21

By Theorem 4, there exist a jump mapping and a PL mapping such that (1) = 1 and =. By Theorem 3, there exist jump mappings and such that =. By Theorem 5, there exist a PL mapping z and a jump mapping such that z() = and = z. By Theorem 3, there exist jump mappings, such that =. Observe that = = = and = = z = z. Hence = and = z. Notice aso that () = ( )() = ( )() = ( )(), (1) = ()(1) = ()(1) = ()(1) and () = ( )() = ( z )() = ( )(). Let = r 1 < r 2 < : : : < r N = 1 be a partition of [; 1] such that, for each i 2 f2; : : : ; Ng and each 2 f ; ; ; z g, we have j(r i ) (r i 1 )j < 1 2( ). Given a nonempty cosed set B R and a point x 2 R, choose the owest (in the order of the rea ine) point (x; B) 2 B such that jx (x; B)j = minfjx yj : y 2 Bg. Let : f1; 2; : : : ; 2Ng! [; 1] be given by (i) = ( 1 (((ri )); f ; ; : : : ; g), if i 2 f1; 2; : : : ; Ng, 1 ((((r 2N i+1 ))); f ; ; : : : ; g), if i 2 fn + 1; : : : ; 2Ng. Let : f1; 2; : : : ; Ng [ f2n + 1; : : : ; 3Ng! [; 1] be given by (i) = ( 1 (((ri )); f ; ; : : : ; g), if i 2 f1; 2; : : : ; Ng, 1 ((z((r i 2N ))); f ; ; : : : ; g), if i 2 f2n + 1; : : : ; 3Ng. Let L = f1; : : : ; 4Ng and et F; G : L! X be the functions de ned by: 8 F (i) = (), if (i) =, 2 f; 1; : : : ; j R g, i 2 f1; : : : ; Ng, >< (2j R ), if (i) =, 2 fj R + 1; : : : ; g, i 2 f1; : : : ; Ng, (j R ), if (i) = >:, 2 f; : : : ; g, i 2 fn + 1; : : : ; 2Ng, F (i 2N), if i 2 f2n + 1; : : : ; 4Ng, 22

and 8 G(i) = (), if (i) =, 2 f; 1; : : : ; j R g, i 2 f1; : : : ; Ng, >< (2j R ), if (i) =, 2 fj R + 1; : : : ; g, i 2 f1; : : : ; Ng, G(2N i + 1), if i 2 fn + 1; : : : ; 2Ng, (j R ), if (i) = >:, 2 f; : : : ; g, i 2 f2n + 1; : : : ; 3Ng, G(6N i + 1), if i 2 f3n + 1; : : : ; 4Ng. In the foowing caim we resume some easy to chec equaities. Caim 1. F (1) = u = G(1); F (N) = (j i ); G(N) = (j i ); F (N + 1) = (j i ); G(N + 1) = (j i ); F (2N) = (j R ), where (2N) = = ((); f ; : : : ; g); G(2N) = u ; F (2N + 1) = u, G(2N + 1) = (j R ); F (3N) = (j i ); G(3N) = (j R 1 ), where 1 = (3N) = ((z((1))); f ; : : : ; g); F (3N + 1) = (j i ); G(3N + 1) = G(3N); F (4N) = (j R ) and G(4N) = (j R ). Let h : L! X be given by h(i) = (F (i); G(i)). From Caim 1, the foowing caim is immediate. Caim 11. h(1) = u ; h(n) = h(n + 1); h(2n) = h(2n + 1); h(3n) = h(3n + 1) and h(4n) = (j R ), for some 2 f; 1; : : : ; g. Caim 12. d(f (i); F (i + 1)) < 3 for each i 2 L fn; 2N; 3N; 4Ng. To prove Caim 12, rst we consider the case that i 2 f1; : : : ; N 1g. Let (i) = 1 and (i + 1) = 2. By the choice of r i and r i+1, 1 j((r i ) ((r i+1 )j < 2( ). This impies that j 1 2 j 1. Anaizing the possibiities for 1 and 2 ( 1 ; 2 j R, 1 j R < 2, 2 j R < 1 and j R 1, 2 ) it can be seen that F (i) = (i 1 ) and F (i + 1) = (i 2 ) for some i 1 ; i 2 2 f1; : : : ; j R g such that ji 1 i 2 j 1. By the choice of (i 1 ) and (i 2 ), there exists T 2 W such that (i 1 ); (i 2 ) 2 T. Therefore, d(f (i); F (i+1)) < 3. The case i 2 fn + 1; : : : ; 2N 1g is simiar. The case i 2 f2n + 1; : : : ; 3N 1g [ f3n + 1; : : : ; 4N 1g foows from the previous ones and the de nition of F. This competes the proof of Caim 12. 23