A continuum is said to be retractable onto near trees (abbreviating RNT) provided that for each > 0 there exists > 0 such that if T is a tree in X and H(T,X) < , then there exists an -retraction r of X onto T. In this paper it is proved that an arcwise connected continuum with the property RNT is a dendroid. We focus our study on dendroids. Dendrites are characterized as: (a) smooth dendroids with property RNT, (b) dendroids with the property RNT hereditarily. Finally, an example of a nonlocally connected dendroid with property RNT is shown

Veronica Martinez-de-la-Vega

English

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GLASNIK MATEMATICKIVol. 37(57)(2002), 375 { 382THE PROPERTY RNT ON DENDROIDSVeronica Martnez-de-la-VegaUniversidad Nacional Autonoma de Mexico, MexicoAbstract. A continuum is said to be retractable onto near trees(abbreviating RNT ) provided that for each " > 0 there exists  > 0 suchthat if T is a tree in X and H(T; X) < , then there exists an "-retraction rof X onto T . In this paper it is proved that an arcwise connected continuumwith the property RNT is a dendroid. We focus our study on dendroids.Dendrites are characterized as: (a) smooth dendroids with property RNT ,(b) dendroids with the property RNT hereditarily. Finally, an example ofa nonlocally connected dendroid with property RNT is shown.1. IntroductionA continuum is a compact, connected metric space. A property of a con-tinuum X is said to be hereditary provided that each subcontinuum of X hasthe property. In particular, a continuum X is said to be hereditarily unicoher-ent provided that the intersection of any two subcontinua of X is connected.A dendroid is an arcwise connected hereditarily unicoherent continuum. Fora dendroid X and points a; b 2 X , let ab denote the only arc in X joininga and b if a 6= b or ab = fag if a = b. A dendrite is dened as a locallyconnected dendroid.A nite graph is a nite union of arcs intersecting only at their end points.A tree is a nite graph that does not contain simple closed curves. Let X bea continuum with a metric d. A retraction from X onto a subcontinuum Yof X is dened as a mapping r : X ! Y such that r(y) = y for every y 2 Y .An "-retraction r : X ! Y is a retraction such that every x 2 X satisesd(x; r(x)) < ".Let C(X) denote the hyperspace of subcontinua of X with the Hausdormetric H .2000 Mathematics Subject Classication. 54F15, 54F50, 54F65, 54C15.Key words and phrases. Continuum, dendrite, dendroid, retraction, tree.375376 V. MARTINEZ-DE-LA-VEGAGiven a continuum X with a metric d, a point p in X , a closed subset Aof X and a positive number ", let B"(p) = fq 2 X : d(p; q) < "g, d(p;A) =minfd(p; a) : a 2 Ag, N("; A) = fq 2 X : there exists a point a 2 A such thatd(q; a) < "g and clX(B) denote the closure of the subset B of X .In the early sixties in Wroc law Higher Topology Seminar of the PolishAcademy of Sciencies, Knaster saw dendroids as those continua that can beretracted onto their subdendrites or even onto their subtrees under smallretractions ("-retractions). The contemporary denition of dendroids (theone given above) was formulated in a more convenient way. It is still an openproblem if for every dendroid X and every positive ", there exist a subtree Tof X and an "-retraction from X onto T . Some partial positive answers tothis question can be found in [2, 3, 4].As we shall see later (Theorem 3.1) each dendrite X has the property thatfor each " > 0 and for each subtree T of X which is close enough to X , thereexists an "-retraction from X onto T . This property motivates the followingdenition.Definition 1.1. A continuum X has the property RNT provided thatfor every " > 0 there exists a  > 0 such that if T is a tree contained in Xand H(T;X) < , then there exists an "-retraction r : X ! T .In this paper we prove the following results.(1) An arcwise connected continuum with the property RNT is a dendroid(Theorem 2.5).(2) Dendrites have the property RNT (Theorem 3.1).(3) A dendroid has property RNT hereditarily if and only if it is a dendrite(Theorem 4.1).(4) A dendroid is smooth and has the property RNT if and only if it is adendrite (Theorem 5.1).(5) There exists a nonlocally connected dendroid with the property RNT .2. The property RNT in arcwise connected continuaThe study of the property RNT was motivated by a question on den-droids. Arcwise connected continua have the property of containing treesclose to X . Therefore it makes sense to study property RNT on these con-tinua. The aim of this section is to prove that arcwise connected continuawith the property RNT are dendroids.Lemma 2.1. Let X be an arcwise connected continuum, T a tree containedin X and F a nite subset of X. Then there exists a tree S such that T [F S  X.Proof. We prove this lemma by induction on the number of elementsof F . First, suppose that F has one element p. We may assume that p =2 T .Since X is arcwise connected, there exists an arc A  X , with end points pTHE PROPERTY RNT ON DENDROIDS 377and t such that T \ A = ftg. Then T1 = T [ A is the required tree. Supposethat the conclusion is true for nite subsets of X with exactly n elements, andtake a subset F of X with n+ 1 elements. Let x be an element of F . By theinduction hypothesis, there exists a tree R such that (F  fxg)[T  R  X .Applying the rst step of the induction to R and the point x, we obtain thedesired tree S. This completes the induction and the proof of the lemma.Lemma 2.2. Let X be an arcwise connected continuum and T a tree con-tained in X. Then for every  > 0 there exists a tree T  X such that T  Tand H(T; X) < .Proof. By compactness of X , there exist a nite number of pointsx1; x2; : : : ; xn of X such that X = B(x1)[B(x2)[   [B(xn). By Lemma2.1, there exists a tree T  X such that T  T and fx1; x2; : : : ; xng  T:Clearly, H(T; X) < .Definition 2.3. A lock L in a continuum X is a subcontinuum of X ofthe form L = A [ B, where A;B 2 C(X), A is an arc with end points p andq, and A \ B = fp; qg.Lemma 2.4. If an arcwise connected continuum is not hereditarily unico-herent, then it contains a lock.Proof. Suppose that X is not hereditarily unicoherent. Then there existM;N 2 C(X) such that M \ N is not connected. Take points p0 2 C1 andq0 2 C2, where C1 and C2 are dierent components of M \ N . Since X isarcwise connected, there exists an arc A0  X with end points p0 and q0.Since A0 * M \N , we may assume that A0 is not contained in M . Let J bea component of A0 nM . So A = clX(J) is an arc that intersects M exactly atits end points. Therefore L = M [ A is a lock in X .Theorem 2.5. Each arcwise connected continuum with the property RNTis a dendroid.Proof. Let X be an arcwise connected continuum with the propertyRNT . We only need to prove that X is hereditarily unicoherent.Suppose that X is not hereditarily unicoherent. By Lemma 2.4 thereexists a lock L in X . Let L = A [ B, where A;B 2 C(X), A is an arc withend points p; q and A\B = fp; qg. Fix a point s 2 AnB. Then " = d(s;B) isa positive number. Take any positive number . By Lemma 2.2 there existsa tree T such that A  T  X and H(X;T) < . Let r : X ! T be anyretraction. Then r(B) is a subcontinuum of the tree T that contains p and q.This implies that A  r(B). Therefore there exists a point b 2 B such thatr(b) = s. Hence d(b; r(b))  ". We have proved that r is not an "-retraction.Thus, X does not have property RNT . This ends the proof of the theorem.378 V. MARTINEZ-DE-LA-VEGA3. The property RNT on dendritesWe know that arcwise connected continua with the property RNT aredendroids. It will be proved that among dendroids, dendrites always have theproperty RNT .Theorem 3.1. Each dendrite has property RNT .Proof. Given a subcontinuum Z of a dendrite X , let r : X ! Z be therst point map, that is, given a point x 2 X , let r(x) be the unique point inZ such that r(x) is a point of any arc in X from x to any point of Z. Theexistence and continuity of r is guaranteed by [7, Lemma 10.24, p. 175]. Notethat r is a retraction. Let " > 0. By compactness of X , there exists a nitecover U of X consisting of open arcwise connected subsets of X such thateach element of U has diameter less than ". Let  be a Lebesgue number forU . Take any subtree T of X such that H(T;X) < . Let r : X ! T be therst point map corresponding to T . Then for each point x 2 X , there existsa point t 2 T such that d(x; t) < . Thus there exists an element U 2 U suchthat x; t 2 U , and therefore xt  U . By denition of r we have r(x) 2 xt  U .So, d(x; r(x)) < ". Therefore, X has the property RNT .4. The hereditary property RNTOn Section 6 of the paper we show an example of a nonlocally connecteddendroid with the property RNT . In this section we characterize dendritesas dendroids which have the property RNT hereditarily.Theorem 4.1. A dendroid has the property RNT hereditarily if and onlyif it is a dendrite.Proof. Suppose that dendroid X is a dendrite. Since every subcontin-uum of X is a dendrite, the suciency follows from Theorem 3.1.In order to prove the necessity, suppose that X has the property RNThereditarily and X is not locally connected. By Theorem 13 of [6], X containsa semi-comb or a semi-broom ([6, Denition 11]).It follows then that X contains:(a) two points a; b 2 X ,(b) two points p 6= q in A = ab,(c) a sequence of points fpng1n=1 in X  A, and(d) a sequence of points fqng1n=1 in A (for the case of semi-broom ([6,Denition 11]), take qn = q for each n) such that:(i) pn ! p, qn ! q,(ii) pnqn \ A = fqng for each n, and(iii) p1q1   fq1g, p2q2   fq2g; : : : are pairwise disjoint.Taking the natural order for ab, where a < b, we may assume that a q < p  b. Since qn ! q, we may also assume that there exists a pointTHE PROPERTY RNT ON DENDROIDS 379c 2 ab such that a  qn  c < p for each n  1. Fix a point e 2 cp  fc; pg.Consider the subcontinuum Y = ae [ clX(Sfpnqn : n  1g) of X . Let " > 0be such that d(p; ae) > " and " < minfd(x; y) : x 2 ac and y 2 epg. Since weare assuming that X has the property RNT hereditarily, Y has the propertyRNT . Let  > 0 be as in the denition of the property RNT for Y and". Take M  1 such that the tree T = ae [ (Sfpnqn : n  Mg) has theproperty that H(T; Y ) < . Since pn 2 Y for each n, p 2 Y . Let r : Y ! Tbe an "-retraction. Since d(p; r(p)) < ", r(p) =2 ae. Therefore, r(p) 2 pmqmfor some m  M . Since e 2 T , we have that r(e) = e. It follows that r(ep)contains qm. Thus there exists a point s 2 ep such that r(s) = qm. Therefored(s; r(s)) = d(s; qm) > ", since s 2 ep and qm 2 ac. This contradicts that r isan "-retraction and proves that X is locally connected.5. The property RNT in smooth dendroidsRecall that a dendroid X is said to be smooth provided that there exists apoint v 2 X such that for each sequence fxng1n=1 in X converging to a pointx 2 X , the sequence of arcs vxn converges to vx (in C(X)) ([1, p. 298]). Thepoint v is called an initial point of X .Theorem 5.1. A dendroid is smooth and has the property RNT if andonly if it is a dendrite.Proof. By [1, Corollary 4, p. 298] dendrites are smooth dendroids. Thenthe suciency follows from Theorem 3.1.In order to prove the necessity suppose that X is a nonlocally connectedsmooth dendroid with an initial point v. By [1, Theorem 1], X is locallyconnected at v. We are going to show that X does not have the propertyRNT . Since X is not locally connected, there exist (see [6, Lemma 12]):(1) three open subsets V , W and U of X ,(2) two dierent points p; q 2 clX(V ),(3) two sequences fpng1n=1 and fqng1n=1 of points of clX(V ),(4) a sequence of distinct components C0; C1; C2; : : : of U , such that(a) clX(V ) W  clX(W )  U ,(b) pq  clX(V ) \ C0,(c) pn ! p, qn ! q,(d) pnqn  clX(V ) \ Cn for each n  1,(e) v =2 U .For each n  1, let En be the component of clX(V ) \ Cn that containspnqn and let Dn be the component of clX(W ) that contains En. Denote bycn the only point in En such that vcn \ En = fcng. Taking a subsequence ifnecesary, we may assume that cn ! c for some c 2 X and En ! K for somecontinuum K of X . Since p 6= q, we may also assume that c 6= q and cn 6= qnfor each n  1. Note that p; q; c 2 K  clX(V ). Let E0 be the component380 V. MARTINEZ-DE-LA-VEGAof clX(V ) that contains p. Then c 2 E0 and E0  C0. Given n  1, by thechoice of cn, we infer that vcn  vqn. By the smoothness of X at v, vcn ! vcand vqn ! vq. Thus vc  vq. Since c 6= q, q =2 vc.Take " > 0 such that " < minfd(x; y) : x 2 clX(V ) and y =2 Wg and3" < d(q; vc). We show that X does not satisfy the denition of propertyRNT for this ". Given  > 0, by Lemma 2.2 there exists a tree T  X suchthat H(T;X) <  and v 2 T .We claim that T intersects only nitely many sets Dn. Let T1; : : : ; Tmbe subtrees of T such that each Ti has diameter less than minfd(x; y) : x 2clX(W ) and y =2 Ug and T = T1 [    [ Tm. If Ti intersects Dn, then Ti  Uand therefore Ti  Cn. So, each Ti can only intersect one set of the formDn. Thus T intersects at most m of the sets Dn. Let N  1 be such thatT \Dn = ; for each n  N .Since v is an initial point of X , vcn ! vc. Thus we may assume thatvcn  N("; vc) and d(q; qn) < " for each n  N .Note that cNqN  EN \ DN  clX(V ) \ DN and cNqN \ T = ;. LetT0 = T [ vcN . Then T0 is a tree, qN =2 T0 and H(T0; X) < .Suppose that there is an "-retraction r : X ! T0. If r(qN ) 2 vcN , bythe choice of N , there is a point y 2 vc such that d(r(qN ); y) < ". Thend(q; y)  d(q; qN ) + d(qN ; r(qN )) + d(r(qN ); y) < 3". This contradicts thechoice of " and proves that r(qN ) =2 vcN . Since r(qN ) 2 B("; qN )\T W \T ,we have that there exists a component D of clX(W ) such that r(qN ) 2 Dand D 6= DN . Since r is a retraction, r(cN ) = cN . Then the set r(cNqN )intersects two dierent components of clX(W ). Thus there exists z 2 cNqNsuch that r(z) =2 W . Since z 2 clX (V ), by the choice of ", d(z; r(z))  ".This contradiction proves that there is not any "-retraction from X onto T0.Therefore X does not have the property RNT . This completes the proof ofthe theorem.6. A nonlocally connected dendroid with the property RNTIn [5] A. Illanes constructed an example of a nonlocally connected den-droid X with the property that, for each subcontinuum A of X , there existsa retraction r : X ! A. The dendroid is constructed as a subset of theEuclidean plane R2. In this section we modify Illanes' dendroid to obtain anonlocally connected dendroid Y with the property RNT .Let us recall the construction of X .Given two points p and q in the Euclidean plane R2, denote by hp; qithe convex segment joining them, if p 6= q , and put hp; qi = fpg, if p = q.For a point p = (x; y) 2 R2, dene p0 = ( x; y). Given a subset B inR2, dene B0 = fp0 2 R2 : p 2 Bg. The origin in R2 is denoted by .Let 1; 2 : R2 ! R be the respective projections on the rst and secondcoordinates. We will dene, inductively, a sequence A0; : : : ; An; : : : of subsetsTHE PROPERTY RNT ON DENDROIDS 381of R2 such that, for each n > 0, An is a polygonal line joining  to a pointan = (un; vn). Let a0 = .Let A0 = fg, A1 =; (1;  14 ). Suppose that A0; : : : ; An have beendened and n > 1. Dene bn = an + a0n 1 and An+1 = An [ (an + A0n 1) [(bn +An). See Figure 1 in [5, p. 803].For each n > 1, dene Cn = (0;12n 2 ) + f( xn ;  y2nvn ) 2 R2 : (x; y) 2 Angand Bn = Cn[(0; 12n 2 );[h; (1; 0)i. Given points p and q in Bn, hhp; qiiwill denote the subarc in Bn joining p and q, if p 6= q and hhp; qii = fpg, ifp = q.Finally, dene X =SfBn : n  2g = h; (0; 1)i [ h; (1; 0)i [ (SfCn :n  2g). Clearly, X is a nonlocally connected dendroid. See Figure 4 in [5,p. 806].Dene Y = X [ h(1; 0); (2; 0)i. We show that Y has the property RNT .In order to do this, we will use the following assertion proved in [5, p. 801].Assertion 6.1. For each q 2 Cn, there exists a retraction (which dependson q) n;q : Bn ! hhq; (1; 0)ii, such that j1(p)  1(n;q(p))j  3n .Let " > 0. Let N  1 be such that 3N + 12N 2 < ". For each 2  n  N ,let cn be the end point of the arc Cn which is dierent from (0;12n 2 ) andlet zn 2 Cn n fcng be such that the diameter of the subarc zncn is less than". Let n > 0 be such that Bn(cn)  zncn. Let 0 > 0 be such thatB0((2; 0)) \X = ; and let  = minf"; 0; 1; : : : ; Ng.Take any tree T of Y such that H(T; Y ) < . Notice that, for each n  2,T \ Cn 6= ;, so there exists tn 2 T such that Cn \ T =(0; 12n 2 ); tnandif n  N , then T \ Bn(cn) 6= ;, so tn 2 zncn. Let t0 2 T be such thatT \ hh; (2; 0)ii = hh; t0ii. Then t0 2 B0((2; 0)).Dene r : Y ! T byr(y) =8>><>>:y, if y 2 T ,tn, if y 2 Cn n T and 2  n  N ,n;tn(y), if y 2 Cn n T and N < n ,t0, if y 2 ht0; (2; 0)i ,where n;tn is chosen as in Assertion 6.1.The continuity of r follows from Assertion 6.1. Therefore r is a retraction.In order to show that r is an "-retraction, take a point y 2 Y n T . Ify 2 Cn n T and 2  n  N , d(y; r(y)) = d(y; tn) < ", since tn; y 2 zncn andthe diameter of zncn is less than ". The case that y 2 ht0; (2; 0)i is similar.Finally, if y 2 Cn n T and N < n, since n;tn(y) 2 Bn, j2(y)   2(r(y))j =j2(y) 2(n;tn(y))j  12n 2 and, by the Assertion 6.1, we know that j1(y) 1(r(y))j = j1(y)   1(n;tn(y))j  3n . This implies that d(y; r(y)) < ".Therefore r is an "-retraction. This completes the proof that Y has theproperty RNT .382 V. MARTINEZ-DE-LA-VEGAAcknowledgements.The author wishes to thank J. J. Charatonik and A. Illanes for fruitfuldiscussions on the topic of this paper.References[1] J. J. Charatonik and C. A. Eberhart, On smooth dendroids, Fund. Math. 67 (1970),297{322.[2] C. A. Eberhart and J. B. Fugate, Approximating continua from within, Fund. Math.72 (1971), 223{231.[3] J. B. Fugate, Retracting fans onto nite fans, Fund. Math. 71 (1971), 113{125.[4] J. B. Fugate, Small retractions of smooth dendroids onto trees, Fund. Math. 71(1971), 255{262.[5] A. Illanes, A retractible nonlocally connected dendroid, Comment. Math. Univ. Car-olinae 39 (1998), 797{807.[6] A. Illanes and V. Martnez-de-la-Vega, Product topology the hyperspace of subcon-tinua, Topology and Its Applications 105 (2000), 305{317.[7] S. B. Nadler, Jr., Continuum Theory, An introduction, Monographs and Textbooksin Pure and Applied Mathematics 158, Marcel Dekker, Inc., New York, Basel & HongKong, 1992.Instituto de Matematicas, UNAMCircuito ExteriorCd. UniversitariaMexico, 04510, D.F.E-mail address: vmvm@matem.unam.mxReceived : 12.10.2000.Revised : 07.11.2001.

The property RNT on dendroids

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GLASNIK MATEMATIČKIVol. 37(57)(2002), 375 – 382THE PROPERTY RNT ON DENDROIDSVerónica Mart́ınez-de-la-VegaUniversidad Nacional Autónoma de México, MéxicoAbstract. A continuum is said to be retractable onto near trees(abbreviating RNT ) provided that for each ε > 0 there exists δ > 0 suchthat if T is a tree in X and H(T, X) < δ, then there exists an ε-retraction rof X onto T . In this paper it is proved that an arcwise connected continuumwith the property RNT is a dendroid. We focus our study on dendroids.Dendrites are characterized as: (a) smooth dendroids with property RNT ,(b) dendroids with the property RNT hereditarily. Finally, an example ofa nonlocally connected dendroid with property RNT is shown.1. IntroductionA continuum is a compact, connected metric space. A property of a con-tinuum X is said to be hereditary provided that each subcontinuum of X hasthe property. In particular, a continuum X is said to be hereditarily unicoher-ent provided that the intersection of any two subcontinua of X is connected.A dendroid is an arcwise connected hereditarily unicoherent continuum. Fora dendroid X and points a, b ∈ X , let ab denote the only arc in X joininga and b if a 6= b or ab = {a} if a = b. A dendrite is defined as a locallyconnected dendroid.A finite graph is a finite union of arcs intersecting only at their end points.A tree is a finite graph that does not contain simple closed curves. Let X bea continuum with a metric d. A retraction from X onto a subcontinuum Yof X is defined as a mapping r : X → Y such that r(y) = y for every y ∈ Y .An ε-retraction r : X → Y is a retraction such that every x ∈ X satisfiesd(x, r(x)) < ε.Let C(X) denote the hyperspace of subcontinua of X with the Hausdorffmetric H .2000 Mathematics Subject Classification. 54F15, 54F50, 54F65, 54C15.Key words and phrases. Continuum, dendrite, dendroid, retraction, tree.375376 V. MARTÍNEZ-DE-LA-VEGAGiven a continuum X with a metric d, a point p in X , a closed subset Aof X and a positive number ε, let Bε(p) = {q ∈ X : d(p, q) < ε}, d(p,A) =min{d(p, a) : a ∈ A}, N(ε, A) = {q ∈ X : there exists a point a ∈ A such thatd(q, a) < ε} and clX(B) denote the closure of the subset B of X .In the early sixties in Wroc law Higher Topology Seminar of the PolishAcademy of Sciencies, Knaster saw dendroids as those continua that can beretracted onto their subdendrites or even onto their subtrees under smallretractions (ε-retractions). The contemporary definition of dendroids (theone given above) was formulated in a more convenient way. It is still an openproblem if for every dendroid X and every positive ε, there exist a subtree Tof X and an ε-retraction from X onto T . Some partial positive answers tothis question can be found in [2, 3, 4].As we shall see later (Theorem 3.1) each dendrite X has the property thatfor each ε > 0 and for each subtree T of X which is close enough to X , thereexists an ε-retraction from X onto T . This property motivates the followingdefinition.Definition 1.1. A continuum X has the property RNT provided thatfor every ε > 0 there exists a δ > 0 such that if T is a tree contained in Xand H(T,X) < δ, then there exists an ε-retraction r : X → T .In this paper we prove the following results.(1) An arcwise connected continuum with the property RNT is a dendroid(Theorem 2.5).(2) Dendrites have the property RNT (Theorem 3.1).(3) A dendroid has property RNT hereditarily if and only if it is a dendrite(Theorem 4.1).(4) A dendroid is smooth and has the property RNT if and only if it is adendrite (Theorem 5.1).(5) There exists a nonlocally connected dendroid with the property RNT .2. The property RNT in arcwise connected continuaThe study of the property RNT was motivated by a question on den-droids. Arcwise connected continua have the property of containing treesclose to X . Therefore it makes sense to study property RNT on these con-tinua. The aim of this section is to prove that arcwise connected continuawith the property RNT are dendroids.Lemma 2.1. Let X be an arcwise connected continuum, T a tree containedin X and F a finite subset of X. Then there exists a tree S such that T ∪F ⊂S ⊂ X.Proof. We prove this lemma by induction on the number of elementsof F . First, suppose that F has one element p. We may assume that p /∈ T .Since X is arcwise connected, there exists an arc A ⊂ X , with end points pTHE PROPERTY RNT ON DENDROIDS 377and t such that T ∩ A = {t}. Then T1 = T ∪ A is the required tree. Supposethat the conclusion is true for finite subsets of X with exactly n elements, andtake a subset F of X with n+ 1 elements. Let x be an element of F . By theinduction hypothesis, there exists a tree R such that (F −{x})∪T ⊂ R ⊂ X .Applying the first step of the induction to R and the point x, we obtain thedesired tree S. This completes the induction and the proof of the lemma.Lemma 2.2. Let X be an arcwise connected continuum and T a tree con-tained in X. Then for every δ > 0 there exists a tree Tδ ⊂ X such that T ⊂ Tδand H(Tδ, X) < δ.Proof. By compactness of X , there exist a finite number of pointsx1, x2, . . . , xn of X such that X = Bδ(x1)∪Bδ(x2)∪ · · ·∪Bδ(xn). By Lemma2.1, there exists a tree Tδ ⊂ X such that T ⊂ Tδ and {x1, x2, . . . , xn} ⊂ Tδ.Clearly, H(Tδ, X) < δ.Definition 2.3. A lock L in a continuum X is a subcontinuum of X ofthe form L = A ∪ B, where A,B ∈ C(X), A is an arc with end points p andq, and A ∩ B = {p, q}.Lemma 2.4. If an arcwise connected continuum is not hereditarily unico-herent, then it contains a lock.Proof. Suppose that X is not hereditarily unicoherent. Then there existM,N ∈ C(X) such that M ∩ N is not connected. Take points p′ ∈ C1 andq′ ∈ C2, where C1 and C2 are different components of M ∩ N . Since X isarcwise connected, there exists an arc A′ ⊂ X with end points p′ and q′.Since A′ * M ∩N , we may assume that A′ is not contained in M . Let J bea component of A′ \M . So A = clX(J) is an arc that intersects M exactly atits end points. Therefore L = M ∪ A is a lock in X .Theorem 2.5. Each arcwise connected continuum with the property RNTis a dendroid.Proof. Let X be an arcwise connected continuum with the propertyRNT . We only need to prove that X is hereditarily unicoherent.Suppose that X is not hereditarily unicoherent. By Lemma 2.4 thereexists a lock L in X . Let L = A ∪ B, where A,B ∈ C(X), A is an arc withend points p, q and A∩B = {p, q}. Fix a point s ∈ A\B. Then ε = d(s,B) isa positive number. Take any positive number δ. By Lemma 2.2 there existsa tree Tδ such that A ⊂ Tδ ⊂ X and H(X,Tδ) < δ. Let r : X → Tδ be anyretraction. Then r(B) is a subcontinuum of the tree Tδ that contains p and q.This implies that A ⊂ r(B). Therefore there exists a point b ∈ B such thatr(b) = s. Hence d(b, r(b)) ≥ ε. We have proved that r is not an ε-retraction.Thus, X does not have property RNT . This ends the proof of the theorem.378 V. MARTÍNEZ-DE-LA-VEGA3. The property RNT on dendritesWe know that arcwise connected continua with the property RNT aredendroids. It will be proved that among dendroids, dendrites always have theproperty RNT .Theorem 3.1. Each dendrite has property RNT .Proof. Given a subcontinuum Z of a dendrite X , let r : X → Z be thefirst point map, that is, given a point x ∈ X , let r(x) be the unique point inZ such that r(x) is a point of any arc in X from x to any point of Z. Theexistence and continuity of r is guaranteed by [7, Lemma 10.24, p. 175]. Notethat r is a retraction. Let ε > 0. By compactness of X , there exists a finitecover U of X consisting of open arcwise connected subsets of X such thateach element of U has diameter less than ε. Let δ be a Lebesgue number forU . Take any subtree T of X such that H(T,X) < δ. Let r : X → T be thefirst point map corresponding to T . Then for each point x ∈ X , there existsa point t ∈ T such that d(x, t) < δ. Thus there exists an element U ∈ U suchthat x, t ∈ U , and therefore xt ⊂ U . By definition of r we have r(x) ∈ xt ⊂ U .So, d(x, r(x)) < ε. Therefore, X has the property RNT .4. The hereditary property RNTOn Section 6 of the paper we show an example of a nonlocally connecteddendroid with the property RNT . In this section we characterize dendritesas dendroids which have the property RNT hereditarily.Theorem 4.1. A dendroid has the property RNT hereditarily if and onlyif it is a dendrite.Proof. Suppose that dendroid X is a dendrite. Since every subcontin-uum of X is a dendrite, the sufficiency follows from Theorem 3.1.In order to prove the necessity, suppose that X has the property RNThereditarily and X is not locally connected. By Theorem 13 of [6], X containsa semi-comb or a semi-broom ([6, Definition 11]).It follows then that X contains:(a) two points a, b ∈ X ,(b) two points p 6= q in A = ab,(c) a sequence of points {pn}∞n=1 in X −A, and(d) a sequence of points {qn}∞n=1 in A (for the case of semi-broom ([6,Definition 11]), take qn = q for each n) such that:(i) pn → p, qn → q,(ii) pnqn ∩ A = {qn} for each n, and(iii) p1q1 − {q1}, p2q2 − {q2}, . . . are pairwise disjoint.Taking the natural order for ab, where a < b, we may assume that a ≤q < p ≤ b. Since qn → q, we may also assume that there exists a pointTHE PROPERTY RNT ON DENDROIDS 379c ∈ ab such that a ≤ qn ≤ c < p for each n ≥ 1. Fix a point e ∈ cp− {c, p}.Consider the subcontinuum Y = ae ∪ clX(⋃{pnqn : n ≥ 1}) of X . Let ε > 0be such that d(p, ae) > ε and ε < min{d(x, y) : x ∈ ac and y ∈ ep}. Since weare assuming that X has the property RNT hereditarily, Y has the propertyRNT . Let δ > 0 be as in the definition of the property RNT for Y andε. Take M ≥ 1 such that the tree T = ae ∪ (⋃{pnqn : n ≤ M}) has theproperty that H(T, Y ) < δ. Since pn ∈ Y for each n, p ∈ Y . Let r : Y → Tbe an ε-retraction. Since d(p, r(p)) < ε, r(p) /∈ ae. Therefore, r(p) ∈ pmqmfor some m ≤ M . Since e ∈ T , we have that r(e) = e. It follows that r(ep)contains qm. Thus there exists a point s ∈ ep such that r(s) = qm. Therefored(s, r(s)) = d(s, qm) > ε, since s ∈ ep and qm ∈ ac. This contradicts that r isan ε-retraction and proves that X is locally connected.5. The property RNT in smooth dendroidsRecall that a dendroid X is said to be smooth provided that there exists apoint v ∈ X such that for each sequence {xn}∞n=1 in X converging to a pointx ∈ X , the sequence of arcs vxn converges to vx (in C(X)) ([1, p. 298]). Thepoint v is called an initial point of X .Theorem 5.1. A dendroid is smooth and has the property RNT if andonly if it is a dendrite.Proof. By [1, Corollary 4, p. 298] dendrites are smooth dendroids. Thenthe sufficiency follows from Theorem 3.1.In order to prove the necessity suppose that X is a nonlocally connectedsmooth dendroid with an initial point v. By [1, Theorem 1], X is locallyconnected at v. We are going to show that X does not have the propertyRNT . Since X is not locally connected, there exist (see [6, Lemma 12]):(1) three open subsets V , W and U of X ,(2) two different points p, q ∈ clX(V ),(3) two sequences {pn}∞n=1 and {qn}∞n=1 of points of clX(V ),(4) a sequence of distinct components C0, C1, C2, . . . of U , such that(a) clX(V ) ⊂W ⊂ clX(W ) ⊂ U ,(b) pq ⊂ clX(V ) ∩ C0,(c) pn → p, qn → q,(d) pnqn ⊂ clX(V ) ∩ Cn for each n ≥ 1,(e) v /∈ U .For each n ≥ 1, let En be the component of clX(V ) ∩ Cn that containspnqn and let Dn be the component of clX(W ) that contains En. Denote bycn the only point in En such that vcn ∩ En = {cn}. Taking a subsequence ifnecesary, we may assume that cn → c for some c ∈ X and En → K for somecontinuum K of X . Since p 6= q, we may also assume that c 6= q and cn 6= qnfor each n ≥ 1. Note that p, q, c ∈ K ⊂ clX(V ). Let E0 be the component380 V. MARTÍNEZ-DE-LA-VEGAof clX(V ) that contains p. Then c ∈ E0 and E0 ⊂ C0. Given n ≥ 1, by thechoice of cn, we infer that vcn ⊂ vqn. By the smoothness of X at v, vcn → vcand vqn → vq. Thus vc ⊂ vq. Since c 6= q, q /∈ vc.Take ε > 0 such that ε < min{d(x, y) : x ∈ clX(V ) and y /∈ W} and3ε < d(q, vc). We show that X does not satisfy the definition of propertyRNT for this ε. Given δ > 0, by Lemma 2.2 there exists a tree T ⊂ X suchthat H(T,X) < δ and v ∈ T .We claim that T intersects only finitely many sets Dn. Let T1, . . . , Tmbe subtrees of T such that each Ti has diameter less than min{d(x, y) : x ∈clX(W ) and y /∈ U} and T = T1 ∪ · · · ∪ Tm. If Ti intersects Dn, then Ti ⊂ Uand therefore Ti ⊂ Cn. So, each Ti can only intersect one set of the formDn. Thus T intersects at most m of the sets Dn. Let N ≥ 1 be such thatT ∩Dn = ∅ for each n ≥ N .Since v is an initial point of X , vcn → vc. Thus we may assume thatvcn ⊂ N(ε, vc) and d(q, qn) < ε for each n ≥ N .Note that cNqN ⊂ EN ∩ DN ⊂ clX(V ) ∩ DN and cNqN ∩ T = ∅. LetT0 = T ∪ vcN . Then T0 is a tree, qN /∈ T0 and H(T0, X) < δ.Suppose that there is an ε-retraction r : X → T0. If r(qN ) ∈ vcN , bythe choice of N , there is a point y ∈ vc such that d(r(qN ), y) < ε. Thend(q, y) ≤ d(q, qN ) + d(qN , r(qN )) + d(r(qN ), y) < 3ε. This contradicts thechoice of ε and proves that r(qN ) /∈ vcN . Since r(qN ) ∈ B(ε, qN )∩T ⊂W ∩T ,we have that there exists a component D of clX(W ) such that r(qN ) ∈ Dand D 6= DN . Since r is a retraction, r(cN ) = cN . Then the set r(cNqN )intersects two different components of clX(W ). Thus there exists z ∈ cNqNsuch that r(z) /∈ W . Since z ∈ clX (V ), by the choice of ε, d(z, r(z)) ≥ ε.This contradiction proves that there is not any ε-retraction from X onto T0.Therefore X does not have the property RNT . This completes the proof ofthe theorem.6. A nonlocally connected dendroid with the property RNTIn [5] A. Illanes constructed an example of a nonlocally connected den-droid X with the property that, for each subcontinuum A of X , there existsa retraction r : X → A. The dendroid is constructed as a subset of theEuclidean plane R2. In this section we modify Illanes’ dendroid to obtain anonlocally connected dendroid Y with the property RNT .Let us recall the construction of X .Given two points p and q in the Euclidean plane R2, denote by 〈p, q〉the convex segment joining them, if p 6= q , and put 〈p, q〉 = {p}, if p = q.For a point p = (x, y) ∈ R2, define p′ = (−x, y). Given a subset B inR2, define B′ = {p′ ∈ R2 : p ∈ B}. The origin in R2 is denoted by Θ.Let π1, π2 : R2 → R be the respective projections on the first and secondcoordinates. We will define, inductively, a sequence A0, . . . , An, . . . of subsetsTHE PROPERTY RNT ON DENDROIDS 381of R2 such that, for each n > 0, An is a polygonal line joining Θ to a pointan = (un, vn). Let a0 = Θ.Let A0 = {Θ}, A1 =〈Θ, (1,− 14 )〉. Suppose that A0, . . . , An have beendefined and n > 1. Define bn = an + a′n−1 and An+1 = An ∪ (an + A′n−1) ∪(bn +An). See Figure 1 in [5, p. 803].For each n > 1, define Cn = (0,12n−2 ) + {( xn ,−y2nvn) ∈ R2 : (x, y) ∈ An}and Bn = Cn∪〈(0, 12n−2 ),Θ〉∪〈Θ, (1, 0)〉. Given points p and q in Bn, 〈〈p, q〉〉will denote the subarc in Bn joining p and q, if p 6= q and 〈〈p, q〉〉 = {p}, ifp = q.Finally, define X =⋃{Bn : n ≥ 2} = 〈Θ, (0, 1)〉 ∪ 〈Θ, (1, 0)〉 ∪ (⋃{Cn :n ≥ 2}). Clearly, X is a nonlocally connected dendroid. See Figure 4 in [5,p. 806].Define Y = X ∪ 〈(1, 0), (2, 0)〉. We show that Y has the property RNT .In order to do this, we will use the following assertion proved in [5, p. 801].Assertion 6.1. For each q ∈ Cn, there exists a retraction (which dependson q) φn,q : Bn → 〈〈q, (1, 0)〉〉, such that |π1(p)− π1(φn,q(p))| ≤ 3n .Let ε > 0. Let N ≥ 1 be such that 3N + 12N−2 < ε. For each 2 ≤ n ≤ N ,let cn be the end point of the arc Cn which is different from (0,12n−2 ) andlet zn ∈ Cn \ {cn} be such that the diameter of the subarc zncn is less thanε. Let δn > 0 be such that Bδn(cn) ⊂ zncn. Let δ0 > 0 be such thatBδ0((2, 0)) ∩X = ∅ and let δ = min{ε, δ0, δ1, . . . , δN}.Take any tree T of Y such that H(T, Y ) < δ. Notice that, for each n ≥ 2,T ∩ Cn 6= ∅, so there exists tn ∈ T such that Cn ∩ T =〈〈(0, 12n−2 ), tn〉〉andif n ≤ N , then T ∩ Bδn(cn) 6= ∅, so tn ∈ zncn. Let t0 ∈ T be such thatT ∩ 〈〈Θ, (2, 0)〉〉 = 〈〈Θ, t0〉〉. Then t0 ∈ Bδ0((2, 0)).Define r : Y → T byr(y) =y, if y ∈ T ,tn, if y ∈ Cn \ T and 2 ≤ n ≤ N ,φn,tn(y), if y ∈ Cn \ T and N < n ,t0, if y ∈ 〈t0, (2, 0)〉 ,where φn,tn is chosen as in Assertion 6.1.The continuity of r follows from Assertion 6.1. Therefore r is a retraction.In order to show that r is an ε-retraction, take a point y ∈ Y \ T . Ify ∈ Cn \ T and 2 ≤ n ≤ N , d(y, r(y)) = d(y, tn) < ε, since tn, y ∈ zncn andthe diameter of zncn is less than ε. The case that y ∈ 〈t0, (2, 0)〉 is similar.Finally, if y ∈ Cn \ T and N < n, since φn,tn(y) ∈ Bn, |π2(y) − π2(r(y))| =|π2(y)−π2(φn,tn(y))| ≤ 12n−2 and, by the Assertion 6.1, we know that |π1(y)−π1(r(y))| = |π1(y) − π1(φn,tn(y))| ≤ 3n . This implies that d(y, r(y)) < ε.Therefore r is an ε-retraction. This completes the proof that Y has theproperty RNT .382 V. MARTÍNEZ-DE-LA-VEGAAcknowledgements.The author wishes to thank J. J. Charatonik and A. Illanes for fruitfuldiscussions on the topic of this paper.References[1] J. J. Charatonik and C. A. Eberhart, On smooth dendroids, Fund. Math. 67 (1970),297–322.[2] C. A. Eberhart and J. B. Fugate, Approximating continua from within, Fund. Math.72 (1971), 223–231.[3] J. B. Fugate, Retracting fans onto finite fans, Fund. Math. 71 (1971), 113–125.[4] J. B. Fugate, Small retractions of smooth dendroids onto trees, Fund. Math. 71(1971), 255–262.[5] A. Illanes, A retractible nonlocally connected dendroid, Comment. Math. Univ. Car-olinae 39 (1998), 797–807.[6] A. Illanes and V. Mart́ınez-de-la-Vega, Product topology the hyperspace of subcon-tinua, Topology and Its Applications 105 (2000), 305–317.[7] S. B. Nadler, Jr., Continuum Theory, An introduction, Monographs and Textbooksin Pure and Applied Mathematics 158, Marcel Dekker, Inc., New York, Basel & HongKong, 1992.Instituto de Matemáticas, UNAMCircuito ExteriorCd. UniversitariaMéxico, 04510, D.F.E-mail address: vmvm@matem.unam.mxReceived : 12.10.2000.Revised : 07.11.2001.

The property RNT on dendroids

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