Boundaries in hyperspaces II

Topology and its Applications 188 (2015) 51–63

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Boundaries in hyperspaces II Claudia G. Domínguez-López Instituto de Matemáticas, Universidad Nacional Autónoma de México, Circuito Exterior, Ciudad Universitaria, México, D.F., 04510, Mexico

a r t i c l e

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Article history: Received 16 April 2014 Received in revised form 17 October 2014 Accepted 17 March 2015 Available online xxxx MSC: 54B20 54F15

a b s t r a c t Let C(X) be the hyperspace of subcontinua of a continuum X, and let A be a proper subcontinuum of X. We continue the study of the boundary Bd(C(A)) of C(A) in C(X). We show properties of continua for which these boundaries are 2-cells. Moreover, we define a natural boundary function of hyperspaces, and we characterize the simple closed curve as the unique arcwise connected continuum for which this function is continuous. Results in this paper, as those in [3], extend the study of semi-boundaries in hyperspaces done by A. Illanes in [4]. © 2015 Elsevier B.V. All rights reserved.

Keywords: Boundary Continuum Dendroid Hyperspace Semi-boundary

1. Introduction A continuum is a nonempty compact connected metric space. Given a continuum X we consider the hyperspace C(X) of subcontinua of X endowed with the Hausdorff metric. This paper follows a common topic in the study of hyperspaces: if we know properties of some hyperspace H(X), then it is possible to infer other properties about the base space X. Particularly, our results are inserted in the context of the following problem: given a property P of continua X, find a property Q such that Bd(C(A)) has Q for all elements A of C(X) − {X} if and only if X has P . Let A be a proper subcontinuum of a continuum X. In [4] A. Illanes introduced and studied the semiboundary SB(A) of C(A) in C(X). We recall that an element B of C(A) belongs to SB(A) if B is arcwise accessible from C(X) − C(A) in the sense given in [8] by S.B. Nadler, Jr. Notice that SB(A) is a subset of the boundary Bd(C(A)) of C(A) in C(X). In [3] we began the study of Bd(C(A)). There, among other results, we have shown that Bd(C(A)) is always an arcwise connected subcontinuum of C(X). Here we E-mail address: [email protected]. http://dx.doi.org/10.1016/j.topol.2015.03.007 0166-8641/© 2015 Elsevier B.V. All rights reserved.

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continue the study of sets of the form Bd(C(A)). We prove that if Bd(A) is connected and Bd(C(A)) is a 2-cell for every proper nondegenerate subcontinuum A of X, then X is indecomposable and has only arcs and one-point sets as proper subcontinua. We also introduce the boundary function F from C(X) − {X} into C(C(X)) defined by F (A) = Bd(C(A)) for each element A of C(X) − {X}, and we characterize the simple closed curve as the unique arcwise connected continuum for which this function F is continuous. This result is analogous to the one given in [4] by using the semi-boundary function. We accomplished this characterization by proving that the boundary function is not upper semi-continuous for any dendroid. 2. Preliminaries Given a subset A of a space X, Cl X (A), Int X (A) and Bd X (A) denote the closure, interior and boundary of A in X, respectively. For simplicity, usually the dependence on the containing space X is suppressed in this notation. The set of singletons of A is F1 (A) = {{x} : x ∈ A}. Let A be a subcontinuum of a continuum X. We say that A is nondegenerate if A contains more than one point, and that A is proper if A is different from X. For a point p in X, we denote C(p, X) = {B ∈ C(X) : p ∈ B}. Given a subset U of X we denote C(U ) = {B ∈ C(X) : B ⊂ U } and D(U ) = {B ∈ C(X) : B ∩ U = ∅}. It is known that if U is an open set of X then C(U ) and D(U ) are open sets of the hyperspace C(X) [9, (0.13)]. An arc is a space homeomorphic to the closed interval [0, 1]. Given an arc A in a continuum X, we say that A is a free arc in X provided that A without its end points is an open set in X. A space homeomorphic to the circumference in the Euclidean plane {(x, y) ∈ R2 : x2 + y 2 = 1} is called a simple closed curve. A continuum is decomposable provided that it can be written as the union of two proper subcontinua. A continuum which is not decomposable is said to be indecomposable. A continuum X is unicoherent provided that whenever A and B are subcontinua of X such that X = A ∪ B, then A ∩ B is connected. A continuum X is hereditarily indecomposable (hereditarily unicoherent) provided that each of its subcontinua is indecomposable (unicoherent). Let A and B be subcontinua of a continuum X such that A ⊂ B and A = B, an order arc in C(X) from A to B is a continuous function α : [0, 1] → C(X) such that α(0) = A, α(1) = B and α(s) ⊂ α(t) = α(s) if 0 ≤ s < t ≤ 1. If the range of α is contained in a given subset H of C(X), then we say that α is an order arc in H. It is known that whenever A and B are subcontinua of a continuum X such that A ⊂ B and A = B, there exists an order arc in C(X) from A to B [9, (1.8) and (1.11)]. Given a proper subcontinuum A of a continuum X, an element B of C(A) belongs to the semi-boundary SB(A) of C(A), if there exists a continuous function α : [0, 1] → C(X) such that α(0) = B and α(t) is not contained in A for every t > 0 [4]. We notice that A belongs to SB(A) and that SB(A) is contained in Bd(C(A)) [4, Theorem 1.2 (c) and (e)]. 3. Boundaries which are 2-cells n Given a positive integer n, an n-cell is a space homeomorphic to the finite product I n = j=1 Ij , where Ij = [0, 1], 1 ≤ j ≤ n. A continuum X is an n-od if X contains a subcontinuum A such that X − A has at least n components. A simple triod is a continuum that is the union of three arcs having exactly one point in common, which is an end point of each one of these arcs. This common point is called the ramification point of the simple triod. Lemma 3.1. Let X be a continuum. If Bd(C(A)) is a 2-cell for each proper nondegenerate subcontinuum A of X, then X does not contain simple triods.

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Proof. Assume that X is a continuum that contains a simple triod T . We assert that Bd(T ) = T . To prove this, let us assume the opposite, equivalently, assume that Int(T ) = ∅. Let r be the ramification point of T . Let A be an arc contained in Int(T ) − {r}. Let p and q be the end points of A. We notice that A − {p, q} is an open set of T contained in Int(T ), which is an open set of X. It follows that A − {p, q} is an open set of X. This means that A is a free arc in X. By [3, Corollary 4.2], we have that Bd(C(A)) is contained in C(p, A) ∪ C(q, A), which is an arc in C(A), see [5, Example 5.1]. Thus, Bd(C(A)) is an arc, in contradiction with the hypothesis. This proves our assertion. Since F1 (Bd(T )) is contained in Bd(C(T )) [3, Remark 3.4], we have that F1 (T ) is contained in Bd(C(T )). By hypothesis Bd(C(T )) is a 2-cell. Let δ(Bd(C(T ))) be the manifold boundary of this 2-cell. We notice that if x ∈ T and U is an open set in C(T ) such that {x} ∈ U, then U ∩ Bd(C(T )) is not homeomorphic to an open ball of the Euclidean plane, [5, Example 5.1]. It follows that F1 (T ) is contained in δ(Bd(C(T ))). Since F1 (T ) is a simple triod and δ(Bd(C(T ))) is a simple closed curve, we have a contradiction. This proves the lemma. 2 Lemma 3.2. Let X be a continuum and let n be a positive integer. If X contains an (n + 1)-od, then X contains a subcontinuum A such that Bd(C(A)) contains an n-cell. Proof. Assume that Y is an (n + 1)-od in X. Let B be a subcontinuum of Y such that Y − B has at least n + 1 components. Let E1 , . . . , En+1 be n + 1 distinct components of Y − B. Notice that B ∪ Ei is a subcontinuum of Y [10, Corollary 5.9]. Let αi be an order arc in C(Y ) from B to B ∪ Ei [9, (1.8) and (1.11)]. Let A = α1 ( 12 ) ∪ E2 · · · ∪ En+1 . Notice that A is a subcontinuum of X. Define β : [0, 1] → C(X) by β(t) = α1 ( t+1 2 ) for every t ∈ [0, 1]. Observe that β is an order arc in C(X) 1 from α1 ( 2 ) to B ∪ E1 . Moreover, if t > 0 then β(t) − α1 ( 12 ) is contained in E1 . Since E1 ∩ Ei = ∅ for each i ∈ {2, . . . , n + 1}, it follows that β(t) is not contained in A for every t > 0. Hence, α1 ( 12 ) belongs to SB(A). Let D = {α1 ( 12 ) ∪ α2 (s2 ) ∪ · · · ∪ αn+1 (sn+1 ) : 0 ≤ si ≤ 1, i ∈ {2, . . . , n + 1}}. Define h : I n → D by h((s1 , . . . , sn )) = α1 ( 12 ) ∪ α2 (s1 ) ∪ · · · ∪ αn+1 (sn ), for each (s1 , . . . , sn ) ∈ I n . In a similar way to the proof of Theorem 1 of [11], we can see that h is a homeomorphism. Thus D is an n-cell. On the other hand, since each element of D is the union of n + 1 subcontinua of A each one of them containing B, we notice that D is contained in C(A). Moreover, since each element of D contains α1 ( 12 ), by [4, Theorem 1.2 (b) and (e)], we have that D is contained in Bd(C(A)). 2 Lemma 3.3. Let X be a continuum and let n be a positive integer. If X contains a subcontinuum A such that Bd(A) has at least n components, then X contains an n-od. Proof. Assume that A is a subcontinuum of X such that Bd(A) has at least n components. Let C1 , . . . , Cn be n distinct components of Bd(A). We analyze two cases: (1) There exists i ∈ {1, . . . , n} such that for every component K of X − A, it holds that Ci ∩ Cl(K) = ∅; and (2) For each i ∈ {1, . . . , n} there exists a component Ki of X − A such that Ci ∩ Cl(Ki ) = ∅. In the case (1) we fix a point x ∈ Ci and a sequence {xm }∞ m=1 in X − A such that lim xm = x. Let Km be the component of X − A that contains xm . If there exists n ∈ N such that Km = KN for each m ≥ N , then x ∈ Cl(KN ). Thus, Ci ∩ Cl(KN ) = ∅, contrary to the hypothesis of this case (1). Therefore, there  exists a sequence {mj }∞ j=1 such that Kmj = Kmj if j = j . Let Y = A ∪ Km1 ∪ · · · ∪ Kmn . Notice that Y is a subcontinuum of X [10, Corollary 5.9]. Moreover, since Y − A = Km1 ∪ · · · ∪ Kmn , we have that Y is an n-od in X. In the case (2) we notice that Ci ∪ Cl(Ki ) is a continuum that contains properly Ci , and we take an order arc αi in C(X) from Ci to Ci ∪ Cl(Ki ), for each i ∈ {1, . . . , n} [9, (1.8) and (1.11)]. Let Ui be an open set in X such that Ci ⊂ Ui and Ui ∩ Uj = ∅ if i, j ∈ {1, . . . , n} and i = j. Observe that C(Ui ) is an open set

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in C(X) such that αi (0) ∈ C(Ui ). By continuity of αi , there exists ti > 0 such that αi (ti ) ∈ C(Ui ), which means that αi (ti ) ⊂ Ui , for each i ∈ {1, . . . , n}. Let Y = A ∪ α1 (t1 ) ∪ · · · ∪ αn (tn ). Since Ci ⊂ A ∩ αi (ti ), we n  have that Y is a subcontinuum of X. Besides, note that Y − A = (αi (ti ) − A); since ti > 0, αi (ti ) − A = ∅; i=1

and, since αi (ti ) − A ⊂ Ui , Y − A has at least n components. Therefore, Y is an n-od in X.

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Lemma 3.4. Let X be a continuum. If Bd(C(A)) is a 2-cell for each proper nondegenerate subcontinuum A of X, then Bd(A) is a locally connected subspace of X for each subcontinuum A of X. Proof. Assume that X contains a subcontinuum A such that Bd(A) is not locally connected. By hypothesis and Lemma 3.2, X contains no a 4-od. Thus, by Lemma 3.3, Bd(A) has at most three components. Since a finite union of locally connected closed sets is a locally connected set [7, Theorem 1, p. 230], Bd(A) has a component C which is not locally connected. Hence, there exist a nondegenerate subcontinuum K of C and a sequence {Kn }∞ n=1 of subcontinua of C such that lim Kn = K and Kn ∩ Km = ∅ if n = m [10, Theorem 5.12]. We fix a point x ∈ K and a sequence {xn }∞ n=1 of points in X such that lim xn = x and xn ∈ Kn for each n ∈ N. Let U be an open set in X such that x ∈ U and K is not contained in Cl(U ). Notice that C(U ) and D(X − Cl(U )) are open sets in C(X) containing {x} and K respectively. Thus, we can assume that {xn } ∈ C(U ) and Kn ∈ D(X − Cl(U )) for each n ∈ N. Since x and xn belong to Bd(A), we have that the one-point sets {x} and {xn } belong to Bd(C(A)) [3, Remark 3.4]. From the hypothesis it follows that Bd(C(A)) is locally connected. Then we can take an open set V of Bd(C(A)) such that {x} ∈ V ⊂ Cl(V) ⊂ C(U ) ∩ Bd(C(A)); and we can assume that {xn } ∈ V for each n ∈ N.  Let M = Cl(V). Since Cl(V) is a subcontinuum of C(X), we know that M is a subcontinuum of X [9, (1.49)]. Since Cl(V) ⊂ C(U ), which means E ⊂ U for each E ∈ Cl(V), we have that M ⊂ U . Let Y = M ∪ K1 ∪ · · · ∪ K4 . Since xi ∈ M ∩ Ki for each i ∈ {1, 2, 3, 4}, we observe that Y is a 4  subcontinuum of X. Moreover, we notice that Y − M = (Ki − M ); and since Ki ∈ D(X − Cl(U )), which i=1

means Ki ∩ (X − Cl(U )) = ∅, we have that Ki − M = ∅. Therefore, Y is a 4-od in X. By Lemma 3.2, it follows that X contains a subcontinuum Z such that Bd(C(Z)) contains a 3-cell, which contradicts the hypothesis. This proves the lemma. 2 Theorem 3.5. Let X be a continuum. If Bd(A) is connected and Bd(C(A)) is a 2-cell for each proper nondegenerate subcontinuum A of X, then X is an indecomposable continuum and each proper nondegenerate subcontinuum of X is an arc. Proof. First we will prove that X is an indecomposable continuum. To do this we suppose that X is decomposable. Thus, X contains a proper nondegenerate subcontinuum A which has nonempty interior [7, Theorem 2, p. 207]. We notice that Bd(A) = A and Bd(A) = Cl(X − A). Since Bd(A) is connected and X = A ∪ Cl(X − A), we observe that Cl(X − A) is a connected set [7, Corollary 5, p. 133], thus it is a subcontinuum of X. Therefore, A and Cl(X − A) are proper subcontinua of X, and Bd(A) is a proper subcontinuum of each one of them. Let U1 and U2 be open sets of A and of Cl(X − A) respectively, such that Bd(A) ⊂ U1 ∩ U2 , Cl(U1 ) = A and Cl(U2 ) = Cl(X − A). Let C1 and C2 be the components of Cl(U1 ) and Cl(U2 ) that contain Bd(A), respectively. We will arrive to a contradiction by proving a series of claims. (1) Bd(C1 ) − Bd(A) = ∅ and Bd(C2 ) − Bd(A) = ∅. To prove (1) we fix a point x ∈ C1 ∩ Bd A (Cl(U1 )) [10, Theorem 5.6]. Let {xn }∞ n=1 be a sequence in ∞ A − Cl(U1 ) such that lim xn = x. Since C1 ⊂ Cl(U1 ), we have that {xn }n=1 is a sequence of points in

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X − C1 that converges to x. Thus, x ∈ Bd(C1 ). On the other hand, we notice that Bd A (Cl(U1 )) ⊂ Bd A (U1 ) and, since U1 is an open set of A, U1 ∩ Bd A (U1 ) = ∅. It follows that U1 ∩ Bd A (Cl(U1 )) = ∅. Therefore, since x ∈ Bd A (Cl(U1 )) and Bd(A) ⊂ U1 , we have that x does not belong to Bd(A). Thus, x ∈ Bd(C1 ) − Bd(A). Hence, Bd(C1 ) − Bd(A) = ∅. Similarly, Bd(C2 ) − Bd(A) = ∅. (2) Bd(A) is a proper subcontinuum of Bd(C1 ). Take a point x ∈ Bd(A). Let {xn }∞ n=1 be a sequence of points in X − A such that lim xn = x. Since ∞ C1 ⊂ A, {xn }n=1 is a sequence of points in X −C1 . It follows that x ∈ Cl(X − C1 ). Besides, since x ∈ Bd(A) and Bd(A) ⊂ C1 , we have that x ∈ C1 . Thus, x ∈ Bd(C1 ). Hence, Bd(A) ⊂ Bd(C1 ). By this and (1) we have proved (2). (3) Bd(C1 ) ∩ Bd(C2 ) is a nonempty subset of Bd(A). From (2), we have that Bd(C1 ) = Bd(A) ∪ Cl(Bd(C1 ) − Bd(A)). By hypothesis, Bd(C1 ) is a connected set. Therefore, we can take a point x ∈ Bd(A) ∩ Cl(Bd(C1 ) − Bd(A)). We notice that x ∈ Bd(C1 ). We will prove that the point x also belongs to Bd(C2 ). To do this we take a sequence {xn }∞ n=1 of points in Bd(C1 ) − Bd(A) such that lim xn = x. We observe that xn ∈ Bd(C1 ) ⊂ C1 ⊂ A, thus xn ∈ A − Bd(A). It follows that xn does not belong to Cl(X − A). Therefore, since C2 ⊂ Cl(X − A), we have that {xn }∞ n=1 is a sequence of points in X − C2 . Thus, x ∈ Cl(X − C2 ). Besides, since x ∈ Bd(C(A)) ⊂ C2 , we have that x ∈ C2 . Hence, x ∈ Bd(C2 ). This proves that Bd(C1 ) ∩ Bd(C2 ) is not the empty set. Moreover, since Bd(C1 ) ⊂ C1 ⊂ A and Bd(C2 ) ⊂ C2 ⊂ Cl(X − A), we have that Bd(C1 ) ∩ Bd(C2 ) ⊂ Bd(A). This proves claim (3). Let Y = Bd(C1 ) ∪ Bd(C2 ). (4) The continuum Y is an arc in X. To prove this we notice that, by hypothesis and (3), Y is a subcontinuum of X. By Lemma 3.4, we have that Bd(C1 ) and Bd(C2 ) are locally connected. It follows that Y is locally connected [7, Theorem 1, p. 230]. By Lemma 3.1, X does not contain simple triods. This implies that Y is either an arc or a simple closed curve [10, 8.40]. Assume that Y is a simple closed curve. By (1) and (3), we have that Bd(C1 ) and Bd(C2 ) are proper nondegenerate subcontinua of Y . Thus, Bd(C1 ) and Bd(C2 ) are arcs. We notice that the end points of Bd(C1 ) must belong to the intersection Bd(C1 ) ∩ Bd(C2 ), otherwise we can find a simple triod in Y , which is not possible by Lemma 3.1. Hence, by (3), it follows that the end points of Bd(C1 ) belong to Bd(A). By this and (2), we have that Bd(A) is a subcontinuum of the arc Bd(C1 ) having the end points of this arc. Therefore, Bd(A) = Bd(C1 ). This is a contradiction, because by (2) Bd(A) is a proper subcontinuum of Bd(C1 ). This proves that Y is an arc. Let a and b be the end points of the arc Y . Let E be a subarc of Y such that Bd(A) ⊂ E ⊂ Y − {a, b} and E ∩ (Bd(Ci ) − Bd(A)) = ∅ for each i ∈ {1, 2}. (5) The arc E does not belong to SB(Y ). To prove (5) we suppose that E belongs to SB(Y ). Assuming this we will prove first that there exists a subcontinuum C of X which satisfies the following conditions: (i) Bd(C) ∩ {a, b} = ∅; (ii) Bd(C) ∩ Y = ∅; and (iii) Bd(C) ∩ (X − Y ) = ∅. To prove that a such subcontinuum exists, we take an open set V of X such that E ⊂ V and Cl(V ) ∩ {a, b} = ∅. Since E ∈ SB(Y ), by [4, Theorem 1.2 (a)] we can take an order arc α in C(X) such that α(0) = E and α(t) is not contained in Y for each t ∈ (0, 1]. We observe that C(V ) is an open set in C(X) containing α(0). Thus, by continuity of α, there exists t0 > 0 such that α(t0 ) ⊂ V . Next we analyze two cases: (5.1) Bd(α(t0 )) ∩ (X − Y ) = ∅; and (5.2) Bd(α(t0 )) ∩ (X − Y ) = ∅. In the case (5.1), since E ⊂ Y ∩α(t0 ), we observe that Y ∩α(t0 ) = ∅. Besides, since {a, b} ⊂ Y ∩(X −α(t0 )), we have that Y ∩ (X − α(t0 )) = ∅. Therefore, since Y is connected, we have that Bd(α(t0 )) ∩ Y = ∅. Thus, in this case, putting C = α(t0 ), we have a subcontinuum of X satisfying the conditions (i), (ii) and (iii). In the case (5.2) we have that Bd(α(t0 )) ⊂ Y . On the other hand, since t0 > 0, we have that α(t0 ) is not contained in Y . We take a point p ∈ α(t0 ) − Y . Note that p ∈ Int(α(t0 )). Let W be an open set of X such that p ∈ W ⊂ α(t0 ) and Cl(W ) ∩ Y = ∅. Since E ⊂ Y ∩ α(t0 ), we have that E ⊂ α(t0 ) − W . Let C

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be the component of α(t0 ) − W that contains E. Since Bd(C) ⊂ C ⊂ α(t0 ) and α(t0 ) ∩ {a, b} = ∅, we have that Bd(C) ∩ {a, b} = ∅. Thus, C satisfies (i). Besides, since E ⊂ Y ∩ C and {a, b} ⊂ Y ∩ (X − C), we have that Y ∩ C = ∅ and Y ∩ (X − C) = ∅. Therefore, by connectedness of Y , we have that Bd(C) ∩ Y = ∅. Hence, C satisfies (ii). Moreover, by [10, Theorem 5.6] we know that C ∩ Bd(α(t0 ) − W ) = ∅. Since Bd(α(t0 ) − W ) ⊂ Cl(W ), it follows that C ∩ Cl(W ) = ∅. Thus, we can take a point x ∈ C and a sequence ∞ {xn }∞ n=1 of points in W such that lim xn = x. Since C ⊂ α(t0 ) − W , we notice that {xn }n=1 is a sequence of points in X − C that converges to x. This implies that x ∈ Bd(C). Since Cl(W ) ∩ Y = ∅, we also notice that x ∈ X − Y . It follows that Bd(C) ∩ (X − Y ) = ∅. Thus, C satisfies (iii). We have proved that there exists a subcontinuum C of X that satisfies the conditions (i), (ii) and (iii) given above. By conditions (ii) and (iii) we can take points p ∈ Bd(C) ∩ Y and q ∈ Bd(C) ∩ (X − Y ). By Lemma 3.4, Bd(C) is a locally connected continuum. Thus, Bd(C) is arcwise connected [10, Theorem 8.23]. Let J be an arc in Bd(C) with end points p and q. By condition (i) we have that J ∩ {a, b} = ∅. This implies that there exists a simple triod in Y ∪ J. By Lemma 3.1, this is a contradiction. This proves (5). (6) The arc E does not belong to Bd(C(Y )). To prove this we suppose that E ∈ Bd(C(Y )). By (4) and [3, Lemma 5.1], there exists a sequence {En }∞ n=1 in C(X) such that lim En = E and En ∩ Y = ∅ for each n ∈ N. By (2), Bd(A) ⊂ Bd(C1 ). Thus Bd(A) ⊂ Y . It follows that En ∩ Bd(A) = ∅. Therefore, En ⊂ (X − A) ∪ Int(A). Since En is connected, we have that either En ⊂ X − A or En ⊂ Int(A). This implies that there exists a sequence {nk }∞ k=1 in N such that one of the two conditions below holds: (6.1) For each k ∈ N, Enk ⊂ X − A; or (6.2) For each k ∈ N, Enk ⊂ Int(A). Suppose that (6.1) holds. Take a point x ∈ E ∩ (Bd(C1 ) − Bd(A)). Since lim En = E, there exists a sequence of points {xnk }∞ k=1 such that lim xnk = x and xnk ∈ Enk for each k ∈ N. Notice that xnk ∈ X − A for each k ∈ N. It follows that x ∈ Cl(X − A). Also notice that x ∈ C1 ⊂ A. Therefore, x ∈ Bd(A). This contradicts the choice of the point x. Suppose that (6.2) holds. Take a point x ∈ E ∩ (Bd(C2 ) − Bd(A)). As in the previous case there exists a sequence {xnk }∞ k=1 that converges to x and xnk ∈ Enk for each k ∈ N. It follows that xnk ∈ A for each k ∈ N. Thus, x ∈ A. On the other hand, since x ∈ Bd(C2 ) ⊂ C2 ⊂ Cl(X − A), it holds that x ∈ Cl(X − A). Hence, x ∈ Bd(A), which contradicts the choice of the point x. These contradictions prove that the arc E does not belong to Bd(C(Y )). (7) Bd(Y ) = Y . To prove this we take a point y ∈ Y . We will show that there exists a sequence {yn }∞ n=1 of points in X − Y such that lim yn = y. In order to do this we analyze three cases: (7.1) y ∈ Bd(C1 ) ∩ Bd(C2 ); (7.2) y ∈ Bd(C1 ) − Bd(C2 ); and (7.3) y ∈ Bd(C2 ) − Bd(C1 ). In the case (7.1), by (3), we observe that y ∈ Bd(A). Thus, we can take a sequence {yn }∞ n=1 of points in X − A such that lim yn = y. Since Bd(C1 ) ⊂ C1 ⊂ A, we have that yn ∈ X − Bd(C1 ) for each n ∈ N. Notice that if there exists N ∈ N such that yn ∈ X − Bd(C2 ) for each n ≥ N , then {yn }∞ n=N is a sequence of points in X − Y that converges to y, and we are done. Therefore, in this case, we can assume that yn ∈ Bd(C2 ) for every n ∈ N. Then, for each n ∈ N we can take a sequence {ym,n }∞ m=1 of points in X − C2 such that lim ym,n = yn . Since X − A is an open set of X and yn ∈ X − A, we can suppose that ym,n ∈ X − A for every m, n ∈ N. Since C1 ⊂ A, it follows that ym,n ∈ X − C1 for every m, n ∈ N. Thus, since Y ⊂ C1 ∪ C2 , we have that {ym,n }∞ m=1 is a sequence of points in X − Y for every n ∈ N. Let d be the metric of the continuum X. For each n ∈ N we fix mn ∈ N such that d(yn , ym,n ) < n1 . We notice that {ymn ,n }∞ n=1 is a sequence of points in X − Y such that lim ymn ,n = y. In the case (7.2), we take a sequence {yn }∞ n=1 of points in X −C1 such that lim yn = y. Since Bd(C2 ) ⊂ C2 , we observe that y ∈ X − C2 . Since X − C2 is an open set of X, we can assume that yn ∈ X − C2 for every n ∈ N. Therefore, since Y ⊂ C1 ∪ C2 , we have that {yn }∞ n=1 is a sequence of points in X − Y that converges to y.

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Fig. 1. An indecomposable continuum.

The case (7.3) is similar to the case (7.2). Thus, in every case we can find a sequence in X − Y that converges to y. By this y ∈ Bd(Y ). Hence, we have proved (7). By (4), (6) y (7), we have that Y is an arc in X such that Bd(C(Y )) = C(Y ) and Bd(Y ) = Y . From [3, Theorem 6.1], it follows that Bd(C(Y )) is not unicoherent. On the other hand, by hypothesis Bd(C(Y )) is unicoherent [7, Examples, p. 163]. Thus, we have arrived to a contradiction. This proves that X is indecomposable. Next, we will prove the second part of the theorem. Let B be a proper nondegenerate subcontinuum of X. Since X is indecomposable, we observe that Bd(B) = B [7, Theorem 2, p. 207]. By Lemma 3.4, we have that Bd(B) is locally connected. Thus, B also is. By Lemma 3.1, X does not contain simple triods. Therefore, neither B. It follows that B is either an arc or a simple closed curve [10, (8.40)]. Let V be an open set of X such that B ⊂ V and Cl(V ) = X. Let K be the component of V that contains B. By similar arguments as given in the last paragraph, we have that Cl(K) is an arc or a simple closed curve. We notice that Cl(K) ∩ Bd(V ) = ∅ [10, Theorem 5.6]. It follows that B is a proper nondegenerate subcontinuum of Cl(K). This implies that B is an arc. 2 Problem 3.6. Can the condition Bd(A) is connected in Theorem 3.5 be removed? Next example shows that the converse in Theorem 3.5 is not true. Example 3.7. Let Y be the indecomposable continuum having exactly two end points, p and q, given in Example 3 of [7, p. 205], this means that each one of p and q is an end point of every arc that contains it. Let X be the continuum obtained from Y by identifying the points p and q to a single point, which we denote by o. In Fig. 1 we show a picture of this continuum X. We notice that X is an indecomposable continuum, and that every proper nondegenerate subcontinuum of X is an arc. We also notice that the continuum X contains a proper subcontinuum A such that Bd(C(A)) is not a 2-cell. In order to see the last assertion in Example 3.7, let A be an arc in X, with end points a and b, such that o ∈ A − {a, b}. Let B be an arc contained in A − {a, b} with end points c and d, such that o ∈ B − {c, d}. We observe that B is not the limit of a sequence of elements in C(X) − C(A). Thus, B does not belong to Bd(C(A)). Hence, Bd(C(A)) = C(A). Since X is indecomposable, we also observe that Bd(A) = A [7, Theorem 2, p. 207]. By [3, Theorem 6.1], it follows that Bd(C(A)) is not unicoherent. Therefore, Bd(C(A)) is not a 2-cell [7, Examples, p. 163]. For next remark we recall that a continuum X, with metric d, is called a Kelley continuum provided that for each ε > 0 there exists δ > 0 such that if a, b ∈ X and d(a, b) < δ and a ∈ A ∈ C(X), then there exists B ∈ C(X) such that b ∈ B and H(A, B) < ε [6, 3.2], where H denotes the Hausdorff metric in C(X). Remark 3.8. Every Kelley indecomposable continuum whose proper nondegenerate subcontinua are arcs lies in the hypothesis of Theorem 3.5. Solenoids different to the circle are examples of such continua [10, 2.8

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and 2.16], [9, (16.26)]. We notice that the hypothesis of Theorem 3.5 does not imply the property of Kelly continua, the continuum X in Fig. 55 (b) of [5, p. 426] is such that Bd(A) is connected and Bd(C(A)) is a 2-cell for each proper nondegenerate subcontinuum A of X, and it is not a Kelley continuum. 4. The boundary function It is a natural strategy in the study of hyperspaces to define functions on hyperspaces of continua, and to analyze relationships between properties of these functions and properties of continua. As examples of this we cite [2], [4, Section 7], and [9, Chapter XV]. Given a continuum X, we know that for every proper nondegenerate subcontinuum A of X, Bd(C(A)) is a subcontinuum of C(X) [3, Corollary 3.2]. Then we define the boundary function F : C(X)−{X} → C(C(X)) given by F (A) = Bd(C(A)) for each A ∈ C(X) − {X}. Let us observe that the element X is left outside of the domain of this function since boundary of C(X) is the empty set, which is not an element of C(C(X)). If the function F is continuous (upper semi-continuous) then we say that the boundary function is continuous (upper semi-continuous) for the continuum X. Our objective in this section is to prove that the simple closed curve is the unique arcwise connected continuum for which this boundary function is continuous, Theorem 4.5, and the motivation of this comes from results in section 7 of [4], where A. Illanes studied the semi-boundary function. Let Z be a space and let Y be a continuum. We recall that a function G : Z → C(Y ) is upper semicontinuous at a point p of Z, written usc at p, provided that for each open set V of Y such that G(p) ⊂ V , there exists an open set U of X such that p ∈ U and G(z) ⊂ V for every point z ∈ U . The function G : Z → C(Y ) is upper semi-continuous, written usc, provided that G is usc at every point z of Z. We notice that if a function G : Z → C(Y ) is continuous at a point p of Z, then G is usc at p. Indeed, if V is an open set of Y such that G(p) ⊂ V , then C(V ) is an open set of C(Y ) and G(p) ∈ C(V ). Hence, there exists an open set U of Z such that p ∈ U and G(z) ∈ C(V ) for every z ∈ U . Thus, G(z) ⊂ V for every z ∈ U . To prove Theorem 4.5, we first prove Lemmas 4.1 and 4.3. We recall that a dendroid is an arcwise connected, hereditarily unicoherent continuum. We notice that given two distinct points x and y in a dendroid X there exists a unique arc in X with end points x and y, which we denote by xy. Lemma 4.1. The boundary function is not usc for any dendroid. Proof. Let X be a dendroid. It is known that X is a decomposable continuum [1, Corollary, p. 17]. Let E and D be proper subcontinua of X such that X = E ∪ D. We take points e ∈ E − D and d ∈ D − E. There exist two points w0 and z0 in X such that ed ⊂ w0 z0 and w0 z0 is a maximal arc in X [1, Lemma 3]. Since every subcontinuum of X is unicoherent, we notice that if {w0 , z0 } ⊂ E or {w0 , z0 } ⊂ D then w0 z0 ⊂ E or w0 z0 ⊂ D; which implies that d ∈ E or e ∈ D. Therefore, we have that neither {w0 , z0 } ⊂ E nor {w0 , z0 } ⊂ D. Hence, we can assume that w0 ∈ E − D and z0 ∈ D − E. Let p be a point in (D−E) ∩w0 z0 −{z0 } and let A = E ∪w0 p. We observe that A is a proper subcontinuum of X. We will prove that the boundary function is not usc at A. In order to do this we first prove the following assertion: ∞ Assertion: There exist a sequence {Bn }∞ n=1 in C(X) − {X} and a sequence {wn }n=1 of points in X such that lim Bn = A, lim{wn } = {w0 } and {wn } belongs to Bd(C(Bn )) for each n ∈ N. We take ε > 0 such that p does not belong to the ε-ball in X around w0 , B(w0 , ε). For each n ∈ N, let ∞  An be the component of A − B(w0 , nε ) that contains p. Clearly An ⊂ A. In fact this inclusion becomes n=1

an equality. Indeed, given x ∈ A − {w0 , p} there exists a point y ∈ w0 z0 such that xy ∩ w0 z0 = {y}. Since w0 z0 is a maximal arc, we notice that y = w0 . Moreover, w0 does not belong to the arc xy ∪ yp. Therefore,

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there exists n ∈ N such that B(w0 , nε ) ∩ (xy ∪ yp) = ∅. Thus, xy ∪ yp ⊂ An . Hence, x ∈ An . This proves ∞  that An = A. Since An ⊂ An+1 , it follows that lim An = A. n=1

For each n ∈ N, we take a point wn ∈ B(w0 , nε ) ∩ (w0 p − {w0 }) and we write Bn = An ∪ wn p. Since p ∈ An ∩ wn p and since w0 does not belong to Bn , we observe that each Bn is a proper subcontinuum of X. Moreover, lim Bn = A ∪ w0 p = A. Besides, we notice that lim wn = w0 , which gives that lim{wn } = {w0 }. We also notice that w0 wn ∩Bn = {wn }, which implies that wn ∈ Bd(Bn ). It follows that {wn } ∈ Bd(C(Bn )) [3, Remark 3.4]. This proves our assertion. Now we observe that X − D is an open set of X that contains the point w0 and it is contained in A. Thus, C(X − D) is an open set of C(X) that contains the singleton {w0 } and it is contained in C(A). It follows that {w0 } belongs to Int(C(A)). Hence, {w0 } does not belong to Bd(C(A)). Let V and W be open disjoint sets of C(X) such that Bd(C(A)) ⊂ V and {w0 } ∈ W. We notice that for each open set U of C(X) − {X} containing A there exists n ∈ N such that Bn ∈ U and {wn } ∈ W. It follows that Bd(C(Bn )) ∩ W = ∅. This implies that Bd(C(Bn )) is not contained in V. This proves that the boundary function is not usc at A. 2 Corollary 4.2. The boundary function is not continuous for any dendroid. Lemma 4.3. If the boundary function is continuous for a continuum X, then each proper subcontinuum of X is unicoherent. Proof. Suppose that X is a continuum for which the boundary function is continuous, and suppose that X contains a proper subcontinuum A which is not unicoherent. Let H and K be proper subcontinua of A such that A = H ∪ K and H ∩ K is not connected. Let P and Q be nonempty closed disjoint sets such that H ∩ K = P ∪ Q. Let M be a subcontinuum of H irreducible respect to intersect both P and Q, this means M ∩ P = ∅, M ∩ Q = ∅ and for each proper subcontinuum N of M , either N ∩ P = ∅ or N ∩ Q = ∅ [7, Theorem 2 p. 54]. Let B = K ∪ M . We will prove claim (1) below. (1) The continuum M belongs to Bd(C(B)). To prove (1) we fix a point p ∈ M ∩ P , and we take an order arc α in C(M ) from {p} to M [9, (1.8) and (1.11)]. Let Bn = K ∪ α(1 − n1 ) for each n ∈ N. Since p ∈ K ∩ α(1 − n1 ), each Bn is a subcontinuum of X. We assert that Bn is not contained in M , and that M is not contained in Bn , for each n ∈ N. To prove this, since α(1 − n1 ) ⊂ M , we observe that Bn ∩ M = (K ∩ M ) ∪ α(1 − n1 ). Besides, since M ⊂ H, we have that K ∩ M = K ∩ (H ∩ M ) = (P ∩ M ) ∪ (Q ∩ M ). It follows that: Bn ∩ M = α(1 − n1 ) ∪ (P ∩ M ) ∪ (Q ∩ M ). Since α(1 − n1 ) ∪ (P ∩ M ) and Q ∩ M are nonempty closed disjoint sets, we obtain that Bn ∩ M is not connected. This proves what we asserted. Let Cn be the component of Bn ∩ M that contains α(1 − n1 ) for each n ∈ N. By [4, Theorem 1.2 (e) and (f)], it follows that Cn belongs to Bd(C(Bn )) for each n ∈ N. We observe that lim Bn = B. Thus, since the boundary function is continuous, lim Bd(C(Bn )) = Bd(C(B)). Besides, since lim α(1 − n1 ) = M and α(1 − n1 ) ⊂ Cn ⊂ M , we have that lim Cn = M . This implies that M belongs to Bd(C(B)). Hence, claim (1) is proved. Let U and V be open sets of X such that P ⊂ U , Q ⊂ V and Cl(U ) ∩ Cl(V ) = ∅. Since M and K are connected and each one intersects both U and V , we notice that M − (U ∪ V ) = ∅ and K − (U ∪ V ) = ∅. Moreover, since M ∩ K ⊂ U ∪ V , we observe that M − (U ∪ V ) and K − (U ∪ V ) are closed disjoint sets. Let W and W  be open disjoint sets of X such that M − (U ∪ V ) ⊂ W and K − (U ∪ V ) ⊂ W  . Let U = C(U ∪ V ∪ W ) ∩ D(U ) ∩ D(V ) ∩ D(W ). We have that U is an open set of C(X) [9, (0.13)]. We notice that M belongs to U. By (1) we can take an order arc β in Bd(C(B)) from M to B [3, Theorem 3.1]. By continuity of β we can fix t0 > 0 such that β(t0 ) belongs to U.

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Since M ⊂ β(t0 ) ⊂ B = K ∪M and M = β(t0 ), we have that β(t0 ) −M is a nonempty set contained in K. We assert that β(t0 ) −M ⊂ U ∪V ; to prove this we assume that there exists a point x ∈ (β(t0 ) −M ) −(U ∪V ). It follows that x ∈ K −(U ∪V ). Thus, x ∈ W  . Since β(t0 ) ∈ C(U ∪V ∪W ), which means β(t0 ) ⊂ U ∪V ∪W , it follows that x ∈ W . Hence, x ∈ W ∩ W  . This is a contradiction because W and W  are disjoint sets. Therefore, we have proved that β(t0 ) − M ⊂ U ∪ V . Then one of the following two conditions holds: (i) (β(t0 ) − M ) ∩ U = ∅; or (ii) (β(t0 ) − M ) ∩ V = ∅. We first assume that (i) holds. We fix a point q ∈ Q ∩ M and we take an order arc γ in C(M ) from {q} to M [9, (1.8) and (1.11)]. Let Bn = K ∪ γ(1 − n1 ) for each n ∈ N. We notice that {Bn }∞ n=1 is a sequence in  C(X) − {X} and that lim Bn = B. Thus, since the boundary function is continuous by hypothesis, we have that lim Bd(C(Bn )) = Bd(C(B)). Since β(t0 ) ∈ Bd(C(B)), it follows that there exists a sequence {En }∞ n=1 in C(X) such that lim En = β(t0 ) and En ∈ Bd(C(Bn )) for each n ∈ N. We observe that U − M is an open set of X and, since we assume (i), β(t0 ) ∩ (U − M ) = ∅. It follows that D(U − M ) is an open set of C(X) and β(t0 ) ∈ D(U − M ). We also recall that U is an open set of C(X) and β(t0 ) ∈ U. Therefore, we can take n ∈ N such that En ∈ U and En ∈ D(U − M ). This means that En ⊂ U ∪ V ∪ W , En ∩ U = ∅, En ∩ V = ∅, En ∩ W = ∅ and En ∩ (U − M ) = ∅. Next we will prove claims (2)-(5) below: (2) En ⊂ (K ∩ Cl(U )) ∪ (γ(1 − n1 ) ∪ Cl(V )). Let x be a point in En . Since En ⊂ Bn = K ∪ γ(1 − n1 ), it follows that x ∈ K ∪ γ(1 − n1 ). If x ∈ γ(1 − n1 ) then there is nothing else to do. Thus, we suppose that x ∈ K −γ(1 − n1 ). We notice that if x does not belong to U ∪ V , then x ∈ K − (U ∪ V ). Thus, since K − (U ∪ V ) ⊂ W  , x ∈ W  . Besides, since En ⊂ U ∪ V ∪ W , x ∈ W . Hence, x ∈ W ∩ W  . This is a contradiction because W and W  are disjoint sets. This proves that x ∈ U ∪ V . Therefore, x ∈ K ∩ Cl(U ) or x ∈ Cl(V ). Hence, we have proved claim (2). (3) K ∩ Cl(U ) and γ(1 − n1 ) ∪ Cl(V ) are disjoint sets. Since Cl(U ) ∩ Cl(V ) = ∅, we have that (K ∩ Cl(U )) ∩ (γ(1 − n1 ) ∪ Cl(V )) = K ∩ Cl(U ) ∩ γ(1 − n1 ). Since γ(1 − n1 ) ⊂ M ⊂ H, K ∩ Cl(U ) ∩ γ(1 − n1 ) = K ∩ H ∩ Cl(U ) ∩ γ(1 − n1 ). Since K ∩ H = P ∪ Q, it follows that K ∩ Cl(U ) ∩ γ(1 − n1 ) = (P ∪ Q) ∩ Cl(U ) ∩ γ(1 − n1 ). We recall that P ⊂ Cl(U ), Q ⊂ Cl(V ) and Cl(U ) ∩ Cl(V ) = ∅. Therefore, (P ∪ Q) ∩ Cl(U ) ∩ γ(1 − n1 ) = P ∩ γ(1 − n1 ). We obtain that K ∩ Cl(U ) ∩ γ(1 − n1 ) = P ∩ γ(1 − n1 ). Now we notice that γ(1 − n1 ) is a proper subcontinuum of M and γ(1 − n1 ) ∩ Q = ∅. By irreducibility of M , it follows that P ∩ γ(1 − n1 ) = ∅. This proves (3). (4) En ∩ K ∩ Cl(U ) = ∅. We take a point x ∈ En ∩ (U − M ). Since En ⊂ Bn = K ∪ γ(1 − n1 ) and γ(1 − n1 ) ⊂ M , it follows that x ∈ K ∩ U , which proves (4). (5) En ∩ (γ(1 − n1 ) ∪ Cl(V )) = ∅. Since En ∈ U ⊂ D(V ), it follows that En ∈ D(V ). It means that En ∩ V = ∅. This implies what (5) says. Since En is connected, with the facts (2)–(5) we have a contradiction. 1 If we assume that (ii) holds then, by using the sequence {Bn }∞ n=1 with Bn = K ∪ α(1 − n ), and the sets 1 K ∩ Cl(V ) and α(1 − n ) ∪ Cl(U ), we obtain a similar contradiction. This proves the theorem. 2 The following example shows that the condition of continuity cannot be replaced by the condition of usc in Lemma 4.3. Example 4.4. Let X be the compactification of the real line R with remainder the circumference S 1 as in Fig. 2. The boundary function is usc for this continuum X. Comparing with Lemma 4.3, we notice that this continuum X contains a nonunicoherent subcontinuum. We will prove what we asserted in Example 4.4. In order to do this we suppose that there exists an element A of C(X) − {X} such that the boundary function is not usc at A. Then there exist an open set V of C(X) and a sequence {An }∞ n=1 in C(X) − {X} such that lim An = A, Bd(C(A)) ⊂ V and Bd(C(An )) is

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Fig. 2. A compactification of the real line.

not contained in V for every n ∈ N. Take an element En in Bd(C(An )) − V for each n ∈ N. Thus, {En }∞ n=1 is a sequence in C(X) − V and, by compactness, we may assume that there exists E ∈ C(X) − V such that lim En = E. Since En ⊂ An , we observe that E ⊂ A. Hence, E ∈ C(A). Moreover, since Bd(C(A)) is contained in V and E does not belong to V, we notice that: (∗) E does not belong to Bd(C(A)). Next we will arrive to a contradiction with (∗) analyzing four cases. (1) A ⊂ S 1 . In this case we notice that there exists a sequence of arcs in the real line R which converges to A. This implies that Bd(C(A)) = C(A). Since E ∈ C(A), we have that E ∈ Bd(C(A)). (2) A ⊂ R. In this case we observe that A is a free arc in X. Let a and b be the end points of A. We know that Bd(C(A)) = C(a, A) ∪ C(b, A) [3, Corollary 4.2]. We can assume that each An is an arc in R with end points an and bn such that lim an = a and lim bn = b. Similarly we have that Bd(C(An )) = C(an , An ) ∪ C(bn , An ) for each n ∈ R. Since En ∈ Bd(C(An )), without loss of generality we may assume that an ∈ En for each n ∈ N. Since lim En = E and lim an = a, it follows that a ∈ E. Thus, E ∈ Bd(C(A)). (3) A = S 1 ∪ [a, ∞) or A = S 1 ∪ (−∞, b], for some points a and b in R. Let us analyze the case A = S 1 ∪ [a, ∞), the other one is similar. We assert that Bd(C(A)) = C(a, A) ∪ C(S 1 , A) ∪ C(S 1 ), where C(S 1 , A) = {B ∈ C(A) : S 1 ⊂ B}. To prove this let B be a subcontinuum of A that does not belong to C(a, A) ∪ C(S 1 , A) ∪ C(S 1 ). Then B is an arc contained in the open set (a, ∞). Thus, C((a, ∞)) is an open set of C(X) and B ∈ C((a, ∞)) ⊂ C(A). Hence, B ∈ Int(C(A)). Therefore, B does not belong to Bd(C(A)). This proves that Bd(C(A)) is contained in C(a, A) ∪ C(S 1 , A) ∪ C(S 1 ). Conversely, given B ∈ C(a, A) we write Bn = B ∪ [a − n1 , a]; and given B ∈ C(S 1 , A) we write Bn = B ∪ (−∞, −n]. In both cases we have that {Bn }∞ n=1 is a sequence in C(X) − C(A) that converges to B. Hence B ∈ Bd(C(A)). This proves that C(a, A) ∪ C(S 1 , A) is contained in Bd(C(A)). If B ∈ C(S 1 ) then there exists a sequence of arcs in the ray (−∞, a] that converges to B. By this B ∈ Bd(C(A)). Therefore C(S 1 ) is contained in Bd(C(A)). This proves our assertion. Since lim An = A = S 1 ∪ [a, ∞), it is sufficient to analyze the following two cases: (3.1) For each n ∈ N, An = S 1 ∪ [an , ∞), where lim an = a; and (3.2) For each n ∈ N, An = S 1 ∪ [an , ∞) ∪ (−∞, bn ], where lim an = a and {bn }∞ n=1 is a decreasing sequence in (−∞, a] which does not have a lower bound. In case (3.1), in a similar way that we obtained Bd(C(A)), we have that Bd(C(An )) = C(an , A) ∪ C(S 1 , An ) ∪ C(S 1 ), where C(S 1 , An ) = {B ∈ C(An ) : S 1 ⊂ B}. Therefore, we can restrict ourselves to study the following three subcases: (3.1.1) For each n ∈ N, En ∈ C(an , An ): In this subcase we obtain that a ∈ E. Thus, E ∈ C(a, A). (3.1.2) For each n ∈ N, En ∈ C(S 1 , An ): In this subcase it follows that S 1 ⊂ E. Hence, E ∈ C(S 1 , A). (3.1.3) For each n ∈ N, En ∈ C(S 1 ): In this subcase we have that E ⊂ S 1 . Then E ∈ C(S 1 ).

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Therefore, in case (3.1), we conclude that E ∈ Bd(C(A)). In case (3.2) we can prove that Bd(C(An )) = C(an , An ) ∪C(bn , An ) for each n ∈ N. By this it is sufficient to consider two subcases: (3.2.1) For each n ∈ N, En ∈ C(an , An ): This condition implies that E ∈ C(a, A). (3.2.2) For each n ∈ N, En ∈ C(bn , An ): In this subcase, by compactness, we can assume that there exists 1 a point b ∈ X such that lim bn = b. By properties of the sequence {bn }∞ n=1 , we notice that b ∈ S . Besides, 1 1 since bn ∈ En , we observe that b ∈ E. Therefore E ∩ S = ∅. It follows that S ⊂ E or E ⊂ S 1 . Thus, E ∈ C(S 1 , A) ∪ C(S 1 ). By this, in case (3.2), we also obtain that E ∈ Bd(C(A)). In any situation into case (3), we have proved that E ∈ Bd(C(A)). (4) A = S 1 ∪ (−∞, a] ∪ [b, ∞), where a < 0 < b. In this case we can prove that Bd(C(A)) = C(a, A) ∪ C(b, A); and without loss of generality we may assume that An = S 1 ∪(−∞, an ] ∪[bn , ∞), where lim an = a and lim bn = b. We also have that Bd(C(An )) = C(an , An ) ∪ C(bn , An ). Thus, for each n ∈ N, an ∈ En or bn ∈ En . This implies that a ∈ E or b ∈ E. Thus, E ∈ C(a, A) ∪ C(b, A). Hence, E ∈ Bd(C(A)). Summarizing, in each case (1)-(4) we obtain that the continuum E belongs to Bd(C(A)), which is a contradiction with (∗). This proves that the boundary function is usc for the continuum X of Example 4.4. Next the main result of this section, it is analogous to the equivalence between (a) and (b) in Theorem 7.8 of [4]. In the last part of the proof of this result we use the notion of Whitney map. We recall that a Whitney map for C(X) is a continuous function μ : C(X) → [0, 1] such that μ({p}) = 0 for each p ∈ X, and μ(A) < μ(B) whenever A ⊂ B = A. It is known that Whitney maps exist for the hyperspace of any continuum [9, (0.50.1)–(0.50.3)]. Theorem 4.5. If X is an arcwise connected continuum, then the boundary function is continuous for X if and only if X is a simple closed curve. Proof. Necessity: First we will prove that X contains a simple closed curve. To do this we assume the contrary. It follows that X is uniquely arcwise connected, it means that for any two distinct points in X there exists a unique arc in X having them as its end points. Using this, with similar arguments that those given in the proof of (b) implies (a) in Theorem 7.8 of [4], it can be proved that each proper subcontinuum of X is arcwise connected. We assert that X is unicoherent. To prove this we assume that there exist two subcontinua A and B of X such that X = A ∪ B and A ∩ B is not connected. Let p and q be points in different components of A ∩ B. We can take an arc α in A, and an arc β in B, joining p and q. It follows that α ∪ β is a subcontinuum of X that contains a simple closed curve, which is a contradiction with our assumption at the beginning of this proof. This proves what we asserted. By Lemma 4.3, each proper subcontinuum of X is unicoherent. It follows that X is hereditarily unicoherent. Since X is arcwise connected, this implies that X is a dendroid. This is a contradiction with Lemma 4.2. This proves that X contains a simple closed curve. Let S be a simple closed curve in X. Since S is not unicoherent, by Lemma 4.3 it follows that S is not a proper subcontinuum of X. Thus, X = S. Sufficiency: Let A be an element of C(X) − {X} and let {An }∞ n=1 be a sequence in C(X) − {X} such that lim An = A. We can suppose that A, and each An is an arc in X. Let a and b be the end points of A; let an and bn be the end points of An . We may assume that lim an = a and lim bn = b. We observe that A, and each An , is a free arc in X. It follows that Bd(C(A)) = C(a, A) ∪ C(b, A) and Bd(C(An )) = C(an , An ) ∪ C(bn , An ) [3, Corollary 4.2]. By compactness, without loss of generality we can assume that the sequence {Bd(C(An ))}∞ n=1 has a limit in C(C(X)). We will prove that this limit coincides with Bd(C(A)).

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Let B be an element of lim Bd(C(An )). There exists a sequence {Bn }∞ n=1 in C(X) such that lim Bn = B and Bn ∈ Bd(C(An )) for each n ∈ N. We have that an ∈ Bn or bn ∈ Bn for each n ∈ N. This implies that a ∈ B or b ∈ B. Thus, B ∈ Bd(C(A)). Hence, lim Bd(C(An )) is contained in Bd(C(A)). Conversely, let B be an element of Bd(C(A)). We first assume that B = A, and we observe that An ∈ Bd(C(An )) [4, (c) and (e)]. Since lim An = A, we obtain that B ∈ lim Bd(C(An )). Next we suppose that B = A. Let μ be a Whitney map for C(X). Let t = μ(B). We notice that t < μ(A). By continuity of μ, we may assume that t < μ(An ) for each n ∈ N. It follows that there exists a subcontinuum Bn of An such that an ∈ Bn and μ(Bn ) = t for each n ∈ N. We observe that Bn ∈ Bd(C(An )) for each n ∈ N. Without loss of generality we may assume that the sequence {Bn }∞ n=1 has a limit in C(X). We will see that this limit is B. Since an ∈ Bn ⊂ An , we have that a ∈ lim Bn ⊂ A. Thus, B and lim Bn are subcontinua of the arc A having in common the end point a. It follows that lim Bn ⊂ B or B ⊂ lim Bn . Besides, since μ(lim Bn ) = lim μ(Bn ) = t, we have that μ(lim Bn ) = μ(B). This implies that lim Bn = B. This proves that B ∈ lim Bd(C(An )). Therefore, Bd(C(A)) is contained in lim Bd(C(An )). We have proved that lim Bd(C(An )) = Bd(C(A)). Thus, the boundary function is continuous for the simple closed curve. 2 Remark 4.6. There exist continua that are not arcwise connected for which the boundary function is continuous. To see this let us recall that a continuum X is said to be C ∗ -smooth provided that the function C ∗ : C(X) → C(C(X)) defined by C ∗ (A) = C(A) for each A ∈ C(X) is continuous [9, (15.5)]. Besides we notice that if X is a Kelley indecomposable continuum then Bd(C(A)) = C(A) for every A ∈ C(X) − {X}, thus for a such continuum the boundary function coincides with the function C ∗ in C(X) − {X}. Therefore, if X is a Kelley indecomposable C ∗ -smooth continuum then the boundary function is continuous for X. It is known that any hereditarily indecomposable continuum is C ∗ -smooth and is a Kelley continuum [9, (15.20) and (16.27)]. It is also known that any solenoid different to the circle is C ∗ -smooth and is a Kelley indecomposable continuum [9, (14.76.6) and (16.26)]. Hence, the boundary function is continuous for any hereditarily indecomposable continuum and for any solenoid different to the circle. Acknowledgements The author wishes to thank Professor Alejandro Illanes for his useful comments given to me during the preparation of this paper. The author also wishes to thank the referee for suggestions which led to the improvement of this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]

K. Borsuk, A theorem on fixed points, Bull. Acad. Pol. Sci. 2 (1954) 17–20. W.J. Charatonik, Some functions on hyperspaces of continua, Topol. Appl. 22 (1986) 211–221. C.G. Domínguez-López, Boundaries in hyperspaces, Topol. Appl. 167 (2014) 10–25. A. Illanes, Semi-boundaries in hyperspaces, Topol. Proc. 16 (1991) 63–87. A. Illanes, S.B. Nadler Jr., Hyperspaces: Fundamentals and Recent Advances, Monogr. Textbooks Pure Appl. Math., vol. 216, Marcel Dekker, Inc., New York, Basel, 1999. J.L. Kelley, Hyperspaces of a continuum, Trans. Am. Math. Soc. 52 (1942) 22–36. K. Kuratowski, Topology, vol. II, Academic Press, New York, London, 1968. S.B. Nadler Jr., Arcwise accessibility in hyperspaces, Diss. Math. 138 (1976). S.B. Nadler Jr., Hyperspaces of Sets, Monogr. Textbooks Pure Appl. Math., vol. 49, Marcel Dekker, Inc., New York, Basel, 1978. S.B. Nadler Jr., Continuum Theory: An Introduction, Monogr. Textbooks Pure Appl. Math., vol. 158, Marcel Dekker, Inc., New York, Basel, Hong Kong, 1992. J.T. Rogers Jr., Dimension of hyperspaces, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 20 (1972) 177–179.