On the n-fold symmetric product suspensions of a continuum

Topology and its Applications 157 (2010) 597–604 Contents lists available at ScienceDirect Topology and its Applications www.elsevier.com/locate/topol On the n-fold symmetric product suspensions of a continuum Franco Barragán 1 Facultad de Ciencias Físico Matemáticas, BUAP, Ave. San Claudio y Rio Verde, Ciudad Universitaria, San Manuel Puebla, Pue. C.P. 72570, Mexico a r t i c l e i n f o Article history: Received 30 October 2009 Accepted 31 October 2009 MSC: 54B20 a b s t r a c t In 1979 Sam B. Nadler, Jr. defined the hyperspace suspension of a continuum. We define the n-fold symmetric product suspensions of a continuum using n-fold symmetric products. We study some properties of this hyperspace: unicoherence, local connectedness, arcwise connectedness.  2009 Elsevier B.V. All rights reserved. Keywords: Arcwise connected Compactification Continuum Locally connected continuum Retraction Symmetric product Unicoherent continuum 1. Introduction In 1979 Sam B. Nadler, Jr. introduced the hyperspace suspension of a continuum [1]. In 2004 S. Macías, defined the n-fold hyperspace suspension of a continuum [2]. We define the n-fold symmetric product suspensions of a continuum, and the purpose of this paper is to present a study of some properties of this hyperspace. The paper is divided into six sections. In Section 2, we give the basic definitions for understanding the paper. In Section 3, we give examples of geometric models for the n-fold symmetric product suspensions of given continua. In Section 4, we prove that the n-fold symmetric product suspension of a continuum is unicoherent if n  3, and we present an example of a unicoherent continuum whose second symmetric product suspension is not unicoherent. In Section 5, we prove that a continuum is locally connected if and only if its n-fold symmetric product suspension is locally connected. In Section 6, we give a class of nonarcwise connected continua for which their n-fold symmetric product suspensions are arcwise connected. Also, we give a class of continua for which their n-fold symmetric product suspensions are not arcwise connected. 2. Definitions The symbols N, R and C will denote the set of positive integers, real numbers and complex numbers, respectively. A continuum is a nonempty compact, connected metric space. A subcontinuum is a continuum contained in a space X . If X is a continuum, then given A ⊂ X and ǫ > 0, the open ball about A of radius ǫ is denoted by Vǫ ( A ), and the closure of A in X by Cl X ( A ). A map means a continuous function. An onto map f : X → Y between continua is said to be monotone provided that for any point y of Y , f −1 ( y ) is a connected subset of X . 1 E-mail address: [email protected]. This research is part of the author’s Dissertation under the supervision of Professors Raúl Escobedo and Sergio Macías. 0166-8641/$ – see front matter doi:10.1016/j.topol.2009.10.017  2009 Elsevier B.V. All rights reserved. 598 F. Barragán / Topology and its Applications 157 (2010) 597–604 The symbol I will denote the closed interval [0, 1]. An arc is any space which is homeomorphic to I . A simple closed curve is a space homeomorphic to S 1 = {e it ∈ C: t ∈ R}. A ray is a space homeomorphic to [0, ∞). A continuum X is unicoherent provided that if X = A ∪ B, where A and B are subcontinua of X , then A ∩ B is connected. For each topological space Y , we define b0 (Y ) = (number of components of Y ) − 1 if this number is finite, and b0 (Y ) = ∞ otherwise. The multicoherence degree, r ( X ), of a continuum X is defined by:   r ( X ) = sup b0 ( H ∩ K ): H and K are subcontinua of X and X = H ∪ K . Given a continuum X and n ∈ N, the product of X with itself n times will be denoted by X n , the symbol F n ( X ) denotes the n-fold symmetric product of X ; that is: F n ( X ) = { A ⊂ X | A has at most n points} topologized with the Hausdorff metric, which is defined as follows:    H( A , B ) = inf ǫ > 0  A ⊂ Vǫ ( B ) and B ⊂ Vǫ ( A ) . Given a finite collection, U 1 , . . . , U m , of subsets of X , U 1 , . . . , U m n , denotes the follows subset of F n ( X ):   m    A ∈ Fn( X )  A ⊂ U k and A ∩ U k = ∅ for each k ∈ {1, . . . , m} . k =1 It is known that the family of all subsets of F n ( X ) of the form U 1 , . . . , U m n , where each U i is an open subset of X , forms a basis for a topology for F n ( X ) (see [3, 0.11]) called the Vietoris topology, and that the Vietoris topology and the topology induced by the Hausdorff metric coincide [3, 0.13]. Given a continuum X and n ∈ N with n  2, we define the n-fold symmetric product suspension of the continuum X , denoted by SF n ( X ), by the quotient space: SF n ( X ) = F n ( X )/ F 1 ( X ) with the quotient topology. The fact that SF n ( X ) is a continuum follows from 3.10 of [4]. Notation 2.1. Given a continuum X , qnX : F n ( X ) → SF n ( X ) denotes the quotient map. Also, let F nX denotes the point qnX ( F 1 ( X )). Remark 2.2. Note that SF n ( X ) \ { F nX } is homeomorphic to F n ( X ) \ F 1 ( X ), using the appropriate restriction of qnX . 3. Examples In this section we present examples of geometric models of n-fold symmetric product suspensions for some given continua. Example 3.1. K. Borsuk and S. Ulam proved that F 2 ( I ) is homeomorphic to I 2 [5, Theorem 6]. It is known actually that there exists a homeomorphism from F 2 ( I ) onto the triangle in the Euclidean plane which has as vertices the points (0, 0), (0, 1) and (1, 1) [6, Example 3.7] such that F 1 ( I ) is homeomorphic to the segment that joins to the points (0, 0) and (1, 1). If we identify F 1 ( I ) to a point, we obtain a space homeomorphic to this triangle. Hence, SF 2 ( I ) is homeomorphic to this triangle. Thus, SF 2 ( I ) is homeomorphic to F 2 ( I ). Therefore, SF 2 ( I ) is homeomorphic to I 2 . A simple triod T is a continuum which is the union of three arcs which have only one point in common. We construct a model for the hyperspace SF 2 ( T ). Example 3.2. By [6, Example 3.8], F 2 ( T ) is homeomorphic to the Fig. 1. Which is a 2-cell D 0 with three 2-cells D 1 , D 2 and D 3 glued in such a way that D 0 ∩ D j is an arc, j ∈ {1, 2, 3}, and D 1 ∩ D 2 ∩ D 3 is a single point p. Note that F 1 ( T ) is contained in the manifold boundaries of D 1 , D 2 , D 3 and F 1 ( T ) ∩ D 0 = { p }. Hence, if we identify F 1 ( T ) to a point in D 1 ∪ D 2 ∪ D 3 , we obtain a space homeomorphic to D 1 ∪ D 2 ∪ D 3 . Therefore, SF 2 ( T ) is homeomorphic to F 2 ( T ). In Examples 3.1 and 3.2, we give continua such that their respective 2-fold symmetric product and 2-fold symmetric product suspension are homeomorphic. In the following example we will prove that F 2 ( S 1 ) and SF 2 ( S 1 ) are topologically different. This example may be suffice to show that it is interesting to study the n-fold symmetric product suspensions of a continuum. F. Barragán / Topology and its Applications 157 (2010) 597–604 599 Fig. 1. Geometric model of the 2-fold symmetric product suspension of the simple triod. Example 3.3. It is known that F 2 ( S 1 ) is homeomorphic to the Möbius strip [5, p. 877], and F 1 ( S 1 ) is homeomorphic to the manifold boundary of the Möbius strip. Since it is well known that if we identify the manifold boundary of the Möbius strip to a point, we obtain the real projective plane RP2 , we have that SF 2 ( S 1 ) is homeomorphic to RP2 . Remark 3.4. Observe that S 1 is an example of a nonunicoherent continuum such that F 2 ( S 1 ) is not unicoherent but SF 2 ( S 1 ) is unicoherent [7, p. 197]. In Example 4.4 we present a unicoherent continuum X such that SF 2 ( X ) is not unicoherent. A Hilbert cube, denoted by Q , is a continuum which is homeomorphic to the countable cartesian product each I i = I , with the product topology [4, 1.4]. ∞ i =1 I i , where Example 3.5. If Q is the Hilbert cube, then SF n ( Q ) is the Hilbert cube, for each n ∈ N with n  2. To see this let n  2. Then F n ( Q ) is homeomorphic to Q [8, Theorem 2.4]. Since Q is contractible, we have that F 1 ( Q ) is contractible. Since F 1 ( Q ) is nowhere dense in F n ( Q ), it follows that F 1 ( Q ) has the shape of a point (in the sense of Borsuk) [9, 5.5, p. 28]. Hence, F n ( Q ) \ F 1 ( Q ) is homeomorphic to Q \ { p }, for some point p of Q [10, 25.2]. Since F n ( Q ) \ F 1 ( Q ) is homeomorphic to SF n ( Q ) \ { F nQ }, we obtain that Q \ { p } is homeomorphic to SF n ( Q ) \ { F nQ }. Therefore, SF n ( Q ) is homeomorphic to Q [11, 2.23]. A retraction is a continuous function, r, from a space, X , into itself such that r is the identity on its range (i.e., r (r (x)) = r (x) for each x ∈ X ). A closed subset A of X is said to be a retract of X provided that there is a retraction of X onto A. A space X is called an absolute retract provided that X is a retract of every space Y containing X as a closed subset. Theorem 3.6. If X is an absolute retract and n ∈ N with n  2, then SF n ( X ) is an absolute retract. Proof. Without loss of generality, we assume that X is embedded in the Hilbert cube Q [12, 1.1.16]. Since X is an absolute retract, there exists a retraction r : Q → X . It is easy to verify that the induced function SF n (r ) : SF n ( Q ) → SF n ( X ), which can be defined by SF n (r )(χ ) =  qnX ( F n (r )((qnQ )−1 (χ ))), if χ = F nQ ; F nX , if χ = F nQ , is a retraction. Since SF n ( Q ) is homeomorphic to Q (Example 3.5), we have that SF n ( X ) is an absolute retract [13, Theorem 7, p. 341]. ✷ Since absolute retracts have the fixed point property [13, Theorem 11, p. 343], we have: Corollary 3.7. Let X be an absolute retract and n ∈ N with n  2. Then SF n ( X ) has the fixed point property. 4. Unicoherence Theorem 4.1. Let X be a continuum and let n  3 be a positive integer. Then SF n ( X ) is a unicoherent continuum. Proof. By [14, Theorem 8], F n ( X ) is unicoherent. We note that qnX is a monotone map. Then, by [4, Corollary 13.34], we have that SF n ( X ) is a unicoherent continuum. ✷ For the case n = 2, it is known that r ( F 2 ( X ))  1 (see [14, p. 181]). Since qnX is a monotone map, by [4, Theorem 13.33], we have that r (SF 2 ( X ))  1. Hence, we have the following: Theorem 4.2. Let X be a continuum. Then r (SF 2 ( X ))  1. 600 F. Barragán / Topology and its Applications 157 (2010) 597–604 Fig. 2. Continuum for which their 2-fold symmetric product suspension is not unicoherent. In Example 4.4 we will present a continuum X such that r (SF 2 ( X )) = 1. In order to give this example, first we will prove the following lemma: Lemma 4.3. Let X be a continuum and let A, B, M be subcontinua of X such that X = A ∪ B and A ∩ B is not connected. If there is a component K of A ∩ B such that K ∩ M = ∅, then the quotient space X / M is not unicoherent. Proof. Let K be a component of A ∩ B such that K ∩ M = ∅. We will consider two cases: (1) Suppose that no component of A ∩ B intersects M. Then M ⊂ X \ ( A ∩ B ). Considering the quotient map, q M : X → X / M, we have that X / M = q M ( A ) ∪ q M ( B ), q M ( A ∩ B ) = q M ( A ) ∩ q M ( B ) and q M ( A ∩ B ) is homeomorphic to A ∩ B. Since A ∩ B is not connected, q M ( A ) ∩ q M ( B ) is not connected. Therefore, X / M is not unicoherent. (2) Suppose that there exists a component L of A ∩ B such that L ∩ M = ∅. It is clear that q M ( A ) and q M ( B ) are subcontinua of X / M such that X / M = q M ( A ) ∪ q M ( B ). We will prove that q M ( A ) ∩ q M ( B ) is not connected. Let N=  {C : C is a component of A ∩ B and C ∩ M = ∅}. We have that L ⊂ N, thus N = ∅. We note that N is a closed subset of X and no connected subset of A ∩ B intersects both N and K . Then, by [4, Theorem 5.2], there exist two closed subsets P and Q of A ∩ B such that A ∩ B = P ∪ Q , K ⊂ P , N ⊂ Q and P ∩ Q = ∅. We have that q M ( A )∩ q M ( B ) = q M ( P )∪ q M ( Q ). Since P ∩ M = ∅, it follows that q M ( P )∩ q M ( Q ) = ∅. Hence, q M ( A )∩ q M ( B ) is not connected. Therefore X / M is not unicoherent. ✷ E. Castañeda [15, Example 2.1] gives a unicoherent continuum X whose F 2 ( X ) is not unicoherent. With very similar techniques, we give another proof of that F 2 ( X ) is not unicoherent. We use our proof to verify that SF 2 ( X ) is not unicoherent. t + 2)e it ∈ C: t ∈ R} and X = S 1 ∪ S 2 ∪ Y . Then X is a unicoherent Example 4.4. Let S 2 = {3e it ∈ C: t ∈ R}, Y = {( 1+| t| continuum (see Fig. 2) such that SF 2 ( X ) is not unicoherent. We define z A = { z, w } ∈ F 2 ( X ): Re w z B = { z, w } ∈ F 2 ( X ): Re w 0 , 0 , where Re( wz ) is the real part of the complex number wz . Since Re( wz )  0 if and only if Re( wz )  0, we have that A and B are well defined. First, we prove that A is compact. We define the following subset of X 2 : A1 = ( z, w ) ∈ X 2 : Re z w 0 . F. Barragán / Topology and its Applications 157 (2010) 597–604 601 We note that A1 is a closed subset of X 2 . We consider the map f 2 : X 2 → F 2 ( X ) given by f 2 ( z, w ) = { z, w }, for each (z, w ) ∈ X 2 (see [16, Lemma 1]). It follows that f 2 (A1 ) = A. This implies that A is a closed subset of F 2 ( X ); and hence, compact. In a similar way, we have that B is compact. To prove that A is connected, we define D = {{ z, w } ∈ A: z, w ∈ S 1 }, E = {{ z, w } ∈ A: z, w ∈ Y } and F = D ∪ E . We see that D is connected. We take the point {1} ∈ D , and let P = { z, w } ∈ D . Then there exist r , s ∈ R such that z = e ir and w = e is . Without loss of generality, we assume that r  s. Since P ∈ A, it follows that Re( wz ) = Re(e i (r −s) )  0. Thus, cos(r − s)  0. Then we have two possibilities: there exists m ∈ N ∪ {0} such that −π ( 4m2+5 )  r − s  −π ( 4m2+3 ) or − π2  r − s  0. We suppose that −π ( 4m2+5 )  r − s  −π ( 4m2+3 ) (similarly, if − π2  r − s  0). Since −π ( 4m2+5 )  −2π (m + 1)  −π ( 4m2+3 ), we can assume that −π ( 4m2+5 )  r − s  −2π (m + 1) (similarly, if −2π (m + 1)  r − s  −π ( 4m2+3 )). Then − π4  r −s+2π2 (m+1)  0. We define M=  e i (r −t ) ,e i (s+t )  ∈ F 2 ( X ): t ∈ r − s + 2π (m + 1) 2 ,0  . Note that M is connected. To prove that M is a subset of A, let t ∈ [ r −s+2π (m+1) , 0]. 2 Then − π4  t  0, and hence, 0  i (r −t ) −2t  π2 . Thus, −π ( 4m2+5 )  −2t + r − s  −π ( 4m2+3 ). This implies that cos(−2t + r − s)  0. It follows that Re( eei(s+t ) )  0. r +s−2π (m+1) r − s + 2 π ( m + 1 ) il Consequently M ⊂ A. Hence, M ⊂ D . Let l = Let N= and Q = {e }. Thus, for t = 2 2 , Q ∈ M. Also, P ∈ M. e i (l+t ) ∈ F 2 ( X ): t ∈ R .    Then N is a connected subset of D such that Q , {1} ∈ N . This implies that M ∪ N is a connected subset of D containing the points P and {1}. Therefore, D is connected. On the other hand, let P = { z, w } ∈ E . To prove that F is connected it is sufficient to verify that there exists a connected r subset P of F containing the point P and P ∩ D = ∅. Let r , s ∈ R such that z = ( 1+| + 2)e ir and w = ( 1+|s s| + 2)e is . We r| assume that r  s. Since P ∈ A, it follows that cos(r − s)  0. Hence, either there exists m ∈ N ∪ {0} such that −π ( 4m2+5 )  r − s  −π ( 4m2+3 ) or − π2  r − s  0. We suppose that −π ( 4m2+5 )  r − s  −π ( 4m2+3 ) (similarly, if − π2  r − s  0). We assume that −π ( 4m2+5 )  r − s  −2π (m + 1) (a similar argument, if −2π (m + 1)  r − s  −π ( 4m2+3 )). Let M= r −t 1 + |r − t | +2 e i (r −t ) , s+t 1 + |s + t | r +s−2π (m+1) +2 e i (s+t ) ∈ F 2 ( X ): t ∈ r − s + 2π (m + 1) l+2π (m+1) l Let l = and Q = {( 1+| + 2)e il , ( 1+|l+2π (m+1)| + 2)e i (l+2π (m+1) )}. Hence, for t = 2 l| is a connected subset of F such that P , Q ∈ M. Let N= l+t 1 + |l + t | + 2 e i (l+t ) , l + 2π (m + 1) + t 1 + |l + 2π (m + 1) + t | 2 ,0  r −s+2π (m+1) , 2 . Q ∈ M. Then M + 2 e i (l+2π (m+1)+t ) : t ∈ R . We note that N is a connected subset of E such that Q ∈ N . For each j ∈ N, let t j = −2 j π and Bj = l +tj 1 + |l + t j | + 2 e i (l+t j ) , l + 2π (m + 1) + t j 1 + |l + 2π (m + 1) + t j | + 2 e i (l+2π (m+1)+t j ) . Hence, { B j }∞ is a sequence contained in N converging to {e il }. Then {e il } ∈ Cl F 2 ( X ) (N ). It follows that, N ∪ {e il } is a j connected subset of F . Thus, P = M ∪ N ∪ {e il } is a connected subset of F , P ∈ P and P ∩ D = ∅. Therefore, F is connected. Note that A = Cl F 2 ( X ) (F ). Since F is connected, we have that A is connected. Hence, A is a subcontinuum of F 2 ( X ). With similar arguments to the ones given to prove that A is connected, we can verify that B is connected. Therefore, B is a subcontinuum of F 2 ( X ). To see that A ∩ B is not connected, we define H = { z, w } ∈ F 2 ( X ): K = { z, w } ∈ F 2 ( X ): z w z w = it , t  0 , = it , t  0 . Note that H and K are closed subsets of F 2 ( X ) such that A ∩ B = H ∪ K and H ∩ K = ∅. Hence, A ∩ B is not connected. Since F 2 ( X ) = A ∪ B , it follows that F 2 ( X ) is not unicoherent. On the other hand, since F 1 ( X ) ∩ B = ∅, by Lemma 4.3, we have that SF 2 ( X ) is not unicoherent. 602 F. Barragán / Topology and its Applications 157 (2010) 597–604 5. Local connectedness In this section we present results about of the n-fold symmetric product suspensions of locally connected continua. Lemma 5.1. Let X be a continuum and let n  2 be a positive integer. If F n ( X ) \ F 1 ( X ) is locally connected, then X is locally connected. Proof. Let x be a point in X and let U be an open subset of X such that x ∈ U . Let δ > 0 be such that the open ball about x of radius δ , V δ (x), is contained in U . By [4, Corollary 5.5], there exists a subcontinuum K of X such that {x}  K ⊂ V δ (x). Let y ∈ K be such that x = y. Let α > 0 be such that V α (x) ∩ V α ( y ) = ∅. We denote A = {x, y } and let ǫ > 0 be such that ǫ < min{δ, α }. Let U 1 = V ǫ (x) and U 2 = V ǫ ( y ). Hence, U 1 , U 2 n is an open subset of F n ( X ). Then U 1 , U 2 n ∩ ( F n ( X ) \ F 1 ( X )) is an open subset of F n ( X ) \ F 1 ( X ) such that A ∈ U 1 , U 2 n ∩ ( F n ( X ) \ F 1 ( X )). Since F n ( X ) \ F 1 ( X ) is locally connected, there exists a connected open 1 ( X )). Then,  subset C of F n ( X ) \ F 1 ( X ) (andhence of F n ( X )) such that A ∈ C ⊂ U 1 , U 2 n ∩ ( F n ( X ) \ F by [17, Lemma 6.1], C is an open subset ofX and ( C ) ∩ U 1 is a connected subset of X . This implies that ( C ) ∩ U 1 is a connected open subset of X such that x ∈ ( C ) ∩ U 1 ⊂ U . It follows that X is locally connected at the point x. Therefore, X is locally connected. ✷ Theorem 5.2. A continuum X is locally connected if and only if SF n ( X ) is locally connected for each positive integer n  2. Proof. We suppose that X is locally connected. By [16, Lemma 2], we have that F n ( X ) is locally connected. Since qnX ( F n ( X )) = SF n ( X ), by [4, Proposition 8.16], it follows that SF n ( X ) is locally connected. If SF n ( X ) is locally connected, then, by [13, Theorem 3, p. 230], SF n ( X ) \ { F nX } is locally connected. Since SF n ( X ) \ { F nX } is homeomorphic to F n ( X ) \ F 1 ( X ), we have that F n ( X ) \ F 1 ( X ) is locally connected. Therefore, by Lemma 5.1, X is locally connected. ✷ 6. Arcwise connectedness Given an arc A, let α : [0, 1] → A be a homeomorphism and let α (0) = p and α (1) = q. In this case, we say that A is an arc from p to q (the end points of A), or we say that A is an arc joining the points p and q. A continuum X is said to be arcwise connected provided that any two points can be joined by an arc contained in X . Let X be a continuum. It is known that if X is an arcwise connected continuum, then F n ( X ) is an arcwise connected continuum [18, Proposition 2.7]. Hence, we have the following: Theorem 6.1. Let X be an arcwise connected continuum and let n  2 be a positive integer. Then SF n ( X ) is an arcwise connected continuum. Theorem 6.2. Let X be a continuum and let n  2 be a positive integer. Then the following are equivalent: 1. X contains an arc. 2. F n ( X ) contains an arc. 3. SF n ( X ) contains an arc. Proof. If X contains an arc, then X contains two disjoint arcs, from which we see that F n ( X ) contains an arc missing F 1 ( X ); clearly, then SF n ( X ) contains an arc. Hence, (1) implies (3). (3) implies that there is an arc Γ in SF n ( X ) \ { F nX }; hence, (qnX )−1 (Γ )  is an arc in F n ( X ). Now we assumethat A is an arc in F n ( X ). Then, by [19, Lemma 2.2], A is a nondegenerate locally connected compact A contains an arc [4, Theorem 8.23]. This proves the theorem. ✷ subset of X . Thus, As an easy consequence of Theorem 6.2, we have the following: Remark 6.3. Let X be a continuum and let n  2. If either X is the pseudo-arc [4, 1.23], or X is hereditarily indecomposable [4, 1.23], or X is constructed as in [4, 2.27] which is a hereditarily decomposable continuum that contains no arc, then SF n ( X ) is not arcwise connected. The following theorem proves that the converse of Theorem 6.1 is not true. Theorem 6.4. Let X be a compactification of [0, ∞) with a nondegenerate arcwise connected continuum L as remainder, and let n  2 be a positive integer. If there exists a retraction r : X → L, then SF n ( X ) is an arcwise connected continuum. F. Barragán / Topology and its Applications 157 (2010) 597–604 603 Proof. Let A ∈ SF n ( X ) \ { F nX }. To prove that SF n ( X ) is arcwise connected, we will check that there exists an arcwise connected subset of SF n ( X ) containing the points A and F nX . Let A ∈ F n ( X ) \ F 1 ( X ) such that qnX ( A ) = A. Using the arc components of X , namely L and X \ L, we can find a point B ∈ F 2 ( X ) \ F 1 ( X ) and an arcwise connected subset Γ1 of F n ( X ) such that A , B ∈ Γ1 . Hence, qnX (Γ1 ) is an arcwise connected subset of SF n ( X ) joining to A and qnX ( B ). Since X have two arc components, we have the following cases: Case (1). B is in a single arc component of X . Let Γ2 be an arcwise connected subset of F n ( X ) joining to B and F 1 ( X ). Hence, qnX (Γ1 ) ∪ qnX (Γ2 ) is an arcwise connected subset of SF n ( X ) joining to A and F nX . Case (2). B ∩ L = ∅ and B ∩ ( X \ L ) = ∅. We assume that B ∩ ( X \ L ) = { p } and B ∩ L = {q}. Let γ : [0, 1) → ( X \ L ) be a homeomorphism. Let Γ2 be an arcwise connected subset of F n ( X ) joining to { p , r ( p )} and B = { p , q}. It follows that qnX (Γ2 ) is an arcwise connected subset of SF n ( X ) joining to qnX ({ p , r ( p )}) and qnX ( B ). We consider the map δ : [0, 1) → F n ( X ) be given by δ(t ) = {γ (t ), r (γ (t ))}. Let M = δ([0, 1)). We note that δ is a homeomorphism onto M such that { p , r ( p )} ∈ M. Let Γ3 = Cl F n ( X ) (M). Then Γ3 is a compactification of [0, 1). Let R = Γ3 \ M be the remainder. Since R ⊂ F 1 ( X ), we have that qnX (R) = { F nX }. Also, since M ∩ F 1 ( X ) = ∅, qnX (M) is a ray. Then qnX (Γ3 ) = qnX (M) ∪ { F nX } is an arc joining the points qnX ({ p , r ( p )}) and F nX . Thus, qnX (Γ1 ) ∪ qnX (Γ2 ) ∪ qnX (Γ3 ) is an arcwise connected subset of SF n ( X ) containing to A and F nX . Therefore, SF n ( X ) is an arcwise connected continuum. ✷ The following corollary is an easy consequence of Theorem 6.4 and [20, p. 30]: Corollary 6.5. Let X be a compactification of [0, ∞) with a nondegenerate locally connected continuum as remainder, and let n  2 be a positive integer. Then SF n ( X ) is an arcwise connected continuum. We define S + = {(1 + e −θ , θ): θ  0}, S − = {(1 + e −θ , −θ): θ  0}, and let W = S 1 ∪ S + ∪ S −. The Waraszkiewicz spirals are subcontinua of W [12, 2.6.35]. However, W is not a Waraszkiewicz spiral. Remark 6.6. Note that Theorem 6.4 implies that the n-fold symmetric product suspensions of the Elsa continua [21, p. 329] and the Waraszkiewicz spirals [22] are arcwise connected continua. Next we will prove that SF n ( W ) is an arcwise connected continuum. Example 6.7. Let n ∈ N be such that n  2. Then SF n ( W ) is an arcwise connected continuum. n To see this let A ∈ SF n ( W ) \ { F W }. Then there exists A ∈ F n ( W ) \ F 1 ( W ) such that qnX ( A ) = A. Let γ : [0, 1) → S + be a homeomorphism, p = γ (0) and q = φ( p ), where φ : W → S 1 is the projection map given by φ(r , θ) = (1, θ). We have the following cases: Case (1). A is in a single arc component of X . Then there exists an arcwise connected subset of SF n ( X ) joining to A and F nX . Case (2). A ∩ S 1 = ∅ y A ∩ ( S + ∪ S − ) = ∅. Since S 1 and S + ∪ S − are arcwise connected subsets of W , we have an arcwise connected subset Γ1 of F n ( W ) such that A , { p , q} ∈ Γ1 . It follows that qnW (Γ1 ) is an arcwise connected subset of SF n ( W ) containing to A and qnW ({ p , q}). Let δ : [0, 1) → F n ( W ) be the map given by δ(t ) = {φ(γ (t )), γ (t )}. Let M = δ([0, 1)). Then δ is a homeomorphism onto M such that { p , q} ∈ M. Let Γ2 = Cl F n (W ) (M). Thus, Γ2 is a compactification of [0, 1). Let R = Γ2 \ M be the remainder. n Note that R ⊂ F 1 ( W ) and M ∩ F 1 ( W ) = ∅. It follows that qnW (Γ2 ) is an arc joining the points qnW ({ p , q}) and F W . Hence, n qnW (Γ1 ) ∪ qnW (Γ2 ) is an arcwise connected subset of SF n ( W ) containing to A and F W . Therefore, SF n ( W ) is an arcwise connected continuum. With the next theorem we give a class of continua, for which their n-fold symmetric product suspensions are not arcwise connected. Theorem 6.8. Let X be a continuum, and let n ∈ N be such that n  2. If there exist two different closed arc components of X , then SF n ( X ) is not arcwise connected. Proof. Let L 1 and L 2 be two different closed arc components of X . Let a1 ∈ L 1 , a2 ∈ L 2 , and let A = {a1 , a2 }. Let A be an arc in F n ( X ) such that A ∈ A. First  we will prove that A ⊂  L 1 , L 2 n . By [19, Lemma 2.2], wehave that A is a compact locally connected subset of X . If A is a subcontinuum of X , then, by [4, Theorem A is an arcwise connected  8.23], A is not a subcontinuum of X . On the subcontinuum of X . This implies that L 1 ∩ L 2 = ∅, this is a contradiction. Hence, 604 F. Barragán / Topology and its Applications 157 (2010) 597–604   other  hand, by [17, Lemma 2.2], A has at most two components. Let A = C 1 ∪ C 2 , where C 1 and C 2 are the components A. We note that C 1 and C 2 are arcwise connected subcontinua of X . Thus, neitherA ⊂ C 1 nor A ⊂ C 2 . Hence, we can of A ⊂ L1 ∪ L2 . assume that a1 ∈ C 1 and a2 ∈ C 2 . This implies that C 1 ⊂ L 1 and C 2 ⊂ L 2 . It follows that Now let B ∈ A. Then B ⊂ L 1 ∪ L 2 . If B ∩ L 1 = ∅, then B ⊂ L 2 . Let b ∈ B. Since F n ( L 2 ) is an arcwise connected subset of F n ( X ) [18, Proposition 2.7], there exists an arc B contained in F n ( X ) joining to B and {b}. It follows that A ∪ B isa locally connected subcontinuum [13, Theorem 1, p. 230] of F n ( X ). By [19, Lemma 2.2] and [17, Lemma 2.2], we have that (A ∪ B ) X is a locally connected subcontinuum of . Hence, by [4, Theorem 8.23], (A ∪ B ) is an arcwise connected subcontinuum  of X such that a1 , a2 ∈ (A ∪ B ). Therefore, we have a contradiction. Hence B ∩ L 1 = ∅. Similarly B ∩ L 2 = ∅. This implies that B ∈  L 1 , L 2 n . Thus, A ⊂  L 1 , L 2 n . Therefore, if A is an arc in F n ( X ) such that A ∈ A, then A ⊂  L 1 , L 2 n . Since  L 1 , L 2 n is closed subset of F n ( X ) and  L 1 , L 2 n ∩ F 1 ( X ) = ∅, let U be an open subset of F n ( X ) such that F 1 ( X ) ⊂ U and U ∩  L 1 , L 2 n = ∅. It follows that qnX (U ) is an open subset of SF n ( X ) such that F nX ∈ qnX (U ). Suppose that SF n ( X ) is arcwise connected. Since A ∈ F n ( X ) \ F 1 ( X ), qnX ( A ) ∈ SF n ( X ) \ { F nX }. Let Γ be an arc in SF n ( X ) joining to qnX ( A ) and F nX , and let α : [0, 1] → Γ be a homeomorphism such that α (0) = qnX ( A ) and α (1) = F nX . We take a point r ∈ [0, 1) such that α (r ) ∈ qnX (U ) \ { F nX }. Let D ∈ F n ( X ) such that qnX ( D ) = α (r ). We consider the subarc α ([0, r ]) of Γ . Then (qnX )−1 (α ([0, r ])) is an arc contained in F n ( X ) such that A , D ∈ (qnX )−1 (α ([0, r ])). Then (qnX )−1 (α ([0, r ])) ⊂  L 1 , L 2 n . On the other hand, D ∈ U = (qnX )−1 (qnX (U )). Thus U ∩  L 1 , L 2 n = ∅. Therefore, we have a contradiction. This implies that SF n ( X ) is not arcwise connected. ✷ Remark 6.9. Let X be the continuum of Example 4.4 and let n  2 be a positive integer. As a consequence of Theorem 6.8, we have that SF n ( X ) is not arcwise connected. On the other hand, related to Theorem 6.4, we note that there exist compactifications of [0, ∞) whose n-fold symmetric product suspensions are not arcwise connected, namely: let Z be a compactification of [0, ∞) with the continuum X of Example 4.4 as remainder. By Theorem 6.8, it follows that SF n ( Z ) is not arcwise connected. 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