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The hyperspace of regular subcontinua
Accepted Manuscript The Hyperspace of Regular Subcontinua
Norberto Ordoñez
PII: DOI: Reference:
S0166-8641(17)30631-4 https://doi.org/10.1016/j.topol.2017.11.038 TOPOL 6340
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Topology and its Applications
Received date: Revised date: Accepted date:
2 December 2016 15 July 2017 29 November 2017
Please cite this article in press as: N. Ordoñez, The Hyperspace of Regular Subcontinua, Topol. Appl. (2017), https://doi.org/10.1016/j.topol.2017.11.038
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The Hyperspace of Regular Subcontinua Norberto Ordo˜ nez Universidad Aut´ onoma del Estado de M´exico, Facultad de Ciencias, Instituto Literario No. 100, Col. Centro, Toluca, Estado de M´exico, M´exico, C. P. 50000.
Abstract Given a metric continuum X, we define the hyperspace of regular subcontinua of X as the collection of all regular closed subcontinua of X. This is a new hyperspace for X. In this paper we study the connectedness, compactness, arcwise connectedness and contractibility of this hyperspace. Also we raise some open questions. Keywords: Continuum, Hyperspaces, Regular Set, Regular Subcontinua. 2010 MSC: Primary: 54B20, Secondary: 54B05, 54F15. 1. Introduction A continuum is a nonempty compact connected metric space. Given a continuum X, by a hyperspace of X we mean a specified collection of subsets of X endowed with the Hausdorff metric. The hyperspaces are a powerful tool to study the continuum for which they are defined and, of course, they provide new problems to solve. Some hyperspaces for a continuum X are 2X the hyperspace of all nonempty closed subsets of X, C(X) the hyperspace of all connected elements of 2X and F1 (X) the hyperspace of all one-point sets of X. A subcontinuum is an element of C(X). These hyperspaces have important properties, for example, 2X and C(X) are always arcwise connected continua [1, Corollary 14.10] and unicoherent [1, Theorem 19.8]. Also if X is locally connected, then 2X and C(X) are absolute retracts [1, Theorem 10.8] and contractible [1, Corollary 20.14]. And F1 (X) is an isometric copy of X. For more information related with these and other hypespaces see [1]. Email address: [email protected] (Norberto Ordo˜ nez) Preprint submitted to Elsevier
November 30, 2017
To introduce the hyperspace that we study in this paper, recall that for a topological space X, we say that a nonempty closed subset A of X is a regular closed provided that closure of the interior of A is equal to A. Thus, given a continuum X, we define: The hyperspace of regular subcontinua of X as D(X) = {A ∈ C(X) : A is regular closed in X}. For the study of D(X), this paper is organized as follow. In Section 2, we introduce general notation and definitions that we will use along the paper. We also enunciate known results in hyperspaces. In Section 3 we study the connectedness of D(X). We show a dendroid X for which D(X) is not connected and we prove that if X is locally connected, then D(X) is locally connected, contractible and a dense subset in C(X). In Section 4, we study the compactness of D(X). The main result that we prove is that if D(X) is infinite, then D(X) is not compact and thus we obtain, as a corollary, that D(X) is compact if an only if D(X) is finite. In Section 5, we show a decomposable continuum X for which D(X) = {X}. Along the paper we also raise some open questions. 2. Preliminaries N will denote the natural numbers. If X is a topological space and Y ⊂ X, X then Y , BdX (Y ) and IntX (Y ) denote the closure, the boundary and the interior of Y in X, respectively. We will write just Y , Bd(Y ) and Int(Y ) when the space where we calculate these is understood. We introduce the concept of the Hausdorff metric. Given a continuum X with metric d, x ∈ X, A ⊂ X and ε > 0, we define the open ball ofradius ε with center x as Bεd (x) = {y ∈ X : d(x, y) < ε} and let N (ε, A) = {Bεd (a) : a ∈ A}. If A and B are two closed subsets of X, the Hausdorff distance between A and B is defined by: H(A, B) = inf{ε > 0 : A ⊂ N (ε, B) and B ⊂ N (ε, A)}. It is known that H is a Metric for 2X when X is a continuum (see [1, Theorem 2.2]) and H is called the Hausdorff metric. Additionally, if we make the convention that N (0, A) = A for every subset A of X, then the following lemmas are easy to prove: 2
Lemma 2.1. Let X be a continuum. If t, s ∈ [0, 1] and A ∈ C(X), then N (t, N (s, A)) = N (t + s, A). Lemma 2.2. Let X be a continuum, A, B ∈ C(X) and ε > 0. Then H(A, B) < ε if and only if A ⊂ N (ε, B) and B ⊂ N (ε, A). A convex metric for a space X is a metric d for X that induces the topology on X, which satisfies that for any x, y ∈ X there exists z ∈ X such that d(x, z) = 12 d(x, y) = d(z, y). It is known that every locally connected continuum admits a convex metric (see [1, Theorem 10.3]). Let X be a continuum with a convex metric and A ∈ 2X . Then N (ε, A) = {x ∈ X : d(x, a) ≤ ε for some a ∈ A} (see [1, Exercise 10.11]) and hence by [3, 0.65.3], we have the following result. Lemma 2.3. Let X be a continuum with a convex metric. The map K : [0, 1] × C(X) → 2X defined by K((t, A)) = N (t, A) is continuous. The proof of the following lemma is easy. Lemma 2.4. Let X be a continuum. If U is a connected open subset of X, then U ∈ D(X). Lemma 2.5. Let X be a continuum with a convex metric and let A ∈ C(X). If ε > 0, then N (ε, A) ∈ D(X). Proof. As a consequence of [1, Proposition 10.4], we have that N (ε, A) is connected and since it is an open subset of X, the result follows from Lemma 2.4. A continuum X is said to be: • connected im kleinen at a point p ∈ X, written cik at p, provided that every neighborhood of p contains a connected neighborhood of p (where neighborhood means a subset of X which has p in its interior). • decomposable provided that X can be written as the union of two proper subcontinua. • indecomposable provided that X is not decomposable. • hereditarily indecomposable provided that every subcontinuum of X is indecomposable. • an arc provided that X is homeomorphic to the unit interval [0, 1]. 3
3. The connectedness of D(X) Unfortunately, the hyperspace D(X) is not always connected. Example 3.1. In the Euclidean plane let L = [0, 1] × {0} and for every 2 n ∈ N, let Ln = {(t, 1−t )) ∈ R : t ∈ [0, 1]}. Define Y = L ∪ ( n∈N Ln ). We n consider the following space: W = Y ∪ {(x, y) ∈ R2 : (−x, −y) ∈ Y }. Notice that W is a dendroid (a hereditarily unicoherent and arcwise connected continuum). Proposition 3.2. The hyperspace D(W ) is not connected. Proof. We shall prove that A = {A ∈ D(W ) : A ⊂ Y } is an open-closed and proper subset of D(W ). Since A = D(W ) ∩ C(Y ) and W ∈ D(W ) − A, we obtain that A is closed and a proper subset of D(W ). We are going to prove that A is open in D(W ). Let A ∈ A, we consider two cases. Case 1. (1, 0) ∈ A. From the facts that the components of Y −{(1, 0)} are {Ln −{(0, 1)} : n ∈ N}∪{L−{(0, 1)}} and Int(L−{(0, 1)}) = ∅, we obtain that A ⊂ Ln −{(1, 0)} for some n ∈ N. Since Ln − {(0, 1)} is an open subset of X, we can choose ε > 0 such that for every B ∈ BεH (A), we have that B ⊂ Ln − {(1, 0)}. Therefore A ∈ BεH (A) ∩ D(W ) ⊂ A. Case 2. (1, 0) ∈ A. H We shall prove that B H 1 (A) ∩ D(W ) ⊂ A. Let B ∈ B 1 (A) ∩ D(W ). Since 4
4
B ⊂ N ( 14 , A) ⊂ X − {(−1, 0)}, we obtain that B ⊂ Y ∪ ((−1, 0] × {0}). Suppose that B ∩ ([−1, 0] × {0}) is not degenerate, then every point x ∈ B ∩ ([−1, 0) × {0}) satisfies that x ∈ Int(B) = B, which is a contradiction. Thus B ∈ A and we have that A ∈ B H 1 (A) ∩ C(W ) ⊂ A. 4 This shows that A is an open-closed and proper subset of D(W ) and we obtain that D(W ) is not connected. From the above example, we consider the following problem. Problem 3.3. For which class of continua X, is the hyperspace D(X) connected? 4
In the rest of this section we will show that for the class of locally connected continua and for indecomposable continua, D(X) is connected. Directly from Lemmas 2.3 and 2.5, we obtain the following proposition. Proposition 3.4. Let X be a locally connected continuum with convex metric. If A ∈ D(X) and t ∈ [0, 1], then α : [0, 1] → D(X) defined by α(s) = N (s · t, A), is well defined and continuous. Theorem 3.5. If X is a locally connected continuum with a convex metric, then D(X) is locally connected. Proof. By (b) of [4, Exercise 5.22], it is enough to show that D(X) is cik at every element. Let A ∈ D(X) and ε > 0. By [4, Exercise 8.48], we have that C(X) is locally connected, then there exists a connected open subset V of ε C(X) such that A ∈ V ⊂ B H and B ∈ V}. ε (A). Let U = {N (t, B) : 0 ≤ t < 2 2 By Lemma 2.5, we have that U ⊂ C(X). Since V ⊂ U , then A ∈ V ∩ D(X) ⊂ U ∩ D(X), which implies that A ∈ IntD(X) (U ∩ D(X)). We shall prove that U ∩ D(X) ⊂ BεH (A) ∩ D(X). Let E ∈ U ∩ D(X), then E = N (t, B) for some B ∈ V and 0 ≤ t < 2ε . ε Since V ⊂ B H . Now, from the fact that t < 2ε , ε (A), we have that H(B, A) < 2 2
we obtain that N (t, B) ⊂ N ( 2ε , B) and since B ⊂ N ( 2ε , N (t, B)), we conclude that H(B, N (t, B)) < 2ε . Thus H(A, E) ≤ H(A, B) + H(B, N (t, B)) < ε. This shows that U ∩ D(X) ⊂ BεH (A) ∩ D(X). We shall prove that U ∩ D(X) is connected. Let E ∈ U ∩ D(X), then E = N (t, B) for some B ∈ V and 0 ≤ t < ε . Since V is a connected open subset in a locally connected space, by [4, 2 Theorem 5.12], there exits an embedding α : [0, 1] → V such that α(0) = A and α(1) = B. We consider two cases. Case 1. t0 > 0. Let β : [0, 1] → U ∩ D(X) be defined by β(s) = N (st0 , α(s)). By Lemma 2.5, we have that β is well defined and since 0 ≤ st0 ≤ t0 < 2ε for every s ∈ [0, 1], by the continuity of α and Lemma 3.4, it follows that β continuous. From the fact that β(0) = N (0, A) = A and β(1) = N (t0 , B) = E, we have that β([0, 1]) is a connected path from A to E in U ∩ D(X). Case 2. t0 = 0. In this case E = N (0, B) = B. Let t1 be such that 0 < t1 < 2ε and let β, γ : [0, 1] → U ∩ D(X) be defined by β(s) = N (st1 , α(s)) and γ(s) = N (st1 , B). Similarly as Case 1, β and γ are well defined continuous maps 5
such that β(s) = A, β(1) = N (t1 , B), γ(0) = E and γ(1) = N (t1 , B). Thus β([0, 1]) ∪ γ([0, 1]) is a connected path from E to A contained in U ∩ D(X). This finishes the proof that U ∩ D(X) is a connected subset. Thus, we have showed that U ∩ D(X) is a connected neighborhood of A contained in BεH (A) ∩ D(X), which shows that D(X) is cik at A and hence D(X) is locally connected. Theorem 3.6. If X is a locally connected continuum, then (1) D(X) is dense in C(X). (2) D(X) is contractible. (3) D(X) is arcwise connected. Proof. By [1, Theorem 10.3], we may assume that X has a convex metric. (1) Let A ∈ C(X) and ε > 0. By Lemma 2.4, we have that N ( 2ε , A) ∈ D(X) and it is clear that H(A, N ( 2ε , A)) < ε. (2) We can suppose that the diameter of X is 1. Let G : D(X) × [0, 1] → D(X) be defined by G((A, t)) = N (t, A). Using Lemma 3.4, it is a routine to show that G is well defined and continuous. Notice that G((A, 0)) = N (0, A) = A and G((A, 1)) = N (1, A) = X, for each A ∈ D(X). Thus D(X) is contractible. (3) Since every contractible space is arcwise connected, the result follows from (2). Finally, the following result is from the fact that a continuum X is indecomposable if and only if every proper subcontinuum of X has empty interior (see [2, Theorem 2]). Theorem 3.7. If X is an indecomposable continuum, then D(X) = {X}. 4. The compactness of D(X) The aim of this section is to show that, if D(X) is infinite, then D(X) is not compact. I wish to thank Prof. C. G. Mouron for providing me the tools to prove the principal result of this section.
6
Regular subsets Recall that a nonempty subset Y of X is regular, if Y is a regular closed subset of X. The following proposition is easy to show. Proposition 4.1. Let X be a continuum and let A and B be nonempty subsets of X. (a) If A is closed, then Int(A ∪ B) ⊂ A ∪ Int(B) and Int(B − A) = Int(B) − A. (b) A is regular if and only if A is regular. (c) If A and B are regular closed, then A ∪ B is regular closed. (d) If A and B are closed and C is a closed subset of B − A, then A ∪ C is closed in X. (e) If A is closed and {A}∞ n=1 is a sequence of closed subsets of X such that An ∩ Int(A) = ∅ for each n ∈ N and lim An = A (with the Hausdorff metric), then A is not regular, even more, we have that Int(A) = ∅. Proposition 4.2. Let B be a regular subset of X. If A is a closed subset of X, then B − A is regular. Proof. Suppose that B −A is nonempty. We only need to show that B − A ⊂ Int(B − A). Let x ∈ B − A and let U be an open subset of X such that x ∈ U . Then there exists b ∈ U such that b ∈ B and b ∈ A. Since A is closed, we have that W = U ∩ (X − A) is a non empty open subset such that b ∈ W ∩ B − A. Since B is regular, we have that W ∩ Int(B) = ∅. Let b1 ∈ W ∩ Int(B). Since b1 ∈ A and by (a) of Proposition 4.1, we have that b1 ∈ Int(B − A) ⊂ Int(B − A). This shows that B − A ⊂ Int(B − A) and we finish the proof of this proposition. Corollary 4.3. If B is a regular subset of X and V is an open subset, then B ∩ V is a regular subset of X. Proposition 4.4. Let Y and C be two subsets of X such that: (1) Y and Y ∪ C are closed, 7
(2) Y ∩ C = ∅, (3) C is not regular. Then Y ∪ C is not regular. Proof. Since C is not regular, there exist x ∈ C − Int(C) and an open subset U of X such that x ∈ U and U ∩ Int(C) = ∅. Let y ∈ C ∩ U . Since Y ∩ C = ∅, we have that y ∈ Y ∪ Int(C). By (a) of Proposition 4.1, we obtain that Int(Y ∪ C) ⊂ Y ∪ Int(C) = Y ∪ Int(C). Therefore, we have that y ∈ Int(Y ∪ C). From the fact that y ∈ Y ∪ C, we obtain that Int(Y ∪ C) Y ∪ C. Thus Y ∪ C is not regular. Regularity and components Given a subset A of a topological space X, we define Com(A) = {C ⊂ A : C is a component of A}. Given A a collection of subsets of a topological space X, we define A∗ =
A∈A
A.
Lemma 4.5. Let X be a continuum. If A is a no connected subset of X and A = U ∪ V is a separation of A, then there exist two nonempty subsets A and B of Com(A); and two disjoint open subsets U and V of X such that: (1) A ∩ B = ∅, (2) A ∪ B = Com(A), (3) A∗ ⊂ U and B ∗ ⊂ V , (4) U ⊂ U and V ⊂ V . Proof. Since X is a completely normal space, there exist two disjoint open subsets U and V of X such that U ⊂ U and V ⊂ V . It is routine to show that A = {C ∈ Com(A) : C ⊂ U }, B = {C ∈ Com(A) : C ⊂ V } and the open sets U and V satisfy the conditions that we need. 8
Proposition 4.6. Let A be a regular subset of X and let A = U ∪ V be a separation of A. If C ∈ Com(A) and A is a subset of Com(A) such that: (1) C ∈ A, (2) A∗ ∩ U = ∅ and A∗ ∩ V = ∅, (3) A∗ is closed in A, (4) A∗ is regular. Then there exists a subset A1 of A such that: (a) C ∈ A1 and A1 is a proper subset of A, (b) A∗1 is closed in A, (c) A∗1 is regular. Proof. Condition (2) implies that A∗ is not connected. By Lemma 4.5, there exist two subsets A1 and B1 of Com(A∗ ) = A and two disjoint open subsets U and V of X such that: (1) A1 ∩ B1 = ∅, (2) A1 ∪ B1 = A, (3) A∗1 ⊂ U and B1∗ ⊂ V . We may assume, without loss of generality, that C ∈ A1 . Since B1 is nonempty and A1 ∩ B1 = ∅, condition (a) holds. From the fact that X − V is closed in X and A∗ is closed in A, we obtain that A∗1 = A ∩ A∗ ∩ (X − V ) is closed in A. Thus condition (b) is true. Finally, since A∗1 is a proper subset of A∗ and U is an open subset of X, by Corollary 4.3, we have that A∗1 = A∗ ∩U is a regular subset of X and condition (c) is proved. The following theorem is known as the Brouwer Reduction Theorem (see [5, Theorem 11.1]). Theorem 4.7. Let Y be a second countable topological space and let Z be a nonempty family of closed subsets of Y . If every decreasing sequence of elements in Z satisfies that its intersection belongs to Z, then Z has minimal elements. 9
Theorem 4.8. Let A be a nonempty regular subset and let C be a component of A. If C is not regular, then there exists a sequence {Bn }∞ n=1 of subsets of Com(A) such that: (1) C ∈ Bn for each n ∈ N, (2) Bn+1 Bn for each n ∈ N, (3) Bn∗ is regular for each n ∈ N, (4) Bn∗ is closed in A for each n ∈ N, ∗ (5) B = ∞ n=1 Bn is a subset of Com(A) such that B is closed in A but it is not regular. Proof. Let Z = {B ∗ ⊂ A : C ∈ B ⊂ Com(A), B ∗ is closed in A and regular in X}. We consider the following cases. Case 1. Every decreasing sequence in Z satisfies that its intersection belongs to Z. By Theorem 4.7, there exists a minimal element B ∗ in Z. Therefore B ∗ is a regular subset of X such that C ∈ Com(B ∗ ) = B, B ∗ is closed in A and B ∗ is a regular subset. Since C ∈ Com(B ∗ ) = B and C is not regular it follows that B ∗ is not connected, therefore there exist two nonempty subsets U and V of B ∗ such that B ∗ = U ∪ V are a separation of B ∗ . By Proposition 4.6, there exists a proper subset B1 of B such that C ∈ B1 and B1 is a proper subset of B; B1∗ is closed in B ∗ and B1∗ is regular. Notice that C ∈ B1 ⊂ B ⊂ Com(A). Since B1∗ is closed in B ∗ , which is closed in A, we obtain that B1∗ in closed in A. Finally we have that B1∗ is regular, which shows that B1∗ ∈ Z. But B1∗ is a proper subset of B ∗ , since B1 is a proper subset of B, which contradicts the minimality of B ∗ in Z. This shows that Case 1 is not possible. 2. There exists a decreasing sequence {Bn∗ }∞ n=1 in Z, such that Case ∗ n∈N Bn ∈ Z. Notice that the sequence {Bn∗ }∞ n=1 must have infinite number of elements ∗ and we may assume, without loss of generality that Bn+1 Bn∗ for each ∗ ∞ n ∈ N. Let {Bn }∞ n=1 be the sequence in Com(A) induced by {Bn }n=1 . We ∞ are going to show that {Bn }n=1 satisfies the theorem. Let n ∈ N. Since Bn∗ ∈ Z, we have that C ∈ Bn ⊂ Com(A), Bn∗ is closed in A and regular in X. Therefore the conditions (1), (3) and (4) are 10
∗ ∗ satisfied. Given n ∈ N, since Bn+1 Bn∗ , there exists y ∈ Bn∗ − Bn+1 . If F is the component of A which has y, then F ∈ Bn − Bn+1 . This shows that condition (2) holds. We prove condition (5). Let B = n∈N Bn . It is clear that B ∗ = n∈N Bn∗ . Notice that C ∈ B and B ∗ is a closed subset of A, since C ⊂ Bn and Bn∗ is subset of A for each n ∈ N. Finally, from the fact that B ∗ = a closed ∗ ∗ n∈N Bn ∈ Z, we have that B is not a regular subset of X. This ends the proof of the theorem.
Proposition 4.9. Let A and B be two elements of D(X). If B − A is a component which is not regular, then D(X) is not closed. Proof. Let C be a non regular component of B − A. By Corollary 4.3, we have that B − A = B ∩ (X − A) is regular and by Theorem 4.8, there exists a sequence {Bn }∞ n=1 of subsets in Com(B − A) such that: (1) C ∈ Bn for every n ∈ N, (2) Bn+1 ⊂ Bn for every n ∈ N, (3) Bn∗ is regular for every n ∈ N, (4) Bn∗ is closed in B − A for every n ∈ N, (5) B = n∈N Bn is a subset of Com(B − A) with the property that B ∗ is closed in B − A but not regular. For every n ∈ N, let Kn = A ∪ Bn∗ y K = A ∪ B ∗ . By [4, Corollary 5.9], we have that K and Kn are connected, for each n ∈ N, and by (d) of Proposition 4.1, it follows that K and Kn are closed in X. Let n ∈ N. Since Bn∗ is regular, we have that Bn∗ is a regular closed, and by (c) of Proposition 4.1, we obtain that Kn = A∪Bn∗ = A∪Bn∗ is regular and by Proposition4.4, we have that K is not regular. Since A ∪ B ∗ = n∈N A ∪ Bn∗ , then K = n∈N Kn and hence lim Kn = K. Therefore we have showed that lim Kn = K where K ∈ D(X), which shows that D(X) is not closed. Minimal limit elements of D(X) An element B ∈ C(X) is called a minimal limit element of D(X) provided that B is a cluster element of D(X) and if A is a cluster element of C(X) such that A ⊂ B, then A = B. 11
Lemma 4.10. Let X be a continuum such that D(X) is compact. If A ∈ D(X) and {An }∞ n=1 is a sequence in D(X) such that: (1) An is a proper subset of A, for each n ∈ N, (2) lim An = A. Then A is not a minimal limit element of D(X). Proof. Since A1 ∈ D(X), we have that Int(A1 ) = ∅, therefore by (e) of Proposition 4.1 there exists N1 ∈ N such that Am ∩ A1 = ∅ for every m ≥ N1 . We consider three cases. Case 1. A1 ∪ Am is a proper subset of A for every m ≥ N1 . Let KN1 be a component of A − (A1 ∪ AN1 ). Since D(X) is closed, by Proposition 4.9, we have that KN1 is regular and by (b) of Proposition 4.1, we have that K N1 is regular. Suppose that there exist natural numbers N1 < N2 < . . . < Nt−1 with the property that for each i ∈ {1, . . . , t − 1} there exists a component KNi of A − (A1 ∪ ANi ), such that K Ni is regular and K Ni = K Nj , for each i, j ∈ {1, . . . , t − 1} with i = j. Claim 1. There exists Nt > Nt−1 such that Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Proof. Notice that Int(K Ni ) is a nonempty open subset of X contained in A for each i ∈ {1, . . . , t − 1}. Since lim An = A, we can choose Nt ∈ N such that Nt > Nt−1 and An ∩ Int(K Ni ) = ∅ for each i ∈ {1, . . . , t − 1}. Let i ∈ {1, . . . , t − 1} and x ∈ ANt ∩ Int(K Ni ). Since ANt ∈ D(X) and x ∈ Int(K Ni ), we have that Int(ANt ) ∩ Int(K Ni ) = ∅. This ends the proof of this claim. Since A1 ∪ ANt is a proper subset of A, there exists a component KNt of A − (A1 ∪ ANt ). By Proposition 4.9, we have that KNt is regular and by (b) of Proposition 4.1, we obtain that K Nt is regular. Claim 2. K Nt = K Ni for each i ∈ {1, . . . , t − 1}. Proof. Let i ∈ {1, . . . , t − 1}. Since KNt ⊂ A − (A1 ∪ ANt ), we have that KNt ⊂ A − Int(ANt ) and thus K Nt ⊂ A − Int(ANt ). By Claim 1, we have that Ui = Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X such that Ui ∩ K Nt = ∅. Since Ui ⊂ K Ni , we obtain that K Nt = K Ni and we finish the proof of this claim. Inductively, there exists a sequence of pairwise disjoints regular subsets {K Ni }∞ i=1 such that KNi is a component of A − (A1 ∪ ANi ) for each i ∈ N. 12
We can suppose, without loss of generality, that lim K Ni = K for some K ∈ C(X). This implies that K is a cluster element of D(X) and by construction we have that K = lim K Ni ⊂ A − Int(A1 ). Since Int(A1 ) = ∅ we obtain that K is proper in A and thus A cannot be a minimal limit point of D(X). This ends Case 1. Case 2. A1 ∪ Ak = A for a finite number of elements in {m ∈ N : m ≥ N1 }. Let N1 = max{m ∈ N : A1 ∪ Am = A}. Then A1 ∪ Am is a proper subset of A for each m ≥ N1 and we apply Case 1. ∞ Case 3. A1 ∪ Ami = A for a subsequence {mi }∞ i=1 of {m}m=N1 . We may assume, without loss of generality, that A1 ∪ Am = A for each m N1 . Since AN1 is a proper subset of A, there exists a component KN1 of A − AN1 ⊂ A1 . Since D(X) is closed, by Proposition 4.9, we have that KN1 is regular and by (b) of Proposition 4.1, we obtain that K N1 is regular. Suppose that there exist natural numbers N1 < N2 < . . . < Nt−1 with the property that for each i ∈ {1, . . . , t − 1} there exists a component KNi of A−ANi such that K Ni is regular and K Ni = K Nj , for every i, j ∈ {1, . . . , t−1} with i = j. Claim 3. There exists Nt > Nt−1 such that Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Proof. Notice that Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Since lim An = A, there exists Nt ∈ N such that Nt > Ni−1 and An ∩ Int(K Ni ) = ∅ for each n ≥ Nt . Let i ∈ {1, . . . , t − 1} and x ∈ ANt ∩ Int(K Ni ). Since ANt ∈ D(X) and x ∈ Int(K Ni ), we have that Int(ANt ) ∩ Int(K Ni ) = ∅. This ends the proof of this claim. Since ANt is a proper subset of A, there exists a component KNt of A−ANt . By Proposition 4.9, we have that KNt is regular and by (b) of Proposition 4.1, we have that K Nt is regular. Claim 4. K Nt = K Ni for each i ∈ {1, . . . , t − 1}. Proof. Let i ∈ {1, . . . , t − 1}. Since KNt ⊂ A − ANt , we have that KNt ⊂ A − Int(ANt ) and thus K Nt ⊂ A − Int(ANt ). By Claim 3, we have that Ui = Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X such that Ui ∩ KNt = ∅. Since Ui ⊂ K Ni , we have that K Nt = K Ni and we finish the proof of this claim. Inductively, there exists a sequence of pairwise disjoints regular subsets {K Ni }∞ i=1 such that KNi is a component of A − ANi for each i ∈ N. We can suppose, without loss of generality, that lim K Ni = K for some K ∈ D(X). Notice that, if i ∈ N, then KNi ⊂ A1 , therefore we have that lim KNi = K ⊂ 13
A1 . Since K is a cluster element of D(X), we obtain that A is not minimal. This ends the proof of this lemma. Lemma 4.11. If D(X) has infinite number of elements, then D(X) has a minimal limit point. Proof. Let A cluster element of D(X) and let Z = {B ∈ C(X) : B ⊂ A and B is a cluster element of D(X)}. Notice that A ∈ Z. Let {Am }∞ m=1 be a decreasing sequence in Z and we may assume, without loss of generality, that ∞ it is strictly decreasing. For each m ∈ N there exists a sequence {Am j }n=1 in m m D(X) where Am n = Ak for every j, k ∈ N and j = k, such that lim An = Am . n→∞
1 Let A1N1 be an element of the sequence {A1n }∞ n=1 such that H(A1 , AN1 ) < 1. Suppose that m ∈ N is such that for every i ∈ {1, . . . , m} there exists 1 i an element AiNi of the sequence {Ain }∞ n=1 such that H(Ai , ANi ) < i and j ANj = AkNk , for every j, k ∈ {1, . . . , m} such that j = k. We can choose an 1 m+1 ∞ }n=1 such that H(Am+1 , Am+1 element Am+1 Nm+1 of the sequence {An Nm+1 ) < m+1 i and Am+1 Nm+1 = ANi , for each i ∈ {1, . . . , m}. j ∞ k Inductively we have constructed a sequence {Am Nm }m=1 where ANj = ANk 1 m ∞ for every j, k ∈ N and j = k such ∞ that H(Am , ANm ) < m . Since {Am }m=1 is a decreasing sequence, if B = m=1 Am , then lim Am = B and therefore it is easy to verify that lim Am Nm = B. This shows that B is a cluster element of D(X), which shows that the intersection of every decreasing sequence in Z belongs to Z. By Theorem 4.7, we obtain that Z has minimal elements. This shows that D(X) has minimal limit elements.
Main theorem Theorem 4.12. If D(X) is not finite, then D(X) is not compact. Proof. Suppose that D(X) is compact. By Lemma 4.11, there exists a minimal limit element Y ∈ D(X). Let {An }∞ n=1 be a sequence in D(X) − Y such that lim An = Y . By Lemma 4.10 and (e) of Proposition 4.1, we may assume that An ⊂ Y and An ∩ Y = ∅ for every n ∈ N. Let K1 be a component of A1 − Y . By Proposition 4.9 and (b) of Proposition 4.1, we have that K 1 is regular. Since K1 ⊂ A1 − Y , we obtain that K 1 ∩ Int(Y ) = ∅.
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Suppose that t ∈ N is such that for every i ∈ {1, . . . , t − 1} there exists a component Kni of Ani − Y such that K ni is regular, K ni ∩ Int(Y ) = ∅ and also K ni = K nj , for every i, j ∈ {1, . . . , t − 1} such that i = j. Claim 1. There exist nt > nt−1 and a component Knt of Ant − Y such that (i) K nt is regular, (ii) K nt ∩ Int(Y ) = ∅, and (iii) K nt = K ni for each i ∈ {i, . . . , t − 1}. Proof. Suppose that for every n > nt−1 , the closure of every component of An − Y do not satisfy some condition that we request. Notice the follow. If n > nt−1 and Kn is a component of An − Y , since D(X) is closed, then by Proposition 4.9 and (b) of Proposition 4.1, we conclude that K n is regular. This means that (i) always holds. Since Kn ⊂ An − Y , then Kn ⊂ An − Int(Yn ) and therefore we have that K n ⊂ An − Int(Y ). This means that (ii) always holds. Hence for every n > nt−1 , the closure of every component of An − Y is equal to K i , for some i ∈ {1, . . . , t−1}. Thus we can suppose, without loss of generality, that for every n > nt−1 , the closure of any component of An − Y is equal to K 1 , which implies that K 1 ⊂ An for every n > nt−1 . Now, notice that K 1 ⊂ lim An = Y , which is a contradiction because K1 ⊂ X − Y . This ends the proof of Claim 1. Inductively, we construct a sequence {K ni }∞ i=1 in D(X) such that K ni ∩ Int(Y ) = ∅ for each i ∈ N and K ni = K nj for every i = j. Since D(X) is closed, we may assume, without loss of generality, that lim K ni = K where K ∈ D(X). Therefore K is a cluster element of D(X). From the fact that K ni ⊂ Ani and lim Ani = Y , we have that K ⊂ Y and since Y is a minimal limit element of D(X), we obtain that K = Y . But K ni ∩ Int(Y ) = ∅ for every i ∈ N, and by (e) of Proposition 4.1, we have that Y ∈ D(X), which is a contradiction. This finish the proof of the theorem. Conclusion As a corollary of Theorem 4.12, we have: Corollary 4.13. Let X be a continuum, then D(X) is compact if and only if D(X) is finite. Thus it is natural to consider the following problem. 15
Problem 4.14. For which n ∈ N does there exist a continuum X such that D(X) has exactly n elements? Related to the above problem, if we recall the fact that in an indecomposable continuum, every proper subcontinuum has empty interior ([2, Theorem 2]), then the following result is easy to obtain. Proposition 4.15. Let n ∈ N, let A1 , . . . , An be indecomposable continua and let pi ∈ Ai be such that pi = pj for every i, j ∈ {1, . . . , n} and i = j. Let X = A1 ∪ . . . ∪ An . (1) If Ai ∩ Aj = {pi } if and only if |i − j| = 1, then D(X) has n + (n − 1) + . . . + 1 = n(n+1) elements. 2 (2) If Ai ∩ Aj = {pi } if and only if |i − j| = 1 or |i − j| = n − 1, then D(X) has n(n − 1) + 1 elements. (3) If Ai ∩ Aj = {p1 } for every i, j ∈ {1, . . . , n} and i = j, then D(X) has n n n k=1 k = 2 − 1 elements. 5. A special space By Theorem 4.12, we have that D(X) is a continuum (degenerate) if and only if D(X) = {X}; and by Theorem 3.7, we have that if X is an indecomposable continuum, then D(X) = {X}. In this section we give an example of a decomposable continuum X for which D(X) = {X}. Let C be the canonical Cantor set contained in [0, 1] and let Y be the Buckethandle continuum contained in {(x, y, 0) ∈ R3 : x, y ∈ R} which is constructed over the Cantor set C and whose end point is (0, 0, 0) (see [1, Example 22.11]). Let Y = {(x, y − 1, 0) ∈ R3 : (x, y, 0) ∈ Y and y ≤ 0} ∪ {(x, y + 1, 0) ∈ R3 : (x, y, 0) ∈ Y and y ≥ 0} ∪ {(c, t, 0) ∈ R3 : c ∈ C, t ∈ [−1, 1]}. Notice that Y is a Buckethandle continuum that contains naturally the cylinder Z = {(c, t, 0) ∈ R3 : c ∈ C, t ∈ [−1, 1]}. For each c ∈ C, let 16
Yc = {(c, y, x) ∈ R3 : (x, y, 0) ∈ Y } and Lc = {(c, t, 0) ∈ R3 : t ∈ [−1, 1]}. Notice that for every c ∈ C, we have that • Yc is a copy of Y contained in the plane x = c. • Lc is an arc. • Y ∩ Yc = Lc . With the previous notation, we define the continuum X in R3 as Yc . X=Y ∪ c∈C
By construction, we obtain the following result. Proposition 5.1. The continuum X satisfies the following properties: (a) X is decomposable. (b) Yc ∩ Yv = ∅, for every c, v ∈ C and c = v. (c) Int(Yc ) = ∅, for every c ∈ C. (d) If {cn }∞ n=1 is a sequence in C such that lim cn = c, for some c ∈ C, then lim Ycn = Yc . (e) X − Y is open but not connected, even more, its components are the sets Yc − Lc , where c ∈ C. We adopt the following notation. Let A ∈ C(X) and c ∈ C be such that A ∩ Yc = ∅ and A ⊂ Yc . If x ∈ A ∩ Yc , then Cxc denotes the component of A ∩ Yc which contains x and Cx denotes the composant of Yc which contains x. Lemma 5.2. Let A ∈ C(X) and let e ∈ C be such that A ⊂ Ye . If x ∈ A∩Ye , then Cxe ∩ Le is a nonempty connected subset.
17
Proof. First we prove that Cxe ∩ Le = ∅. Suppose that Cxe ∩ Le = ∅. Then Cxe ⊂ X − Y . By [1, Theorem 14.6], there exists an order arc α : [0, 1] → C(X) from Cxe to A. Let t > 0 be such that α(t) ∩ Y = ∅. By (e) of Proposition 5.1, there exists c ∈ C such that α(t) ⊂ Yc − Lc . Since x ∈ α(t) ∩ Ye , we have that α(t) ⊂ Ye . Now, by definition of Cxe , we have that α(x) ⊂ Cxe . Which contradicts the fact that Cxe = α(0) α(t) ⊂ Cxe . Thus Cxe ∩ Le = ∅. We shall prove that Cxe ∩ Le is connected. We only need to consider the case when Ye ⊂ Cxe . In this case, notice that Cxe is an arc of Cx , since Le ⊂ Cx and Cxe ∩ Le = ∅, we have that Cxe ∪ Le is an arc. Therefore Cxe ∩ Le is an arc and we end the proof of this lemma. Proposition 5.3. If A is a subcontinuum of X such that Y ∩ A = ∅, then A ∩ Y is a subcontinuum of Y . Proof. Let A ∈ C(X) be such that Y ∩ A = ∅. It is enough to prove that A ∩ Y is connected in the case when A ⊂ Y . Suppose that there exist two nonempty disjoint closed subsets H and K of X such that A ∩ Y = H ∪ K. Notice that, if x ∈ A−Y , then there exists c ∈ C such that x ∈ Yc . By Lemma 5.2, we have that Cxc ∩ Lc is nonempty and connected, therefore Cxc ∩ Lc ⊂ H or Cxc ∩ Lc ⊂ K. Let: H = {x ∈ A : there exists c ∈ C such that x ∈ Yc and Cxc ∩ H = ∅} ∪ H and K = {x ∈ A : there exists c ∈ C such that x ∈ Yc and Cxc ∩ K = ∅} ∪ K. It is routine to prove that H and K are two disjoint closed subset of X such that A = K ∪ H , which can not happen. Thus, we finish the proof of this lemma. Proposition 5.4. If A ∈ C(X) and c ∈ C, then A ∩ Yc is a subcontinuum of Yc . Proof. Let A ∈ C(X) and c ∈ C be such that A∩Yc is nonempty and A ⊂ Yc . We are going to prove that A ∩ Yc is connected. By Lemma 5.2, every component of A ∩ Yc intersects Lc . Let a ∈ A ∩ Lc . If there exists a component C of A ∩ Yc such that C ⊂ Ca , then C ∩ Lc = ∅. Which implies that C is subcontinuum of an indecomposable continuum Yc , which intersects two different composants of a Yc , therefore A ∩ Yc = Yc . If every component of A ∩ Yc is a subset of Ca , then there exists an arc pq with end points p and q contained in Ca such that A ∩ Yc ⊂ pq and 18
it is minimal with this property. Using the fact that A ∩ Y is connected (Proposition 5.3) and Y is the Buckethandle continuum, we have that (A ∩ Y ) ∩ Lc is an arc contained in A ∩ LC ⊂ A ∩ Yc ⊂ pq, and we conclude that A ∩ Yc = pq. This ends the proof of the proposition. Proposition 5.5. If A is a regular subcontinuum of X, then Y ⊂ A. Proof. By Proposition 5.3, B = A ∩ Y is asubcontinuum of Y , therefore B is a subcontinuum of X. If B ∩ (Y − Yc ) = ∅, then A is a proper c∈C
subcontinuum of Y or A ⊂ Yc for some c ∈ C. In both cases we have that Int(A) = ∅ which can not happen. Let x ∈ B ∩ (Y − Yc ). Since Yc c∈C
c∈C
is closed in X, there exists an open subset U of X such that x ∈ U and U ∩ Yc = ∅. c∈C
Since x ∈ A = Int(A), there exists y ∈ U ∩ Int(A). Thus W = U ∩ Int(A) is a nonempty open set of X such that W ⊂ B. Therefore B is subcontinuum of Y with nonempty interior in Y , which implies that Y = B. Proposition 5.6. Let A be a subcontinuum of X and c ∈ C. If Int(A) ∩ (Yc − Lc ) = ∅, then Yc ⊂ A. Proof. By Proposition 5.4, B = A ∩ Yc is a subcontinuum of Yc . Let x ∈ Int(A) ∩ (Yc − Lc ) and let U be an open subset in X such that x ∈ U ⊂ A and U ∩ Y = ∅. Then U ∩ Yc is a nonempty open subset of Yc contained in B, which shows that B is a subcontinuum with nonempty interior in Yc , therefore we have that Yc = B. Proposition 5.7. If A is a regular subcontinuum of X, then D = {c ∈ C : Yc ⊂ A} is dense in C. Proof. By Proposition 5.5, we havethat Y ⊂ A. Let c ∈ C and ε > 0. Let x ∈ Lc be such that x ∈ Int( Ye ). Since A is regular, there exists e∈C y ∈ Int(A) ∩ Bεd (x) ∩ Int( Ye ). e∈C Since y ∈ Int( Ye ), there exists c1 ∈ C such that y ∈ Yc1 . Therefore e∈C
Yc1 is a subcontinuum of X such that Int(A) ∩ (Yc1 − Lc1 ) = ∅. And hence, by Proposition 5.6, we have that Yc1 ⊂ A. Let t, s ∈ [0, 1] be such that x = (c, t, 0) and y = (c1 , s, 0). Since y ∈ Bεd (x), we have that |c − c1 | d(x, y) < ε. This shows that D is dense in C. 19
Theorem 5.8. The continuum X does not contain proper regular subcotinua, in other words D(X) = {X}. Proof. Let A be a regular subcontinuun of X. By Proposition 5.5, we have that Y ⊂ A; and by Proposition 5.7, we have that {c ∈ C : Yc ⊂ A} is dense in C, which implies that A is dense in X. Thus A = X. Problem 5.9. Does there exist a hereditarily decomposable continuum X for which D(X) = {X}? Acknowledgement. The author wishes to thank Prof. A. Illanes by his motivation to conclude this paper. References [1] A. Illanes and S. B. Nadler, Jr, Hyperspaces, Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, New York, Basel, 1999. [2] K. Kuratowsky, Topology, Vol. 2, Academic Press and PWN, New York, London and Warszawa, 1968. [3] S. B. Nadler, Jr., Hyperspaces of sets, Monographs and Textbooks in Pure and Applied Math., Vol 49, Marcel Dekker, New York, Basel, 1978. [4] S. B. Nadler, Jr., Continuum Theory: an Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [5] G. T. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ., vol. 28, Providence, RI, 1942, reprinted with corrections 1971.
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Norberto Ordoñez
PII: DOI: Reference:
S0166-8641(17)30631-4 https://doi.org/10.1016/j.topol.2017.11.038 TOPOL 6340
To appear in:
Topology and its Applications
Received date: Revised date: Accepted date:
2 December 2016 15 July 2017 29 November 2017
Please cite this article in press as: N. Ordoñez, The Hyperspace of Regular Subcontinua, Topol. Appl. (2017), https://doi.org/10.1016/j.topol.2017.11.038
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The Hyperspace of Regular Subcontinua Norberto Ordo˜ nez Universidad Aut´ onoma del Estado de M´exico, Facultad de Ciencias, Instituto Literario No. 100, Col. Centro, Toluca, Estado de M´exico, M´exico, C. P. 50000.
Abstract Given a metric continuum X, we define the hyperspace of regular subcontinua of X as the collection of all regular closed subcontinua of X. This is a new hyperspace for X. In this paper we study the connectedness, compactness, arcwise connectedness and contractibility of this hyperspace. Also we raise some open questions. Keywords: Continuum, Hyperspaces, Regular Set, Regular Subcontinua. 2010 MSC: Primary: 54B20, Secondary: 54B05, 54F15. 1. Introduction A continuum is a nonempty compact connected metric space. Given a continuum X, by a hyperspace of X we mean a specified collection of subsets of X endowed with the Hausdorff metric. The hyperspaces are a powerful tool to study the continuum for which they are defined and, of course, they provide new problems to solve. Some hyperspaces for a continuum X are 2X the hyperspace of all nonempty closed subsets of X, C(X) the hyperspace of all connected elements of 2X and F1 (X) the hyperspace of all one-point sets of X. A subcontinuum is an element of C(X). These hyperspaces have important properties, for example, 2X and C(X) are always arcwise connected continua [1, Corollary 14.10] and unicoherent [1, Theorem 19.8]. Also if X is locally connected, then 2X and C(X) are absolute retracts [1, Theorem 10.8] and contractible [1, Corollary 20.14]. And F1 (X) is an isometric copy of X. For more information related with these and other hypespaces see [1]. Email address: [email protected] (Norberto Ordo˜ nez) Preprint submitted to Elsevier
November 30, 2017
To introduce the hyperspace that we study in this paper, recall that for a topological space X, we say that a nonempty closed subset A of X is a regular closed provided that closure of the interior of A is equal to A. Thus, given a continuum X, we define: The hyperspace of regular subcontinua of X as D(X) = {A ∈ C(X) : A is regular closed in X}. For the study of D(X), this paper is organized as follow. In Section 2, we introduce general notation and definitions that we will use along the paper. We also enunciate known results in hyperspaces. In Section 3 we study the connectedness of D(X). We show a dendroid X for which D(X) is not connected and we prove that if X is locally connected, then D(X) is locally connected, contractible and a dense subset in C(X). In Section 4, we study the compactness of D(X). The main result that we prove is that if D(X) is infinite, then D(X) is not compact and thus we obtain, as a corollary, that D(X) is compact if an only if D(X) is finite. In Section 5, we show a decomposable continuum X for which D(X) = {X}. Along the paper we also raise some open questions. 2. Preliminaries N will denote the natural numbers. If X is a topological space and Y ⊂ X, X then Y , BdX (Y ) and IntX (Y ) denote the closure, the boundary and the interior of Y in X, respectively. We will write just Y , Bd(Y ) and Int(Y ) when the space where we calculate these is understood. We introduce the concept of the Hausdorff metric. Given a continuum X with metric d, x ∈ X, A ⊂ X and ε > 0, we define the open ball ofradius ε with center x as Bεd (x) = {y ∈ X : d(x, y) < ε} and let N (ε, A) = {Bεd (a) : a ∈ A}. If A and B are two closed subsets of X, the Hausdorff distance between A and B is defined by: H(A, B) = inf{ε > 0 : A ⊂ N (ε, B) and B ⊂ N (ε, A)}. It is known that H is a Metric for 2X when X is a continuum (see [1, Theorem 2.2]) and H is called the Hausdorff metric. Additionally, if we make the convention that N (0, A) = A for every subset A of X, then the following lemmas are easy to prove: 2
Lemma 2.1. Let X be a continuum. If t, s ∈ [0, 1] and A ∈ C(X), then N (t, N (s, A)) = N (t + s, A). Lemma 2.2. Let X be a continuum, A, B ∈ C(X) and ε > 0. Then H(A, B) < ε if and only if A ⊂ N (ε, B) and B ⊂ N (ε, A). A convex metric for a space X is a metric d for X that induces the topology on X, which satisfies that for any x, y ∈ X there exists z ∈ X such that d(x, z) = 12 d(x, y) = d(z, y). It is known that every locally connected continuum admits a convex metric (see [1, Theorem 10.3]). Let X be a continuum with a convex metric and A ∈ 2X . Then N (ε, A) = {x ∈ X : d(x, a) ≤ ε for some a ∈ A} (see [1, Exercise 10.11]) and hence by [3, 0.65.3], we have the following result. Lemma 2.3. Let X be a continuum with a convex metric. The map K : [0, 1] × C(X) → 2X defined by K((t, A)) = N (t, A) is continuous. The proof of the following lemma is easy. Lemma 2.4. Let X be a continuum. If U is a connected open subset of X, then U ∈ D(X). Lemma 2.5. Let X be a continuum with a convex metric and let A ∈ C(X). If ε > 0, then N (ε, A) ∈ D(X). Proof. As a consequence of [1, Proposition 10.4], we have that N (ε, A) is connected and since it is an open subset of X, the result follows from Lemma 2.4. A continuum X is said to be: • connected im kleinen at a point p ∈ X, written cik at p, provided that every neighborhood of p contains a connected neighborhood of p (where neighborhood means a subset of X which has p in its interior). • decomposable provided that X can be written as the union of two proper subcontinua. • indecomposable provided that X is not decomposable. • hereditarily indecomposable provided that every subcontinuum of X is indecomposable. • an arc provided that X is homeomorphic to the unit interval [0, 1]. 3
3. The connectedness of D(X) Unfortunately, the hyperspace D(X) is not always connected. Example 3.1. In the Euclidean plane let L = [0, 1] × {0} and for every 2 n ∈ N, let Ln = {(t, 1−t )) ∈ R : t ∈ [0, 1]}. Define Y = L ∪ ( n∈N Ln ). We n consider the following space: W = Y ∪ {(x, y) ∈ R2 : (−x, −y) ∈ Y }. Notice that W is a dendroid (a hereditarily unicoherent and arcwise connected continuum). Proposition 3.2. The hyperspace D(W ) is not connected. Proof. We shall prove that A = {A ∈ D(W ) : A ⊂ Y } is an open-closed and proper subset of D(W ). Since A = D(W ) ∩ C(Y ) and W ∈ D(W ) − A, we obtain that A is closed and a proper subset of D(W ). We are going to prove that A is open in D(W ). Let A ∈ A, we consider two cases. Case 1. (1, 0) ∈ A. From the facts that the components of Y −{(1, 0)} are {Ln −{(0, 1)} : n ∈ N}∪{L−{(0, 1)}} and Int(L−{(0, 1)}) = ∅, we obtain that A ⊂ Ln −{(1, 0)} for some n ∈ N. Since Ln − {(0, 1)} is an open subset of X, we can choose ε > 0 such that for every B ∈ BεH (A), we have that B ⊂ Ln − {(1, 0)}. Therefore A ∈ BεH (A) ∩ D(W ) ⊂ A. Case 2. (1, 0) ∈ A. H We shall prove that B H 1 (A) ∩ D(W ) ⊂ A. Let B ∈ B 1 (A) ∩ D(W ). Since 4
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B ⊂ N ( 14 , A) ⊂ X − {(−1, 0)}, we obtain that B ⊂ Y ∪ ((−1, 0] × {0}). Suppose that B ∩ ([−1, 0] × {0}) is not degenerate, then every point x ∈ B ∩ ([−1, 0) × {0}) satisfies that x ∈ Int(B) = B, which is a contradiction. Thus B ∈ A and we have that A ∈ B H 1 (A) ∩ C(W ) ⊂ A. 4 This shows that A is an open-closed and proper subset of D(W ) and we obtain that D(W ) is not connected. From the above example, we consider the following problem. Problem 3.3. For which class of continua X, is the hyperspace D(X) connected? 4
In the rest of this section we will show that for the class of locally connected continua and for indecomposable continua, D(X) is connected. Directly from Lemmas 2.3 and 2.5, we obtain the following proposition. Proposition 3.4. Let X be a locally connected continuum with convex metric. If A ∈ D(X) and t ∈ [0, 1], then α : [0, 1] → D(X) defined by α(s) = N (s · t, A), is well defined and continuous. Theorem 3.5. If X is a locally connected continuum with a convex metric, then D(X) is locally connected. Proof. By (b) of [4, Exercise 5.22], it is enough to show that D(X) is cik at every element. Let A ∈ D(X) and ε > 0. By [4, Exercise 8.48], we have that C(X) is locally connected, then there exists a connected open subset V of ε C(X) such that A ∈ V ⊂ B H and B ∈ V}. ε (A). Let U = {N (t, B) : 0 ≤ t < 2 2 By Lemma 2.5, we have that U ⊂ C(X). Since V ⊂ U , then A ∈ V ∩ D(X) ⊂ U ∩ D(X), which implies that A ∈ IntD(X) (U ∩ D(X)). We shall prove that U ∩ D(X) ⊂ BεH (A) ∩ D(X). Let E ∈ U ∩ D(X), then E = N (t, B) for some B ∈ V and 0 ≤ t < 2ε . ε Since V ⊂ B H . Now, from the fact that t < 2ε , ε (A), we have that H(B, A) < 2 2
we obtain that N (t, B) ⊂ N ( 2ε , B) and since B ⊂ N ( 2ε , N (t, B)), we conclude that H(B, N (t, B)) < 2ε . Thus H(A, E) ≤ H(A, B) + H(B, N (t, B)) < ε. This shows that U ∩ D(X) ⊂ BεH (A) ∩ D(X). We shall prove that U ∩ D(X) is connected. Let E ∈ U ∩ D(X), then E = N (t, B) for some B ∈ V and 0 ≤ t < ε . Since V is a connected open subset in a locally connected space, by [4, 2 Theorem 5.12], there exits an embedding α : [0, 1] → V such that α(0) = A and α(1) = B. We consider two cases. Case 1. t0 > 0. Let β : [0, 1] → U ∩ D(X) be defined by β(s) = N (st0 , α(s)). By Lemma 2.5, we have that β is well defined and since 0 ≤ st0 ≤ t0 < 2ε for every s ∈ [0, 1], by the continuity of α and Lemma 3.4, it follows that β continuous. From the fact that β(0) = N (0, A) = A and β(1) = N (t0 , B) = E, we have that β([0, 1]) is a connected path from A to E in U ∩ D(X). Case 2. t0 = 0. In this case E = N (0, B) = B. Let t1 be such that 0 < t1 < 2ε and let β, γ : [0, 1] → U ∩ D(X) be defined by β(s) = N (st1 , α(s)) and γ(s) = N (st1 , B). Similarly as Case 1, β and γ are well defined continuous maps 5
such that β(s) = A, β(1) = N (t1 , B), γ(0) = E and γ(1) = N (t1 , B). Thus β([0, 1]) ∪ γ([0, 1]) is a connected path from E to A contained in U ∩ D(X). This finishes the proof that U ∩ D(X) is a connected subset. Thus, we have showed that U ∩ D(X) is a connected neighborhood of A contained in BεH (A) ∩ D(X), which shows that D(X) is cik at A and hence D(X) is locally connected. Theorem 3.6. If X is a locally connected continuum, then (1) D(X) is dense in C(X). (2) D(X) is contractible. (3) D(X) is arcwise connected. Proof. By [1, Theorem 10.3], we may assume that X has a convex metric. (1) Let A ∈ C(X) and ε > 0. By Lemma 2.4, we have that N ( 2ε , A) ∈ D(X) and it is clear that H(A, N ( 2ε , A)) < ε. (2) We can suppose that the diameter of X is 1. Let G : D(X) × [0, 1] → D(X) be defined by G((A, t)) = N (t, A). Using Lemma 3.4, it is a routine to show that G is well defined and continuous. Notice that G((A, 0)) = N (0, A) = A and G((A, 1)) = N (1, A) = X, for each A ∈ D(X). Thus D(X) is contractible. (3) Since every contractible space is arcwise connected, the result follows from (2). Finally, the following result is from the fact that a continuum X is indecomposable if and only if every proper subcontinuum of X has empty interior (see [2, Theorem 2]). Theorem 3.7. If X is an indecomposable continuum, then D(X) = {X}. 4. The compactness of D(X) The aim of this section is to show that, if D(X) is infinite, then D(X) is not compact. I wish to thank Prof. C. G. Mouron for providing me the tools to prove the principal result of this section.
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Regular subsets Recall that a nonempty subset Y of X is regular, if Y is a regular closed subset of X. The following proposition is easy to show. Proposition 4.1. Let X be a continuum and let A and B be nonempty subsets of X. (a) If A is closed, then Int(A ∪ B) ⊂ A ∪ Int(B) and Int(B − A) = Int(B) − A. (b) A is regular if and only if A is regular. (c) If A and B are regular closed, then A ∪ B is regular closed. (d) If A and B are closed and C is a closed subset of B − A, then A ∪ C is closed in X. (e) If A is closed and {A}∞ n=1 is a sequence of closed subsets of X such that An ∩ Int(A) = ∅ for each n ∈ N and lim An = A (with the Hausdorff metric), then A is not regular, even more, we have that Int(A) = ∅. Proposition 4.2. Let B be a regular subset of X. If A is a closed subset of X, then B − A is regular. Proof. Suppose that B −A is nonempty. We only need to show that B − A ⊂ Int(B − A). Let x ∈ B − A and let U be an open subset of X such that x ∈ U . Then there exists b ∈ U such that b ∈ B and b ∈ A. Since A is closed, we have that W = U ∩ (X − A) is a non empty open subset such that b ∈ W ∩ B − A. Since B is regular, we have that W ∩ Int(B) = ∅. Let b1 ∈ W ∩ Int(B). Since b1 ∈ A and by (a) of Proposition 4.1, we have that b1 ∈ Int(B − A) ⊂ Int(B − A). This shows that B − A ⊂ Int(B − A) and we finish the proof of this proposition. Corollary 4.3. If B is a regular subset of X and V is an open subset, then B ∩ V is a regular subset of X. Proposition 4.4. Let Y and C be two subsets of X such that: (1) Y and Y ∪ C are closed, 7
(2) Y ∩ C = ∅, (3) C is not regular. Then Y ∪ C is not regular. Proof. Since C is not regular, there exist x ∈ C − Int(C) and an open subset U of X such that x ∈ U and U ∩ Int(C) = ∅. Let y ∈ C ∩ U . Since Y ∩ C = ∅, we have that y ∈ Y ∪ Int(C). By (a) of Proposition 4.1, we obtain that Int(Y ∪ C) ⊂ Y ∪ Int(C) = Y ∪ Int(C). Therefore, we have that y ∈ Int(Y ∪ C). From the fact that y ∈ Y ∪ C, we obtain that Int(Y ∪ C) Y ∪ C. Thus Y ∪ C is not regular. Regularity and components Given a subset A of a topological space X, we define Com(A) = {C ⊂ A : C is a component of A}. Given A a collection of subsets of a topological space X, we define A∗ =
A∈A
A.
Lemma 4.5. Let X be a continuum. If A is a no connected subset of X and A = U ∪ V is a separation of A, then there exist two nonempty subsets A and B of Com(A); and two disjoint open subsets U and V of X such that: (1) A ∩ B = ∅, (2) A ∪ B = Com(A), (3) A∗ ⊂ U and B ∗ ⊂ V , (4) U ⊂ U and V ⊂ V . Proof. Since X is a completely normal space, there exist two disjoint open subsets U and V of X such that U ⊂ U and V ⊂ V . It is routine to show that A = {C ∈ Com(A) : C ⊂ U }, B = {C ∈ Com(A) : C ⊂ V } and the open sets U and V satisfy the conditions that we need. 8
Proposition 4.6. Let A be a regular subset of X and let A = U ∪ V be a separation of A. If C ∈ Com(A) and A is a subset of Com(A) such that: (1) C ∈ A, (2) A∗ ∩ U = ∅ and A∗ ∩ V = ∅, (3) A∗ is closed in A, (4) A∗ is regular. Then there exists a subset A1 of A such that: (a) C ∈ A1 and A1 is a proper subset of A, (b) A∗1 is closed in A, (c) A∗1 is regular. Proof. Condition (2) implies that A∗ is not connected. By Lemma 4.5, there exist two subsets A1 and B1 of Com(A∗ ) = A and two disjoint open subsets U and V of X such that: (1) A1 ∩ B1 = ∅, (2) A1 ∪ B1 = A, (3) A∗1 ⊂ U and B1∗ ⊂ V . We may assume, without loss of generality, that C ∈ A1 . Since B1 is nonempty and A1 ∩ B1 = ∅, condition (a) holds. From the fact that X − V is closed in X and A∗ is closed in A, we obtain that A∗1 = A ∩ A∗ ∩ (X − V ) is closed in A. Thus condition (b) is true. Finally, since A∗1 is a proper subset of A∗ and U is an open subset of X, by Corollary 4.3, we have that A∗1 = A∗ ∩U is a regular subset of X and condition (c) is proved. The following theorem is known as the Brouwer Reduction Theorem (see [5, Theorem 11.1]). Theorem 4.7. Let Y be a second countable topological space and let Z be a nonempty family of closed subsets of Y . If every decreasing sequence of elements in Z satisfies that its intersection belongs to Z, then Z has minimal elements. 9
Theorem 4.8. Let A be a nonempty regular subset and let C be a component of A. If C is not regular, then there exists a sequence {Bn }∞ n=1 of subsets of Com(A) such that: (1) C ∈ Bn for each n ∈ N, (2) Bn+1 Bn for each n ∈ N, (3) Bn∗ is regular for each n ∈ N, (4) Bn∗ is closed in A for each n ∈ N, ∗ (5) B = ∞ n=1 Bn is a subset of Com(A) such that B is closed in A but it is not regular. Proof. Let Z = {B ∗ ⊂ A : C ∈ B ⊂ Com(A), B ∗ is closed in A and regular in X}. We consider the following cases. Case 1. Every decreasing sequence in Z satisfies that its intersection belongs to Z. By Theorem 4.7, there exists a minimal element B ∗ in Z. Therefore B ∗ is a regular subset of X such that C ∈ Com(B ∗ ) = B, B ∗ is closed in A and B ∗ is a regular subset. Since C ∈ Com(B ∗ ) = B and C is not regular it follows that B ∗ is not connected, therefore there exist two nonempty subsets U and V of B ∗ such that B ∗ = U ∪ V are a separation of B ∗ . By Proposition 4.6, there exists a proper subset B1 of B such that C ∈ B1 and B1 is a proper subset of B; B1∗ is closed in B ∗ and B1∗ is regular. Notice that C ∈ B1 ⊂ B ⊂ Com(A). Since B1∗ is closed in B ∗ , which is closed in A, we obtain that B1∗ in closed in A. Finally we have that B1∗ is regular, which shows that B1∗ ∈ Z. But B1∗ is a proper subset of B ∗ , since B1 is a proper subset of B, which contradicts the minimality of B ∗ in Z. This shows that Case 1 is not possible. 2. There exists a decreasing sequence {Bn∗ }∞ n=1 in Z, such that Case ∗ n∈N Bn ∈ Z. Notice that the sequence {Bn∗ }∞ n=1 must have infinite number of elements ∗ and we may assume, without loss of generality that Bn+1 Bn∗ for each ∗ ∞ n ∈ N. Let {Bn }∞ n=1 be the sequence in Com(A) induced by {Bn }n=1 . We ∞ are going to show that {Bn }n=1 satisfies the theorem. Let n ∈ N. Since Bn∗ ∈ Z, we have that C ∈ Bn ⊂ Com(A), Bn∗ is closed in A and regular in X. Therefore the conditions (1), (3) and (4) are 10
∗ ∗ satisfied. Given n ∈ N, since Bn+1 Bn∗ , there exists y ∈ Bn∗ − Bn+1 . If F is the component of A which has y, then F ∈ Bn − Bn+1 . This shows that condition (2) holds. We prove condition (5). Let B = n∈N Bn . It is clear that B ∗ = n∈N Bn∗ . Notice that C ∈ B and B ∗ is a closed subset of A, since C ⊂ Bn and Bn∗ is subset of A for each n ∈ N. Finally, from the fact that B ∗ = a closed ∗ ∗ n∈N Bn ∈ Z, we have that B is not a regular subset of X. This ends the proof of the theorem.
Proposition 4.9. Let A and B be two elements of D(X). If B − A is a component which is not regular, then D(X) is not closed. Proof. Let C be a non regular component of B − A. By Corollary 4.3, we have that B − A = B ∩ (X − A) is regular and by Theorem 4.8, there exists a sequence {Bn }∞ n=1 of subsets in Com(B − A) such that: (1) C ∈ Bn for every n ∈ N, (2) Bn+1 ⊂ Bn for every n ∈ N, (3) Bn∗ is regular for every n ∈ N, (4) Bn∗ is closed in B − A for every n ∈ N, (5) B = n∈N Bn is a subset of Com(B − A) with the property that B ∗ is closed in B − A but not regular. For every n ∈ N, let Kn = A ∪ Bn∗ y K = A ∪ B ∗ . By [4, Corollary 5.9], we have that K and Kn are connected, for each n ∈ N, and by (d) of Proposition 4.1, it follows that K and Kn are closed in X. Let n ∈ N. Since Bn∗ is regular, we have that Bn∗ is a regular closed, and by (c) of Proposition 4.1, we obtain that Kn = A∪Bn∗ = A∪Bn∗ is regular and by Proposition4.4, we have that K is not regular. Since A ∪ B ∗ = n∈N A ∪ Bn∗ , then K = n∈N Kn and hence lim Kn = K. Therefore we have showed that lim Kn = K where K ∈ D(X), which shows that D(X) is not closed. Minimal limit elements of D(X) An element B ∈ C(X) is called a minimal limit element of D(X) provided that B is a cluster element of D(X) and if A is a cluster element of C(X) such that A ⊂ B, then A = B. 11
Lemma 4.10. Let X be a continuum such that D(X) is compact. If A ∈ D(X) and {An }∞ n=1 is a sequence in D(X) such that: (1) An is a proper subset of A, for each n ∈ N, (2) lim An = A. Then A is not a minimal limit element of D(X). Proof. Since A1 ∈ D(X), we have that Int(A1 ) = ∅, therefore by (e) of Proposition 4.1 there exists N1 ∈ N such that Am ∩ A1 = ∅ for every m ≥ N1 . We consider three cases. Case 1. A1 ∪ Am is a proper subset of A for every m ≥ N1 . Let KN1 be a component of A − (A1 ∪ AN1 ). Since D(X) is closed, by Proposition 4.9, we have that KN1 is regular and by (b) of Proposition 4.1, we have that K N1 is regular. Suppose that there exist natural numbers N1 < N2 < . . . < Nt−1 with the property that for each i ∈ {1, . . . , t − 1} there exists a component KNi of A − (A1 ∪ ANi ), such that K Ni is regular and K Ni = K Nj , for each i, j ∈ {1, . . . , t − 1} with i = j. Claim 1. There exists Nt > Nt−1 such that Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Proof. Notice that Int(K Ni ) is a nonempty open subset of X contained in A for each i ∈ {1, . . . , t − 1}. Since lim An = A, we can choose Nt ∈ N such that Nt > Nt−1 and An ∩ Int(K Ni ) = ∅ for each i ∈ {1, . . . , t − 1}. Let i ∈ {1, . . . , t − 1} and x ∈ ANt ∩ Int(K Ni ). Since ANt ∈ D(X) and x ∈ Int(K Ni ), we have that Int(ANt ) ∩ Int(K Ni ) = ∅. This ends the proof of this claim. Since A1 ∪ ANt is a proper subset of A, there exists a component KNt of A − (A1 ∪ ANt ). By Proposition 4.9, we have that KNt is regular and by (b) of Proposition 4.1, we obtain that K Nt is regular. Claim 2. K Nt = K Ni for each i ∈ {1, . . . , t − 1}. Proof. Let i ∈ {1, . . . , t − 1}. Since KNt ⊂ A − (A1 ∪ ANt ), we have that KNt ⊂ A − Int(ANt ) and thus K Nt ⊂ A − Int(ANt ). By Claim 1, we have that Ui = Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X such that Ui ∩ K Nt = ∅. Since Ui ⊂ K Ni , we obtain that K Nt = K Ni and we finish the proof of this claim. Inductively, there exists a sequence of pairwise disjoints regular subsets {K Ni }∞ i=1 such that KNi is a component of A − (A1 ∪ ANi ) for each i ∈ N. 12
We can suppose, without loss of generality, that lim K Ni = K for some K ∈ C(X). This implies that K is a cluster element of D(X) and by construction we have that K = lim K Ni ⊂ A − Int(A1 ). Since Int(A1 ) = ∅ we obtain that K is proper in A and thus A cannot be a minimal limit point of D(X). This ends Case 1. Case 2. A1 ∪ Ak = A for a finite number of elements in {m ∈ N : m ≥ N1 }. Let N1 = max{m ∈ N : A1 ∪ Am = A}. Then A1 ∪ Am is a proper subset of A for each m ≥ N1 and we apply Case 1. ∞ Case 3. A1 ∪ Ami = A for a subsequence {mi }∞ i=1 of {m}m=N1 . We may assume, without loss of generality, that A1 ∪ Am = A for each m N1 . Since AN1 is a proper subset of A, there exists a component KN1 of A − AN1 ⊂ A1 . Since D(X) is closed, by Proposition 4.9, we have that KN1 is regular and by (b) of Proposition 4.1, we obtain that K N1 is regular. Suppose that there exist natural numbers N1 < N2 < . . . < Nt−1 with the property that for each i ∈ {1, . . . , t − 1} there exists a component KNi of A−ANi such that K Ni is regular and K Ni = K Nj , for every i, j ∈ {1, . . . , t−1} with i = j. Claim 3. There exists Nt > Nt−1 such that Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Proof. Notice that Int(K Ni ) is a nonempty open subset of X for each i ∈ {1, . . . , t − 1}. Since lim An = A, there exists Nt ∈ N such that Nt > Ni−1 and An ∩ Int(K Ni ) = ∅ for each n ≥ Nt . Let i ∈ {1, . . . , t − 1} and x ∈ ANt ∩ Int(K Ni ). Since ANt ∈ D(X) and x ∈ Int(K Ni ), we have that Int(ANt ) ∩ Int(K Ni ) = ∅. This ends the proof of this claim. Since ANt is a proper subset of A, there exists a component KNt of A−ANt . By Proposition 4.9, we have that KNt is regular and by (b) of Proposition 4.1, we have that K Nt is regular. Claim 4. K Nt = K Ni for each i ∈ {1, . . . , t − 1}. Proof. Let i ∈ {1, . . . , t − 1}. Since KNt ⊂ A − ANt , we have that KNt ⊂ A − Int(ANt ) and thus K Nt ⊂ A − Int(ANt ). By Claim 3, we have that Ui = Int(ANt ) ∩ Int(K Ni ) is a nonempty open subset of X such that Ui ∩ KNt = ∅. Since Ui ⊂ K Ni , we have that K Nt = K Ni and we finish the proof of this claim. Inductively, there exists a sequence of pairwise disjoints regular subsets {K Ni }∞ i=1 such that KNi is a component of A − ANi for each i ∈ N. We can suppose, without loss of generality, that lim K Ni = K for some K ∈ D(X). Notice that, if i ∈ N, then KNi ⊂ A1 , therefore we have that lim KNi = K ⊂ 13
A1 . Since K is a cluster element of D(X), we obtain that A is not minimal. This ends the proof of this lemma. Lemma 4.11. If D(X) has infinite number of elements, then D(X) has a minimal limit point. Proof. Let A cluster element of D(X) and let Z = {B ∈ C(X) : B ⊂ A and B is a cluster element of D(X)}. Notice that A ∈ Z. Let {Am }∞ m=1 be a decreasing sequence in Z and we may assume, without loss of generality, that ∞ it is strictly decreasing. For each m ∈ N there exists a sequence {Am j }n=1 in m m D(X) where Am n = Ak for every j, k ∈ N and j = k, such that lim An = Am . n→∞
1 Let A1N1 be an element of the sequence {A1n }∞ n=1 such that H(A1 , AN1 ) < 1. Suppose that m ∈ N is such that for every i ∈ {1, . . . , m} there exists 1 i an element AiNi of the sequence {Ain }∞ n=1 such that H(Ai , ANi ) < i and j ANj = AkNk , for every j, k ∈ {1, . . . , m} such that j = k. We can choose an 1 m+1 ∞ }n=1 such that H(Am+1 , Am+1 element Am+1 Nm+1 of the sequence {An Nm+1 ) < m+1 i and Am+1 Nm+1 = ANi , for each i ∈ {1, . . . , m}. j ∞ k Inductively we have constructed a sequence {Am Nm }m=1 where ANj = ANk 1 m ∞ for every j, k ∈ N and j = k such ∞ that H(Am , ANm ) < m . Since {Am }m=1 is a decreasing sequence, if B = m=1 Am , then lim Am = B and therefore it is easy to verify that lim Am Nm = B. This shows that B is a cluster element of D(X), which shows that the intersection of every decreasing sequence in Z belongs to Z. By Theorem 4.7, we obtain that Z has minimal elements. This shows that D(X) has minimal limit elements.
Main theorem Theorem 4.12. If D(X) is not finite, then D(X) is not compact. Proof. Suppose that D(X) is compact. By Lemma 4.11, there exists a minimal limit element Y ∈ D(X). Let {An }∞ n=1 be a sequence in D(X) − Y such that lim An = Y . By Lemma 4.10 and (e) of Proposition 4.1, we may assume that An ⊂ Y and An ∩ Y = ∅ for every n ∈ N. Let K1 be a component of A1 − Y . By Proposition 4.9 and (b) of Proposition 4.1, we have that K 1 is regular. Since K1 ⊂ A1 − Y , we obtain that K 1 ∩ Int(Y ) = ∅.
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Suppose that t ∈ N is such that for every i ∈ {1, . . . , t − 1} there exists a component Kni of Ani − Y such that K ni is regular, K ni ∩ Int(Y ) = ∅ and also K ni = K nj , for every i, j ∈ {1, . . . , t − 1} such that i = j. Claim 1. There exist nt > nt−1 and a component Knt of Ant − Y such that (i) K nt is regular, (ii) K nt ∩ Int(Y ) = ∅, and (iii) K nt = K ni for each i ∈ {i, . . . , t − 1}. Proof. Suppose that for every n > nt−1 , the closure of every component of An − Y do not satisfy some condition that we request. Notice the follow. If n > nt−1 and Kn is a component of An − Y , since D(X) is closed, then by Proposition 4.9 and (b) of Proposition 4.1, we conclude that K n is regular. This means that (i) always holds. Since Kn ⊂ An − Y , then Kn ⊂ An − Int(Yn ) and therefore we have that K n ⊂ An − Int(Y ). This means that (ii) always holds. Hence for every n > nt−1 , the closure of every component of An − Y is equal to K i , for some i ∈ {1, . . . , t−1}. Thus we can suppose, without loss of generality, that for every n > nt−1 , the closure of any component of An − Y is equal to K 1 , which implies that K 1 ⊂ An for every n > nt−1 . Now, notice that K 1 ⊂ lim An = Y , which is a contradiction because K1 ⊂ X − Y . This ends the proof of Claim 1. Inductively, we construct a sequence {K ni }∞ i=1 in D(X) such that K ni ∩ Int(Y ) = ∅ for each i ∈ N and K ni = K nj for every i = j. Since D(X) is closed, we may assume, without loss of generality, that lim K ni = K where K ∈ D(X). Therefore K is a cluster element of D(X). From the fact that K ni ⊂ Ani and lim Ani = Y , we have that K ⊂ Y and since Y is a minimal limit element of D(X), we obtain that K = Y . But K ni ∩ Int(Y ) = ∅ for every i ∈ N, and by (e) of Proposition 4.1, we have that Y ∈ D(X), which is a contradiction. This finish the proof of the theorem. Conclusion As a corollary of Theorem 4.12, we have: Corollary 4.13. Let X be a continuum, then D(X) is compact if and only if D(X) is finite. Thus it is natural to consider the following problem. 15
Problem 4.14. For which n ∈ N does there exist a continuum X such that D(X) has exactly n elements? Related to the above problem, if we recall the fact that in an indecomposable continuum, every proper subcontinuum has empty interior ([2, Theorem 2]), then the following result is easy to obtain. Proposition 4.15. Let n ∈ N, let A1 , . . . , An be indecomposable continua and let pi ∈ Ai be such that pi = pj for every i, j ∈ {1, . . . , n} and i = j. Let X = A1 ∪ . . . ∪ An . (1) If Ai ∩ Aj = {pi } if and only if |i − j| = 1, then D(X) has n + (n − 1) + . . . + 1 = n(n+1) elements. 2 (2) If Ai ∩ Aj = {pi } if and only if |i − j| = 1 or |i − j| = n − 1, then D(X) has n(n − 1) + 1 elements. (3) If Ai ∩ Aj = {p1 } for every i, j ∈ {1, . . . , n} and i = j, then D(X) has n n n k=1 k = 2 − 1 elements. 5. A special space By Theorem 4.12, we have that D(X) is a continuum (degenerate) if and only if D(X) = {X}; and by Theorem 3.7, we have that if X is an indecomposable continuum, then D(X) = {X}. In this section we give an example of a decomposable continuum X for which D(X) = {X}. Let C be the canonical Cantor set contained in [0, 1] and let Y be the Buckethandle continuum contained in {(x, y, 0) ∈ R3 : x, y ∈ R} which is constructed over the Cantor set C and whose end point is (0, 0, 0) (see [1, Example 22.11]). Let Y = {(x, y − 1, 0) ∈ R3 : (x, y, 0) ∈ Y and y ≤ 0} ∪ {(x, y + 1, 0) ∈ R3 : (x, y, 0) ∈ Y and y ≥ 0} ∪ {(c, t, 0) ∈ R3 : c ∈ C, t ∈ [−1, 1]}. Notice that Y is a Buckethandle continuum that contains naturally the cylinder Z = {(c, t, 0) ∈ R3 : c ∈ C, t ∈ [−1, 1]}. For each c ∈ C, let 16
Yc = {(c, y, x) ∈ R3 : (x, y, 0) ∈ Y } and Lc = {(c, t, 0) ∈ R3 : t ∈ [−1, 1]}. Notice that for every c ∈ C, we have that • Yc is a copy of Y contained in the plane x = c. • Lc is an arc. • Y ∩ Yc = Lc . With the previous notation, we define the continuum X in R3 as Yc . X=Y ∪ c∈C
By construction, we obtain the following result. Proposition 5.1. The continuum X satisfies the following properties: (a) X is decomposable. (b) Yc ∩ Yv = ∅, for every c, v ∈ C and c = v. (c) Int(Yc ) = ∅, for every c ∈ C. (d) If {cn }∞ n=1 is a sequence in C such that lim cn = c, for some c ∈ C, then lim Ycn = Yc . (e) X − Y is open but not connected, even more, its components are the sets Yc − Lc , where c ∈ C. We adopt the following notation. Let A ∈ C(X) and c ∈ C be such that A ∩ Yc = ∅ and A ⊂ Yc . If x ∈ A ∩ Yc , then Cxc denotes the component of A ∩ Yc which contains x and Cx denotes the composant of Yc which contains x. Lemma 5.2. Let A ∈ C(X) and let e ∈ C be such that A ⊂ Ye . If x ∈ A∩Ye , then Cxe ∩ Le is a nonempty connected subset.
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Proof. First we prove that Cxe ∩ Le = ∅. Suppose that Cxe ∩ Le = ∅. Then Cxe ⊂ X − Y . By [1, Theorem 14.6], there exists an order arc α : [0, 1] → C(X) from Cxe to A. Let t > 0 be such that α(t) ∩ Y = ∅. By (e) of Proposition 5.1, there exists c ∈ C such that α(t) ⊂ Yc − Lc . Since x ∈ α(t) ∩ Ye , we have that α(t) ⊂ Ye . Now, by definition of Cxe , we have that α(x) ⊂ Cxe . Which contradicts the fact that Cxe = α(0) α(t) ⊂ Cxe . Thus Cxe ∩ Le = ∅. We shall prove that Cxe ∩ Le is connected. We only need to consider the case when Ye ⊂ Cxe . In this case, notice that Cxe is an arc of Cx , since Le ⊂ Cx and Cxe ∩ Le = ∅, we have that Cxe ∪ Le is an arc. Therefore Cxe ∩ Le is an arc and we end the proof of this lemma. Proposition 5.3. If A is a subcontinuum of X such that Y ∩ A = ∅, then A ∩ Y is a subcontinuum of Y . Proof. Let A ∈ C(X) be such that Y ∩ A = ∅. It is enough to prove that A ∩ Y is connected in the case when A ⊂ Y . Suppose that there exist two nonempty disjoint closed subsets H and K of X such that A ∩ Y = H ∪ K. Notice that, if x ∈ A−Y , then there exists c ∈ C such that x ∈ Yc . By Lemma 5.2, we have that Cxc ∩ Lc is nonempty and connected, therefore Cxc ∩ Lc ⊂ H or Cxc ∩ Lc ⊂ K. Let: H = {x ∈ A : there exists c ∈ C such that x ∈ Yc and Cxc ∩ H = ∅} ∪ H and K = {x ∈ A : there exists c ∈ C such that x ∈ Yc and Cxc ∩ K = ∅} ∪ K. It is routine to prove that H and K are two disjoint closed subset of X such that A = K ∪ H , which can not happen. Thus, we finish the proof of this lemma. Proposition 5.4. If A ∈ C(X) and c ∈ C, then A ∩ Yc is a subcontinuum of Yc . Proof. Let A ∈ C(X) and c ∈ C be such that A∩Yc is nonempty and A ⊂ Yc . We are going to prove that A ∩ Yc is connected. By Lemma 5.2, every component of A ∩ Yc intersects Lc . Let a ∈ A ∩ Lc . If there exists a component C of A ∩ Yc such that C ⊂ Ca , then C ∩ Lc = ∅. Which implies that C is subcontinuum of an indecomposable continuum Yc , which intersects two different composants of a Yc , therefore A ∩ Yc = Yc . If every component of A ∩ Yc is a subset of Ca , then there exists an arc pq with end points p and q contained in Ca such that A ∩ Yc ⊂ pq and 18
it is minimal with this property. Using the fact that A ∩ Y is connected (Proposition 5.3) and Y is the Buckethandle continuum, we have that (A ∩ Y ) ∩ Lc is an arc contained in A ∩ LC ⊂ A ∩ Yc ⊂ pq, and we conclude that A ∩ Yc = pq. This ends the proof of the proposition. Proposition 5.5. If A is a regular subcontinuum of X, then Y ⊂ A. Proof. By Proposition 5.3, B = A ∩ Y is asubcontinuum of Y , therefore B is a subcontinuum of X. If B ∩ (Y − Yc ) = ∅, then A is a proper c∈C
subcontinuum of Y or A ⊂ Yc for some c ∈ C. In both cases we have that Int(A) = ∅ which can not happen. Let x ∈ B ∩ (Y − Yc ). Since Yc c∈C
c∈C
is closed in X, there exists an open subset U of X such that x ∈ U and U ∩ Yc = ∅. c∈C
Since x ∈ A = Int(A), there exists y ∈ U ∩ Int(A). Thus W = U ∩ Int(A) is a nonempty open set of X such that W ⊂ B. Therefore B is subcontinuum of Y with nonempty interior in Y , which implies that Y = B. Proposition 5.6. Let A be a subcontinuum of X and c ∈ C. If Int(A) ∩ (Yc − Lc ) = ∅, then Yc ⊂ A. Proof. By Proposition 5.4, B = A ∩ Yc is a subcontinuum of Yc . Let x ∈ Int(A) ∩ (Yc − Lc ) and let U be an open subset in X such that x ∈ U ⊂ A and U ∩ Y = ∅. Then U ∩ Yc is a nonempty open subset of Yc contained in B, which shows that B is a subcontinuum with nonempty interior in Yc , therefore we have that Yc = B. Proposition 5.7. If A is a regular subcontinuum of X, then D = {c ∈ C : Yc ⊂ A} is dense in C. Proof. By Proposition 5.5, we havethat Y ⊂ A. Let c ∈ C and ε > 0. Let x ∈ Lc be such that x ∈ Int( Ye ). Since A is regular, there exists e∈C y ∈ Int(A) ∩ Bεd (x) ∩ Int( Ye ). e∈C Since y ∈ Int( Ye ), there exists c1 ∈ C such that y ∈ Yc1 . Therefore e∈C
Yc1 is a subcontinuum of X such that Int(A) ∩ (Yc1 − Lc1 ) = ∅. And hence, by Proposition 5.6, we have that Yc1 ⊂ A. Let t, s ∈ [0, 1] be such that x = (c, t, 0) and y = (c1 , s, 0). Since y ∈ Bεd (x), we have that |c − c1 | d(x, y) < ε. This shows that D is dense in C. 19
Theorem 5.8. The continuum X does not contain proper regular subcotinua, in other words D(X) = {X}. Proof. Let A be a regular subcontinuun of X. By Proposition 5.5, we have that Y ⊂ A; and by Proposition 5.7, we have that {c ∈ C : Yc ⊂ A} is dense in C, which implies that A is dense in X. Thus A = X. Problem 5.9. Does there exist a hereditarily decomposable continuum X for which D(X) = {X}? Acknowledgement. The author wishes to thank Prof. A. Illanes by his motivation to conclude this paper. References [1] A. Illanes and S. B. Nadler, Jr, Hyperspaces, Fundamentals and Recent Advances, Monographs and Textbooks in Pure and Applied Math., Vol. 216, Marcel Dekker, New York, Basel, 1999. [2] K. Kuratowsky, Topology, Vol. 2, Academic Press and PWN, New York, London and Warszawa, 1968. [3] S. B. Nadler, Jr., Hyperspaces of sets, Monographs and Textbooks in Pure and Applied Math., Vol 49, Marcel Dekker, New York, Basel, 1978. [4] S. B. Nadler, Jr., Continuum Theory: an Introduction, Monographs and Textbooks in Pure and Applied Math., Vol. 158, Marcel Dekker, New York, Basel, Hong Kong, 1992. [5] G. T. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ., vol. 28, Providence, RI, 1942, reprinted with corrections 1971.
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