PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 89(103) (2011), 49–56 DOI:

HEREDITARILY INDECOMPOSABLE HAUSDORFF CONTINUA HAVE UNIQUE HYPERSPACES 2^{X} AND C_{n}(X)

Alejandro Illanes

Communicated by Miloš Kurilić

Abstract. Let X be a Hausdorff continuum (a compact connected Hausdorff space). Let 2^{X} (respectively, C_{n} (X)) denote the hyperspace of nonempty closed subsets of X (respectively, nonempty closed subsets of X with at most n components), with the Vietoris topology. We prove that if X is hereditarily indecomposable, Y is a Hausdorff continuum and 2^{X} (respectively C_{n} (X)) is homeomorphic to 2^{Y} (respectively, C_{n} (Y)), then X is homeomorphic to Y.

1. Introduction

A Hausdorff continuum is a compact connected Hausdorff space X with more than one point. A subcontinuum A of X is a closed connected subset of X.

For a Hausdorff continuum X and a positive integer n, define the hyperspaces

2^{X} = {A ⊂ X : A is closed and nonempty},
C_{n}(X) = {A ∈ 2^{X} : A has at most n components},
C(X) = C_{1}(X),
F_{n}(X) = {A ∈ 2^{X} : A has at most n points}.

The hyperspace 2^{X} is endowed with the Vietoris topology. That is, the basis for the topology of 2^{X} is the family B = {⟨U_{1}, …, U_{n}⟩ : n is a positive integer and U_{1}, …, U_{n} are open subsets of X}, where ⟨U_{1}, …, U_{n}⟩ = {A ∈ 2^{X} :A ⊂ U_{1}∪⋯ ∪U_{n} and A ∩ U_{i} ≠ ∅, for each i ∈ {1, …, n}}.

The Hausdorff continuum X is hereditarily indecomposable provided that given A, B ∈ C(X), either A ∩ B = ∅ or A ⊂ B or B ⊂ A and X is said to have unique hyperspace 2^{X} (resp., C_{n}(X)), provided that, if Y is a Hausdorff continuum and 2^{X} is homeomorphic to 2^{Y} (resp., C_{n}(X) is homeomorphic to C_{n}(Y)), then X is homeomorphic to Y.

2010 Mathematics Subject Classification: Primary 54B20.

Key words and phrases: generalized arc, Hausdorff continuum, hereditarily indecomposable, hyperspace, unique hyperspace, Vietoris topology.

Uniqueness of hyperspaces has been widely studied (see [2], [3, p. 285] and [4]). Macías proved in [7] and [8, Theorem 6.1] that if X is a metric hereditarily indecomposable Hausdorff continuum, then X has unique hyperspaces 2^{X} and C_{n}(X) (for each n). Lončar [6, Theorem 2.4] proved that rim metrizable hereditarily indecomposable continua have unique hyperspace C(X). Using techniques of inverse limits, Peláez generalized these results by proving that if X is a rim metrizable (X has a basis of neighborhoods with metrizable boundary) hereditarily indecomposable Hausdorff continuum, then X has unique hyperspaces 2^{X} and C_{n}(X) (for each n).

In this paper we generalize these results by proving:

Theorem 1.1. If X is a hereditarily indecomposable Hausdorff continuum, then X has unique hyperspaces 2^{X} and C_{n}(X) (for each n).

2. Generalized arcs

The proof of Theorem 1.1 is based in an analysis of the generalized arcs (see below) in the hyperspace 2^{X}.

A generalized arc joining p and q in a topological space Z is a subcontinuum α of Z such that p, q ∈ α and each point z ∈ α − {p, q}, separates p and q in α. Given z ∈ α − {p, q}, let α − {z} = U_{z} ∪ V_{z}, where U_{z} and V_{z} are disjoint open subsets of α such that p ∈ U_{z} and p ∈ V_{z}. Let U_{p} = ∅ = V_{q}, V_{p} = α − {p} and U_{q} = α − {q}. If z, w ∈ α and z ≠ w, define z < w if and only if z ∈ U_{w}.

The following lemma summarizes the basic facts about generalized arcs. For the proof of (b), see Theorem 6.16 of [10]. The rest of Lemma 2.1 is easy to prove.

Lemma 2.1. Let α be a generalized arc joining points p and q in a topological space Z. Then:

(a) the relation < is a well defined linear order,

(b) α has the topology induced by <,

(c) given z, w ∈ α, the intervals [z, w), (z, w), (z, w] and [z, w], defined in the natural way, are connected subsets of α,

(d) (α, <) has the property of the supremum (every nonempty subset of α has a supremum in α),

(e) the relation ∽ defined in Z by z ∽ w if an only if there exists a generalized arc joining z and w in Z, is an equivalence relation; the equivalence classes are called g-arcwise components.

Let X be a Hausdorff continuum. Given A, B ∈ 2^{X}, with A ⊊ B, an order arc in 2^{X}, from A to B is a subcontinuum α of 2^{X} such that A ⊂ C ⊂ B for each C ∈ α and, for every C, D ∈ α, either C ⊂ D or D ⊂ C. The fundamental theorem for order arcs in hyperspaces is the following.

Theorem 2.2. [5, Theorem 15.3] Let X be a Hausdorff continuum and let A, B ∈ 2^{X} be such that A ⊊ B. Then there exists an order arc in 2^{X}, from A to B if and only if each component of B intersects A.

UNIQUE HYPERSPACES 2^{X} AND C_{n} (X) 51

Lemma 2.3. Let X be a Hausdorff continuum, A, B ∈ 2^{X} and let α be and order arc in 2^{X}, from A to B. Then α is a generalized arc joining A and B in 2^{X}.

Proof. It is enough to show that for each C ∈ α − {A, B}, C separates A and B in α. Let A = {D ∈ α : D ⊂ C} and B = {D ∈ α : C ⊂ D}. It is easy to show that A and B are closed subsets of α. By the definition of order arc, we have α = A ∪ B, A ∈ A − {C}, B ∈ B − {C} and A ∩ B = {C}. This proves that C separates A and B in α. □

3. Property of Kelley

A Hausdorff continuum X is said to have the Property of Kelley (see [1, p. 115]), provided that for every p ∈ X, A ∈ C(X) and open subset U of C(X) such that p ∈ A ∈ U, there exists an open subset T of X such that p ∈ T and, for each q ∈ T, there exists B ∈ C(X) such that q ∈ B ∈ U.

For metric Hausdorff continua, property of Kelley has been widely studied, see ([5, pp. 167 and 405] and [9, Chapter XIV]). In [9, Theorem 16.27] it is proved that hereditarily indecomposable Hausdorff metric continua have property of Kelley. Next, we extend this result to Hausdorff continua.

Theorem 3.1. Let X be a hereditarily indecomposable Hausdorff continuum. Then X has property of Kelley.

Proof. Suppose that X does not have the property of Kelley. Then there exist p ∈ X, A ∈ C(X) and V = ⟨V_{1}, …, V_{m}⟩ ∩ C(X), a basic set in C(X), such that p ∈ A ∈ V and for each open subset T of X such that p ∈ T there exists a point q ∈ T for which there is no element B ∈ C(X) with the property that q ∈ B ∈ V. Note that A ≠ X.

Let W be an open subset of X such that A ⊂ W ⊂ cl_{X} (W) ⊂ V_{1}∪ ⋯∪ V_{m} and W ≠ X. Let T = {T ⊂ X : T is an open subset of X such that p ∈ T ⊂ W }. For each T ∈ T , choose a point q_{T} ∈ T for which there is no element B ∈ C(X) with the property that q_{T} ∈ B ∈ V and let D_{T} be the component of cl_{X} (W) such that q_{T} ∈ D_{T}. Since D_{T} ⊂ V_{1}∪ ⋯∪ V_{m} and D_{T} ∉ V, there exists i ∈ {1, …, m} such that D_{T} ∩ V_{i} = ∅. Let E = {q_{T} ∈ X : T ∈ T }. Then p ∈ cl_{X} (E). For each i ∈ {1, …, m}, let T_{i} = {T ∈ T : D_{T} ∩ V_{i} = ∅} and E_{i} = {q_{T} ∈ X : T ∈ T_{i}}. Then T = T_{1} ∪ ⋯∪ T_{m} and E = E_{1} ∪ ⋯∪ E_{m}. Thus, there exists i ∈ {1, …, m} such that p ∈ cl_{X} (E_{i}). We may assume that p ∈ cl_{X} (E_{1}).

For each Q ∈ T , let F (Q) = cl_{X} ( ⋃ {D_{T} : T ∈ T_{1} and q_{T} ∈ Q}). Since p ∈ cl_{X} (E_{1}), it follows that p ∈ F (Q). Thus, F (Q) is a nonempty compact subset of X. Let F = ⋂ {F (Q) : Q ∈ T } and let G be the component of F such that p ∈ G.

We need to show that Fr_{X} (W)∩G ≠ ∅. Suppose to the contrary that Fr_{X} (W)∩ G = ∅. By [5, Theorem 12.9], there exist disjoint compact subsets K and L of X such that F = K ∪ L, G ⊂ K and K ∩ Fr_{X} (W) = ∅. Let R and S be disjoint open subsets of X such that K ⊂ R and Fr_{X} (W) ∪ L ⊂ S. Since F ⊂ R ∪ S, there exist k ⩾ 1 and elements Q_{1}, …, Q_{k} ∈ T of X such that F ⊂ F (Q_{1})∩⋯ ∩ F (Q_{k}) ⊂ R∪S. Let Q_{0} = Q_{1}∩⋯ ∩Q_{k}. Then Q_{0} is an open subset of X such that p ∈ Q_{0} ⊂ W and F (Q_{0}) ⊂ R ∪ S. Since p ∈ R ∩ Q_{0} and p ∈ cl_{X} (E_{1}), there exists T ∈ T_{1} such that q_{T} ∈ R ∩ Q_{0} ∩ D_{T}. Then D_{T} ⊂ F (Q_{0}) ⊂ R ∪ S. Since D_{T} is connected, D_{T} ⊂ R, so Fr_{X} (W) ∩ D_{T} = ∅, this contradicts [5, Theorem 12.10]. We have shown that Fr_{X} (W) ∩ G ≠ ∅.

Recall that G ∈ C(X) and p ∈ G ∩ A. Since X is hereditarily indecomposable, G ⊂ A or A ⊂ G. Since Fr_{X} (W) ∩ G ≠ ∅ and Fr_{X} (W) ∩ A = ∅, we obtain that A ⊂ G. Since A ∩ V_{1} ≠ ∅, we can choose a point x ∈ A ∩ V_{1} ⊂ F ∩ V_{1}. Then x ∈ V_{1}∩ F (W). Thus, there exists T ∈ T_{1} such that V_{1}∩ D_{T} ≠ ∅. This contradicts the definition of T_{1} and ends the proof of the theorem. □

4. Main results

Lemma 4.1. Let X be a hereditarily indecomposable Hausdorff continuum. Suppose that A, B ∈ 2^{X}, C ∈ C(X), A ⊂ C, B ⊈ C and α is a generalized arc joining A and B in 2^{X}. Then C ∈ α.

Proof. Let D = {D ∈ α : D ⊂ C}. Then D is a nonempty proper closed subset of α. Since α is connected, there exists D_{0} ∈ Fr_{α} (D). Since D is closed in α, D_{0} ⊂ C. Let < be the order defined, as in the paragraph previous to Lemma 2.1, for α satisfying A < B. We claim that D_{0} = C. Suppose to the contrary that D_{0} ≠ C. Fix a point x_{0} ∈ C − D_{0}. Since ⟨X − {x_{0}}⟩ ∩ α is an open subset of α containing D_{0}, by Lemma 2.1(b), we may assume that there exists an element D_{1} ∈ α, such that D_{0} < D_{1} such that [D_{0}, D_{1}] ⊂ ⟨X − {x_{0}}⟩ and [D_{0}, D_{1}] ⊈ D. Let E = ⋃ {F : F ∈ [D_{0}, D_{1}]}. Notice that D_{0} ⊂ E, x_{0} ∉ E and E ⊈ C. We check that E is closed in X. Let x ∈ X − E. Then {⟨X − cl_{X} (U)⟩ : x ∈ U and U is an open subset of X} is an open cover of the compact set [D_{0}, D_{1}]. Thus, there exists an open subset U_{0} of X such that x ∈ U_{0} and [D_{0}, D_{1}] ⊂ ⟨X − cl_{X} (U_{0})⟩. Hence, U_{0}∩ E = ∅. This proves that E is closed in X. Fix a point y_{0} ∈ E − C.

We claim that the component G of E that contains y_{0} intersects C. Suppose to the contrary that G ∩ C = ∅. By [5, Theorem 12.9], there exists an open and closed subset K of E such that y_{0} ∈ K and K ∩ C = ∅. Let L = E − K. Then L is a compact subset of X and D_{0} ⊂ L. Let E = {F ∈ [D_{0}, D_{1}] : F ⊂ L} and F = {F ∈ [D_{0}, D_{1}] : F ∩ K ≠ ∅}. Clearly, E and F are closed disjoint subsets of [D_{0}, D_{1}], D_{0} ∈ E, F ≠ ∅ and [D_{0}, D_{1}] = E ∪ F. This contradicts the connectedness of [D_{0}, D_{1}] and completes the proof that G ∩ C ≠ ∅.

Since X is hereditarily indecomposable, G and C are subcontinua of X and G ∩ C ≠ ∅, we have G ⊂ C or C ⊂ G. Since y_{0} ∈ G − C and G ⊂ E ⊂ X − {x_{0}}, we obtain a contradiction. This ends the proof that C = D_{0} ∈ α. □

The following lemma can be proved imitating the proof of Theorem 11.3 of [9] and using Theorem 2.2.

Lemma 4.2. Let Y be a Hausdorff continuum, n ⩾ 1 and A ∈ 2^{Y} − C(Y). Then 2^{Y} − {A} and C_{n}(Y) − {A} are arcwise connected.

Lemma 4.3. Let X be a hereditarily indecomposable continuum and n ⩾ 1. Let A ∈ C(X) − ({X} ∪ F_{1}(X)). Let E be the g-arcwise component of C_{n}(X) − {A} that contains X. Then E is the only dense g-arcwise component of C_{n}(X) − {A}.

Proof. Let F be a g-arcwise component of C_{n}(X) − {A} such that F ≠ E. Given B ∈ C_{n}(X) − {A} such that B ⊈ A, taking an order arc from B to X, we obtain that B ∈ E. This implies that F ⊂ C_{n}(A). Since C_{n}(A) is compact and C_{n}(A) ≠ C_{n}(X), we conclude that F is not dense in C_{n}(X) − {A}.

Now we check that E is dense in C_{n}(X) − {A}. Let B ∈ C_{n}(X) − {A} and U = ⟨U_{1}, …, U_{m}⟩ ∩ C_{n}(X) be a basic set in C_{n}(X) such that B ∈ U, where each U_{i} is a nonempty open subset of X. We may assume that B ∉ E. Then B ⊊ A. Let B_{1}, …, B_{k} be the components of B. Then k ⩽ n. Fix points p_{1} ∈ B_{1}, …, p_{k} ∈ B_{k}. For each i ∈ {1, …, k}, let F_{i} = {j ∈ {1, …, m} : B_{i} ∩ U_{j} ≠ ∅} and U_{i} = {C ∈ C(X) : C ⊂ ⋃ {U_{j} : j ∈ F_{i}} and C ∩ U_{j} ≠ ∅ for each j ∈ F_{i}}. Then U_{i} is open in C(X) and B_{i} ∈ U_{i}. By Theorem 3.1, there exists an open subset V_{i} of X such that p_{i} ∈ V_{i} and, if u ∈ V_{i}, then there exists Q ∈ C(X) such that u ∈ Q ∈ U_{i}.

Fix a point x ∈ X −A. Let α be an order arc in C(X) joining {x} to X. Notice that α ⊂ E. Let E_{0} ∈ α be the first element of α containing A, that is E_{0} ∈ α, A ⊂ E_{0} and, if E ⊊ E_{0} and E ∈ α, then A ⊈ E. Since the set ⟨V_{1}, …, V_{k}, X⟩∩C(X) is an open subset of C(X) containing E_{0}, there exists E ∈ α such that E ⊊ E_{0} and E ∈ ⟨V_{1}, …, V_{k}, X⟩. For each i ∈ {1, …, k}, choose a point u_{i} ∈ E ∩ V_{i}. By the choice of V_{i}, there exists Q_{i} ∈ C(X) such that u_{i} ∈ Q_{i} ∈ U_{i}. Let D = {u_{1}, …, u_{k}}.

Since x ∈ E, E ⊈ A. Since X is hereditarily indecomposable and E ∈ C(X), E ∩ A = ∅. Thus, D ∩ A = ∅. Let Q = Q_{1}∪ ⋯∪ Q_{k}, since D ⊂ Q, we have that Q ∈ E. It is easy to show that Q ∈ U. This completes the proof of the lemma. □

Theorem 4.4. If X is a hereditarily indecomposable Hausdorff continuum, then X has unique hyperspaces 2^{X} and C_{n}(X), for each n.

Proof. Let Y be a Hausdorff continuum. Let K(X) denote one of the hyperspaces 2^{X} or C_{n}(X) of X and let K(Y) be the correspondig hyperspace of Y. Suppose that K(X) is homeomorphic to K(Y). Let h : K(X) → K(Y) be a homeomorphism. Since F_{1}(X) (respectively F_{1}(Y)) is homeomorphic to X (respectively Y) it is enough to show that h(F_{1}(X)) = F_{1}(Y).

Using Proposition 6.3 of [9], it is easy to show that every proper subcontinuum of X has empty interior.

Claim 1. Let A ∈ C(X) − ({X} ∪ F_{1}(X)). Let E be the g-arcwise component of K(X)−{A} that contains X. Then E is the only dense g-arcwise component of K(X) − {A}.

We prove Claim 1. By Lemma 4.3, Claim 1 holds for the case that K(X) = C_{n}(X). Thus, in the proof of Claim 1, we assume that K(X) = 2^{X}. Let F be a g-arcwise component of 2^{X} − {A} such that F ≠ E. Given B ∈ 2^{X} − {A} such that B ⊈ A, taking an order arc from B to X, we obtain that B ∈ E. This implies that F ⊂ 2^{A}. Since 2^{A} is compact and 2^{A} ≠ 2^{X}, we conclude that F is not dense in 2^{X} − {A}. Now we check that E is dense in 2^{X} − {A}. Let B ∈ 2^{X} − {A} and let U = ⟨U_{1}, …, U_{m}⟩ be a basic set in 2^{X} such that B ∈ U, where each U_{i} is a nonempty open subset of X. Since int_{X} (A) = ∅, for each i ∈ {1, …, m}, U_{i} ⊈ A.

Given i ∈ {1, …, m}, fix a point p_{i} ∈ U_{i} − A. Let G = {p_{1}, …, p_{m}}. Then G ∈ U ∩ E. This completes the proof of Claim 1.

Claim 2. h(C(X)) ⊂ C(Y).

In order to prove Claim 2, take an element A ∈ C(X) − ({X} ∪ F_{1}(X)). By Lemma 4.1, K(X) − {A} is not g-arcwise connected. By Lemma 4.2, if h(A) is not connected, then K(Y) − {h(A)} is g-arcwise connected. Since h is a homeomorphism, we conclude that h(A) is connected. That is, h(A) ∈ C(Y). Therefore, h(C(X) − ({X} ∪ F_{1}(X))) ⊂ C(Y). Since cl_{2^{X}} (C(X) − ({X} ∪ F_{1}(X))) = C(X) and C(Y) is compact, we conclude that h(C(X)) ⊂ C(Y).

Claim 3. Let A ∈ C(X) − {X} and let B ∈ K(X) be such that B ⊂ A.

Then h(B) ⊂ h(A).

We prove Claim 3. In the case that A ∈ F_{1}(X), B = A and h(B) = h(A). Thus, suppose that A ∉ F_{1}(X). By Claim 2, h(A) is connected. We can suppose that h(A) ≠ Y. Let E (respectively, F) be the g-arcwise component of K(X) − {A} (respectively, K(Y) − {h(A)}) that contains X (respectively, Y). By Lemma 4.1, B ∉ E. If h(B) ⊈ h(A), then taking an order arc from h(B) to Y in K(Y) we can prove that h(B) ∈ F. Since h is a homeomorphism, h(E) ≠ F, so h(E) ∩ F = ∅. Since K(Y) − (2^{h(A)} ∩ K(Y)) ⊂ F, h(E) ⊂ 2^{h(A)} ∩ K(Y). Since h(A) ≠ Y, 2^{h(A)} ∩ K(Y) is a proper compact subset of K(Y). Thus, h(E) is not dense in K(Y)−{h(A)}. This implies that E is not dense in K(X) − {A}, contrary to Claim 1. This proves that h(B) ⊂ h(A).

Claim 4. h(F_{1}(X)) ⊂ F_{1}(Y).

Suppose, contrary to Claim 4, that there exists a point p ∈ X such that h({p}) ∉ F_{1}(Y). Since F_{1}(Y) is compact, there exists an open subset U of X such that p ∈ U and ⟨U⟩ ⊂ K(X) − h^{−1} (F_{1}(Y)). Taking an order arc from {p} to X in C(X), it is possible to find two nondegenerate proper subcontinua A and B of X such that p ∈ A ⊊ B ⊂ U. Let E be the g-arcwise connected component of K(X) − {B} such that {p}, A ∈ E. By Lemma 4.1, E ⊂ K(X) ∩ 2^{B} ⊂ ⟨U⟩. Thus h(E) ⊂ K(Y) − F_{1}(Y). Since h is a homeomorphism, h(E) is a g-arcwise component of K(Y) − {h(B)}. By Claims 2 and 3, h(A) ∈ C(Y) and h(A) ⊊ h(B). Fix a point y ∈ h(A) and take an order arc β from {y} to h(A) in C(Y). Notice that β ⊂ K(Y) − {h(B)}. Since h(A) ∈ h(E), β ⊂ h(E). Hence, {y} ∈ h(E) ∩ F_{1}(Y), a contradiction. Claim 4 is proved.

Claim 5. Let A ∈ C(X). Then h(A) = ⋃ {h({p}) ∈ K(Y) : p ∈ A}.

Let B = ⋃ {h({p}) ∈ K(Y) : p ∈ A}. By [1, Lemma 2.1], B is a subcontinuum of Y. In the case that A ≠ X, by Claim 3, B ⊂ h(A). Now, we see that, in the case that A = X, we also have that B ⊂ h(A). It is enough to show that, if p ∈ X, then h({p}) ⊂ h(X). Suppose to the contrary that there exists a point y ∈ h({p}) − h(X). Let W = h^{−1} (⟨Y − {y}⟩). Then W is an open subset of K(X) such that X ∈ W. Let λ be an order arc from {p} to X in C(X). Then there exists E ∈ λ such that E ≠ X and E ∈ W. By Claim 3, h({p}) ⊂ h(E). Thus, y ∈ h(E) ∈ ⟨Y − {y}⟩. This contradiction proves that, in every case, B ⊂ h(A).

Suppose that B ≠ h(A). Let V = K(Y) − (2^{B} ∩ K(Y)) and U = h^{−1} (V). Then U is an open subset of K(X) such that A ∈ U. Fix a point p_{0} ∈ A and let α be an order arc from {p_{0}} to A in C(X). Then there exists C ∈ α − {A, {p_{0}}} such that C ∈ U. Then {p_{0}} ⊊ C ⊊ A. Fix a point q_{0} ∈ A − C. Since h({p_{0}}), h({q_{0}}) ⊂ B, taking order arcs in C(Y), it is possible to construct a generalized arc β joining h({p_{0}}) and h({q_{0}}) in 2^{B} ∩ K(Y). Thus, h^{−1} (β) is a generalized arc in K(X) joining {p_{0}} and {q_{0}}. By Lemma 4.1, C ∈ h^{−1} (β). Thus, h(C) ∈ β ⊂ 2^{B} ∩ K(Y). Hence, h(C) ∈ 2^{B} ∩ K(Y) ∩ V, a contradiction. We have shown that B = h(A). This ends the proof of Claim 5.

Claim 6. Let A ∈ K(X). Then h(A) ⊂ h(X).

In order to prove claim 6, we may assume that A ≠ X. By Claim 2, h(X) is connected. In order to prove Claim 6, since 2^{h(X)} ∩ K(Y) is compact, it is enough to show that h(A) ∈ cl_{2^{Y}} (2^{h(X)} ∩ K(Y)). Let V be an open subset of K(Y) such that h(A) ∈ V. Let U = ⟨U_{1}, …, U_{m}⟩ ∩ K(X) be a basic open subset of K(X) such that A ∈ U ⊂ h^{−1} (V) and X ∉ U, where each U_{i} is open in X. Note that X ⊈ U_{1}∪ ⋯∪ U_{m}. Fix a point p ∈ X and let α be an order arc from {p} to X in C(X). We analyze two cases:

Case 1. K(X) = 2^{X}. Since X ∈ ⟨U_{1}, …, U_{m}, X⟩, there exists B ∈ α ∩ ⟨U_{1}, …, U_{m}, X⟩ such that B ≠ X. For each i ∈ {1, …, m}, choose a point p_{i} ∈ B ∩ U_{i}. Let C = {p_{1}, …, p_{m}}. Then C ⊂ B, C ∈ U and h(C) ∈ V. By Claim 3, h(C) ⊂ h(B). By Claim 5, h(B) = ⋃ {h({b}) ∈ 2^{Y} : b ∈ B} ⊂ ⋃ {h({x}) ∈ 2^{Y} : x ∈ X} = h(X). Hence, h(C) ∈ 2^{h(X)} . Hence, h(A) ∈ cl_{2^{Y}} (2^{h(X)} ∩ K(Y)) = 2^{h(X)} ∩ K(Y).

Case 2. K(X) = C_{n}(X). Let A_{1}, …, A_{k} be the components of A. Then k ⩽ n. For each i ∈ {1, …, k}, choose a point a_{i} ∈ A_{i}, let F_{i} = {j ∈ {1, …, m} : A_{i} ∩ U_{j} ≠ ∅} and U_{i} = {C ∈ C(X) : C ⊂ ⋃ {U_{j} : j ∈ F_{i}} and C ∩ U_{j} ≠ ∅ for each j ∈ F_{i}}. Then U_{i} is open in C(X) and A_{i} ∈ U_{i}. By Theorem 3.1, there exists an open subset V_{i} of X such that a_{i} ∈ V_{i} and, if u ∈ V_{i}, then there exists Q ∈ C(X) such that u ∈ Q ∈ U_{i}. Since X ∈ ⟨V_{1}, …, V_{k}, X⟩ ∩ K(X), there exists B ∈ (α − {X}) ∩ ⟨V_{1}, …, V_{k}, X⟩. For each i ∈ {1, …, k}, fix a point b_{i} ∈ B ∩ V_{i} and let Q_{i} ∈ C(X) be such that b_{i} ∈ Q_{i} ∈ U_{i}. Let Q = Q_{1}∪ ⋯∪ Q_{k}. Then Q ∈ C_{n}(X) and Q ∈ U. Since Q ⊂ U_{1} ∪ ⋯∪ U_{m}, Q ≠ X. Since X is hereditarily indecomposable and B ≠ X, we obtain that B ∪ Q ≠ X. Notice that B ∪ Q is a subcontinuum of X. By Claim 3, h(Q) ⊂ h(B ∪ Q) and, by Claim 5, h(B ∪Q) = ⋃ {h({b}) ∈ 2^{Y} : b ∈ B ∪Q} ⊂ ⋃ {h({x}) ∈ 2^{Y} : x ∈ X} = h(X). Thus, h(Q) ⊂ h(X) and h(Q) ∈ V. Therefore, h(A) ∈ cl_{2^{Y}} (2^{h(X)} ∩K(Y)) = 2^{h(X)} ∩K(Y). Claim 6 is proved.

We are ready to show that h(F_{1}(X)) = F_{1}(Y). Let y ∈ Y and B = h^{−1} ({y}). By Claim 6, {y} = h(B) ⊂ h(X). By Claim 5, h(X) = ⋃ {h({x}) ∈ 2^{Y} : x ∈ X}. Thus, there exists x ∈ X such that y ∈ h({x}). By Claim 4, h({x}) is a onepoint set. Hence, {y} = h({x}) ∈ h(F_{1}(X)). We have shown that F_{1}(Y) ⊂ h(F_{1}(X)). Thus, Claim 4 implies that h(F_{1}(X)) = F_{1}(Y). This ends the proof of the theorem. □

Question 1. (see [3, Problem 37]) Let X be a hereditarily indecomposable Hausdorff continuum. Is it true that X has a unique hyperspace F_{n}(X), for each n ⩾ 2? That is, suppose that Y is a Hausdorff continuum such that F_{n}(X) is homeomorphic to F_{n}(Y), does it follows that X is homeomorphic to Y?

The answer to this question is not known even for the case that X is a metric continuum and n = 2.

Acknowledgment. The author wishes to thank the participants in the Third Workshop of Research in Continua and Hyperspaces (Toluca City, July, 2009) for useful discussions on the topic of this paper, particularly, Gerardo Acosta, Mauricio E. Chacón, Jorge M. Martínez, Rodrigo Hernández, Antonio Peláez and Alicia Santiago. The author also thanks Marianela Vélez for her technical help during the preparation of this paper.

References

1. J. J. Charatonik and A. Illanes, Local connectedness in hyperspaces, Rocky Mountain J. Math. 36 (2006), 811–856.

2. R. Hernández-Gutiérrez, A. Illanes and V. Martínez-de-la-Vega, Uniqueness of hyperspaces for Peano continua, to appear in Rocky Mountain J. Math.

3. A. Illanes, Hyperspaces of Continua; in: Elliott Pearl (ed.), Open Problems in Topology II, Elsevier, 2007, 279–288.

4. A. Illanes, Dendrites with unique hyperspace C_{2}(X), II, Topology Proc. 34 (2009), 77–96.

5. A. Illanes and S. B. Nadler, Jr., Hyperspaces: Fundamentals and Recent Advances, Monographs and Textbooks Pure Appl. Math. 216, Marcel Dekker, New York and Basel, 1999.

6. I. Lončar, Nonmetric rim-metrizable continua and unique hyperspace, Publ. Inst. Math., Nouv. Sér. 73(87) (2003), 97–113.

7. S. Macías, Hereditarily indecomposable continua have unique hyperspace 2^{X} , Bol. Soc. Mat. Mexicana 5(3) (1999), 415–418.

8. S. Macías, On the hyperspaces C_{n} (X) of a continuum X, II, Topology Proc. 25 (2000), 255–276.

9. S. B. Nadler, Jr., Hyperspaces of sets: A text with research questions, Monographs and Textbooks Pure Appl. Math. 49, Marcel Dekker, New York and Basel, 1978.

10. S. B. Nadler, Jr., Continuum Theory: An Introduction, Monographs and Textbooks Pure Appl. Math. 158, Marcel Dekker, New York, Basel and Hong Kong, 1992.

11. A. Peláez, On the uniqueness of the hyperspaces 2^{X} and C_{n} (X) of rim-metrizable continua, Topology Proc. 30 (2006), 565–576.

Instituto de Matemáticas (Received 22 04 2010) Universidad Nacional Autónoma de México Area de la Investigación Científica Circuito Exterior Ciudad Universitaria México, 04510, D.F. México

 illanes@matem.unam.mx