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Reflexive families of closed sets Zhongqiang Yang and Dongsheng Zhao Abstract. Let S(X) denote the set of all closed subsets of a topological space X, and C(X) denote the set of all continuous mappings f : X → X. A family A ⊆ S(X) is called reflexive if there exists C ⊆ C(X) such that A = {A ∈ S(X) : f (A) ⊆ A for every f ∈ C}. In this paper, we investigate the conditions for a family of closed sets to be reflexive. Recall [3] that a collection A of closed subspaces of a Hilbert space H is called reflexive if there exists a collection F of continuous operators on H such that A = Lat(F) = {A| A is a closed subspace of H with T (A) ⊆ A, ∀T ∈ F}. Reflexive families of continuous operators is defined in a dual way. See [2][3] [4] [5] [6][7] for more about the characterizations of such families. In [9], the second author considered the reflexive families in concrete categories. For the category SET of sets, a complete characterization for both reflexive 1991 Mathematics Subject Classification. 47A46,54C08,54D05,54E35. Key words and phrases. reflexive families of closed sets, s-reflexive topological space, strongly zero-dimensional metric space, countable metric space, hereditarily disconnected space, reflexive closed set. The first author was supported for this work by National Natural Science Foundation of China (No. 10471084) and by Guangdong Provincial Natural Science Foundation (No. 04010985). 1 2 Z. YANG AND D. ZHAO families of sets and reflexive families of mappings were obtained in [9]. In the present paper we investigate the reflexive families in the context of topological spaces. Given a topological space X, let S(X) be the set of all closed sets in a topological space X, and C(X) the set of all continuous mappings f : X → X. For any A ⊆ S(X) and F ⊆ C(X) define Alg(A) = {f ∈ C(X) : f (A) ⊆ A for every A ∈ A} and Lat(F) = {A ∈ S(X) : f (A) ⊆ A for every f ∈ F}. A family A ⊆ S(X) is reflexive if there exists F ⊆ C(X) such that A = Lat(F), or equivalently if Lat(Alg(A)) = A. A topological space X is s-reflexive whenever every family A of closed subsets of X satisfying conditions (a) X, ∅ ∈ A, (b) B ⊆ A implies T B ∈ A, and S (c) B ⊆ A implies cl( B) ∈ A is reflexive. The main results in this paper are: every strongly zero-dimensional complete metric space is s-reflexive; every countable metric space is s-reflexive; every Hausdorff s-reflexive space is hereditarily disconnected. From these it is deduced that a locally compact metric space is s-reflexive if and only if it is zero-dimensional. We end the paper by listing some problems for possible further work. Let S(X) be the set of all closed subsets in a topological space X, and C(X) be the set of all continuous mappings f : X → X. For any A ⊆ S(X) and F ⊆ C(X), define Alg(A) = {f ∈ C(X) : f (A) ⊆ A, ∀A ∈ A}, Lat(F) = {A ∈ S(X) : f (A) ⊆ A, ∀f ∈ F}. REFLEXIVE FAMILIES OF CLOSED SETS 3 The two mappings Alg : P(S(X)) → P(C(X)) and Lat : P(C(X)) → P(S(X)) form a Galois connection between P(S(X)) and P(C(X)). Thus as in the general case, for any A ⊆ S(X) and F ⊆ C(X) we have (i) Lat(Alg(A)) ⊇ A, Alg(Lat(F)) ⊇ F; (ii) Alg(Lat(Alg(A))) = Alg(A), Lat(Alg(Lat(F))) = Lat(F). A family A ⊆ S(X) is called reflexive if A = Lat(Alg(A)). Similarly, F ⊆ C(X) is reflexive if F = Alg(Lat(F)). As in the general case [9], A ⊆ S(X) is reflexive if and only if there exists F ⊆ C(X) such that A = Lat(F). Also by the above condition (i), A is reflexive if and only if Lat(Alg(A)) ⊆ A. Lemma 1. If A ⊆ S(X) is reflexive, then the following conditions are satisfied: (a) X, ∅ ∈ A. (b) B ⊆ A implies T B ∈ A. S (c) B ⊆ A implies cl( B) ∈ A. (d) If D is a connect component of A ∈ A and B ⊆ D for some nonempty B in A, then D ∈ A. Proof. Only (d) needs verification. For any f ∈ Alg(A), f (B) ⊆ B because B ∈ A. Also f (B) ⊆ f (D) ⊆ f (A) ⊆ A, and f (D) ∩ D ⊇ f (B) ∩ B = f (B) 6= ∅. Thus f (D) is a connected set contained in A and has non-empty intersection with the connected component D of A, hence f (D) ⊆ D. Therefore D ∈ Lat(Alg(A)) = A. 4 Z. YANG AND D. ZHAO If X is a space with the discrete topology, then S(X) = P(X) and C(X) is the set of all mappings from X to X. By Theorem 1 of [9], a family A ⊆ P(X) is reflexive if and only if A is closed under arbitrary union and intersection, so every family A of closed subsets of a discrete space satisfying conditions (a),(b) and (c) in Lemma 1 is reflexive. A natural question is: beside the discrete topological spaces, which else spaces also have this property? Definition 1. A topological space X is called s-reflexive if every family A of closed subsets of X satisfying the conditions (a),(b) and (c) in Lemma 1 is reflexive. Lemma 2. For every topological space X, the following conditions are equivalent: (1) For any two points a and b in X, the following mapping fa,b : X → X is continuous: fa,b (x) = x, if x 6= a, b, if x = a. (2) For any finite set D ⊆ X and any element b ∈ X the following mapping fD,b : X → X is continuous: fD,b (x) = x, if x 6∈ D, b, if x ∈ D. (3) For any non-empty proper closed set B and finite set D, both B \ D and B ∪ D are closed. REFLEXIVE FAMILIES OF CLOSED SETS 5 Proof. (1) implies (2). This follows from that fD,b is actually the composition of mappings in {fa,b : a ∈ D}. (2) implies (3). Let B be a non-empty proper closed set and D be a −1 finite set. Choose any b ∈ B, then fD,b (B) = B ∪ D, so B ∪ D is closed. −1 (B) = B \ D is also closed. Choose any c ∈ X \ B, then fD,c (3) implies (1). If a = b, then fa,b = idX , the identity mapping, which is continuous. If a 6= b, for any non-empty proper closed set B, B ∪ {a}, if b ∈ B, −1 fa,b (B) = B \ {a}, if b 6∈ B. −1 (B) is always closed and hence fa,b is continuous. So fa,b Corollary 1. Let X satisfy the equivalent conditions in Lemma 2 with |X| = 6 2 . If there exists one closed singleton {a}, then X is T1 . In fact, assume that X is not a singleton. For any point x ∈ X, {x, a} = {a} ∪ {x} is a proper closed set. Thus {x, a} \ {a} = {x} is also closed. The set {a, b} with the non-T1 topology satisfies the conditions in Lemma 2. Thus the requirement |X| = 6 2 is needed. Let A ⊆ S(X) be a collection of closed subsets of X satisfying conditions T (a),(b) and (c)in Lemma 1. For each Y ∈ S(X), let φA (Y ) = {A ∈ A : Y ⊆ A}. If no confusion occurs, we simply write φ(Y ) for φA (Y ). On this mapping, we have the following simple lemma. Lemma 3. Let A ⊆ S(X) satisfy conditions (a),(b) and (c). (1) For each Y ⊆ X, φ(Y ) ∈ A. And Y ∈ A if and only if Y = φ(Y ). (2) For any B ∈ S(X), B ∈ A if and only if φ({x}) ⊆ B for all x ∈ B. S (3) For any B ∈ S(X), B ∈ A if and only if B = {φ({x}) : x ∈ B}. 6 Z. YANG AND D. ZHAO Lemma 4. Let A ⊆ S(X) satisfy conditions (a),(b) and (c). For each U ⊆ X define κ(U ) = {x ∈ X : φ({x}) ∩ U 6= ∅}. Then (1) κ(V ) ⊇ V for all V ⊆ X; S S (2) κ( i∈I Ui ) = i∈I κ(Ui ); (3) if U is an open set of X, then κ(U ) is also open. Proof. (1) and (2) are trivial. We only prove (3). Let U be open and (xλ )λ∈D be a net in X \ κ(U ) which converges to the point a ∈ X. For each S λ ∈ D, φ({xλ }) ∩ U = ∅, Thus cl λ∈D φ({xλ }) ∩ U = ∅. By condition (c), S the set B = cl λ∈D φ({xλ }) is in A. Now a ∈ B, so φ({a}) ⊆ B ⊆ X \ U . Hence φ({a}) ∩ U = ∅, which implies a ∈ X \ κ(U ). Therefore, κ(U ) is open. Theorem 1. If X satisfies the equivalent conditions in Lemma 2, then X is s-reflexive. Proof. Suppose A ⊆ S(X) satisfies conditions (a),(b) and (c). Let B be a closed set not in A. By Lemma 3 (2), there is a ∈ B such that φ({x}) 6⊆ B. Choose a point b ∈ φ({a}) \ B and define the mapping f : X → X by f (x) = x, if x 6= a, b, if x = a. By the assumption on X, f is continuous. For every A ∈ A, if a ∈ A then φ({a}) ⊆ A, so f (a) = b ∈ φ({a}) ⊆ A, from this it follows that f (A) ⊆ A. If a 6∈ A, then f (A) = A. Thus f ∈ Alg(A). But f (B) 6⊆ B, so A ⊇ Lat(Alg(A)) and hence A = Lat(Alg(A)). So A is reflexive. Example 1. (1) Every discrete and indiscrete space satisfies the conditions in Lemma 2, so they are s-reflexive. REFLEXIVE FAMILIES OF CLOSED SETS 7 (2) Let λ be an infinite cardinal and X be any set. Define τ to be the topology consisting all subsets whose complements are either X or has cardinal less than λ. Then (X, τ ) satisfies the conditions in Lemma 2, it is therefore s-reflexive. (3) Suppose E is a subset of X. Define the topology τE such that a subset A of X is closed if and only if either A = X, or A\E is finite. Then (X, τE ) obviously satisfies the conditions in Lemma 2. So (X, τE ) is s-reflexive. (4) The Euclidean interval X = [−1, 1] is not s-reflexive. As a matter of fact, let A = {A ∈ S(X) : A ⊆ [0, 1] or A 3 1}. Then it is easy to verify that A satisfies conditions (a),(b) and (c) in Lemma 1. For every f ∈ Alg(A) and every x ∈ X, since {x, 1} ∈ A, f ({x, 1}) ⊆ {x, 1}, so f (x) = x or f (x) = 1. Let A = {x ∈ [−1, 1) : f (x) = x} and B = {x ∈ [−1, 1) : f (x) = 1}. Then A and B are closed in [−1, 1) and are disjoint, and A∪B = [−1, 1). Hence A = ∅ or B = ∅ because [−1, 1) is connected. For every x ∈ [0, 1], since {x} ∈ A we have f (x) = x. It follows that A ⊃ [0, 1) 6= ∅. This implies A = [−1, 1). Notice that f (1) = 1 for every f ∈ Alg(A) because {1} ∈ A. Thus Alg(A) = {idX }. But Lat({idX }) = S(X) 6= A. Hence A is not reflexive. Since the interval [−1, 1] equipped with the discrete topology is s-reflexive and the Euclidean interval is a continuous image of it, so the continuous image of s-reflexive spaces need not be s-reflexive. In the following we investigate the s-reflexivity of some special spaces . Lemma 5. [1, Theorem 7.3.1(iii)] If (X, d) is a strongly zero-dimensional metric space, then there is a sequence W1 , W2 , · · · , of open covers of the 8 Z. YANG AND D. ZHAO space X such that Wi consists of clopen sets of X, diam(W ) < 1/i for each W ∈ Wi , and Wi+1 is a refinement of Wi . Lemma 6. Let U be an open cover of a strongly zero-dimensional metric space X. Then there is a refinement V of U consisting of pairwise disjoint clopen sets in X. This lemma is a corollary of [8, Lemma 3 ]. The following direct proof was provided by the referee. Proof. Since X is paracompact, we can assume that the family U = {Uα : α < λ} is locally finite and cl(Uα ) ⊆ U for every α < λ. At first we construct a family {Wα : α < λ} of clopen sets of the subspace U by transfinite induction such that, for every α < λ, (i) Wα ⊆ Uα ; (ii) {Wξ : ξ ≤ α} ∪ {Uξ : α < ξ < λ} is a cover of X. In fact, if Wβ is constructed for every β < α in such a way that the conditions (i) and (ii) are fulfilled for {Wβ : β < α}, we consider the closed set [ [ Aα = X \ ( {Wξ : ξ < α} ∪ {Uξ : α < ξ < λ}). It follows from the local finiteness of the family U that Aα ⊆ Uα . Since X is strongly zero-dimensional, we can choose a clopen subset Wα such that Aα ⊆ Wα ⊆ Uα . Then the family {Wβ : β ≤ α} satisfies the conditions (i) and (ii). The transfinite induction is completed. Since the family {Wα : α < λ} is locally finite, the family V = {Vα : α < λ}, where, V0 = W0 and Vα = Wα \ is as required. [ {Vβ : β < α} for every 0 < α < λ, REFLEXIVE FAMILIES OF CLOSED SETS 9 Lemma 7. Let U be an open cover of a strongly zero-dimensional metric space X. Then for any > 0, there is a refinement V of U such that V consists of pairwise disjoint clopen sets and diam(V ) < for all V ∈ V. Proof. By Lemma 6, there exists a refinement W of U consisting of pairwise disjoint clopen sets in X. It follows from Lemma 5 that, for every W ∈ W, there exists an open cover U(W ) of W such that diam(A) < for each A ∈ U(W ). Using again Lemma 6, there exists a refinement V(W ) of U(W ) consisting of pairwise disjoint clopen sets in X. Then the family V= [ {V(W ) : W ∈ W} is as required. Theorem 2. Every strongly zero-dimensional complete metric space (X, d) is s-reflexive. Proof. Let A be a family of closed sets in X satisfying conditions (a),(b) and (c) of Lemma 1. We show A = Lat(Alg(A)). To this end, it is enough to show that if B 6∈ A then there is f ∈ AlgA so that f (B) 6⊆ B. Suppose B ∈ S(X) and B 6∈ A. Then there is b ∈ B such that φ({b}) 6⊆ B. Choose an element c ∈ φ({b})\B and a clopen set U0 with diam(U0 ) < 1 and c ∈ U0 ⊆ X \ B. By Lemma 4(3), κ(U0 ) is open and b ∈ κ(U0 ), there is a clopen set V0 such that diam(V0 ) < 1, b ∈ V0 ⊆ κ(U0 ) and V0 ∩ U0 = ∅. ∞ Now we construct two sequences {Ui }∞ i=1 and {Vi }i=1 of collections of pairwise disjoint non-empty clopen sets, and a mapping αn : Vn → Un for each n, such that the following conditions are satisfied: S S (i) Un = U0 , Vn = V0 for every n; 10 Z. YANG AND D. ZHAO (ii) diam(U ) ≤ 1 n for each U ∈ Un ; (iii) Un+1 is a refinement of Un , and Vn+1 is a refinement of Vn ; (iv) V ⊆ κ(αn (V )) for every V ∈ Vn ; (v) if m ≤ n and V ∈ Vm , W ∈ Vn then either W ⊆ V or V ∩ W = ∅; (vi) if m ≤ n, W ∈ Vn , V ∈ Vm and W ⊆ V , then αn (W ) ⊆ αm (V ). First, let U1 = {U0 }, V1 = {V0 } and α1 (V0 ) = U0 . Then the above six conditions are satisfied. Now suppose Un , Vn and αn : Vn → Un have been defined and satisfy conditions (i)-(vi). For any U ∈ Un , by Lemma 7 there exists a family S Un+1 (U ) of pairwise disjoint clopen sets in X such that Un+1 (U ) = U , S 1 diam(W ) ≤ n+1 for each W ∈ Un+1 (U ). Let Un+1 = {Un+1 (U ) : U ∈ Un }. Obviously Un+1 is a refinement of Un . Next, for each V ∈ Vn , by (iv) we have V ⊆ κ(αn (V )). Note that κ preserves union by Lemma 4(2), it follows that {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))} is an open cover of V . Again, by Lemma 7, there is a cover Vn+1 (V ) of V consisting of pairwise disjoint clopen sets and Vn+1 (V ) is finer than {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))}. Put S Vn+1 = {Vn+1 (V ) : V ∈ Vn }. Then Vn+1 is a refinement of Vn . To define αn+1 , for each A ∈ Vn+1 , there is a unique V ∈ Vn such that A ∈ Vn+1 (V ). Then A ⊆ κ(E) ∩ V for some E ∈ Un+1 (αn (V )) ⊆ Un+1 . Such a set E need not be unique. Choose any of them and let αn+1 (A) = E. Thus we defined a mapping αn+1 : Vn+1 → Un+1 . By induction we defined the sequences {Ui }, {Vi } and mapping αn for each n. The conditions (i),(ii),(iii) and (iv) follow immediately from the construction of these objects. REFLEXIVE FAMILIES OF CLOSED SETS 11 To show that condition (v) is also satisfied, let m ≤ n, then Vn is a refinement of Vm . If V ∈ Vm , W ∈ Vn and W 6⊆ V , then W ⊆ V 0 for some V 0 ∈ Vm , where V 0 6= V , thus W ∩ V ⊆ V 0 ∩ V = ∅. To prove (vi) is valid it is enough to check the case where n = m + 1. Let V ∈ Vm with W ∈ Vm+1 and W ⊆ V . By the definition of αm+1 , αm+1 (W ) = E for some E ∈ Vm+1 (αm (V )), hence αm+1 (W ) = E ⊆ αm (V ). Define the mapping f : X → X as follows: {x}, ∞ {f (x)} = T {αn (Vn ) : x ∈ Vn ∈ Vn }, if x ∈ X\V0 , if x ∈ V0 . n=1 First, f is well defined. As a matter of fact, if x ∈ V0 , then for each n, there is a unique Vn ∈ Vn with x ∈ Vn . If x ∈ Vn ∈ Vn , x ∈ Vm ∈ Vm and m ≤ n, then it follows from the condition (v) that ∅ 6= αn (Vn ) ⊆ αm (Vm ). Thus {αn (Vn ) : n ∈ N} is a sequence of closed sets whose every finite subfamily has a non-empty intersection. Furthermore, diam(αn (Vn )) ≤ T for each n and X is complete, so the set ∞ n=1 αn (Vn ) is a singleton. 1 n The mapping f is clearly continuous on X \ V0 . For any x ∈ V0 and every > 0, there exist n and Vn such that x ∈ Vn ∈ Vn and αn (Vn ) ⊆ B(f (x), ). For each y ∈ Vn , by the definition of f , f (y) ∈ αn (Vn ), hence f (y) ∈ B(f (x), ). This shows that f is also continuous on X \ V0 . For any x ∈ X, we show f (x) ∈ φ({x}). If x 6∈ V0 , f (x) = x ∈ φ({x}). If x ∈ V0 , then for each n, there is a unique Vn ∈ Vn such that x ∈ Vn . But Vn ⊆ κ(αn (Vn )), so φ({x}) ∩ αn (Vn ) 6= ∅. Choose any xn ∈ φ({x}) ∩ αn (Vn ). Then the sequence {xn } converges to f (x), so f (x) ∈ φ({x}) because φ({x}) is closed. 12 Z. YANG AND D. ZHAO Now for each A ∈ A, if x ∈ A and x ∈ X \ V0 , then f (x) = x ∈ A; if x ∈ A ∩ V0 , then f (x) ∈ φ({x}) ⊆ A, again f (x) ∈ A. Therefore f ∈ Alg(A). However, b ∈ V0 ∩ B, f (b) ∈ α1 (V0 ) = U0 , and U0 ∩ B = ∅, so f (b) 6∈ B. Hence f (B) 6⊆ B, which implies B 6∈ Lat(Alg(A)). The proof is complete. ∞ Note that in constructing the sequences {Ui }∞ i=1 , {Vi }i=1 and mapping αn we did not make use of the completeness of X. By [1, Corollary 6.2.8] every countable metric space is strongly zerodimensional but is not necessarily complete. Theorem 3. Every countable metric space is s-reflexive. Proof. Let X be a countable metric space. Then X is strongly zerodimensional. We show that X is s-reflexive. The proof is similar to that of Theorem 2. Again let A be a family of closed sets in X satisfying conditions (a), (b) and (c), and B 6∈ A. Let b ∈ B such that φ({b}) 6⊆ B. Choose an element c ∈ φ({b}) \ B and a clopen set U0 with diam(U0 ) < 1 and c ∈ U0 ⊆ X \ B. As κ(U0 ) is open and b ∈ κ(U0 ), there is a clopen set V0 such that diam(V0 ) < 1, b ∈ V0 ⊆ κ(U0 ) and V0 ∩ U0 = ∅. Arrange V0 as {x1 , x2 , · · · }, where x1 = b. ∞ We now define by induction two sequences {Un }∞ n=1 and {Vn }n=1 of pair- wise disjoint clopen sets of X, a map αn : Vn → Un for each n, and f (xn ) (n ∈ N), such that the conditions (i)-(vi) in the proof of Theorem 2 and the following (vii)-(viii) hold: (vii) for any 1 ≤ i < j ≤ n, if xi ∈ V ∈ Vn and xj ∈ V 0 ∈ Vn , then V ∩ V 0 = ∅; (viii) if i ≤ n and xi ∈ V ∈ Vn , then f (xi ) ∈ αn (V ) ∩ φ(xi ). REFLEXIVE FAMILIES OF CLOSED SETS 13 For n = 1 we let U1 = {U0 }, V1 = {V0 }, α1 (V0 ) = U0 , and f (x1 ) = c. Suppose for each i ≤ n, Ui , Vi , αi and f (i) have been defined and satisfy the conditions. To define these objects for n + 1, for each U ∈ Un choose Un+1 (U ) as a collection of pairwise disjoint clopen sets with diameter less S S 1 and Un+1 (U ) = U . Put Un+1 = {Un+1 (U ) : U ∈ Un }. Let than n+1 V ∈ Vn+1 . We consider three cases: Case A. V ∩ {x1 , x2 , · · · , xn , xn+1 } = ∅ or V ∩ {x1 , x2 , · · · , xn , xn+1 } = {xn+1 }. Then Vn+1 (V ) and the restriction of αn+1 on Vn+1 (V ) are defined in the same way as in the proof of Theorem 2. When the intersection is {xn+1 }, there is W ∈ Vn+1 (V ) with xn+1 ∈ W . Then define f (xn+1 ) be any point in αn+1 (W ) ∩ φ(xn+1 ). Note as xn+1 ∈ W ⊆ κ(αn+1 (W )), such a point f (xn+1 ) exists. Case B. V ∩ {x1 , x2 , · · · , xn , xn+1 } = {xi , xn+1 } for some i ≤ n. Choose two disjoint clopen sets C and D such that xi ∈ C, xn+1 ∈ D, and C ⊆ κ(U 0 ) ∩ V for some U 0 ∈ Un+1 (αn (V )) with f (xi ) ∈ U 0 ( note: by induction assumption xi ∈ V ∈ Vn imply f (xi ) ∈ αn (V ) ∩ φ(xi ), so there is U 0 ∈ Un+1 (αn (V )) with f (xi ) ∈ U 0 ∩ φ(xi ), hence xi ∈ κ(U 0 )). Also D ⊆ κ(W 0 ) ∩ V for some W 0 ∈ Un+1 (αn (V )). Now choose a refinement Vn+1 (V ) of {κ(W ) ∩ V : W ∈ Un+1 (αn (V ))} consisting of pairwise disjoint clopen sets and containing both C and D as members. Define αn+1 on Vn+1 (V ) by letting αn+1 (C) = U 0 , and αn+1 (F ) as before if F 6= C. Also define f (xn+1 ) be any point in αn+1 (D)∩φ(xn+1 ) (note: αn+1 (D)∩φ(xn+1 ) 6= ∅ because xn+1 ∈ D ⊆ κ(αn+1 (D))). . Case C. V ∩ {x1 , x2 , · · · , xn , xn+1 } = {xi } for some i ≤ n. Then the construction of Vn+1 (V ) is defined in the same way as in the proof of theorem 2. 14 Z. YANG AND D. ZHAO At last let Vn+1 = S {Vn+1 (V ) : V ∈ Vn }. Since there is a unique V ∈ Vn that contains xn+1 and satisfies the condition in either Case A or Case B, so f (xn+1 ) is defined. Let g : X → X be the mapping defined by g(x) = x for x 6∈ V0 and g(xi ) = f (xi )(i = 1, 2, · · · ). From the construction it is clear that g(x) ∈ φ(x) for all x ∈ X, thus g(A) ⊆ A for all A ∈ A. In addition g(b) = f (b) = c 6∈ B, g(B) 6⊆ B. To complete the proof we only need to verify that g is continuous, and it is enough to show that f is continuous on V0 . For any x = xn ∈ V0 and any > 0, choose m ≥ n, m1 < . Let xn ∈ V ∈ Vm . Then f (xn ) ∈ αm (V ). If i ≥ m, xi ∈ V , then there is W ∈ Vi with xi ∈ W ⊆ V . Then f (xi ) ∈ αi (W ) ⊆ αm (V ). Thus d(f (xn ), f (xi )) < diam(V ) < Hence f is continuous at xn . 1 m < . A space X is called hereditarily disconnected if X does not contain any connected subset of cardinality larger than one. Lemma 8. For every T1 connected space X with more than two elements, there is a connected subset B ⊆ X such that 1 < |B| and B 6= X. Proof. Suppose the only connected subsets of X are X and singletons. Choose any a ∈ X. Then X \ {a} is not connected and hence it is a union of two disjoint non-empty open sets U and V in X. Then cl(U ) = U ∪ {a}, otherwise, cl(U ) = U , which implies U is clopen and so X is not connected. Similarly, cl(V ) = V ∪ {a}. By the assumptions, cl(U ) is not connected, so cl(U ) is a union of disjoint non-empty closed sets E and F in X. Assume a 6∈ E, then E is a non-empty proper clopen subset of X as E = X \ (F ∪ cl(V )). But this contradicts the connectedness of X. The contradiction completes the proof. REFLEXIVE FAMILIES OF CLOSED SETS 15 Lemma 9. Every s-reflexive Hausdorff space is hereditarily disconnected. Proof. If X has at most two elements, it is clearly hereditary disconnected. Now assume X has more than two elements. Suppose X is not hereditarily disconnected. If X is not connected, then as it is not hereditarily disconnected, one of its connected components, say B, is a proper non-singleton connected subset. If X is connected, by Lemma 8 there exists also a proper non-singleton connected subset B. Choose b1 , b2 ∈ B with b1 6= b2 and x0 ∈ X \ B. Let F = {F ∈ S(X) : F = {b1 } or F 3 x0 } ∪ {∅}. Then F satisfies conditions (a),(b) and (c)in Lemma 1, and {b2 } 6∈ F. However, {b2 } ∈ Lat(Alg(F)) which implies that X is not s-reflexive. In fact, if f ∈ Alg(F) and b ∈ B, from {b, x0 } ∈ F it follows that f (b) = b or f (b) = x0 . Thus B is the union of the disjoint closed sets E = {b ∈ B : f (b) = b} and K = {b ∈ B : f (b) = x0 }. Trivially, b1 ∈ E, thus E = B since B is connected. In particular, f (b2 ) = b2 . We are done. Theorem 4. Let X be a locally compact metric space. Then X is sreflexive if and only X is zero-dimensional. Proof. Noting that every metrizable locally compact space is completely metrizable (see [1, p609]) and the property of s-reflexive is topological, the theorem follows from Theorem 2, Lemma 5 and the fact that hereditarily disconnectedness, zero-dimensional and strong zero-dimensional properties are equivalent for locally compact paracompact spaces (see [1,Theorem 6.2.9]). 16 Z. YANG AND D. ZHAO In the following we construct a locally compact countable complete metric space which has a one point extension that is not s-reflexive. We shall define a non-reflexive family of closed sets in the space that satisfies all conditions (a),(b) and (c), as well as (d) in Lemma 1. Example 2. Let 1 Y1 = {( , m) : n, m = 1, 2, · · · } n and 1 Y0 = {( , 0) : n = 1, 2, · · · } ∪ {(0, 0)}. n As a subspace of R2 , Y = Y0 ∪ Y1 is a locally compact countable complete space. Let X = Y ∪ {p} where p is an element not in Y , and define a local base at p as follows: for every map g : N \ D → N ∪ {0}, where D is a finite subset of N, let 1 U (g) = {p} ∪ {( , m) : n ∈ N \ D, m > g(n)}. n Thus ( n1 , m) 6∈ U (g) for any n ∈ D and m ∈ N ∪ {0}. Note that U (g) ∩ U (h) = U (max{g, h}), where g : N \ D1 → N ∪ {0}, h : N \ D2 → N ∪ {0} and max{g, h} : N \ (D1 ∪ D2 ) → N ∪ {0} defined by max{g, h}(n) = max{f (n), g(n)} for every n ∈ N \ (D1 ∪ D2 ). Thus all U (g)’s form a local base at p. Assume Y is an open subspace of X, we thus define a topology on X. Now we show that X is not s-reflexive. Define a map q : X → Y0 by q(( n1 , m)) = ( n1 , 0), q(p) = q(0, 0) = (0, 0) for any n ∈ N and m ∈ {0} ∪ N. Clearly, q is continuous. The family A = {q −1 (A) : A is closed in Y0 } REFLEXIVE FAMILIES OF CLOSED SETS 17 satisfies conditions (a),(b) and (c). In fact, (a) and (b) are clearly valid. To see that (c) is also satisfied, consider any family {Ai : i ∈ I} of closed S S S subsets of Y0 . Then cl( i∈I q −1 (Ai )) = cl(q −1 ( i∈I Ai )) = q −1 (cl( i∈I Ai )). But A is not reflexive. In fact, {(0, 0)} 6∈ A. For any f ∈ Alg(A), as {(0, 0), p} = q −1 ({(0, 0)}) is in A, we have f ((0, 0)) = (0, 0) or f ((0, 0)) = p. If the later holds, then p = lim f (( n1 , 0)). By the definition of A, f (( n1 , 0)) ∈ n→∞ q −1 ({( n1 , 0)}), suppose f (( n1 , 0)) = ( n1 , g(n)), n ∈ N. Then we have a mapping g : N → N ∪ {0}. But then f (( n1 , 0)) 6∈ U (g) for all n ∈ N, that contradicts p = lim f (( n1 , 0)). This contradiction indicates f ((0, 0)) = (0, 0), n→∞ hence {(0, 0)} ∈ Lat(Alg(A)). Thus A is not reflexive. Note that the only non-empty connected subsets of X are singletons, so the family A constructed above also satisfies condition (d). A closed set A of X is called reflexive if {∅, A, X} is reflexive. Since {∅, X} is reflexive for every X, so both ∅ and X are reflexive. Lemma 10. A closed set A of X is reflexive if and only if for each B ∈ S(X) with A ⊂ B 6= X, there exists f ∈ C(X) such that f (A) ⊆ A but f (B) 6⊆ B. Proof. Assume A 6= ∅ and A 6= X. The necessity is trivial. To show the sufficiency, suppose B ∈ S(X) and B 6∈ {∅, A, X}. We verify that there exists f ∈ C(X) such that f (A) ⊆ A but f (B) 6⊆ B. If A 6⊂ B, choose a ∈ A \ B and define f ∈ C(X) by f (x) = a for all x ∈ X. Then f satisfies the requirement. If A ⊂ B, it follows from the assumption that there exists f ∈ C(X) such that f (A) ⊆ A but f (B) 6⊆ B. Thus {∅, A, X} is reflexive. 18 Z. YANG AND D. ZHAO Proposition 1. If every path-connected component is dense in a Tychonoff space X, then every closed set of X is reflexive. Proof. We only consider non-empty proper closed sets. Let A, B ∈ S(X) with ∅ 6= A ⊂ B 6= X. Choose a ∈ A and b ∈ B \ A. By the assumption there exists x0 ∈ X \ B such that a and x0 are in the same path-connected component. Define a continuous mapping g : X → [0, 1] such that g(x) = 0 for x ∈ A and g(b) = 1. Moreover, there exists a continuous mapping h : [0, 1] → X such that h(0) = a and h(1) = x0 since a and x0 are in the same path-connected component. Now f = h ◦ g ∈ C(X) satisfying f (A) ⊆ A. However, f (b) = x0 6∈ B, so f (B) 6⊆ B. Thus A is reflexive. Let X = [0, 1] ∪ [2, 3] be the subspace of R consisting of two disjoint closed intervals. Then [0, 1] is a path-connect component which is not dense in X. But one still can prove, using Proposition 1, that every closed set of X is reflexive. Proposition 2. Every closed set of a zero-dimensional space is reflexive. Proof. Let X be a zero-dimensional space, and A, B ∈ S(X) with ∅ 6= A ⊂ B 6= X. Choose a ∈ A and x0 ∈ X \ A. There exists a clopen set U 3 x0 such that U ∩ A = ∅. Consider f : X → X defined by f (x) = a if x 6∈ U and f (x) = x0 if x ∈ U . Then f ∈ C(X) and f (A) ⊆ A but f (B) 6⊆ B. Remark 1. We end the paper by listing some problems. 1). If X and Y are both s-reflexive, is the cartesian product X × Y still s-reflexive? REFLEXIVE FAMILIES OF CLOSED SETS 19 2). In what spaces, every (finite) chain of closed sets is reflexive? We still do not know whether every chain in R is reflexive. 3). In Example 2, we constructed a family that satisfies all conditions (a)-(d) and is not reflexive. It would be more interesting and rich full to investigate spaces in which every family satisfying conditions (a)-(d) is reflexive. 4). Another possible future work is to find conditions characterizing reflexive families of continuous endomappings. Such work would be much more complicated than that on reflexive families of closed sets. Acknowledgement 1. The authors would like to express their sincere thanks to the referees for their valuable comments and suggestions. References [1] R.Engelking, General Topology, Polish Scientific Publishers, Warszawa,1977. [2] P.R.Halmos, Ten problems in Hilbert space, Bull.Amer.Math.Soc. 76(1970) 887-933. [3] P.R.Halmos, Reflexive lattices of subspaces, J.London Math. Soc. 4(1971) 257-263. [4] K.J.Harrison and W.E.Longstaff, Automorphic images of commutative subspace lattices, Proc. Amer. Math. Soc.296(1986),217-228. [5] W.E.Longstaff, Strongly reflexive subspace lattices, J.London Math. Soc.(2)11(1975),491-498. [6] W.E.Longstaff, On lattices whose every realization on Hilbert space is reflexive, J.London Math. Soc.(2) 37(1988), 499-508. [7] W.E.Longstaff and P.Oreste, On the ranks of single elements of reflexive operator algebras, Proc.Amer.Math.Soc.125(10)(1997),2875-2882. [8] Z.Yang, Dieudonné-Hahn-Tong theorem for complete chains, Houston Journal of Mathematics 29(2003), 949-960. [9] D.Zhao, On reflexive subobject lattices and reflexive endomorphism algebras, Comment. Math.Univ.Carolinae 44(2003), 23-32. 20 Z. YANG AND D. ZHAO (Z. Yang) Department of Mathematics, Shantou University, Shantou, Guangdong, 515063, China P.R. E-mail address: [email protected] (D. Zhao) Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, 1 Nanyang Walk,Singapore 637616 E-mail address: [email protected]