LazaarEchiAGT.dvi @ Applied General Topology c© Universidad Politécnica de Valencia Volume 10, No. 2, 2009 pp. 227-237 Quasihomeomorphisms and lattice equivalent topological spaces Othman Echi and Sami Lazaar Abstract. This paper deals with lattice-equivalence of topological spaces. We are concerned with two questions: the first one is when two topological spaces are lattice equivalent; the second one is what additional conditions have to be imposed on lattice equivalent spaces in order that they be homeomorphic. We give a contribution to the study of these questions. Many results of Thron [Lattice-equivalence of topological spaces, Duke Math. J. 29 (1962), 671-679] are recovered, clarified and commented. 2000 AMS Classification: 54B30, 54D10, 54F65. Keywords: Quasihomeomorphism; lattice equivalence; sober space; TD- space. 1. Introduction Let X be a topological space; we denote by Γ(X) the lattice of closed sets of X. Two topological spaces X and Y are said to be lattice equivalent if there is a bijective map from Γ(X) to Γ(Y ) which together with its inverse is order-preserving [8]. The question of characterizing when two topological spaces are lattice equiv- alent is still open. A lattice equivalence ϕ : Γ(X) −→ Γ(Y ) is said to be induced by a homeo- morphism if there is a homeomorphism f : X −→ Y such that ϕ(C) = f (C), for each C ∈ Γ(X). In [8], Thron was concerned in the problem of determining what additional conditions have to be imposed on lattice equivalent spaces in order that they be homeomorphic. A complete answer to this problem is still very far off. 228 O. Echi and S. Lazaar It is worth noting that, over the years, several researchers dealt with the concept of lattice equivalent topological spaces and representations of an ab- stract lattice as the family of closed sets on a topological space (see for instance [4], [8], [9]). In 1966, Finch [4] proved that a lattice equivalence ϕ : Γ(X) −→ Γ(Y ) is induced by a homeomorphism if and only if the following conditions are satisfied: (i) For each x ∈ X, there exists yx ∈ Y such that ϕ({x}) = {yx}. (ii) For each y ∈ Y , there exists xy ∈ X such that ϕ −1({y}) = {xy}. (iii) Let x ∈ X and y ∈ Y . Set Xx = {t ∈ X : {x} = {t}} and Yy = {t ∈ Y : {y} = {t}}. Then ϕ−1({y}) = {x} implies that |Xx| = |Yy| (where |Xx| denotes the cardinality number of the set Xx). In 1972, Yip Kai Wing [9] was interested in quasi-homeomorphisms and lattice-equivalences of topological spaces. This work seems to be very close to our paper; but it is important to announce that none of our results is in Yip’s paper [9]. In this connection, the first section will be entirely devoted to quasi-homeomorphisms and comments on Yip’s results in his note [9]. The present paper is devoted to shed some light on lattice equivalent topo- logical spaces. Let us first recall some notions which were introduced by the Grothendieck school (see for example [5] and [6]), such as quasihomeomorphisms and strongly dense subsets. If X is a topological space, we denote by O(X) the set of all open subsets of X. Recall that a continuous map q : X −→ Y is said to be a quasihomeo- morphism if U 7−→ q−1(U ) defines a bijection O(Y ) −→ O(X). We say that a subset S of a topological space X is locally closed if it is an intersection of an open set and a closed set of X. A subset S of a topological space X is said to be strongly dense in X, if S meets every nonempty locally closed subset of X. Thus a subset S of X is strongly dense if and only if the canonical injection S →֒ X is a quasihomeomorphism. It is well known that a continuous map q : X −→ Y is a quasihomeomorphism if and only if the topology on X is the inverse image by q of that on Y and the subset q(X) is strongly dense in Y [5]. Let X be a topological space. If f : X −→ Y is continuous, then we define Γ(f ) : Γ(Y ) −→ Γ(X), by Γ(f )(C) = f −1(C). In particular, if q : X −→ Y is a quasihomeomorphism. Then the map Γ(q) : Γ(Y ) −→ Γ(X) is a lattice equivalence (see for example [1, Proposition 1.9]). The following definition is natural. Definition 1.1. A lattice equivalence ϕ : Γ(X) −→ Γ(Y ) between two topolog- ical spaces is said to be induced by a quasihomeomorphism if there is either a quasihomeomorphism q : Y −→ X such that ϕ = Γ(q) or a quasihomeomor- phism p : X −→ Y such that ϕ−1 = Γ(p). Our main result is a characterization of lattice equivalences induced by a quasihomeomorphism (see Theorem 3.4). This result is very close to the one Lattice equivalent topological spaces 229 done by Finch [4] characterizing lattice equivalences induced by a homeomor- phism. As a consequence, many results of Thron are recovered, clarified and commented. 2. Grothendieck’s Quasihomeomorphisms and Yip’s Quasihomeomorphisms As we have said in the introduction the concept of quasihomeomorphisms was introduced in Algebraic geometry by Grothendieck ([6], [5]). Also, it was shown that this concept arises naturally in the theory of some foliations asso- ciated to closed connected manifolds( see the papers [2] and [3]). The following definition is given in [9]. Definition 2.1. A continuous map q : X −→ Y between topological spaces is said to be a quasihomeomorphism if the following conditions are satisfied: (i) For any closed set C in X, q−1[q(C)] = C. (ii) For any closed set F in Y , q[q−1(F )] = F . Fortunately, the two notions “Grothendieck’s quasihomeomorphism” and “Yip’s quasihomeomorphism” coincides, as it is shown in the following. Proposition 2.2. Let q : X −→ Y be a continuous map between topolog- ical spaces. Then q is a Yip’s quasihomeomorphism if and only if it is a Grothendieck’s quasihomeomorphism. Proof. Suppose that q is a Grothendieck’s quasihomeomorphism. Then Γ(q) : Γ(Y ) −→ Γ(X) is a lattice isomorphism, by [1, Proposition 1.9]. Hence q is a quasihomeomorphism in the sense of Yip, by [9, Theorem 1]. Conversely, if q is a Yip’s quasihomeomorphism, then Γ(q) : Γ(Y ) −→ Γ(X) is a lattice isomorphism, by [9, Theorem 1]; so that Γ(q) is bijective; proving that q is a Grothendieck’s quasihomeomorphism. � Remark 2.3. Let q : X −→ Y be a quasihomeomorphism between topological spaces. In [9, Theorem 2], Yip proved that q(X) is dense in Y . In fact, it follows from [5, Chapter 0, Proposition 2.7.1] that q(X) is even strongly dense in Y . In [9, Theorem 3], the author investigated closed quasihomeomorphisms be- tween T0-spaces. The following result is more precise. Proposition 2.4. Let q : X −→ Y be a quasihomeomorphism. Then the following statements are equivalent: (i) q is a surjective mapping; (ii) q is a closed mapping; (iii) q is an open mapping. Proof. (ii) =⇒ (i) and (iii) =⇒ (i). Let q be closed (resp.open). Then, q(X) is a closed (resp.open) subset of Y . Now, since q−1(q(X)) = q−1(Y ), we get q(X) = Y , by the definition of a quasihomeomorphism. Thus q is onto. 230 O. Echi and S. Lazaar (i) =⇒ (ii) and (iii). Let C be a closed (resp. an open) subset of X. By definition, there is a closed (resp. an open) subset D of Y such that C = q−1(D). But, since q is onto, we get q(C) = D, proving that q(C) is closed (resp. open). � Example 2.5. Let us construct a quasihomeomorphism that is not a sur- jection. Consider an infinite set X and a point α not belonging to X. Set Y := X ∪ {α}. Equip Y with the topology whose closed sets are Y and the finite subsets of X. Hence X is a strongly dense subspace of Y ; so that the canonical embedding i : X →֒ Y is a quasihomeomorphism. Since i is not onto, it is not closed and not open. Before stating the next result, we need to recall the notion of sober space (also introduced by the school of Grothendieck [5] or [6]). A subspace Y of X is called irreducible, if each nonempty open subset of Y is dense in Y (equivalently, if C1 and C2 are two closed subsets of X such that Y ⊆ C1 ∪ C2, then Y ⊆ C1 or Y ⊆ C2). Let C be a closed subset of a space X; we say that C has a generic point if there exists x ∈ C such that C = {x}. Recall that a topological space X is said to be sober if any nonempty irreducible closed subset of X has a unique generic point. Let X be a topological space and S(X) the set of all nonempty irreducible closed subsets of X [5]. Let U be an open subset of X; set Ũ = {C ∈ S(X) | U ∩ C 6= ∅}; then the collection {Ũ | U is an open subset of X} provides a topology on S(X) and the following properties hold [5]: (i) The map ηX : X −→ S(X), which takes x to {x}, is a quasihomeomor- phism. (ii) S(X) is a sober space. (iii) Let f : X −→ Y be a continuous map. Let S(f ) : S(X) −→ S(Y ) be the map defined by S(f )(C) = f (C), for each irreducible closed subset C of X. Then S(f ) is continuous. (iv) The topological space S(X) is called the soberification of X, and the assignment S, defines a functor from the category TOP of topological spaces to itself. Proposition 2.6. Let q : X −→ Y be a quasihomeomorphism. Then the following properties hold. (1) If X is a T0-space, then q is injective. (2) If X is sober and Y is a T0-space, then q is a homeomorphism. Proof. (1) Let x1, x2 be two distinct points of X with q(x1) = q(x2). Then there exists an open subset U of X such that, for example, x1 ∈ U and x2 /∈ U . Since there exists an open subset V of Y satisfying q−1(V ) = U , we get q(x1) ∈ V and q(x2) /∈ V , which is impossible. It follows that q is injective. (2) Firstly, it is obviously seen that if S is a closed subset of Y , then S is irreducible if and only if so is q−1(S). Lattice equivalent topological spaces 231 Now, let us prove that q is surjective. For this end, let y ∈ Y . According to the above observation, q−1({y}) is a nonempty irreducible closed subset of X. Hence q−1({y}) has a generic point x. Thus we have the containments {x} ⊆ q−1({q(x)}) ⊆ q−1({y}) = {x}. So that q−1({q(x)}) = q−1({y}). It follows, from the fact that q is a quasi- homeomorphism, that {q(x)} = {y}. Since Y is a T0-space, we get q(x) = y. This proves that q is a surjective map, and thus q is bijective. But it is easily seen that bijective quasihomeomorphisms are homeomorphisms. � Remark 2.7. Let q : X −→ Y be a quasihomeomorphism. If Y is sober and X is a T0-space, then q need not be a homeomorphism. To see this, it suffices to consider a T0-space X which is not sober. Then the canonical embedding ηX : X −→ S(X) is a quasihomeomorphism which is not a homeomorphism. 3. Lattice equivalence A Brouwerian lattice is a complete lattice L for which x ∨ ( ∧ C) = ∧ {x ∨ y | y ∈ C} for all x ∈ L and all C ⊆ L. A morphism of Brouwerian lattices is a mapping f : L −→ M that preserves all infima and all finite suprema. Let CBL denotes the category of Brouwerian lattices and Brouwerian lattice maps. Then Γ : TOP −→ CBL is a contravariant functor. Remark 3.1. Let f : X −→ Y be a continuous map. Then f is rendered invertible by the functor Γ (i.e., Γ(f ) is a lattice equivalence) if and only if f is a quasihomeomorphism. The following example shows that a lattice equivalence that is induced by a quasihomeomorphism is not necessarily induced by a homeomorphism. Example 3.2. Let X be a topological space which is not sober. Then the canonical quasihomeomorphism ηX : X −→ S(X) induces a lattice equivalence between S(X) and X which is not induced by a homeomorphism. In order to give a complete characterization of lattice equivalence of topolog- ical spaces induced by a quasihomeomorphism, we give the following definition. Definition 3.3. Let X, Y be two topological spaces and ϕ : Γ(X) −→ Γ(Y ) a lattice equivalence. We say that ϕ is a point-closure lattice equivalence if one of the following properties is satisfied: (i) For each x ∈ X, there exists yx ∈ Y such that ϕ({x}) = {yx}. (ii) For each y ∈ Y , there exists xy ∈ X such that ϕ −1({y}) = {xy}. Our main result is the following. 232 O. Echi and S. Lazaar Theorem 3.4. Let ϕ : Γ(X) −→ Γ(Y ) be a lattice equivalence of topological spaces. Then the following statements are equivalent: (i) ϕ is induced by a quasihomeomorphism; (ii) ϕ is a point-closure lattice equivalence. Proof. (i) =⇒ (ii) Suppose that there is a quasihomeomorphism q : Y −→ X such that ϕ = Γ(q). Then for each D ∈ Γ(Y ), we have ϕ−1(D) = C with C ∈ Γ(X) and q−1(C) = D. Hence D ⊆ q−1(q(D)) ⊆ q−1(C) = D. Thus ϕ−1(D) = q(D). Therefore, for each y ∈ Y , we have ϕ−1({y}) = q({y}). But q({y}) = {q(y)}, since q is continuous; so that ϕ−1({y}) = {q(y)}. Now, if we suppose that there is a quasihomeomorphism p : X −→ Y such that ϕ−1 = Γ(p), then we get ϕ({x}) = {p(x)}, for each x ∈ X. It follows that ϕ is a point-closure lattice equivalence. (ii) =⇒ (i) Suppose, for instance, that for each x ∈ X, there exists yx ∈ Y such that ϕ({x}) = {yx}. For each x ∈ X choose q(x) ∈ Y such that ϕ({x}) = {q(x)}. This allows us to define a mapping q : X −→ Y . We are aiming to prove that q is a quasihomeomorphism and ϕ−1 = Γ(q). It is enough to prove that ϕ−1(G) = q−1(G), for each G ∈ Γ(Y ). Let x ∈ ϕ−1(G). Then {x} ⊆ ϕ−1(G). Thus {q(x)} ⊆ G; in particular q(x) ∈ G; therefore x ∈ q−1(G). Conversely, let x ∈ q−1(G); then q(x) ∈ G. Hence {q(x)} ⊆ G; consequently, {x} ⊆ ϕ−1(G). Therefore, x ∈ ϕ−1(G). It follows that ϕ−1 = Γ(q). If we suppose that for each y ∈ Y , there exists a xy ∈ X such that ϕ−1({y}) = {xy}, then by the above argument, there is a quasihomeomor- phism p : Y −→ X such that ϕ = Γ(p). � Corollary 3.5. If X is a topological space and Y is a T1-space, then each lattice equivalence between them is induced by a quasihomeomorphism. Proof. It is easy to check that condition (ii) of Definition 3.3 is fulfilled. Then we can apply Theorem 3.4. � The following result establishes some links between lattice equivalences in- duced by a homeomorphism and those induced by a quasihomeomorphism. For the proof of the next theorem, we need a lemma. Lemma 3.6. Let E, F be two topological spaces such that F is a T0-space. If f, g : E −→ F are two continuous maps such that Γ(f ) = Γ(g), then f = g. Proof. Let x ∈ E. Then f −1({f (x)}) = g−1({f (x)}) and g−1({g(x)}) = f −1({g(x)}). This yields g(x) ∈ {f (x)} and f (x) ∈ {g(x)}. Thus {f (x)} = {g(x)}; so that f (x) = g(x), since Y is a T0-space. � Lattice equivalent topological spaces 233 Theorem 3.7. Let X, Y be two T0-spaces and ϕ : Γ(X) −→ Γ(Y ) a lattice equivalence of topological spaces. Then the following statements are equivalent: (i) ϕ is induced by a homeomorphism; (ii) There are two quasihomeomorphisms q : Y −→ X and p : X −→ Y such that ϕ = Γ(q) and ϕ−1 = Γ(p). Proof. [(i) =⇒ (ii)]. Straightforward. [(ii) =⇒ (i)]. Let q : Y −→ X and p : X −→ Y be two quasihomeomorphisms such that ϕ = Γ(q) and ϕ−1 = Γ(p). We have ϕϕ−1 = 1Γ(Y ) = Γ(1Y ) and ϕ −1ϕ = 1Γ(X) = Γ(1X ). Hence Γ(pq) = Γ(1Y ) and Γ(qp) = Γ(1X ). Thus, according to Lemma 3.6, we get qp = 1X and pq = 1Y . Therefore, ϕ is induced by the homeomorphism p. � The following example shows that the T0 axiom cannot be deleted in Theo- rem 3.7. Example 3.8. Let X, Y be two sets with distinct cardinalities. We equip X and Y with the indiscrete topology. The unique lattice equivalence between X and Y is ϕ : Γ(X) −→ Γ(Y ), defined by ϕ(∅) = ∅ and ϕ(X) = Y . It is easily seen that for any two continuous maps q : Y −→ X and p : X −→ Y , we have ϕ = Γ(q) and ϕ−1 = Γ(p). Moreover, p, q are quasihomeomorphisms. But since the two sets have distinct cardinalities, they cannot be homeomorphic. Recall that a topological space X is said to be a TD-space if for each x ∈ X, {x} is locally closed. It is easily seen that TD is between T0 and T1. The following result will be used in the next corollary. Proposition 3.9. Every quasihomeomorphism between two TD-spaces is a homeomorphism. Proof. Let q : X −→ Y be a quasihomeomorphism between two TD-spaces. Hence q is injective, by Proposition 2.6. On the other hand, q(X) is strongly dense in Y and every point-set is locally closed; so that q(X) = Y . Thus q is a bijective quasihomeomorphism. Therefore, q is a homeomorphism. � For the proof of the next corollary, we need a lemma. Lemma 3.10. Let X, Y be two topological spaces and ϕ : Γ(X) −→ Γ(Y ) a lattice equivalence. Let G be a closed subset of Y . Then the following statements are equivalent: (i) G is irreducible in Y ; (ii) ϕ−1(G) is irreducible in X. 234 O. Echi and S. Lazaar Proof. Since ϕ−1 is also a lattice equivalence, it is enough to show the impli- cation (i) =⇒ (ii). Let F1, F2 be two closed subsets of X such that ϕ −1(G) ⊆ F1 ∪ F2. Hence G ⊆ ϕ(F1 ∪ F2). But ϕ(F1 ∪ F2) = ϕ(F1) ∪ ϕ(F2). Thus G ⊆ ϕ(F1) ∪ ϕ(F2). Now, since G is irreducible in Y , G ⊆ ϕ(F1) or G ⊆ ϕ(F2). This yields ϕ−1(G) ⊆ F1 or ϕ −1(G) ⊆ F2, proving that ϕ −1(G) is irreducible in X. � Corollary 3.11. Let X, Y be two T0-spaces and ϕ : Γ(X) −→ Γ(Y ) a lattice equivalence. (1) If X or Y is a TD-space (resp. sober space), then ϕ is induced by a quasihomeomorphism q : X −→ Y or p : Y −→ X. (2) If X and Y are TD-spaces (resp. sober spaces), then the mapping q of (1) is a homeomorphism (thus ϕ is induced by a homeomorphism). Proof. (1) Suppose for example, that X is a TD-space and Y is a T0-space. Thron showed in [8] that for each x ∈ X, there exists a unique yx ∈ Y such that ϕ({x}) = {yx}. Then ϕ is a point-closure lattice equivalence and according to the proof of Theorem 3.4, ϕ is induced by a quasihomeomorphism q : X −→ Y ( q takes x to yx). Now, suppose that X is sober and Y is T0. Let y ∈ Y ; then, since {y} is an irreducible closed subset of Y , ϕ−1({y}) is an irreducible closed subset of X, by Lemma 3.10. Hence ϕ−1({y}) has a unique generic point. Thus ϕ is a point-closure lattice equivalence; so that there is a quasihomeomorphism p : Y −→ X which induces ϕ (p takes y ∈ Y to the unique generic point of ϕ−1({y})). (2) Every quasihomeomorphism between two TD-spaces is a homeomor- phism, by Proposition 3.9. Also, every quasihomeomorphism between two sober spaces is a homeomorphism, by Proposition 2.6. � Corollary 3.12 ([8, Theorem 2.1 ]). Every lattice equivalence between two TD-spaces is induced by a homeomorphism. Corollary 3.13 ([8, Corollary 2.1 ]). Every lattice equivalence between a T0- space and a T2-space is induced by a homeomorphism. Proof. Let X be a T0-space, Y a T2-space (thus a TD-space) and ϕ : Γ(Y ) −→ Γ(X) a lattice equivalence. Then by Corollary 3.11, there is a quasihomeomor- phism q : Y −→ X such that ϕ−1 = Γ(q). Now, Y is a sober space (since it is T2) and X is T0. This forces q to be a homeomorphism, by Proposition 2.6. Therefore, ϕ is induced by a homeomor- phism. � Example 3.14. A lattice equivalence between a T1-space (thus a TD- space) and a sober space which is not induced by a homeomorphism. For, let Y be an infinite set equipped with the cofinite topology. Let α /∈ Y , and X = Y ∪ {α}. We equip X with the topology whose closed sets are X and Lattice equivalent topological spaces 235 the finite subsets of Y . Hence Y is a strongly dense subspace of X; so that the canonical embedding Y →֒ X is a quasihomeomorphism; thus it induces a lattice equivalence ϕ. Clearly, ϕ is not induced by a homeomorphism, since Y is a T1-space and X is not. Note that Example 3.18 provides nontrivial examples of lattice equivalences between a T1-space and a sober space which are not induced by a homeomor- phism. It is worth noting that the TD-axiom is the weakest requirement under which [8, Theorem 2.1] is true, as shown by Thron in the following Theorem 3.15 ([8, Theorem 2.2]). If X is not a TD-space, then there exists a lattice equivalence between X and some other space Y , which is not induced by a homeomorphism. Looking carefully at the proof of the above theorem, we remark that the lattice equivalence given by the author is not induced by a homeomorphism in both cases, when X is T0 or not; nevertheless, it is induced by a quasi- homeomorphism. This rises the natural question whether a lattice equivalence is always induced by a quasihomeomorphism. Unfortunately, the answer is negative, as shown by the following nice example. Example 3.16. A lattice equivalence of topological spaces that is not induced by a quasihomeomorphism. Let X and Y be two disjoint infinite sets equipped with the cofinite topology. Let α, β /∈ X ∪ Y and α 6= β. Set X ′ = X ∪ {α} and Y ′ = Y ∪ {β}. We equip X ′ (resp. Y ′ ) with the topology whose closed sets are X ′ (resp. Y ′ ) and the finite subset of X (resp. of Y ). Recall that the free union E + F of disjoint spaces E and F is the set E ∪ F with the topology: U ⊆ E + F is open if and only if U ∩ E is open in E and U ∩ F is open in F . Now, consider Λ = X ′ + Y and ∆ = Y ′ + X. It is clear that there exists a unique morphism of lattices ϕ : Γ(Λ) −→ Γ(∆) which satisfies the following properties: – (i) ϕ(X ′ ) = X, ϕ(C) = C, for all finite subset C of X. – (ii) ϕ(Y ) = Y ′ , ϕ(D) = D, for all finite subset D of Y. Clearly, ϕ is a lattice equivalence of topological spaces. Suppose that ϕ is induced by a quasihomeomorphism. Without loss of gen- erality, we may suppose that there is a quasihomeomorphism q : ∆ −→ Λ such that ϕ = Γ(q). Hence q−1(Y ) = ϕ(Y ) = Y ′ ; so that q(β) ∈ Y . On the other hand, {β} = Y ′ . The continuity of q implies that q(Y ′ ) = q({β}) ⊆ q({β}) = {q(β)}. Thus Y ′ ⊆ q−1({q(β)}) = ϕ({q(β)}) = {q(β)}, a contradiction, since Y is infinite. Therefore, ϕ is not induced by a quasihomeomorphism. 236 O. Echi and S. Lazaar In [8, Corollary 2.1], Thron has written that “If X is a T0-space and Y is a T2- space, then they are homeomorphic if and only if they are lattice-equivalent”. The following result shows that in [8, Corollary 2.1], “T2-space” cannot be replaced by “T1-space”. We need to recall the notion of Jacobson space [5]. A topological space X is said to be a Jacobson space if the subset of all closed points of X is strongly dense in X. Theorem 3.17. (1) If a T0-space X is lattice equivalent to a T1-space Y , then X is a Jacobson space. (2) There exist a T0-space X and a T1-space Y which are lattice equivalent but not homeomorphic (hence any lattice equivalence between them is not induced by a homeomorphism). Proof. (1) By Corollary 3.5, the lattice equivalence between X and Y is induced by a quasihomeomorphism q : Y −→ X. According to Proposition 2.6, the induced quasihomeomorphism q1 : Y −→ q(Y ) is bijective. Hence Y is homeomorphic to the subspace q(Y ) of X and q(Y ) is strongly dense in X. It suffices to prove that q(Y ) is the set X0 of all closed points of X. Indeed, X0 ⊆ q(Y ), since q(Y ) is strongly dense in X. On the other hand, let y ∈ Y ; then {q(y)} is closed in q(Y ) since Y is homeomorphic to q(Y ). Hence {q(y)} ∩ q(Y ) = {q(y)}. Let z ∈ {q(y)}; then {z} ∩ q(Y ) 6= ∅. Thus {z} ∩ q(Y ) = {q(y)}. It follows that {q(y)} = {z}. Therefore, z = q(y), since X is a T0-space. (2) It suffices to take a Jacobson T0-space X which is not T1. Let Y be the subspace of X whose elements are the closed points of X. Hence the canonical embedding of Y into X is a quasihomeomorphism. Thus X and Y are lattice equivalent; however, they are not homeomorphic. � Example 3.18. It is easy to give explicit examples of Jacobson T0-spaces which are not T1. Let R be a Hilbert ring which is not a field; i.e., a ring such that the intersection with R of a maximal ideal of the polynomial ring R[t] is maximal (take for example R = K[t1, ..., tn] the polynomial ring on n indeterminates over a field K). Let X = Spec(R) equipped with the hull-kernel topology. Then X is a Jacobson space which is not T1. Here, if we let Y := M ax(R) be the set of all maximal ideals of R, then Y is a T1 strongly dense subspace of X. Thus the canonical quasihomeomorphism i : Y −→ X induces a lattice equivalence between the topological spaces X and Y . On the other hand, the space X is sober by [7, Proposition 4]. This yields a lattice equivalence between a T1-space and a sober space which is not induced by a homeomorphism. Lattice equivalent topological spaces 237 Acknowledgements. The authors wish to thank the DGRST 03/U R/15− 03 for its partial support. References [1] K. Belaid, O. Echi and R. Gargouri, A-spectral spaces, Topology Appl. 138 (2004), 315–322. [2] E. Bouacida, O. Echi and E. Salhi, Foliations, spectral topology, and special morphisms, Advances in commutative ring theory (Fez, 1997), 111–132, Lecture Notes in Pure and Appl. Math. 205, Dekker, New York, 1999. [3] E. Bouacida, O. Echi and E. Salhi, Feuilletages et topologie spectrale, J. Math. Soc. Japan 52 (2000), 447–464. [4] P. D. Finch, On the lattice-equivalence of topological spaces, J. Austral. Math. Soc. 6 (1966), 495–511. [5] A. Grothendieck and J. Dieudonné, Eléments de Géométrie Algébrique, Die Grundlehren der mathematischen Wissenschaften, 166, Springer-Verlag, New York, 1971. [6] A. Grothendieck and J. Dieudonné, Eléments de Géométrie Algébrique. I. Le langage des schémas, Inst. Hautes Etudes Sci. Publ. Math. No. 4, 1960. [7] M. Hochster, Prime ideal structure in commutative rings, Trans. Amer. Math. Soc. 142 (1969), 43–60. [8] W. J. Thron, Lattice-equivalence of topological spaces, Duke Math. J. 29 (1962), 671– 679. [9] K. W. Yip, Quasi-homeomorphisms and lattice-equivalences of topological spaces, J. Austral. Math. Soc. 14 (1972), 41–44. Received April 2008 Accepted April 2009 Othman Echi (othechi@yahoo.com,othechi@math.com) King Fahd University of Petroleum and Minerals, Department of Mathematics & Statistics, P.O. Box 5046, Dhahran 31261, Saudi Arabia. Sami Lazaar (salazaar72@yahoo.fr) Department of Mathematics, University Tunis-El Manar, Faculty of Sciences of Tunis“Campus Universitaire”, 2092 Tunis, Tunisia