Closure operators on sets and algebraic lattices

Transcription

1 Closure operators on sets and algebraic lattices Sergiu Rudeanu University of Bucharest Romania Closure operators are abundant in mathematics; here are a few examples. Given an algebraic structure, such as group, ring, field, lattice, vector space, etc., taking the substructure generated by a set, i.e., the least substructure which includes that set, is a closure operator. Given a binary relation, taking the relation with certain properties, such as reflexive, transitive, equivalence, etc., generated by the given relation, i.e., the least relation with the desired properties which includes the original relation, is a closure operator. The consequence operators in mathematical logic are closure operators. Topological spaces can be defined by Kuratowski closure operators. In this survey paper we have selected those facts about closure operators that we consider to be the most important ones. We present them in the Ockham razor spirit, i.e., under no more hypotheses than necessary. We have thus detected three levels of generality: A) closure operators on posets, B) closure operators on complete lattices, and C) closure operators on fields of sets P(A), which we tackle in this order. Level A, due to Bourbaki, is less known. However the fundamental property used in the applications of closure operators to various fields of mathematics is valid for arbitrary posets: the existence of the closure x of each element x of the underlying set, that is, the least closed element which includes x. Besides, closure operators are in bijection with the associated sets of closed elements, known as closure systems or Moore families. In the case of complete lattices, every (finite or not) meet of closed elements is closed, so that every Moore family is itself a complete lattice, but not a sublattice of the original one. Level C is the framework of algebraic closures, whose associated closure systems describe an algebraic structure, or equivalently, a deductive structure. We have incorporated this equivalence and two intrinsic characterizations of algebraic closures, in a theorem which is emblematic for the unifying power of closure operators. The last part of the paper is devoted to compactly generated complete lattices, known as algebraic lattices. They are related to level C by a representation theorem. For properties of closure operators not included in this paper see e.g. [2, 4.4, 4.5]. * * * 1

2 In the sequel we will use the abbreviation poset for a partially ordered set (P, ) and we will read x y as x is included in y, or y includes x. Definition 1 If P is a poset, a map : P P is called a closure operator or simply a closure, if for every x, y P the following properties hold: x x (extensivity), x = x (idempotence), x y = x ȳ (isotony). The elements x P satisfying x = x are said to be closed. We denote by C the set of closed elements: C = {x P x = x}. We have mentioned above several examples of closures. Besides, note that the identity map x x is a closure operator. Proposition 1 If is a closure on the poset P, then for each x P, x is the least closed element which includes x. Proof: x is closed by idempotence, x x by extensivity, and if y is a closed element such that x y then x ȳ = y, so that x y. Notation For every poset P, every subset S P and every x P, we set S(x] = {y S x y}. Instead of P (x] we use the simplified notation (x] = {y P x y}. Definition 2 If P is a poset, by a closure system or Moore family we mean any non-empty subset C P such that for each x P, the set C(x] has least element. With this definition, Proposition 1 can be reformulated: Proposition 1 The set of closed elements of a closure operator is a closure system. The converse is also true : Proposition 2 Let P be a poset and C a closure system of P. The assignment is a closure on P. x x = least element of C(x] Proof: It is well known that the bottom element of a poset, whenever it exists, is unique. Therefore the above assigment is indeed a map. We have x C(x], hence the extensivity x x holds and x C. The latter membership together with x x show that x C( x], hence the least element x satisfies x x. But extensivity implies x x, therefore the idempotence x = x holds. If x y then C(y] C(x], therefore the (least) element x of C(x] is included in particular in all the elements of C(y] and since ȳ C(y] it follows that x ȳ, proving isotony. 2

3 Theorem 1 The maps constructed in Propositions 1 and 2 establish a bijection between the closure operators and the closure systems of P. Proof: We shall prove that by composing the two maps in the two possible orders one obtains the identity maps, i.e., Prop1 C Prop2 C Prop2? =, Prop1 C? = C. First composition: x is the least element of C(x]. On the other hand x = x, hence x C, and moreover, x C(x] because x x. Besides, if y C(x] then y C and x y, hence x ȳ = y. Therefore x is also the first element of C(x], hence x= x. Second composition: x C x = x x is the least element of C(x] = x C = x C(x] = x is bottom of C(x] x C. In the case of complete lattices, Moore families have a simpler characterization. Proposition 3 A subset C of a complete lattice L is a closure system if and only if every meet of elements from C belongs to C. Comment: The above statement includes the case of the empty meet, which is the greatest element 1. If you don t like this convention, replace the above by 1 C and the meet of every non-empty subset.... Anyway the proof must distinguish the two cases. Proof: Suppose C is a closure system and let be the associated closure operator. Then 1 C because 1 1 and 1 1 as greatest element. Now suppose X C and set a = infx. Then for every x X we have a x hence ā x = x, proving that a is a lower bound of all x X, therefore ā a and since a ā, it follows that a = ā C. Conversely, suppose C satisfies the above condition. Then 1 C, as remarked in Comment, so C. For every x L set a = infc(x]; to prove that a is least element, it remains to check that a C(x]. Since C(x] C, it follows that a C. Besides, since x is a lower bound of C(x], it follows that x a, therefore a C(x]. Proposition 3 can be extended to an important property. Theorem 2 Every closure system C of a complete lattice (L, inf, sup, 0, 1) is a complete lattice (C, inf,, 0, 1) under the same meet operation as L, while X = supx for every X C. Proof: The property established in Proposition 3 implies that C is a complete lattice under the same meet operation as L because, according to a well-known remark, the join of an arbitrary subset X C is obtained as the meet of the set of all upper bounds of X. Clearly 0 is the least element of C and it was shown in the proof of Proposition 3 that 1 C. 3

4 The other construction of expresses the fact that supx is the least upper bound in C of all the elements of the subset X C In the sequel we take an arbitrary but fixed non-empty set A and we concentrate on the closure operators of the set P(A) of all subsets of A. Since (P(A),,,, A) is a complete lattice, the previous results apply to this case, but there are also many specific results of greatest interest. Recall that an algebra A with underlying set A is a structure A = (A, (f i ) i I ) where each f i is an operation f i : A ni A of arity n i on A. The subalgebras of A (properly speaking, of A) are defined in a natural way and every intersection of subalgebras is a subalgebra, therefore the set SubA of all subalgebras of A is a Moore family. Definition 3 A closure system C on P(A) is called algebraic if there exists an algebra A such that C = SubA. Recall also that an ideal of a join semilattice (L, ) is a non-empty subset I L such that x y I = x I and x, y I = x y I. Proposition 4 The ideals of a join semilattice with least element (L,, 0) form an algebraic closure system. Proof: Let (L,, 0) be a join semilattice with zero. For each a L define { a, if a x, ϕ a (x) = x, if a x. The subalgebras of the algebra A = (L,, 0, (ϕ a ) a L ) form a Moore family SubA. To prove that the ideals of L form an algebraic closure system, we will prove that the ideals coincide with the subalgebras of A. If S is a subalgebra then 0 S. If x, y S then x y S. If x S and y x then y = ϕ y (x) S. If I is an ideal then 0 I and if x, y I then x y I. For every x I and every a L, if a x then ϕ a (x) = a I, while if a x then ϕ a (x) = x I again. The following structures are more general than algebras. Definition 4 A deductive structure or a formal proof system is a tuple D = (A, (R i ) i I ) where R i A n(i) A, n(i) 0, for each i I. The R i s are called deduction rules if n(i) > 0 and axiom schemes if n(i) = 0, in which case R i A. If X A and y A, by a formal proof of y from X is meant a sequence (y 1,..., y m = y) of elements from A such that for each k = 1,..., m, either y k X or there is i I and a sequence of indices k 1,..., k n(i) < k such that (y k1,..., y kn(i), y k ) R i. If this happens we say that y is deducible from X and we denote this fact by X D y or simply X y. We set X = {y A X y}. In particular the elements of are called theses. A subset S A is said to be deductively closed if S = S. We say that the map X X is the consequence operator of D. 4

5 Proposition 5 The consequence operator of a deductive structure D is a closure operator on the set P(A) of subsets of the underlying set A of D. Proof: If x X then the singleton sequence (x) is a formal proof of x from X, therefore X X. For idempotence we will prove that X X. Indeed, if y X then there is a formal proof y 1,..., y m = y of y from X. If in the sequence y 1,..., y m we replace each y k by a formal proof of y k from X, we obtain a formal proof of y from X and therefore y X. Finally suppose X Y and take z X. Then there is a formal proof of z from X, say z 1,..., z m = z. For each k = 1,..., m for which z k X, we have z k Y ; this shows that z Y. Therefore X Y = X Y. Definition 5 A closure operator on (P(A), ) is said to be deductive if it is the consequence operator of a deductive structure defined on A. A closure system C is said to be deductive if the closure operator associated with C is deductive. Definition 6 A poset is said to be directed if every two elements of it have a common upper bound. A family (A t ) t T of subsets of A is said to be directed if the index set T is a directed poset and for every s, t T, s t = A s A t. Theorem 3 The following conditions are equivalent for a closure system C on (P(A), ) : (i) C is algebraic; (ii) C is deductive: (iii) if (A t ) t T C is a directed family then t T A t C; (iv) the closure operator associated with C satisfies (*) X = {Y Y X, Y finite} X P(A). Comment: The equivalence (i) (iv) is known as the Birkhoff-Frink theorem. Proof: (i)= (ii): Let C = SubA, where A = (A, (f i ) i I ), and let be the closure operator associated with C. We can identify each operation f i : A ni A with its graph f i A ni A, so that A becomes a deductive structure as well. Moreover, as shown in universal algebra, for every X A, each element y X has finitely many generators from X and the construction of y is the same in the algebra (see e.g. [3], Proposition 6.5.5) and in the deductive structure, that is, y X X y. This shows that C is also the closure system associated with the deductive structure A. (ii)= (iii): Let (A t ) t T C be a directed family and B = t T A t. If B y then there is a formal proof x 1,..., x m = x of x from B. Let {k 1,..., k r } {1,..., m} be the set of those indices k m for which x k B. For each of these indices k j there is t j T such that x kj A tj. Let s T be an upper bound of t 1,..., t r ; it follows that x k1,..., x kr A s. For the other indices k {1,..., m}, the elements x k are obtained by applying deduction rules. This 5

6 shows that A s x and since A s is deductively closed, it follows that x A s B. Thus B x = x B, showing that B C. (iii)= (iv): Let X A and denote by X # the right side of (*). We shall prove that X # = X. We have # = {Y Y } = {Y Y = } =. Now we suppose X and we shall prove that X # is the least set from C which includes X. In view of (iii), in order to prove that X # C, it suffices to prove that X = {Y Y X, Y finite} is a directed family if it is directed by X itself. Indeed, take Y 1, Y 2 X, where Y 1 and Y 2 are finite subsets of X. Then Y 1 Y 2 is a finite subset of X, hence Y 1, Y 2 Y 1 Y 2 X. If x X then x {x} {x} X #. Therefore X X #. Now suppose X Z C. For every Y X # where Y is a finite subset of X we have Y X Z = Z, hence Y Z. It follows that X # Z. (iv)= (i): For every finite subset Z A and every a Z, define f Za : A n A, where n = cardz, as follows: { a, if {x1,..., x f ZA (x 1,..., x n ) = n } = Z, x 1, if {x 1,..., x n } Z; in particular if a then f a = a. Let < X > denote the subalgebra of A = (A, (f Za ) Za ) generated by a subset X A. We shall prove that < X >= X, which will show that C = SubA. It is known from universal algebra that < X >= p 0 X p where X 0 = X and X p+1 = X p Y p where Y p is the set of all the elements f Za (x 1,..., x n ) with arguments x 1,..., x n X p (see e.g. [3], Proposition 6.5.5). In order to show that < X > X we will prove by induction that X p X for all p. We have X 0 = X X. Now we suppose X p X and must prove that X p+1 X; it suffices to show that Y p X. Indeed, take f Za (x 1,..., x n ) Y p ; then x 1,..., x n X p X. If {x 1,..., x n } Z then f Za (x 1,..., x n ) = x 1 X. If {x 1,..., x n } = Z then f Za (x 1,..., x n ) = a Z = {x 1,..., x n } X = X. Thus we have proved that < X > X and it remains to prove that X < X >, which we will do in three steps. For X = there are two posibilities. If = then < >. If then for every a we have a = f a < > because the constant operations belong to any subalgebra, therefore again < >. If X = {x 1,..., x n } then for every A X we have a = f Xa (x 1,..., x n ) < X >, therefore X < X >. We have thus proved that Y < Y > for every finite set Y, therefore in formula (*) we have Y < Y > < X >, so that X < X > for any subset X A. In the sequel we discuss algebraic lattices. The inclusion of this topic within closure operators on sets is justified by the fact that the closure system (C, ) from Theorem 3 is an algebraic lattice, as shown in Theorem 4 below. Moreover, any algebraic lattice can be represented as an algebra of sets, namely the ideal lattice of a join semilattice with 0, as shown in Theorem 5. 6

7 Definition 7 Let S be a complete join semilattice, i.e., a poset S such that supx exists for every non-empty subset X S. An element c S is called compact if for every non-empty subset X S, (**) c sup X = c sup X 0 for some finite subset X 0 X. A complete join semilattice with bottom element 0 is a complete lattice because 0 = sup, so that all joins exist, hence all meets exist, too. Note that 0 is a trivial example of a compact element. Definition 8 A complete lattice is said to be algebraic if every element is a supremum of compact elements. Remark 1 A complete lattice is algebraic iff every element x satisfies x = sup K(x] = sup {c K c x}, where K stands for the set of compact elements. Recall that we denote by C and a closure system and the associated closure operator. Theorem 4 Every algebraic closure (C, ) on (P(A), ) is an algebraic lattice in which the compact elements are the closures X of finite sets X. Proof: We already know that C is a complete lattice by Theorem 2 applied to (P(A),,,, A). Having in view Theorem 3, it remains to prove that the compact elements coincide with the closures of finite sets. The set is the bottom element of the lattice C, hence it is compact, as was noted above. Now take X = {x 1,..., x n }; to prove that X is compact, suppose X sup {F t t T }, where F t C for all t T. Then, using Theorem 3, we get X X sup {F t t T } = F t = {Y Y F t, Y finite}, t T hence for each i {1,..., n} there is Y i t T F t, Y i finite, such that x i Y i. Set Y = n i=1 Y i. Then Y is a finite set, Y t T F t, and since all Y i Y, it follows that Y i Y (i = 1,..., n), hence X n i=1 Y i Y. If Y = then X F t for each t T. If Y = {y 1,..., y p } then for each k {1,..., p} there is t k T such that y k F tk, hence Y p k=1 F t k, therefore X Y t T p F tk = sup {F t1,..., F tp }. k=1 Conversely, let F be a compact element of C. Then F = F = {Y Y F, Y finite} sup {Y Y F, Y finite}, because X X = supx by Theorem 2 applied with := sup and sup :=. Therefore there exists a finite subset Y {Y Y F, Y finite} such that 7

8 F supy. If Y = then supy is the least element of C, so that F and in fact F =, where is a finite set. Otherwise Y = {Y 1,..., Y p }, where Y 1,..., Y p are finite subsets of F. Then Y = p k=1 Y k is a finite subset of F and Y F. On the other hand, Y k Y implies Y k Y (k = 1,..., p), hence F sup {Y 1,..., Y p } Y. Therefore F = Y, where Y is a finite set. Theorems 3 and 4 enable us to strengthen Proposition 4, with a quite different proof. Proposition 6 The set IdS of ideals of a join semilattice with zero (S,, 0) is an algebraic lattice, in which the compact elements are the principal ideals. Proof: Clearly every non-empty intersection of ideals is an ideal, while S is the improper ideal, therefore IdS is a Moore family. It is also clear that IdS satisfies condition (iii) from Theorem 3, therefore IdS is an algebraic closure. It follows by Theorem 4 that IdS is an algebraic lattice and the compact elements are the ideals generated by finite sets. The ideal generated by is the bottom ideal {0} = (0] and the ideal generated by a set {x 1,..., x n } is the principal ideal (x 1... x n ]. Exercise 1 In the lattice IdS the following hold: the ideal generated by a non-empty subset X is X = {y S y x 1... x n ; x 1,..., x n X, n 1} and in particular, if I 1,..., I n IdS then I 1... I n = {y S y x 1... x n ; x k I k (k = 1,..., n)}, while every ideal I satisfies I = x I (x] = x I (x]. Exercise 2 The set K(L) of compact elements of an algebraic lattice (L,,, 0, 1) is a join semilattice with zero (K(L),, 0). Our last aim in this paper is an isomorphism theorem. We need a preparation about poset isomorphisms. Let Pos denote the category whose objects are all the posets, while the morphisms are the functions f which satisfy x y = f(x) f(y), known as isotone functions. Lemma 1 A function f : P Q between two posets P,Q is an isomorphism in Pos if and only if it satisfies the conditions (α) f is surjective, (β) x 1, x 2 P : x 1 x 2 f(x 1 ) f(x 2 ). Proof: The conditions are obviously necessary. Conversely, suppose f satisfies (α) and (β). If f(x 1 ) = f(x 2 ) then f(x 1 ) f(x 2 ) and f(x 2 ) f(x 1 ), hence x 1 x 2 and x 2 x 1, therefore x 1 = x 2, showing that f is an injection, therefore f is a bijection. The implication = in (β) shows that f is isotone, and since f is a bijection, the implication = in (β) shows that f 1 is isotone. 8

9 Lemma 2 An isomorphism f : P Q in Pos preserves all existing infima and suprema. Proof: Let us prove that f(supx) = supf(x) provided there exists a = supx. If X = then a = 0 x for every x P, hence f(a) f(x) where f(x) is an arbitrary element of Q because f is surjective, so that f(a) = 0 (bottom of Q), while f( ) = Q, therefore supf( ) = 0 as well. If X then for every x X we have x a, hence f(x) f(a). This proves that f(a) is an upper bound of f(x); to prove it is the least one, take an arbitrary upper bound z of f(x) in Q. Then z = f(b) for some b P and f(x) z = f(b) for all x X, therefore by applying (β) from Lemma 1 it follows that x b for all x X, hence a b by the definition of a, and finally f(a) f(b) = z. This proves that f(a) is indeed the l.u.b. of f(x), that is, f(a) = supf(x). It is commonplace that lattices have two equivalent definitions, as algebras and as posets with a special property. This is not exactly so if we refer to morphisms. Indeed, a lattice homomorphism is naturally defined as a map which preserves and, that is, f(x y) = f(x) f(y) and f(x y) = f(x) f(y). It is well known that a lattice morphism is isotone but the converse fails. Lemma 2 shows that the equivalence is recaptured at the level of isomorphisms. To state this fact properly, we must extend the isomorphisms of Pos beyond Pos, in the sense of the following definition. Definition 9 Consider a map f : P Q where P and Q are sets endowed with structures that include a partial order. We say that f is an order isomorphism if f is a bijection and both f and f 1 are isotone. Noting that 0 and 1 are sup s (or that they are inf s), from Lemma 2 we obtain the following Corollary 1 In each of the following categories: meet semilattices, join semilattices, lattices, with or without 0, with or without 1, isomorphisms coincide with order isomorphisms. It is clear that this corollary can be extended to other similar categories. At first glance complete lattices don t fall exactly within the similar categories mentioned above. A complete lattice may be regarded either as a poset in which every subset has inf and sup, or as a lattice which has this supplementary property, as we have understood in this paper. In either variant the category of complete lattices is defined by taking as morphisms the maps that preserve all meets and joins. Corollary 2 In the category of complete lattices isomorphisms coincide with order isomorphisms. It is natural to define the category of algebraic lattices as the full subcategory of the category of complete lattices determined by algebraic lattices. 9

10 In the last theorem we use the semilattice (K(L),, 0) of compact elements of an algebraic lattice L and the algebraic lattice IdS of ideals of a semilattice S, which appear in Exercise 2 and Proposition 6, respectively. Theorem 5 Every algebraic lattice L is isomorphic to the algebraic lattice IdK(L) of ideals of the join semilattice K(L). Proof: For every x L, the set k(x) = {c K(L) c x} = K(L)(x] is the principal ideal defined by x in the semilattice K(L), therefore we have defined a map k : L IdK(L). In view of Corollary 2, it suffices to prove that k is an order isomorphism. We will do this by using Lemma 1. To prove (α), that is surjectivity, take I IdK(L). Then I K(L) L and let a = sup I. For every x I we have x K(L) and x a, hence x k(a). This proves that I k(a). Conversely, take x k(a). Then x K(L) and x a = sup I, hence there is a finite subset I 0 I such that x sup I 0. If I 0 = then x 0, therefore x = 0 I. If I 0 = {i 1,..., i n } then x i 1... i n I, hence x I again. This proves that k(a) I, therefore I = k(a). To prove (β), take x 1, x 2 L. If x 1 x 2 then clearly k(x 1 ) k(x 2 ). Conversely, suppose k(x 1 ) k(x 2 ). Then, using Remark 1, we get x 1 = sup k(x 1 ) sup k(x 2 ) = x 2. Exercise 3 Every isomorphism f : L L of algebraic lattices preserves compact elements, i.e., f(k(l)) K(L ). Exercise 4 Every sub-complete-lattice of an algebraic lattice is an algebraic lattice. References [1] N. Bourbaki, Eléments de Mathématiques. Théorie des Ensembles. Chap.3. Hermann, Paris [2] S. Rudeanu, Sets and Ordered Structures. Bentham ebooks, Bentham Sci, Publ., 2012 (online). [3] S. Rudeanu, Chapters VI, VII in C.P. Popovici, S. Rudeanu, H. Georgescu, Bazele informaticii, vol. II. Univ. Bucureşti, Facultatea de Matematică, Bucureşti [4] W. Wechler, Universal Algebra for Computer Scientists. Springer-Verlag, Berlin/Heidelberg/New York,