Cantor-Bendixson, socle, and atomicity H. Simmons

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1 Cantor-Bendixson, socle, and atomicity H. Simmons The University, Manchester, England manchester.ac.uk This document is the second in the series [1] [7], concerned with the use of lattices as a tool for analysing modules and spaces. In the following Preamble I try to explain the motivation behind this idea and give some idea of what is to come in this document. As with all the documents in the series [1] [7] this one is written as a teaching document rather than a research document. In other words, the development is quite slow, and no doubt there are some parts than you can omit because you already know that material. However, as mentioned above, there will be places where I refer to one of [1, 2] so that certain details of some proofs need not be repeated. Of course, the word I refers to myself, but when I say we I mean me and you. Contents 1 Basic notions Kernel structure Inflators, nuclei, and the assembly Frames of sets of intervals Base frame Congruence frame Division frame How to generate nuclei Two fundamental derivatives The socle derivative The Cantor-Bendixson derivative Atomicity properties Strong atomicity Weak atomicity Feeble atomicity Bibliography Preamble Consider any module M (over some ring R) and any topological space S. Each of these has an associated lattice Sub(M) OS the lattice of submodules of M and the topology on S, the lattice of open sets. Each of these is a complete lattice, but in a different way. In Sub(M) infima are computed as intersections but suprema are not just unions. In OS suprema are computed as unions but infima are not just intersection. Nevertheless these lattices do have a lot in common, and it is possible to develop a common analysis of both kinds. Here we look at one aspect of such an analysis. 1

2 For each kind there is a certain measure which can indicate just how complicated the parent module or space is. For the module lattice there is the socle length, and for the spatial lattice there is the Cantor-Bendixson rank. (Depending on the context the words length, rank, and dimension are used to name this kind of measure.) It is true that in the spatial case the Cantor-Bendixson rank is usually attached to the lattice of closed sets. However, by taking the dual complement it can be attached to the topology. What is not so obvious is that each of these measures can be used on the other side, and can give useful information. In this document I show how this is done. More generally I show how each such measure can be attached to a particular kind of lattice, a modular idiom, for which Sub(M) and OS are particular examples. Given such a lattice Λ we attach two inflators (inflationary and monotone functions) soc cbd and iterate these to closure. The length of that iteration is the measure given by that inflator. In the module case a use of soc gives the socle length, and in the spatial case a use of cbd gives the Cantor-Bendixson rank. However, the important observation is that each can be used on any complete lattice Λ of a certain kind, and then we find there is a fairly close interaction between the two inflators. In this document I first set up the general machinery and then develop the two particular inflators. Finally I consider when the two are the same or almost the same. This is concerned with the existence of atoms in parts of the parent lattice. In particular, for the two concrete examples the two inflators are always the same. Some of this material was first published in [34], but I do not recommend reading that. The paper [35] also contain some earlier material. There are also other kinds of measure that can be developed in the same way. These include the Krull, Gabriel, and the Boyle measures. These are not dealt with here, but will be in the later documents. 1 Basic notions In this section we set down all the basic notions needed in this document. There are few proofs. More details can be found in [1] especially Section DEFINITION. A lattice is a structure (Λ,,,,, ) where (Λ, ) is a partially ordered set with a top and a bottom, and for each pair of elements a, b A there is a meet a b join a b within A. A lattice morphism Λ f Γ 2

3 between two lattices is a function f which is monotone ( x, y Λ)[x y = f(x) f(y)] with f( ) = f( ) = and respects the two binary operations, that is f(x y) = f(x) f(y) f(x y) = f(x) f(y) for each x, y A. As suggested by this definition each lattice we consider is bounded since it has a top and a bottom, and a lattice morphism must respect these selected elements. We also follow the common convention and do not distinguish between a lattice and its carrier. We write Λ for both. There are several textbooks on lattices, but perhaps the most comprehensive is [22]. I suggest you have that handy. There are two standard properties that a lattice may or may not have. These play a part here. 1.2 DEFINITION. Let Λ be a lattice. The lattice Λ is modular if for all a, x, b Λ with a b. The lattice Λ is distributive if (a x) b = a (x b) (a x) b = (a b) (x b) (a x) b = (a b) (x b) for all a, x, b Λ. Eventually every lattice we look at is modular. Later in Lemma 4.3 we set down an important consequence of modularity. We are concerned here with a certain kind of complete lattice. Thus each subset X of the parent lattice has both a supremum and an infimum. X X However, these play very different roles. Infinite Distributive Law Given a complete lattice there is a possible (IDL) a X = {a x x X} for certain elements a and certain subsets X. Recall that a subset X of a poset is directed if it is non-empty and for each x, y X there is some z X with x, y z. 3

4 1.3 DEFINITION. An idiom is a complete lattice Λ for which (IDL) holds for all directed sets X. A frame is a complete lattice Λ for which (IDL) holds for all sets X. An idiom morphism Λ f Γ between two idioms is a lattice morphism that also preserves arbitrary suprema, that is f( X) = f[x] for each subset X of Λ. (Here f[x] is the direct image of X across f.) A frame morphism between two frames is an idiom morphism. An idiom is sometimes said to be upper continuous and sometimes -continuous. I find the name idiom quite convenient. It is a way of talking about certain concrete situations. The two books [14] and [28] are concerned with the use of modular idioms to analyse certain aspect of modules. The two book [23, 30] and the article [32] are concerned with frames. It is important to note that an idiom morphism need not preserve arbitrary infima. We will see an example shortly. Take a closer look at (IDL). The comparison a X {a x x X} holds in every complete lattice. Thus the content of the law is the converse comparison. 1.4 LEMMA. Let Λ be an idiom. Then Λ is a frame precisely when it is distributive (in the finitary sense). Proof. Trivially, if Λ is a frame then it is distributive. Thus we need the converse. Suppose Λ is distributive and consider an arbitrary subset Y Λ. Consider any a Λ. We require a Y {a y y Y } since the converse comparison always holds. Let X be the set of all the joins of finite subsets of Y. Thus a typical element x X has the form y 1 y n for y 1,..., y n Y. This set X is closed under joins and hence is directed. Also Y X so that Y X. Thus a Y a X {a x x X} where a use of the idiom distributive law gives the second comparison. But x = y 1 y n so that a x = a y 1 a y n {a y y Y } 4

5 since Λ is distributive. Thus {a x x X} {a y y Y } to give the required result. There is another way of distinguishing frames within idioms. At first this looks a little strange, but the associated technique is important. 1.5 DEFINITION. Let Λ be any lattice. An implication on Λ is a 2-placed operation ( ) such that x (b a) b x a for all a, b, x A. Trivially, any lattice can carry at most one implication. The standard example of an implication is that carried by a boolean algebra. There are also other examples as the following shows. 1.6 THEOREM. Let Λ be a complete lattice. Then Λ carries an implication precisely when Λ is a frame. Proof. Suppose first that Λ is a frame. We check that for a, b A defines an implication. The condition (b a) = {x b x a} x (b a) b x a is immediate, and the converse is a simple consequence of the frame distributive law. Conversely, suppose Λ carries an implication operation. We require b X {b x x X} for b Λ and X Λ. Let a be the supremum on the right hand side. Then x X = b x a = x (b a) so that X (b a) to give the required comparison. We will use implications and partial implications quite a lot. Observe that when there is an implication ( ) and a bottom then we have a negation = ( ) characterized by x b b x = for all elements b, x. To conclude this section let s look at the two concrete examples which motive the study of modular idioms and frames. 5

6 Consider any module M (over some ring R) and any topological space S. Each of these has an associated lattice Sub(M) OS the lattice of submodules of M and the topology on S, the lattice of open sets. Each of these is a complete lattice. Infima in Sub(M) are computed as intersections, and suprema in OS are computed as unions. 1.7 EXAMPLES. For each module M the lattice Sub(M) of submodules is a modular idiom. For each space S the topology OS is a frame. You might like to work out what the implication is on OS. As a hint you should not forget the closed subsets of S. As an example of a frame morphism, consider a continuous map T φ S between two topological spaces. We know that each of the topologies OS and OT is a frame, and a simple exercise shows that the inverse image function OS φ OT is a frame morphism. However, in general, a module morphism does not give an idiom morphism between the two idioms of submodules. 2 Kernel structure In any category of algebras and morphisms there is a notion of a kernel of a morphism, a kind of gadget that controls quotients of algebras. Usually when the algebras are lattices each such kernel is some kind of congruence. Manipulating with those can be quite messy. For idioms there is a much neater, and quite unusual, gadget. In this section we look at these. A more general analysis is given in Section 3 of [1]. However, here we will concentrate on the idiom setting. 2.1 DEFINITION. Let Λ be an arbitrary idiom. An inflator on Λ is a function f : Λ Λ that is inflationary and monotone, that is x f(x) x y = f(x) f(y) for x, y Λ. A closure operation on Λ is an inflator f that is also idempotent, that is f 2 = f. A nucleus on Λ is a closure operation j such that j(a) j(b) j(a b) for all a, b Λ. 6

7 Since a nucleus j is inflationary the characterizing comparison can be improved to j(a) j(b) = j(a b) for a, b Λ. Notice also that we may compare nuclei j, k by the pointwise comparison, that is j k ( a Λ)[j(a) k(a)] so that the set of all nuclei on Λ forms at least a poset. In fact, it forms a frame, but that is for later. In this section we obtain the following result. Of course, the terminology and notation will be explained as we move along. 2.2 THEOREM. Let Λ be an idiom. ( ) Each nucleus j on Λ is the kernel of a canonical quotient a surjective idiom morphism from Λ. (!) Each idiom morphism from Λ Λ Λ j f Λ j Γ has an associated nucleus k which is the kernel of the morphism. ( ) Consider the combined situation of (,!), and suppose j k. Then there is a unique idiom morphism f such that Λ j f Γ f Λ j commutes. Our first job is to define the notion of the kernel of a morphism, and show how to produce it. We do the production first. 2.3 THEOREM. Let Λ f Γ be an idiom morphism. There is a unique function k on Λ such that x k(a) f(x) f(a) for x, a Λ. For this function k we have f k = f, and k is a nucleus. 7

8 Proof. We first observe that there is at most one such function k. For suppose k 1 and k 2 are two such functions. For arbitrary a Λ let x = k 1 (a). Then x k 1 (a), so that f(x) f(a), to give k 1 (a) = x k( 2 ). A similar argument gives k 2 (a) k 1 (a), and hence k 1 = k 2. To produce such a function we let k(a) = {x Λ f(x) f(a)} for arbitrary a Λ. The required implication is trivial. For the converse a use of the morphism property of f gives so that f(k(a)) = f ( {x Λ f(x) f(a)} ) = {f(x) Λ f(x) f(a)} f(a) as required. This argument gives x k(a) = f(x) f(k(a)) f(a) f(k(a)) f(a) for a Λ. Also f(a) f(a), so that a k(a), and hence f(a) f(k(a)). Thus f k = f as required. We have just observed that k is inflationary. Almost trivially k is monotone. For a Λ we have f(k 2 (a)) = f(k(a)) = f(a) so that k 2 (a) k(a), to show that k is idempotent. Finally, for a, b Λ let x = k(a) k(b) so that to give and hence to show that k is a nucleus. f(x) f(a) f(y) f(b) f(x) f(a) f(b) = f(a b) k(a) k(b) = x k(a b) This attaches a nucleus to each idiom morphism, so now we give it a name. 2.4 DEFINITION. For each idiom morphism Λ f Γ the kernel of f is that nucleus k on Λ given by Theorem

9 This proves part (!) of Theorem 2.2. Of course, this is nothing to shout about since it is little more than naming a particular function. Our main jobs are to show why nuclei as kernels are useful. To do that we look at ( ). 2.5 DEFINITION. Let j be a nucleus on the idiom Λ. We let Λ j = {x Λ j(x) = x} = j[λ] to obtain the fixed set of j. Since j is a closure operation the fixed set A j is just the set j[λ] of outputs of j. Since Λ j is a subset of Λ it is partially ordered by the restriction of the comparison on Λ. The proof of the following is not difficult and can be found in Section 3 of [1]. 2.6 LEMMA. Let j be a nucleus on the idiom Λ. The fixed set Λ j is closed under arbitrary infima as calculated in Λ. In particular, Λ j is complete. For each subset X Λ j the element j X = j ( X) is the supremum of X in Λ j. We are now ready to prove part ( ) of Theorem THEOREM. Let j be a nucleus on the idiom Λ. The complete lattice Λ j is an idiom, and the assignment Λ a is an idiom morphism with kernel is j. j Λ j j(a) Proof. We know that Λ j is a complete lattice. We must show that it satisfies (IDL). To this end consider any directed subset X Λ j. Observe that X Λ and X is directed in Λ. Recall also that a meet calculated in Λ j is the same as that calculated in Λ. We require for arbitrary a Λ j. This is a j X = j {a x x X} a j ( X ) = j ( {a x x X} ) by the construction of the supremum in Λ j. But a Λ j, so that a = j(a), and hence j(a) j ( X ) = j ( {a x x X} ) is the requirement. Since j is a nucleus this is j ( a X ) = j ( {a x x X} ) 9

10 which holds by the corresponding property in Λ. We certainly have a function j from Λ to Λ j. We must show this is an idiom morphism. Trivially, this assignment j is monotone. Next we show that it respects suprema, that is for each subset X Λ. This unravels to j ( X) = j j [X] j( X) = j ( j[x] ) and the comparison is more or less immediate. We look at the converse comparison. For each x X we have X x and hence j ( X ) j(x) since j is monotone. Thus j( X) j[x] by releasing x. Since j is monotone and idempotent this gives j( X) = j 2 ( X) j ( j[x] ) as required. Since j is a nucleus the function j passes across meets, and hence the function is an idiom morphism. Finally, let k be the kernel of j. We have for all x, a A. With x = k(a) we have to give k j. With x = j(a) we have x k(a) j (x) j (a) j(x) j(a) k(a) = x j(x) j(a) j(x) = j 2 (a) = j(a) so that to give j k. Thus k = j. j(a) = x k(a) Observe that although they have the same behaviour we must distinguish between the two functions j and j. The function j is a morphism from Λ, but the function j is a closure operation on Λ. Our final job is to prove part ( ) of Theorem

11 2.8 THEOREM. Consider any nucleus j on the idiom Λ and any idiom morphism Λ f Γ with kernel k. Suppose j k. Then there is a unique idiom morphism f such that Λ j f Γ f Λ j commutes. Proof. Since j is surjective there can be at most one such morphism f. Thus is suffices to exhibit an example of such a morphism. For a Λ we have f(a) f(j(a)) f(k(a)) = f(a) where the last equality holds by Theorem 2.7. Thus these three elements of Γ are equal, and so f j = f. Since, as sets, we have Λ j Λ, we may set f (a) = f(a) for a Λ j. This gives a function from Λ j to Γ. Also, by the previous observation, for each a Λ we have (f j )(a) = f(j(a)) = f(a) so the triangle commutes as the function level. Trivially, f is monotone, so it suffices to show is respects suprema and meets. For the suprema case we require for arbitrary X Λ j. This is f ( j X ) = f [X] f ( j( X) ) = f[x] by unravelling the constructions involved. Since f is a morphism, this holds by the first observation. Since f is given as an idiom morphism, it is immediate that f passes across meets. These results show why nuclei on idioms are interesting. In the following section we show how nuclei also get involved in the other parts of idiom life. 11

12 3 Inflators, nuclei, and the assembly Let Λ be an arbitrary idiom. We have seen in Section 2 how nuclei on Λ are the kernels of idiom morphisms from Λ. In this section we consider how a nucleus can be generated, and then we look at the structure of the set of all nuclei on Λ. We give this set a name. 3.1 DEFINITION. For each idiom Λ the assembly is the set NA of all nuclei on Λ. In due course we show that NΛ is a frame. The first part of this section is developed in more detail in Section 2 of [1]. We have met some of the following notions before. 3.2 DEFINITION. Let Λ be an arbitrary idiom. An inflator on Λ is a function f : Λ Λ that is inflationary and monotone, that is x f(x) x y = f(x) f(y) for x, y Λ. A closure operation on Λ is an inflator f that is also idempotent, that is f 2 = f. A nucleus on Λ is a closure operation f such that f(a) f(b) f(a b) for all a, b Λ. A binary pre-nucleus on Λ is an inflator f such that f(a) f(b) f(a b) for all a, b Λ. A unary pre-nucleus on Λ is an inflator f such that f(a) b f(a b) for all a, b Λ. A derivative on Λ is an inflator f such that the closure f is a nucleus. In this document we are concerned mostly with unary pre-nuclei, but binary pre-nuclei are important for the larger picture. The idea of a derivative and the notation f needs a bit more explanation. It is concerned with the ordinal iterates of an inflator. Let Ord be the set of ordinals, or as many of them as we need to analyse the idiom Λ. The composite of two inflators on Λ is itself an inflator. We extend that idea by iteration. 3.3 DEFINITION. Let Λ be an arbitrary idiom, and let f be an inflator on Λ. The ordinal iterates (f α α Ord) 12

13 of f are generated by f 0 = id f α+1 = f f α f λ = {f α α < λ} for each ordinal α and limit ordinal λ. At the limit jump the pointwise supremum is use, that is f λ (a) = {f α (a) α < λ} for each a A. Notice how the completeness of Λ is used to pass across the limit jumps. This kind of construction works for any complete poset, but that generality is not needed here. A simple exercise (or a look at [1]) gives the proof of the following. 3.4 LEMMA. Let Λ be an arbitrary idiom, and let f be an inflator on Λ. (a) Each ordinal iterate of f is an inflator, and the whole family of iterates is an ascending chain of inflators. (b) If f is a binary pre-nucleus then each ordinal iterate f α is also a binary pre-nucleus. In particular, the closure f is a nucleus. (c) If f is a unary pre-nucleus then each ordinal iterate f α is also a unary pre-nucleus, and each limit ordinal iterate f λ is a binary pre-nucleus. In particular, the closure f is a nucleus. (d) Each nucleus is a binary pre-nucleus, each binary pre-nucleus is a unary prenucleus, and each unary pre-nucleus is a derivative. Inflators on Λ are partially ordered by the pointwise comparison. Thus f g ( x A)[f(x) g(x)] for inflators f and g. Starting with any inflators f the ordinals iterates generate an ascending chain of inflators. id = f 0 f = f 1 f 2 f α α Ord On set-theoretic grounds at some stage this chain will stabilize. There is some ordinal θ such that f θ = f α for all larger ordinals α. The value of θ depends on the parent idiom Λ and the particular inflator f. Different examples can give very different values. Here we are not concerned with the value of θ, so we write f for the stable limit of the ascending chain. This is an inflator since it is in the chain, and it is idempotent by the stability. Thus f is a closure operation. A few moment s thought shows that f is the least closure operation above f. This closure operation f may or may not be a nucleus, but often we want it to be. By definition, an inflator f is a derivative precisely when its closure f is a nucleus. The two kinds of pre-nuclei are a simple way of recognizing derivatives. As indicated, the set NΛ of all nuclei on Λ is partially ordered by the pointwise comparison. To analyse that we look at the poset UΛ of all unary pre-nuclei on Λ. The missing details for the rest of this section can be found in Section 4 of [1]. 13

14 3.5 DEFINITION. For an arbitrary idiom Λ we let IΛ UΛ BΛ NΛ be the poset of all inflators unary pre-nuclei binary pre-nuclei nuclei on Λ (under the pointwise comparison). This gives us a short chain NΛ BΛ UΛ IΛ of posets associated with Λ. Here we are concerned mostly with UΛ and NΛ, but it does no harm to include the other two as well. 3.6 DEFINITION. Let Λ be an arbitrary idiom, and let F be any set of inflators on Λ. The pointwise infimum F of F is the function on Λ given by ( F ) (a) = {f(a) f F } for each a A. Observe that, almost trivially, this function F is an inflator on Λ, and is a lower bound for F in IΛ. Of course, we can do better than that. 3.7 LEMMA. Let Λ be an arbitrary idiom, and let F be any set of inflators on Λ. The pointwise infimum F is the actual infimum of F in IΛ. If F UΛ then F UΛ and is the actual infimum in UΛ. If F BΛ then F BΛ and is the actual infimum in BΛ. If F NΛ then F NΛ and is the actual infimum in NΛ. This gives us four complete lattices. Here we are interested in UΛ and NΛ. 3.8 DEFINITION. Let Λ be an idiom. We let id tp be the two functions on Λ gives by id(a) = a tp(a) = for each a Λ. Almost trivially each of id, tp is a nucleus on Λ. We see that id is the bottom of the four complete lattices and tp is the top of the four complete lattices. Suprema in three of these lattices are easy to compute. 14

15 3.9 DEFINITION. Let Λ be an arbitrary idiom, and let F be any non-empty set of inflators on Λ. The pointwise supremum F of F is the function on Λ given by ( F ) (a) = {f(a) f F } for each a A. Note that we use this construction only on a non-empty set of inflators. The pointwise supremum of the empty set of inflators is given by ( ) (a) = = which does not produce an inflator. The least inflator is id, the identity function LEMMA. Let Λ be an arbitrary idiom, and let F be any non-empty set of inflators on Λ. Then the pointwise supremum F is the actual supremum of F in IΛ. Now suppose F is directed. If F UΛ then F UΛ and is the actual supremum in UΛ. If F BΛ then F BΛ and is the actual supremum in BΛ. This result does not extend to sets of nuclei. The pointwise supremum of a directed set of nuclei is certainly a binary pre-nucleus, but it need not be idempotent. To analyse NΛ we first look at UΛ THEOREM. For each idiom Λ the complete lattice UΛ is itself an idiom. Proof. We must show that UΛ satisfies (IDL). Thus we require f G = {f g g G} for each f UΛ and each directed G UΛ. We recall that these two suprema and all infima are computed pointwise. We evaluate at an arbitrary element a A. Thus ( f G ) (a) = f(a) ( G ) (a) = f(a) {g(a) g G} = {f(a) g(a) g G} = {(f g)(a) g G} = ( {f g g G} ) (a) to give the required result. The second step holds since G is directed, and this produces a directed subset of Λ. The third step holds by (IDL). The final step holds since the involved subset of UΛ is directed. Consider any unary pre-nucleus f on Λ. We know that the stable limit f is the least nucleus above f. A few moment s thought shows that the operation ( ) is a closure operation on UΛ. And there is more THEOREM. Let Λ be an arbitrary idiom. The closure operation ( ) on the idiom UΛ is actually a nucleus. It fixed set is precisely the assembly NA of Λ. 15

16 Proof. Since ( ) is a closure operation on UΛ it suffices to show it is a unary prenucleus on UΛ, that is f g (f g) for arbitrary f, g UΛ. To do that us we show f α g (f g) α for each ordinal α, and then take α sufficiently large. To help with this let so that our objective is h = f g f α (a) g(a) h α (a) for arbitrary a A. We proceed by induction on α with allowable variation of a. The base case, α = 0, is immediate. For the induction step, α α + 1, we have f α+1 (a) g(a) = f ( f α (a) ) g(a) f ( f α (a) g(a) ) f ( h α (a) ) where the second step holds since f is a unary pre-nucleus, and the third holds by the induction hypothesis. This gives f α+1 (a) g(a) f ( h α (a) ) g(a) f ( h α (a) ) g ( h α (a) ) = h α+1 (a) to conclude this step. For the induction leap to a limit ordinal λ, for arbitrary a A we have f λ (a) g(a) = {f α (a) α < λ} g(a) = {f α (a) g(a) α < λ} {h α (a) α < λ} = h λ (a) as required. The second step uses (IDL) on Λ, and the third uses the induction hypothesis. Finally, the fixed set property f = f f is a nucleus is immediate. By Theorem 2.7 the last two results give the following COROLLARY. For each idiom Λ each of UΛ, NA is an idiom and UΛ f NA f is a canonical quotient with ( ) as the kernel nucleus. This result can be improved but to do that it seems that we need to use a different tactic. We invoke Theorem 1.6. To do that we need a preliminary. The following result is a generalization of Lemma 3.1 of [34]. 16

17 3.14 LEMMA. Let Λ be an arbitrary idiom, let j be a nucleus, and let f be any inflator on Λ. Let G be the set of unary pre-nuclei g with f g j. The following hold. (i) G is closed under composition. (ii) G is directed. (iii) G has a unique maximum member. (iv) This maximum member is a nucleus. Proof. (i). Consider g 1, g 2 G and let g = g 1 g 2. For each x Λ we have g(x) f(x) = g 1 (g 2 (x)) f(x) g 1 (g 2 (x) f(x)) g 1 (j(x)) where the second step holds since g 1 is unary, and the third step holds since g 2 G. But now, since g 1 G we have to show that g G. g(x) f(x) g 1 (j(x)) f(x) g 1 (j(x)) f(j(x)) j 2 (x) = j(x) (ii). For g 1, g 2 G we have g 1, g 2 g 1 g 2 G, to give the required result. (iii). Consider the pointwise supremum of G given by h = G h(a) = {g(a) g G} for each a Λ. Since G is directed we see that h is an inflator. A use of upper continuity shows that h is a unary pre-nucleus. Thus it suffices to show that h F. For each x, y A we have h(x) y = {g(x) g G} y = {g(x) y g G} where the second step holds by a use of the upper continuity (since the supremum is directed). In particular, we have h(x) f(x) = {g(x) f(x) g G} j(x) to show that h G. (iv). We have h G and hence h 2 G by part (i). But now h 2 h to show that h is a nucleus. With this we can obtain the main result of this section THEOREM. Let Λ be an idiom. Then the assembly NΛ is a frame. 17

18 Proof. We know that NΛ is a complete lattice so, by Theorem 1.6 it suffices to show that NΛ carries an implication. Consider any nuclei j, k NΛ. By Lemma 3.14 there is a nucleus l NΛ such that k g j f l for each inflator g. In particular, this is holds for all g NΛ to show that l is the required implication (k j). A long term project is to understand the structure of this assembly NΛ. Believe me, some weird things happen. It seems that in general UΛ need not be a frame. However, Lemma 3.14 gives the following LEMMA. Let Λ be an idiom and let j be a nucleus and let f be a unary pre-nucleus on Λ. Then there is a nucleus (f j) on Λ such that for each unary pre-nucleus g. g (f j) f g j This is saying that the idiom UΛ of unary pre-nuclei on Λ carries a partial implication. We know that the assembly NΛ is a frame and carries a (full) implication which we also write as ( ). This notation doesn t lead to much confusion. The frame NΛ is a quotient of UΛ via the nucleus ( ) on UΛ. In particular, for each unary pre-nucleus f the closure f is the least nucleus above f LEMMA. Let Λ be an idiom and let j be a nucleus and let f be a unary pre-nucleus on Λ. Then (f j) = (f j) where the left hand side is computed in UΛ and the right hand side is computed in NΛ. Proof. Let to obtain two nuclei. Since f f we have k = (f j) l = (f j) f l f l j and hence l k by Lemma By the construction of k we have so a use of the nucleus ( ) gives f k j f k (f k) j = j and hence k l. Observe that this result ensures that each unary pre-nucleus f has a negation f, and this is a nucleus. 18

19 4 Frames of sets of intervals In this section we eventually attach to each modular idiom Λ three frames B(Λ) C(Λ) D(Λ) Cng Dvs and two frame quotients given by nuclei Cng and Dvs on B(Λ). The two frames B(Λ) and C(Λ) are spatial, but D(Λ) need not be. The structure of D(Λ) tells us a lot about the parent idiom Λ. We will see that it is canonically isomorphic to the assembly NΛ of Λ. The frames B(Λ) and C(Λ) can be produced for any bounded lattice, so for a while we work in that setting. Much of the next two blocks can be found in [22] but expressed in a different way. However, you should note that a couple of following results do require modularity. In the final block of this section we restrict our attention to modular idioms. 4.1 Base frame 4.1 DEFINITION. Let Λ be a bounded lattice. For each a, b Λ with a b we let [a, b] = {x Λ a x b} to obtain the interval between a and b. In particular [a, a] is the trivial interval {a}, and [, ] = Λ is the improper interval. Let O(Λ) be the set of all trivial intervals, singletons, of Λ, and let I(Λ) be the set of all intervals of Λ. We look at certain sets of intervals, and various collections of these form the elements of the frames B(Λ), C(Λ), D(Λ). To defined these sets we need some notation. 4.2 DEFINITION. Let Λ be a bounded lattice, and consider I, J I(Λ). We write J I and say J is a sub-interval of I if J I, that is if J = [x, y] and I = [a, b] for some a x y b from A. We write J I and say J and I are similar if there are l, r Λ with associated intervals L = [l, l r] [l r, r] = R where I, J are L, R in some order. Observe that, by definition, the relation is reflexive and symmetric on I(Λ). It is not transitive, and its transitive closure can be quite complicated. Later we assume that Λ is modular and that makes similarity easier to deal with. 19

20 4.3 LEMMA. Let Λ be a modular lattice and consider L = [l, l r] R = [l r, r] be any pair of similar intervals. Then the two assignments x L = [l, l r] l y x r [l r, r] = R y form an inverse pair of isomorphisms (where L, R are viewed as bounded lattices). We now return to an arbitrary bounded lattice Λ. For each I I(Λ) and a, b Λ we have I [a, a] = I O(Λ) and [a, a] [a b, a b] [b, b] so that O(Λ) is the whole of one block of the equivalence relation generated by similarity. We look at other sets of intervals, all closed under similarity. 4.4 DEFINITION. Let Λ be an arbitrary lattice. (a) A set of intervals A is abstract if it is non-empty and closed under, that is J I A = J A for intervals I, J. Let A(Λ) be the collection of all abstract sets of intervals. (b) A set of intervals B is basic if it is abstract and closed under taking subintervals, that is J I B = J B for intervals I, J. Let B(Λ) be the collection of all basic sets of intervals. (c) A set of intervals C is a congruence set if it is basic and closed under abutting, that is [a, b], [b, c] C = [a, c] C for elements a b c. Let C(Λ) be the collection of all congruence sets of intervals. Trivially, each of A(Λ) B(Λ) C(Λ) is a poset under inclusion. A few moment s thought shows that each is closed under arbitrary intersections, and so each is a complete lattice. Furthermore, in A(Λ) and B(Λ) suprema are unions. However, suprema in C(Λ) are more complicated. The lattice A(Λ) is not too important. It is there just to get us going. Calculations in B(Λ) are easy. Since it is closed under arbitrary intersections and unions it is a frame. (It s actually a topology, but of a rather pathetic kind.) The following is a simple exercise. 4.5 LEMMA. Let Λ be a lattice and consider a non-empty set B of intervals. Then B is basic precisely when it is closed under translations, that is for each interval [a, b] and element x. [a, b] B = [a x, b x], [a x, b x] B 20

21 Since B(Λ) is a frame it carries an implication operation (B, A) (B A) which is an important tool. We need an explicit description of this operation. We describe this for the modular case. 4.6 LEMMA. Let Λ be a modular lattice and consider B, A B(Λ). Then for each interval I I(Λ). I (B A) ( K I)[K B = K A] Proof. For basic sets B, A let M be the set of interval given on the right. We show that M is abstract. To this end consider intervals J I M so that J M is required. Consider any L B with L J. Since Λ is modular a flip across the associated isomorphisms gives L K I for some interval K. But B is abstract so that K B and hence K A since I M. Since A is abstract this gives L A, to show that J M. Trivially, M is closed under taking sub-intervals, and hence M is basic. We have M B A for consider any I M B and K = I. This gives M (B A) and it remains to obtain the converse inclusion. For intervals I (B A) K B K I we have to give I M, as required. K B (B A) A Notice that this description of the implication on B(Λ) makes use of the assumed modularity of Λ. There is a description for arbitrary Λ, but that is not needed here. 4.2 Congruence frame We now look at the poset C(Λ) and in due course explain the name congruence set. The following observation is immediate. 4.7 LEMMA. For each lattice Λ the poset C(Λ) is closed under arbitrary intersections and directed unions. 21

22 Since C(Λ) has all infima, in the form of intersections, it is a complete lattice, and so has all suprema. For arbitrary A C(Λ) we have A B(Λ) but this union need not be in C(Λ). We describe how to convert A into the supremum A in C(Λ) later in this block. Observe that O(Λ) I(Λ) are closed under abutting (rather trivially) and so are the bottom and top of C(Λ). Congruence sets have other useful properties. 4.8 LEMMA. For any lattice Λ consider C C(Λ). Then C is closed under pivoting, that is ( ) [a, x], [a, y] C = [a, x y] C ( ) [x, b], [y, b] C = [x y, b] C for all a, b, x, y Λ. With this we can justify our terminology. The following result is essentially the same as [22] (Lemma 8 and page 20, and Theorem 2 on page 131). The proof is routine. 4.9 THEOREM. For each lattice Λ there is a bijective correspondence given by for a, b Λ. Cong(Λ) C(Λ) < > C ( C) a b and a b [a, b] C ( C) a b [a b, a b] C Since C(Λ) is closed under arbitrary intersections we know that each basic set sits inside a smallest congruence set. We now see how to produce that DEFINITION. Let Λ be an arbitrary lattice and consider any basic set B B(Λ). We let Cng(B) be the set of all intervals [a, b] which can be partitioned into B. In other words there is a finite sequence a = x 0 x i x m+1 = b with [x i, x i+1 ] B for each 0 i m. This is the crucial operation that takes us from B(Λ) to C(Λ). 22

23 4.11 LEMMA. Let Λ be an arbitrary lattice. (a) For each B B(Λ) the set Cng(B) is the least congruence set that includes B. (b) The operation Cng is a nucleus on B(Λ). (c) The fixed family of Cng is precisely the members of C(Λ), that is for each B B(Λ). Cng(B) = B B C(Λ) Proof. (a) Trivially, we have B Cng(B) since each interval is a partition of itself. We check that Cng(B) is closed under translation. Suppose [a, b] Cng(B) and consider any y Λ. Suppose the partition a = x 0 x i x m+1 = b witnesses the membership of Cng(B), that is [x i, x i+1 ] B for each part. Then we have [x i y, x i+1 y] B, so that the partition a y = x 0 y x i y x m+1 y = b y ensures that [a y, b y] Cng(B). A similar argument shows that [a y, b y] Cng(A). This shows that Cng(B) is a basic set. By construction Cng(B) is closed under partitions, and hence is a congruence set. We have observed that B Cng(B). Consider any congruence set C with B C. Consider any interval [a, b] Cng(B). This interval can be partitioned into members of B, and each of these subintervals is in C, so that whole interval is in C since C C(Λ). Thus Cng(B) C. (b) We have B Cng(B) for each B B(Λ). A trivial observation shows that Cng is monotone. To show that Cng is idempotent consider any B B(Λ) and any [a, b] Cng(Cng(B)). This interval can be partitioned into members of Cng(B). Each of these parts can be partitioned into members of B. This gives a partition of [a, b] into members of B, and hence [a, b] Cng(B). To show that Cng is a nucleus consider any [a, b] Cng(A) Cng(B). Since we have [a, b] Cng(A) there is a partition a x y b where [x, y] A for each part. We have [x, y] [a, b] Cng(B), so that [x, y] Cng(B) and this part [x, y] itself has a partition x u v y where [u, v] B for each part. We also have [u, v] [x, y] A so that [u, v] A B. Thus, by combining all the partitions we see that [a, b] Cng(A B), as required. (c) Consider any B B(Λ) with Cng(B) = B. Since Cng(B) C(Λ) this gives B C(Λ). Conversely, consider any B C(Λ) and any [a, b] Cng(B). This interval can be partitioned into members of B. But this B is closed under abutting, so that whole interval is in B. Thus Cng(B) = B. This result show that we have a quotient frame B(Λ) B C(Λ) Cng(B) 23

24 given by the nucleus Cng. This also illustrates why the base frame B(Λ) is useful. We may do various calculations in B(Λ) and then hit the result with Cng. The result also shows that A = Cng( A) for each A C(Λ). The following is a standard frame theoretic consequence COROLLARY. For each lattice Λ we have for each B B(Λ) and C C(Λ). (B C) C(Λ) Proof. Let X = (B C) so that an inclusion Cng(X ) X will give the required result. But X B C so that Cng(X ) B Cng(X ) Cng(B) Cng(X B) Cng(C) = C and hence Cng(X ) (B C) = X as required. The following is not directly relevant here but it is worth noting THEOREM. For each lattice Λ the frame C(Λ) is spatial. It is isomorphic to the topology of a topological space. This space is compact and has a base of compact open sets. This result is essentially due to Garret Birkhoff, although he didn t state it this way. It was stated explicitly as Theorem 5.4 of [34]. 4.3 Division frame So far in this sections we have looked mostly at arbitrary finitary lattices. We now restrict our attention to modular idioms DEFINITION. Let Λ be a modular idiom. A basic set B B(Λ) is -closed if ( x X) [ [a, x] B = [a, X] B ] for each a Λ and X [a, ]. A division set for Λ is a congruence set D C(Λ) which is -closed. Let D(Λ) be the family of all division sets. 24

25 Notice that for a basic set B to be -closed we only need the supremum X of those sets X which are bounded below, and a particular bound is involved in the notion. So perhaps the name -closed is a bit too general. I don t know a terminology for this kind of supremum. Perhaps I can invent one. Our aim is to show that D(Λ) is a quotient frame of C(Λ), and hence of B(Λ). In Section 5 we explain the significance of D(Λ) in a different way. Almost trivially the family D(Λ) of division sets is closed under arbitrary intersection, and so is a complete lattice (with infima given by intersections). We will invoke Theorem 1.6 to show that D(Λ) is a frame. To do that we refine Corollary For this we need to assume the idiom is modular LEMMA. For each modular idiom Λ we have for each B B(Λ) and D D(Λ). Proof. Let (B D) D(Λ) X = (B D) for the given B B(Λ) and D D(Λ). By Corollary 4.12 we have X C(Λ), so it suffices to show that X is closed under suprema. To this end consider any a Λ and any X Λ with for each x X. Thus [a, x] X [a, X] X is the requirement. Since we know that X is a congruence set it is closed under pivoting, so we may assume that X is directed. Consider any elements a y z X with [y, z] B. By Lemma 4.6 it suffices to show that [y, z] D. Consider an arbitrary x X. We have (since [y, z] B), and hence since [a, x] X. Thus where, remember, we have a y x z x x with [y x, z x] B [y x, z x] D [y, y x z] D (y x) z = y (x z) by modularity. Since the set X is directed, we have {y x z x X} = y ( X) z = z 25

26 by a use of upper continuity. Thus [y, z] = [y, y ( X) z] D as required. This show that the complete lattice D(Λ) carries an implication, and hence a use of Theorem 1.6 gives the following THEOREM. Let Λ be any modular idiom. Then D(Λ) is a frame. We want to improve this by showing that D(Λ) is a quotient frame of B(Λ). To do that we look for the controlling nucleus. Since D(Λ) is closed under arbitrary intersection the following is not problematic DEFINITION. Let Λ be any modular idiom. For each B B(Λ) let Dvs(B) be the smallest division set that includes B. This is the operation we want LEMMA. Let Λ be any modular idiom. The operation Dvs is a nucleus on B(Λ). Proof. Trivially, Dvs is a closure operation. Thus, for arbitrary B 1, B 2 we require Dvs(B 1 ) Dvs(B 1 ) Dvs(B 1 B 2 ) since the converse inclusion is trivial. Let D = Dvs(B 1 B 2 ) and remember that D D(Λ). We have B 1 B 2 D so that B 1 (B 2 D) and the right hand set is in D(Λ) by Lemma From this we have Dvs(B 1 ) (B 2 D) so that Dvs(B 1 ) B 2 D to give by a repeat of the argument. B 2 (Dvs(B 1 ) D) and hence Dvs(B 1 ) Dvs(B 2 ) D Trivially, we have Cng Dvs, and hence we have a pair of frame quotients as indicated at the beginning of this section. We know that B(Λ) is a frame, and so carries an implication ( ), as described by Lemma 4.6. We know that UΛ is an idiom and carries a partial implication ( ), as given by Lemma We know that NΛ is a frame and carries a (full) implication ( ). These last two implications do agree as shown by Lemma But how do these match with the implication on UΛ? Quite nicely! 26

27 4.19 LEMMA. Let Λ be a modular idiom and consider B B(Λ), D D(Λ) which correspond to f UΛ, j NΛ in the sense that [a, b] B b f(a) [a, b] D b j(a) for each interval [a, b]. Then the division set (B D) corresponds to the nucleus (f j), that is [a, b] (B D) b (f j) for each interval [a, b]. Proof. For convenience let k = (f j) and let l be the nucleus corresponding to the division set (B D). We require l = k. To obtain that we make use of the characterization of (B D) given by Lemma 4.6. Consider any element a Λ. We have a f(a) l(a) l(a) a f(a) l(a) f(a) so that [a, l(a)] (B D) [a, f(a) l(a)] B and hence [a, f(a) l(a)] D by the characterization of (B D). Thus f(a) l(a) j(a) and hence f l j since a is arbitrary. This gives l k by the construction of k. For the converse comparison consider any a Λ. We require [a, k(a)] (B D), for then k(a) l(a), to give k l. To this end consider any intermediate interval a x y k(a) with [x, y] B. Then y f(x) f(a) so that y f(a) k(a) j(a) j(x) to give [x, y] D. Thus [a, k(a)] (B D), as required. Of course, by Lemma 3.17 we have (f j) = (f j) so our three uses of ( ) do agree where they can. 5 How to generate nuclei We now restrict our attention to an arbitrary modular idiom Λ. In this section we show that the frame D(Λ) is canonically isomorphic to the assembly NΛ of Λ and eventually we obtain an explicit description of the division nucleus Dvs. To do that we show how to generate nuclei from basic sets, 27

28 5.1 DEFINITION. Let Λ be a modular idiom. For arbitrary B B(Λ) and a Λ we set {B }(a) = X where x X [a, x] B} to obtain the associated inflator {B } of B. Trivially, this function f = {B } is inflationary. Consider any a b in Λ and any x Λ with [a, x] B. Then we have so that [b, b x] = [a b, b x] B x b x f(b) to show that f is monotone. This shows that {B } is an inflator. It is always better than that, and sometimes a lot better. 5.2 THEOREM. Let Λ be a modular idiom. (d) For each D D(Λ) the inflator {D } is a nucleus. (c) For each C C(Λ) the inflator {C } is a binary pre-nucleus. (b) For each B B(Λ) with C = Cng(B) C(Λ), the inflator {B } and the binary pre-nucleus {C } have the same closure. In particular, {B } is a derivative. (a) For each B B(Λ) where B is -closed, the inflator {B } is a unary pre-nucleus. Proof. (d) Consider any D D(Λ) and for convenience let h = {D }. Consider any a Λ. We have [a, h(a)] D since D is -closed. Thus [a, h(a)], [h(a), h 2 (a)] D by two applications of this observation. But now since D is a congruence set, and hence [a, h 2 (a)] D h 2 (a) h(a) by the construction of h. This shows that h is a closure operation, and part (b) ensures that h is a nucleus (once that part has been verified). (c). For convenience let g = {C }. We know that g is an inflator. Consider a, b Λ. We show g(a) g(b) g(a b) and remember that the converse comparison is trivial. Consider the two subsets X, Y Λ given by respectively. Thus x X [a, x] C g(a) = X y Y [b, y] C g(b) = Y 28

29 are the two corresponding values of g. Observe that since C is a congruence set the two sets X, Y are directed. Thus by two uses of the upper continuity of Λ. Again since C C(Λ) we have for each x X and y Y. Thus for such x and y, to give g(a) g(b) = {x y x X, y Y } [a b, x y] C x y g(a b) g(a b) g(a) g(b) as required. (b) For convenience let f = {B } and g = {C }. By part (c) we know that g is a pre-nucleus. Consider any a Λ. We have f(a) g(a) since B C. We show there is a tighter relationship between f and g. Consider any [a, x] C = Cng(B). By the construction of C we have a finite chain a = a 0 < a 1 < < a l = x with [a i, a i+1 ] B for each 0 i < l. But then for each such i, so that to give a i+1 f(a i ) x f n (a) f ω (a) f(a) g(a) f ω (a) for this arbitrary a Λ. From this we see that the two ascending chains (f α α Ord) (g α α Ord) interlace, and hence have the same stable limit, as required. (a) Finally, suppose that B is -closed and consider the derivative f = {B }. For each element a Λ we have [a, f(a)] B by the construction of f. Thus for each pair of elements a, b Λ we have since B is basic, to give as required. [a b, f(a) b] B f(a) b f(a b) The following observation is worth extracting. It will be improved later. 29

30 5.3 COROLLARY. Let Λ be a modular idiom, let B B(Λ) be -closed, and let f = {B }. Then [a, b] B b f(a) for each interval [a, b]. Proof. The implication holds for every B B(Λ) (whether it is -closed or not). The implication holds by the construction of f since B is -closed. For each basic set B the associated inflator {B } is a derivative, but it seems that in general it need not be unary. We can now obtain the fundamental correspondence. 5.4 THEOREM. Let Λ be a modular idiom. There is an isomorphism NΛ j < D(Λ) > D given by for a, b Λ with a b. j D [a, b] D b j(a) j D j = {D } Proof. We show first that the indicated constructions set up a bijective correspondence. As is usual for this kind of result, the proof is in several phases. For the first phase suppose we are given D D and consider j = {D }. By Lemma 5.2 we know that j is a nucleus, and Corollary 5.3 gives for each a b from Λ. [a, b] D b j(a) For the second phase suppose j is a given nucleus and consider the associated set D of intervals. We show that D D(Λ). Since j is a binary pre-nucleus, for each p, q Λ we have [p q, q] D q j(p q) q j(p) f(q) q j(p) p q j(p) [p, p q] D to show that D is abstract. For each a x y b with [a, b] D we have y b j(a) j(x) so that [x, y] D, and hence D B(Λ). To show that D C(Λ) consider any [a, b], [b, c] D. Then b j(a) c j(b) 30

31 so that c j 2 (a) = j(a) (since j is idempotent), to give [a, c] D. Finally, for this phase, to show that D D consider and a Λ and X Λ with [a, x] D for each x X. Then for each such x, so that x j(a) X j(a) and hence [a, X] D, as required. A simple consequence of the observation at the end of the first phase shows that that these two assignments form an inverse pair of functions Each of the two functions is monotone, and each of the two lattices is complete. Thus a simple calculation shows that each function is a complete morphism. In Section 4 we saw there is a nucleus Dvs of B(Λ) which produces D(Λ) as a quotient frame. We can now describe Dvs in a different way. 5.5 THEOREM. Let Λ be an idiom, consider any B B(Λ), and let f = {B }. Then [a, b] D b f (a) produces the smallest division set D that includes B, that is D = Dvs(B). Proof. By Lemma 5.2 the closure f is a nucleus. Thus, by Theorem 5.4 the set D is a division set. For each interval [a, b] we have [a, b] B = b f(a) f (a) = [a, b] D to show that B D. Consider any division set E with B E. We require D E. To achieve that we make a preliminary observation. Since E is -closed we have ( ) ( a Λ) [ [a, f(a)] E ] by the construction of f. We use this to show ( ) ( a Λ) [ [a, f α (a)] E ] for each ordinal α. We proceed by a base-step-leap induction. Fix an arbitrary element a Λ. The base case, α = 0, is trivial. For the induction step, α α + 1, we have [a, f α (a)], [f α (a), f α+1 ] E 31

32 by the induction hypothesis and a particular case of ( ). Thus [a, f α+1 ] E since E is a congruence set. For the leap to a limit ordinal λ the induction hypothesis gives for each ordinal α < λ. But [a, f α (a)] E f λ (a) = {f α (a) α < λ} so that [a, f λ (a)] E since E is -closed. Finally, for each a Λ, a particular case of ( ) gives [a, f (a)] E which leads to D E, as required. With this we can give an explicit description of the behaviour of Dvs. 5.6 THEOREM. Let Λ be a modular idiom. Consider any basic set B B. Then we have [a, b] Dvs(B) ( a x < b)( x < y b) [ [x, y] B ] for each interval [a, b]. Proof. Let C = Cng(B) be the congruence set generated by B and let f = {C } be the binary pre-nucleus associated with C. By Lemma 5.2 we know that {B } and f have the same idempotent closure, so that Theorem 5.5 gives [a, b] Dvs(B) b f (a) for each interval [a, b]. We use this to prove the two components of the required equivalence. ( ). Consider any and any element a x < b. We have [a, b] Dvs(B) x f(x) b and we first check that this comparison is strict. Suppose x = f(x) b by way of contradiction. We show x = f α (x) b 32