The ordered set of principal congruences of a countable lattice

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1 The ordered set of principa congruences of a countabe attice Gábor Czédi To the memory of András P. Huhn Abstract. For a attice L, et Princ(L) denote the ordered set of principa congruences of L. In a pioneering paper, G. Grätzer characterized the ordered set Princ(L) of a finite attice L; here we do the same for a countabe attice. He aso showed that every bounded ordered set H is isomorphic to Princ(L) of a bounded attice L. We prove a reated statement: if an ordered set H with a east eement is the union of a chain of principa ideas (equivaenty, if 0 H and H has a cofina chain), then H is isomorphic to Princ(L) of some attice L. 1. Introduction 1.1. Historica background. A cassica theorem of Diworth [1] states that every finite distributive attice is isomorphic to the congruence attice of a finite attice. Since Diworth s resut, the congruence attice representation probem has attracted many researchers, and dozens of papers beonging to this topic have been written. The progress is mie-stoned by Huhn [12] and Schmidt [14], reached its summit in Wehrung [15] and Růžička [13], and was summarized in Grätzer [6]; see aso Czédi [3] and Grätzer [10] for some additiona, recent references. In [7], Grätzer started an anaogous topic of Lattice Theory. Namey, for a attice L, et Princ(L) = Princ(L), denote the ordered set of principa congruences of L. A congruence is principa if it is generated by a pair a, b of eements. Ordered sets and attices with 0 and 1 are caed bounded. Ceary, if L is a bounded attice, then Princ(L) is a bounded ordered set. The pioneering theorem in Grätzer [7] states the converse: each bounded ordered set P is isomorphic to Princ(L) for an appropriate bounded attice L. Actuay, the attice he constructed is of ength 5. Up to isomorphism, he aso characterized finite bounded ordered sets as ordered sets Princ(L) of finite attices L Mathematics Subject Cassification: 06B10. Version: March 18, Key words and phrases: principa congruence, attice congruence, ordered set, order, poset, quasi-coored attice, preordering, quasiordering, cofina chain. This research was supported by the European Union and co-funded by the European Socia Fund under the project Teemedicine-focused research activities on the fied of Mathematics, Informatics and Medica sciences of project number TÁMOP A- 11/1/KONV , and by NFSR of Hungary (OTKA), grant number K83219.

2 2 G. Czédi Agebra univers Terminoogy. Uness otherwise stated, we foow the standard terminoogy and notation of Lattice Theory; see, for exampe, Grätzer [8]. Our terminoogy for weak perspectivity is the cassica one taken from Grätzer [5]. Ordered sets are nonempty sets equipped with orderings, that is, with refexive, transitive, antisymmetric reations. Note that an ordered set is often caed a partiay ordered set, or a poset, or an order Our resut. Motivated by Grätzer s theorem mentioned above, our goa is to prove the foowingtheorem. A set X is countabe if it is finite or countaby infinite, that is, if X ℵ 0. An ordered set P is directed if each two-eement subset of P has an upper bound in P. Nonempty down-sets of P and subsets c = {x P : x c} are caed order ideas and principa (order) ideas, respectivey. Theorem 1.1. (i) An ordered set P = P; is isomorphic to Princ(L) for some countabe attice L if and ony if P is a countabe directed ordered set with zero. (ii) If P is an ordered set with zero and it is the union of a chain of principa ideas, then there exists a attice L such that P = Princ(L) and L P + ℵ 0. An aternative way of formuating the condition in part (ii) is to say that 0 P and there is a cofina chain in P; see the first paragraph of Section 5 for the definition of cofinaity. For a pair a, b L 2 of eements, the east congruence coapsing a and b is denoted by con(a, b) or con L (a, b). As it was pointed out in Grätzer [7], the rue con(a i, b i ) con(a 1 b 1 a 2 b 2, a 1 b 1 a 2 b 2 ) for i {1, 2} (1.1) impies that Princ(L) is aways a directed ordered set with zero. Therefore, the first part of the theorem wi easiy be concuded from the second one. To compare part (ii) of our theorem to Grätzer s resut, note that a bounded ordered set P is aways a union of a (one-eement) chain of principa ideas. Of course, no bounded attice L can represent P by P = Princ(L) if P has no greatest eement Basic idea. Let Q; be the ordered set given in Step (d) of Figure 1. Choose a cofina chain {c 0 < c 1 <... } in Q. In our case, this chain is {c 0 < c 1 }. The figure shows how to construct a attice M in severa steps (in our case, four steps) such that Princ(M) = Q;. In the figure, each ordered set on the eft is isomorphic to the ordered set of principa congruences of the corresponding attice on the right. Note that the attice obtained in Step (b) is an interva of M. Beow, we have a coser ook at our mysterious steps eading to M. In genera, Step (a) of Figure 1 is the foowing. If H; ν is a moduar attice of ength 2, then it is isomorphic to the ordered set Princ(L) for the

3 Vo. 00, XX Principa congruences of a countabe attice 3 Figure 1. An exampe for our construction bounded attice L in Figure 5. The thin edges are abeed by the eements of H \ {0 H, 1 H }, whie the thick edges of Figure 5 and the rest of the nontrivia intervas by 1 H. In this way, the abeing provides an isomorphism between

4 4 G. Czédi Agebra univers. Princ(L) and H; ν. Note that most of the argest abes and previous abes are not indicated in Figure 1. Next, assume that H; ν is a bounded ordered set and that L is the bounded attice constructed such that Princ(L) is isomorphic to H; ν. According to Lemma 5.3, if ν is an ordering and ν ν H 2, then we can extend L to a bounded attice L such that H; ν = Princ(L ). This construction consists of repeated appications of singe steps, which are caed horizonta extensions; see Figures 7 and 8, or see Steps (b) and (d) in Figure 1. The transition from L to L is caed a muti-step horizonta extension. The first muti-step horizonta extension is practicay the same as that in Grätzer [7]. However, in genera, we sha aso perform vertica extensions; see a few ines beow. Generay, after infinitey many vertica extensions, neither the attice L, nor its interva [b p, 1] (in Figure 7) is of finite ength. Thus, horizonta extensions become much more compicated than those in Grätzer [7]; the eements x and y in Figure 1 indicate why. In particuar, M in Figure 1 without x and y woud not work, and the attice in Figure 8 woud be inappropriate if d pq 1 were a coatom. The compexity of vertica extensions makes it necessary to introduce severa auxiiary concepts. Since the ordered set in Theorem 1.1 has no argest eement in genera, we aso need vertica extensions. Assume that Princ(L) = H; ν and that H; ν extends to a bounded ordered set H ; ν such that H is an order idea of H ; ν, 0 H = 0 H, and, except for 1 H, each new eement is incomparabe with any other eement distinct from 0 H and 1 H. (Note that L is not necessariy bounded and so H need not have a argest eement.) It is not difficut to extend L to a arger attice L such that Princ(L ) = H, ν, see Figure 6. We refer to this L as a vertica extension. For exampe Step (c) in Figure 1 is a vertica extension. Note that neither our treatment for horizonta extensions, nor that for vertica ones uses the fact that the orderings in question are antisymmetric. Thus, without extra work, we dea with these extensions in a sighty more genera setting. Next, et {c 0 < c 1 < c 2 <... } = {c ι : ι < κ} be a cofina chain in the ordered set P; of Theorem 1.1, and assume that we have aready constructed a attice L ι such that Princ(L ι ) is isomorphic to the principa idea c ι of P;. In order to extend L ι to a attice L ι+1 such that Princ(L ι+1 ) is isomorphic to c ι+1, we perform a vertica extension foowed by a muti-step horizonta extension. Finay, at imit ordinas, we form directed unions Method. First of a, we need the key idea from Grätzer [7]. However, whie [7] is based on an 11-eement gadget attice, we need a gadget consisting of more eements; see Figure 2. Second, we fee that without the quasi-cooring technique deveoped in Czédi [3], the investigations eading to this paper woud have not even begun. As opposed to coorings, the advantage of quasi-coorings is that we have joins (equivaenty, the possibiity of generation) in their range sets. This aows

5 Vo. 00, XX Principa congruences of a countabe attice 5 us to decompose our construction into a sequence of eementary steps. Each step is accompanied by a quasiordering. If severa steps, possiby infinitey many steps, are carried out, then the join of the corresponding quasiorderings gives a satisfactory insight into the construction. Even if it is the cooring versions of some emmas that we ony use at the end, it is worth aowing their quasi-cooring versions since in this way the proofs wi be simper and the emmas become more genera. Third, the idea of using appropriate auxiiary structures is taken from Czédi [2]. Their roe is to accumuate a the assumptions our induction steps wi need Outine. The rest of the paper is devoted to the proof of Theorem 1.1. Quasi-coored attices, zigzags and auxiiary structures, which are the basic concepts we need in the proof, are introduced in Section 2. Vertica and horizonta extensions, which are our main constructive steps, are discussed in Sections 3 and in the ongest section, Section 4, respectivey. Finay, Section 5 competes the proof by transfinite induction. 2. Quasi-coorings and auxiiary structures 2.1. Quasi-coored attices. A quasiordered set is a structure H; ν where H is a set and ν H 2 is a refexive, transitive reation on H. Quasiordered sets are aso caed preordered ones. Instead of x, y ν, we usuay write x ν y. Aso, we write x < ν y and x ν y for the conjunction of x ν y and y ν x, and that of x, y / ν and y, x / ν, respectivey. Simiary, x = ν y wi stand for the conjunction of x ν y and y ν x. If g H and x ν g for a x H, then g is a greatest eement of H; east eements are defined duay. They are not necessariy unique; if they are, then they are denoted by 1 H and 0 H. If for a x, y H, there exists a z H such that x ν z and y ν z, then H; ν is a directed quasiordered set. Given H, the set of a quasiorderings on H is denoted by Quord(H). It is a compete attice with respect to set incusion. For X H 2, the east quasiorder on H that incudes X is denoted by quo(x). We write quo(x, y) instead of quo({ x, y }). Let L be an ordered set or a attice. For x, y L, x, y is caed an ordered pair of L if x y. The set of ordered pairs of L is denoted by Pairs (L). If X L, then Pairs (X) wi stand for X 2 Pairs (L). Note that we sha often use the fact that Pairs (S) Pairs (L) hods for subsets S of L; this expains why we work with ordered pairs rather than intervas. Note aso that a, b is an ordered pair iff b/a is a quotient; however, the concept of ordered pairs fits better to previous work with quasi-coorings. By a quasi-coored attice we mean a structure L = L; γ, H, ν where L is a attice, H; ν is a quasiordered set, γ: Pairs (L) H is a surjective map, and for a u 1, v 1, u 2, v 2 Pairs (L), (C1) if γ( u 1, v 1 ) ν γ( u 2, v 2 ), then con(u 1, v 1 ) con(u 2, v 2 );

6 6 G. Czédi Agebra univers. Figure 2. Our gadget, L g7 (p, q) (C2) if con(u 1, v 1 ) con(u 2, v 2 ), then γ( u 1, v 1 ) ν γ( u 2, v 2 ). This concept is taken from Czédi [3]. By the antichain variant of (Ci) we mean the condition obtained from (Ci) by substituting the equaity sign for ν and. Prior to [3], the name cooring was used for surjective maps satisfying the antichain variant of (C2) in Grätzer, Lakser, and Schmidt [11], and for surjective maps satisfying the antichain variant of (C1) in Grätzer [6, page 39]. However, in [3], [6], and [11], γ( u, v ) was defined ony for covering pairs u v. To emphasize that con(u 1, v 1 ) and con(u 2, v 2 ) beong to the ordered set Princ(L), we usuay write con(u 1, v 1 ) con(u 2, v 2 ) rather than con(u 1, v 1 ) con(u 2, v 2 ). It foows easiy from (C1), (C2), (1.1), and the surjectivity of γ that if L; γ, H, ν is a quasi-coored attice, then H; ν is a directed quasiordered set with a east eement; possiby with many east eements. The quasi-coored attice L g7 = L g7 (p, q) = L g7 ; γ g7, H g7, ν g7 depicted in Figure 2 is the basic gadget of the paper. In this notation, the subscript g comes from gadget whie 7 comes from (A15 ), see ater. (Note that 7 is sufficienty arge to have reasonaby convenient proofs; smaer vaues might cause probems or at east inconvenience.) The gadget L g7 consists of a 19- eement attice L g7, a quasiordered set H g7 ; ν g7, which is actuay a chain, and γ g7 is defined by the figure as foows: for x, y Pairs (L g7 ), p, if x, y is a p-coored edge in the figure, q, if x, y is a q-coored edge or x, y = c pq 4 γ g7 ( x, y ) =, dpq 4, 0 Hg7, if x = y, 1 Hg7, otherwise (if the interva [x, y] contains a thick edge). It is straightforward to see that L g7 is a quasi-coored attice; this task is particuary easy if one uses the description of congruences given in Grätzer [9].

7 Vo. 00, XX Principa congruences of a countabe attice Horizonta distance and zigzags. For a attice or ordered set L, the interior ordered set of L is L 01 = L 01 ;, where is inherited from L and L 01 = L \ {0 L, 1 L } in the sense that {0 L, 1 L } denotes the set, possiby the empty set, consisting of the east and greatest eements of L. In particuar, L 01 = L iff L has neither a east, nor a greatest eement. The interior comparabiity graph or, in short, the graph of L is G icg (L) = L 01 ; ; its vertex set is L 01, and x, y is an edge of this graph iff x y or y x. For X, Y L 01, if z 0 X, z n Y, z 0 z 1 z n, and {z 0,..., z n } = n + 1, (2.1) then z 0,..., z n is a G icg (L)-path from X to Y, or between X and Y, of ength n. The (horizonta) distance δ(x, Y ) N 0 of X, Y L 01 is the minimum of engths of G icg (L)-paths between X and Y ; it is if there is no path from X to Y. In the most important case of (2.1), X = {x} and Y = {y}; then we write x, y, and δ(x, y) instead of X, Y, and δ(x, Y ). Note that δ: L 01 L 01 N 0 = {0, 1, 2,...} is a distance function. That is, δ(x, y) = 0 x = y, δ(x, y) = δ(y, x), and δ(x, y) + δ(y, z) δ(x, z) hod for a x, y, z L 01. Next, consider a 9-tupe Figure 3. The ordered set Z 7 Z 7 = c 2, d 2 ; c 3, d 3 ; c 4, e, d 4 ; c 5, d 5 (2.2) of eements beonging to L 01. If the ordering of L 01 (equivaenty, the ordering of L) restricted to Z7 set := {c 2, d 2, c 3, d 3, c 4, e, d 4, c 5, d 5 } is the one given by Figure 3, then we ca Z 7 a zigzag of G icg (L). To expain the mysterious subscript 7, note that the abeing provides a canonica embedding of Z7 set into L g7, see Figure 2. The subsets {c 2, d 2 } and {c 4, e, d 4 } are caed the ower fibers of Z 7 given in (2.2) whie {c 3, d 3 } and {c 5, d 5 } are its upper fibers. Note the terminoogica difference: athough the fibers of a zigzag are chains, it has ony four fibers but much more chains. If Z 7 is zigzag of G icg (L) such that its ower fibers are order ideas and its upper fibers are order fiters in L 01, then Z 7 is caed a tight zigzag of G icg (L). We say that a G icg (L)-path (2.1) goes through the tight zigzag Z 7 if each fiber of Z 7 contains at east one of the eements z 0,..., z n. For a subset X of L 01 and n N = {1, 2,...}, the neighborhood with radius n of X is Nbh n (X) = {y L : δ(x, y) n for some x X}.

8 8 G. Czédi Agebra univers. Ceary, the incusions X Nbh 1 (X) Nbh 2 (X)... hod. Figure 4. The attice N 6 Next, et L be a (not necessariy bounded) attice. We say that a quadrupe a 1, b 1, a 2, b 2 L 4 is an N 6 -quadrupe of L if {b 1 b 2 = a 1 a 2, a 1 < b 1, a 2 < b 2, a 1 a 2 = b 1 b 2 } is a six-eement subattice, see Figure 4. If, in addition, b 1 b 2 = 0 L and a 1 a 2 = 1 L, then we speak of a spanning N 6 -quadrupe. For a subset X of L 2, the east attice congruence incuding X is denoted by con(x). In particuar, con({ a, b }) = con(a, b). The east and the argest congruence of L are denoted by L and L, respectivey Auxiiary structures and their substructures. Now, we are in the position to define the key concept we need. In the present paper, by an auxiiary structure we mean a structure such that the foowing eight properties hod: L = L; γ, H, ν, δ, ε, Z (2.3) (A1) L; γ, H, ν is a quasi-coored attice and L 3. (A2) The quasiordered set H; ν has exacty one east eement, 0 H, at most one greatest eement, and at east three eements. (A3) δ and ε are H L maps such that δ(0 H ) = ε(0 H ) and, for a x H \ {0 H }, δ(x) ε(x); note that we usuay write a x and b x instead of δ(x) and ε(x), respectivey. (A4) For a p H, γ( δ(p), ε(p) ) = p, that is, γ( a p, b p ) = p. (A5) If p and q are distinct eements of H \{0 H }, then δ(p), ε(p), δ(q), ε(q), aso denoted by a p, b p, a q, b q, is an N 6 -quadrupe of L. (A6) For a p H \ {0 H }, the subsets D p := {x L : 0 L x a p } and U p := {x L : b p x 1 L } are subattices. (The notation comes from down and up. If L has no greatest eement 1 L, then x 1 L means no restriction and U p = b p, the principa fiter generated by b p. Simiary, D p = a p if L has no east eement 0 L.) (A7) For a p H and x, y Pairs (D p ) Pairs (U p ), if x y, then p ν γ( x, y ). (A8) Z is a set of tight zigzags of G icg (L). (Note that Z need not contain a tight zigzags of G icg (L). In particuar, Z can be the empty set.)

9 Vo. 00, XX Principa congruences of a countabe attice 9 We say that L in (2.3) is a strong auxiiary structure if it is an auxiiary structure and the foowing five additiona properties hod. (A9) H has a (unique) greatest eement 1 H, and L is a bounded attice. (A10) The set {x L : 0 L x 1 L } consists of at east three eements. (A11) con ( { a r, b r : r H and r 1 H } ) L. (A12) For a p H 01 and Z 7 Z, we have Nbh 1 ({a p, b p }) Z7 set =. (A13) If p, q H 01 such that p q and a p, b p, a q, b q is a spanning N 6 -quadrupe, then each G icg (L)-path from {a p, b p } to {a q, b q } goes through at east one tight zigzag from Z. The conjunction of (A1), (A2), and (A10) impy (A9); we wi not rey on this observation. Next, we mention three additiona properties of strong auxiiary structures. The first one, (A14 ), is a straightforward consequence of the fact that if x beongs to the set mentioned in (A10), then x is a compement of a eements in L \ {0 L, x, 1 L }. The next one foows from (A12) and (A13), and the third one from the second and Nbh 1 (D r U r ) Nbh 2 ({a r, b r }). (A14 ) if Ψ is a congruence of L distinct from L, then {0 L } and {1 L } are singeton Ψ-bocks. (A15 ) For a p, q H \ {0 H } such that a p, b p, a q, b q is a spanning N 6 - quadrupe, δ({a p, b p }, {a q, b q }) 7. (A16 ) For a p, q H \ {0 H } such that a p, b p, a q, b q is a spanning N 6 - quadrupe, if x Nbh 1 (D p U p ) and y Nbh 1 (D q U q ), then the eements x and y are compementary, that is, x y = 0 L and x y = 1 L. If H; ν is a quasiordered set, then Θ ν = ν ν 1 is known to be an equivaence reation, and the definition [x]θ ν [y]θ ν x ν y turns the quotient set H/Θ ν into an ordered set H/Θ ν ;. The importance of our auxiiary structures is first shown by the foowing emma. Lemma 2.1. If L in (2.3) is an auxiiary structure, then the ordered set Princ(L) is isomorphic to H/Θ ν ;. In particuar, if ν is an ordering, then Princ(L) is isomorphic to the ordered set H; ν. Proof. Ceary, Princ(L) = {con(x, y) : x, y Pairs (L)}. Consider the map ϕ: Princ(L) H/Θ ν, defined by con(x, y) [γ( x, y )]Θ ν. If con(x 1, y 1 ) = con(x 2, y 2 ), then [γ( x 1, y 1 )]Θ ν = [γ( x 2, y 2 )]Θ ν foows from (C2). Hence, ϕ is a map. It is surjective since so is γ. Finay, it is bijective and an order isomorphism by (C1) and (C2). We say that an auxiiary structure L = L; γ, H, ν, δ, ε, Z is countabe if L ℵ 0. In this case, by the surjectivity of γ, H ℵ 0 aso hods. Next, we give an exampe. Exampe 2.2. Let H be a set, finite or infinite, such that 0 H, 1 H H and H 3. Let us define ν = quo ( ({0 H } H) (H {1 H }) ) ; note that H; ν is an ordered set (actuay, a moduar attice of ength 2). Let L be the attice depicted in Figure 5, where {h, g, p, q,...} is the set H 01 = H \ {0 H, 1 H }.

10 10 G. Czédi Agebra univers. For x y, γ( x, y ) is defined by the abeing of edges ike in case of L g7. In particuar, γ( z, z ) = 0 H for a z L, and [x, y] incudes a thick (unabeed) edge iff γ( x, y ) = 1 H. Let δ(0 H ) = ε(0 H ) = x 0 and Z =. For s H \{0 H }, we define δ(s) = a s and ε(s) = b s. Now, obviousy, L = L; γ, H, ν, δ, ε, Z is a strong auxiiary structure. If H ℵ 0, then L is countabe. Note that the back-fied eements form a simpe, sefdua subattice, which is usuay denoted by M 3,3. Hence, L is a sefdua attice. Figure 5. The auxiiary structure in Exampe 2.2 Substructures are defined in the natura way; note that ν = ν H 2 wi not be required beow. Namey, Definition 2.3. Let L = L; γ, H, ν, δ, ε, Z and L = L ; γ, H, ν, δ, ε, Z be auxiiary structures. We say that L is a substructure of L if (i) L is a subattice of L, H H, ν ν, and 0 H = 0 H ; (ii) γ is the restriction of γ to Pairs (L), δ is the restriction of δ to H, and ε is the restriction of ε to H. If, in addition, (iii) L is strong, Z Z, 0 L = 0 L, 1 L = 1 L, 1 H H, (iv) for a x L, if 0 L x 1 L, then x L, and (v) for each Z 7 Z, if Z set 7 L, then Z 7 Z, then L is a tight substructure of L Assume that L is a tight substructure of L. Since L = con L (0 L, 1 L ) Princ(L), H; ν has a unique argest eement by (C1), (C2), and (A2), and we have 1 H = 1 H. It is straightforward to see that L is strong; for exampe, (A11) foows by restricting the corresponding congruence of L to L, whie (A13) is a consequence of 2.3(v). Hence, we can emphasize that if L is a tight substructure of L, then L,L are strong and 1 H = 1 H. (2.4) If L is a substructure (resp., tight substructure) of L, then we say L is an extension (resp., tight extension) of L. Ceary, if L, L, and L are auxiiary

11 Vo. 00, XX Principa congruences of a countabe attice 11 structures such that L is a substructure of L and L is a substructure of L, then L is a substructure of L ; the same hods for strong auxiiary structures and their tight substructures. This transitivity wi often be used in Section 5; sometimes impicity. The next two sections indicate how easiy and efficienty we can work with auxiiary structures. 3. Vertica extensions Figure 6. The auxiiary structure L Generaizing the idea behind Exampe 2.2, this section captures, in terms of extensions of auxiiary structures, how to add an antichain and a new top eement to the quasiordered set H of coors (even if H has no top eement). For an auxiiary structure L = L; γ, H, ν, δ, ε, Z and an arbitrary (possiby empty) set K, we define the foowing objects. Let H be the disjoint union H K {1 H }, et Z =, and et 0 H = 0 H. Define ν Quord(H ) by ν = quo ( ν ({0 H } H ) (H {1 H }) ). Consider the attice L defined by Figure 6, where u, v,... denote the eements of K. The thick dotted ines indicate but not necessariy ; they are edges ony if L is bounded. Note that a new attice eements distinct from 0 L and 1 L, that is, a eements of L \(L {0 L, 1 L }), are compements of a od eements. Extend δ and ε to maps δ, ε : H L by etting δ (w) = a w and ε (w) = b w for w K {1 H }. Define γ : Pairs (L ) H by γ( x, y ), if x, y Pairs (L), w, if x = a w, y = b w, and w K, γ ( x, y ) = 0 H, if x = y, 1 H, otherwise. As usua, we use thick edges in Figure 6 instead of abeing them by 1 H. Finay, et L = L ; γ, H, ν, δ, ε, Z. The proof of the foowing emma is based on Nbh 1 (L) = L; the straightforward detais wi be omitted.

12 12 G. Czédi Agebra univers. Lemma 3.1. If L is an auxiiary structure, then L is a strong auxiiary structure. Furthermore, L is a substructure of L and L L + K + ℵ 0. Since a new bottom eement and a new top eement are added, we say that L is obtained from L by a vertica extension; this motivates the triange aiming upwards in its notation. Note that if L is a sefdua attice, then so is L. 4. Horizonta extensions of auxiiary structures The purpose of this section is to capture, in terms of tight extensions of auxiiary structures, how to increase the quasiorder ν of H = H; ν by a singe step in case H has a argest eement. If x L 01 g7 beongs to the boundary of the panar attice L g7, then we have δ(e pq, x) = δ(c pq 4, x) 3 for x cpq 6 and δ(e pq, x) = δ(c pq 4, x) = 2 for x = cpq 6, where δ is understood in the graph L 01 g7 ;. This expains that athough the panar attice L g7 is not sefdua, the eements c pq 4 and epq behave simiary in most of our considerations. That is, we can often treat L g7 as if it were a sefdua attice with c pq 4 = epq. When doing so, we wi refer to quasi-duaity. Athough the ast two components beow have not yet been defined, note that, for i {1,..., 6}, the two rows of the foowing matrix ( ap b p a q b q c pq i b q a q b p a p d pq 7 i correspond to each other via quasi-duaity. d pq ) i D p U p Up q Dp q c pq 7 i U q D q Dp q Up q Figure 7. Starting from L,... Assume that L = L; γ, H, ν, δ, ε, Z is a strong auxiiary structure, p, q H 01, and a p, b p, a q, b q is a spanning N 6 -quadrupe. (4.1)

13 Vo. 00, XX Principa congruences of a countabe attice 13 We define a structure L = L (p, q) as foows, and it wi take a ot of work to prove that it is a strong auxiiary structure. We ca L a horizonta extension of L; this expains the horizonta triange in the notation. The construction of L from L is iustrated in Figures 7 and 8. Note that our attices can contain much more eements and in a more compicated way than depicted in these two figures. The convex subsets (actuay, convex subattices) of L defined in (A6) are indicated with ight grey shapes. The soid ines represent the covering reation but the dotted ines, which sti stand for the ordering, are not necessariy edges. For exampe, if b g 1 L, then U g = {b g } and the respective dotted ine denotes the covering b g 1 L ; however, the dotted ine is not a covering if U g has no argest eement. As usua, the thick (unabeed) edges are coored by 1 H or 1 H. First, we change the N 6 -subattice {0 L, a p, b p, a q, b q, 1 L } into an L g7, depicted in Figure 2, that is, we insert the back-fied circe-shaped eements into L. Next, to each x U p \ {b p }, indicated by an empty-fied itte square in Figure 8, we add a new upper cover x + of x, which is indicated by a backfied itte square. The set of these new upper covers pus d pq 1 is denoted by U q p. The eements d pq 1 and cpq 1 wi aso be denoted by b+ p and a + p, respectivey. For x 1, x 2 U p, we et x + 1 x+ 2 iff x 1 x 2. This means that the ordered subset U p Up q of L is isomorphic to the direct product of U p C 2, where C 2 is the 2-eement chain. Finay (and simiary), to each y D q \ {a q }, we add a new ower cover y of y (indicated by a back-fied itte square). We et a q = c pq 6 and b q = d pq 6. For y 1, y 2 D q, y 1 y 2 y 1 y 2. In this way, we have obtained an ordered set denoted by L ; see aso (4.9) ater for more exact detais. We wi prove soon that L is a attice and L is a subattice in it; then it wi be cear that x + = x c pq 1 for x {a p } U p and y = y d pq 6 for y {b q } D q. (4.2) Note that whie Grätzer [7] constructed a attice of ength 5, here even the interva, say, [b p, 1 L ] can be of infinite ength. Next, set H = H and Z pq 7 = c pq 2, dpq 2 ; cpq 3, dpq 3 ; cpq 4, epq, d pq 4 ; cpq 5, dpq 5. Let Z = Z {Z pq 7 }. In Quord(H ), we define ν = quo ( ν { p, q } ). Since ν is refexive and transitive, we have that r 1, r 2 ν r 1 ν p and q ν r 2, or r 1 ν r 2, (4.3) for arbitrary r 1, r 2 H. Hence, it foows easiy from the vaidity of (A2) and (A9) in L that H ; ν has a unique argest eement 1 H, a unique east eement 0 H, and we have 1 H = 1 H and 0 H = 0 H. We extend γ to a map

14 14 G. Czédi Agebra univers. Figure 8...., we obtain L γ : Pairs (L ) H by γ( x, y ), if x, y Pairs (L), γ( z, t ), if x, y = z +, t + Pairs (Up q {c pq 1 }), γ( z, t ), if x, y = z, t Pairs (Dp q {dpq 6 }), γ ( x, y ) = p, if x, y { c pq 2, dpq 2, cpq 3, dpq 3, epq, d pq 4 }, q, if x, y { c pq 4, dpq 4, cpq 5, dpq 5, cpq 4, epq }, 0 H, if x = y, 1 H, otherwise. Finay, after etting δ = δ, and ε = ε, we define L (p, q) = L as L ; γ, H, ν, δ, ε, Z. (4.4) Lemma 4.1. If L satisfies (4.1), then L is a bounded attice. Furthermore, S pq, given by (4.5) beow, and L are {0, 1}-subattices of L. Proof. First, we describe the ordering of L more precisey; this description is the rea definition of L. Consider the foowing subsets of L : N pq = {c pq 1,..., cpq 6, dpq 1,..., dpq 6, epq } Up q Dp q (new eements), B pq = {0 L = 0 L, a p, b p, 1 L } D p U p (eft boundary), Br pq = {0 L, a q, b q, 1 L = 1 L } D q U q (right boundary), B pq = B pq R pq R pq r B pq r (boundary), = D p U p U q p {cpq 1 } (eft region), = D q U q Dp q {d pq 6 } (right region), S pq = N pq B pq, L pq and g7 = {0 L, a p, b p, a q, b q, c pq 1,..., cpq 6, epq, d pq 1,..., dpq 6, 1 L }. (4.5)

15 Vo. 00, XX Principa congruences of a countabe attice 15 In case of four sets above, we ca these sets and their eements eft or right simpy because of their positions in our figures. The definitions of some of these sets above are redundant; for exampe, d pq 1 Uq p. The ordering within B pq, which is a subset of L, is inherited from L. By definition, L pq g7 = L g7 = L g7 (p, q). The ordering within U p {a p } Up q {c pq 1 } and that within D q {b q } Dp q {d pq 6 } are aready cear; for exampe, if x U p {a p } and y + Up q {c pq 1 }, then x y+ iff x L y. Thus, since a p is the top eement of and that within R pq r D p and b q is the bottom of U q, the ordering within R pq are defined. (A6) impies that R pq ; and Rpq r ; are attices. (4.6) The above facts, together with L pq g7 = L g7 and even L pq g7 = L g7(p, q), define the ordering within S pq. A routine argument verifies that S pq = S pq ; S pq is a attice; the detais are omitted. Observe that 1 L / R pq for a x R pq R pq r and and y R pq r, x S pq y = 0 L. (4.7) Therefore, for x N pq, there is a unique east eement x of B pq such that x S pq x. Simiary, for x N pq, there is a unique argest eement x of B pq such that x S pq x. If x L, then we et x = x = x. In this way, x x and x x are maps from L to L. Note that (x ) = x and (y ) = y for x D q {b q } and y Dp q {d pq 6 }; (x + ) = x and (y ) + = y for x U p {a p } and y Up q {cpq 1 }. (4.8) Using these maps, the exact definition of the ordering in L is described as foows: for x, y L, x L y, if x, y L, or x S pq y, if x, y S pq, or x L y (4.9) x L y, if x L \ S pq and y N pq, or x L y, if x N pq and y L \ S pq. Observe that for u 1, u 3 B pq and u 2 N pq, the conjunction of u 1 S pq u 2 and u 2 S pq u 3 impies {0 L, 1 L } {u 1, u 3 }. Hence, it is straightforward to see that L is an ordering and L is the restriction of L to L. Note that, for x L, x = 1 L x and x = 0 L x ; (4.10) that is, x is the greatest eement of L x, and duay for x. Uness otherwise specified,,,, e pq, etc. wi be understood in L. Next, we define a mapping u û from L to S pq. For u L \ {0 L }, either u D p, or u b p U p {1 L } is the smaest eement u p of B pq u. Simiary, Br pq u has a smaest eement u q. If u 0 L, then it foows from (A16 ) that 1 L {u p, u q }. Hence, for each u L, B pq u has a smaest eement;

16 16 G. Czédi Agebra univers. we denote it by û. For u N pq, we et û = u. Note that, for every u L, û is the smaest eement of S pq u, and u U p \ a p impies û = u L b p. (4.11) For ater reference, we quasi-duaize (4.11). For u L\ {1 L }, either u U q, or u a q D q {0 L } is the argest eement u q of Br pq u. Simiary, B pq u has a argest eement u p. Hence, for each u L, B pq u has a argest eement; we denote it by ŭ. For u N pq, we et ŭ = u. Note that, for every u L, ŭ is the argest eement of S pq u, and u D q \ b q impies ŭ = u L a q. (4.12) Next, for x y L, we want to show that x and y has a join in L. There are severa cases to consider. The order idea generated by U q p wi be denoted by U q p. Since U q p is directed, U q p wi turn out to be a attice idea of L. Case 4.2. We caim that if {x, y} L, then L, x L y is the join of x and y in L. To prove this, we can assume that {x, y} U q p, since otherwise {x, y} has no upper bound outside L. Let z N pq be an upper bound of {x, y}. We obtain from (4.10) that x L z (even if x S pq ) and y L z. Hence, x L y L z z, proving x L y = x L y. Case 4.3. We caim that if {x, y} S pq, then x S pq y is the join of x and y in L. (We have aready mentioned that S pq ; is a attice.) For {x, y} R pq or {x, y} R pq r, this foows from (4.8) and (4.10) in a straightforward way by considering severa cases. Next, assume that x R pq and y R pq r, and suppose, for a contradiction, that z L \ {1 L } is an upper bound of {x, y}. We have x L z and y L z by (4.10). Since x x z < 1 L, the eement x R pq has a nontrivia upper bound in L. Thus, x = x D p U p Nbh 1 (D p U p ). Since y R pq r, we have y D q U q Nbh 1 (D q U q ). Hence, the vaidity of (A16 ) in L yieds that 1 L = x L y L z < L 1 L, a contradiction. Thus, we concude the vaidity of x S pq y = x L y for the case {x, y} R pq R pq r. The same hods in the remaining case {x, y} R pq R pq r, because then a upper bounds of {x, y} beong to S pq. Case 4.4. For x L \ B pq = L \ S pq and y N pq = S pq \ L, we caim that x L y = { x S pq y, if x U q p (equivaenty, if x U p), x L y, if x / U q p (equivaenty, if x / U p ). (4.13) To prove this, first we assume that x Up. q We concude from (A16 ) that x D p U p and x Nbh 1 (D p U p ). Suppose, for a contradiction, that {x, y} has an upper bound z in L \ {1 L }. We have y L z by (4.10). Since y D q U q Nbh 1 (D q U q ), (A16 ) yieds z = 1, contradicting z L\{1 L }. Hence, a upper bounds of {x, y} beong to S pq. This proves the first haf of (4.13). The second haf is obvious, because x L \ Up q has no upper bound outside L.

17 Vo. 00, XX Principa congruences of a countabe attice 17 Cases prove that L is a join-semiattice. By quasi-duaity, it is a attice. Cases 4.2, 4.3, and their quasi-duas aso prove that L and S pq are {0, 1}-subattices of L. This competes the proof of Lemma 4.1. We need a emma from Diworth [4], see aso Grätzer [5, Theorem III.1.2]. Lemma 4.5. If L is a attice and u 1, v 1, u 2, v 2 Pairs (L), then the foowing three conditions are equivaent. (i) con(u 1, v 1 ) con(u 2, v 2 ); (ii) u 1, v 1 con(u 2, v 2 ); (iii) there exists an n N and there are x i L for i {0,..., n} and y ij, z ij Pairs (L) for i, j {1,..., n} {0,..., n} such that the foowing equaities and inequaities hod: u 1 = x 0 x 1 x n 1 x n = v 1 y i0 = x i 1, y in = u 2, z i0 = x i, and z in = v 2 for 1 i n, y i,j 1 = z i,j 1 y ij and z i,j 1 z ij for j odd, i, j {1,..., n}, z i,j 1 = y i,j 1 z ij and y i,j 1 y ij for j even, i, j {1,..., n}. (4.14) The situation of Lemma 4.5 is outined in Figure 9; note that the eements depicted do not form a subattice in genera and they are not necessariy distinct. The second haf of (4.14) says that, in terms of Grätzer [5], y i,j 1, z i,j 1 is weaky up or down perspective into y ij, z ij ; up for j odd and down for j even. Besides weak perspectivity, we reca that x 1, y 1 is perspective to x 2, y 2 if there are i, j {1, 2} such that i j, x i = x j y i, and y j = x j y i. Projectivity is the refexive transitive cosure of perspectivity. Figure 9. Iustrating Lemma 4.5 for n = 4 For a quasiordered set H; ν, we say that p H is a join of the eements q 1,..., q n H, in notation, p = ν n i=1 q i, if q i ν p for a i and, for every r H, the conjunction of q i ν r for i = 1,..., n impies p ν r. Even if a join exists, it need not be unique in the usua sense, but it is unique moduo = ν. Lemma 4.6 ( Chain Lemma for quasi-coored attices). If L; γ, H, ν is a quasi-coored attice and {u 0 u 1 u n } is a finite chain in L, then γ( u 0, u n ) = ν n i=1 γ( u i 1, u i ) hods in H; ν. (4.15)

18 18 G. Czédi Agebra univers. Proof. Let p = γ( u 0, u n ) and q i = γ( u i 1, u i ). Since con(u i 1, u i ) con(u 0, u n ), (C2) yieds q i ν p for a i. Next, assume that r H such that q i ν r for a i. By the surjectivity of γ, there exists a v, w Pairs (L) such that γ( v, w ) = r. It foows by (C1) that u i 1, u i con(u i 1, u i ) con(v, w). Since con(v, w) is transitive and coapses the pairs u i 1, u i, it coapses u 0, u n. Hence, con(u 0, u n ) con(v, w), and (C2) impies p ν r. Now, we are in the position to prove the main emma of the paper. Lemma 4.7. The structure L = L (p, q), which is defined in (4.4) with assumption (4.1), is a strong auxiiary structure, and L is a tight substructure of L. Furthermore, L ℵ 0 + L. Proof. Since we work both in L and L, reations, operations and maps are often subscripted by the reevant structure; in the absence of subscripts, we are in L. By Lemma 4.1, L is a bounded attice. We obtain from (4.3) that (A2) hods for L. It foows triviay from the construction and Lemma 4.1 that L satisfies (A3), (A4), (A5), (A9), and (A10). Next, we dea with (A6). Let r H \ {0 H } and {x, y} U r ; we have to prove that x y U r. Equivaenty, we have to prove that x y 1 L. For {x, y} L, this foows from the vaidity of (A6) in L. If {x, y} N pq, then both x and y beong to {c pq 1 } Uq p, since b r x y and b r L. Hence, x y 1 L, because R pq is subattice by (4.6). Therefore, we can assume that x L \ S pq and y N pq. We obtain from (4.10) that b r y and y {c pq 1 } Uq p. If we had y = a p, then b r a p woud contradict either (A5), if r p, or (A3), if r = p. Hence, y U p, and for ater reference, we note that for a r H, b r a p. (4.16) Since L satisfies (A6) and {x, y } U r, we obtain x y 1 L. Ceary, x y U p. Using that R pq is a subattice of L, it foows that (x y ) y R pq. Thus, x y = x (y y) = (x y ) y 1 L. Consequenty, U r is a subattice of L, and L satisfies (A6) by quasi-duaity. The members of Pairs (L )\ ( Pairs (L) Pairs (S pq ) ) wi be caed mixed pairs. In other words, a pair is caed mixed if exacty one of its components beongs to L. By the definition of γ, if x, y is a mixed pair, then γ ( x, y ) = 1 H. (4.17) In order to verify (A7), assume that r H, x, y Pairs (U r ), and x y. If x, y Pairs (L), in other words, x, y is an od pair, or x, y is a mixed pair, or γ ( x, y ) = 1 H, then r ν γ ( x, y ) foows from ν ν, (4.17), and the vaidity of (A7) in L. Hence, we can assume that x, y Pairs (N pq ). Since 0 L b r x and b r L, we obtain that x Up q {cpq 1 }. Actuay, we have x Up q, since x = cpq 1, combined with (4.10) and (cpq 1 ) = a p, woud contradict (4.16). Simiary, y Up q. Ceary, x, y U p L. Hence, using the definitions, the vaidity of (A7) in L, and (4.8), we obtain that

19 Vo. 00, XX Principa congruences of a countabe attice 19 r ν γ( x, y ) = γ ( (x ) +, (y ) + ) = γ ( x, y ). Thus, by quasi-duaity, L satisfies (A7). Next, we prove that, for x, y Pairs (L ), γ ( x, y ) = 1 H = con L (x, y) = L. (4.18) First, assume that x, y Pairs (L) and γ ( x, y ) = 1 H, that is, γ( x, y ) = 1 H. Since (C1) hods in L and γ( 0 L, 1 L ) ν 1 H = γ( x, y ), we obtain L = con L (0 L, 1 L ) con L (x, y). Hence, 0 L, 1 L con L (x, y). Using Lemma 4.5 and the fact that L is a subattice of L by Lemma 4.1, we obtain 0 L, 1 L con L (x, y), and (4.18) hods in this case. Second, assume that x, y Pairs (L ) is a mixed pair in the sense of (4.17), and keep in mind that (A14 ), which is a consequence of (A10), hods in L. Figure 8 shows (and it is straightforward to prove) that there exist x 1, y 1 B pq such that x x 1 y 1 y and x 1, y 1 is perspective to 0 L, u or u, 1 L for some u (L ) 01. Hence, we concude con L (x, y) = 1 H from (A14 ). Third, we are eft with the case x, y Pairs (N pq ) and γ ( x, y ) = 1 H. If [x, y] contains a thick edge x 1 y 1 of Figure 8, then the previous case appies. Otherwise, either x, y Pairs (Up q {c pq 1 }), or x, y Pairs (Dp q {d pq 6 }). By quasi-duaity, we can assume the first aternative. Using (4.8), we have that x, y = (x ) +, (y ) +, which is perspective to x, y. Thus, con L (x, y) = con L (x, y ), γ ( x, y ) = γ ( x, y ), and the first case appies since x, y beongs to Pairs (L). This proves (4.18). Let Θ denote the congruence of L described in (A11). Let Ψ = { x +, y + : x, y U p {a p } and x, y Θ}, Γ = { x, y : x, y D q {b q } and x, y Θ} and Φ = {e pq, c pq 4, dpq 4 }2 {c pq 2, dpq 2 }2 {c pq 3, dpq 3 }2 {c pq 5, dpq 5 }2. Here, Ψ and Γ are equivaence reations on the sets {c pq 1 } Uq p and {dpq 6 } Dq p, respectivey, and Φ is the equivaence on {e pq, c pq 2,..., cpq 5, dpq 2,..., dpq 5 } whose bocks are the fibers of Z pq 7. Let Θ = Θ Ψ Γ Φ; its bocks are the Θ-bocks, the Ψ-bocks, the Γ-bocks, and the Φ-bocks. The restriction of Θ to subset X L wi be denoted by Θ X. Since the Θ-bocks, the Θ {ap } U p -bocks, and the Θ {bq} D q -bocks are convex subattices, so are the Θ -bocks. To prove that Θ is a congruence, assume that x, y Θ Pairs (L ), x y, and z L \ {0 L, 1 L } such that x z; we caim that x z, y z Θ. (4.19) Since Θ is taken from (A11), (A14 ) gives that {x, y} {0 L, 1 L } =. By the convexity of Θ -bocks, we can assume that y z. Based on Grätzer [8, Lemma 11], a tedious but straightforward argument shows that Θ S pq is a congruence. Since so is Θ = Θ L, we can assume that {x, y, z} S pq and {x, y, z} L. Since both L and N pq are unions of Θ -bocks, there are two cases to consider.

20 20 G. Czédi Agebra univers. First, assume that x, y Pairs (L) and z N pq = L \ L. It foows from 0 L < x < z that c pq 1 z. If z Uq p, then x b p, y b p Θ Up 2 = Θ S pq Up, 2 x z, y z = x b p z, y b p z, and the fact that Θ S pq is a congruence give (4.19). Hence, we consider z = c pq 1. If we have y b p, then y z = c pq 1, y dpq 1, and cpq 1 dpq 1 yied x z, y z = cpq 1, dpq 1 Θ. Thus, we assume that y b p, that is, b p < b p y. Since x c pq 1, we have x (c pq 1 ) = a p b p by (4.10). Hence, b p, b p y = b p L x, b p L y Θ L. Thus, (A14 ) yieds b p y 1 L = 1 L, impying y U p. Using (4.11) and y a p, we have that ŷ = y b p. Therefore, y z = ŷ S pq c pq 1 ; either by (4.13), if y L \ S pq, or because y = ŷ S pq. Since ŷ = y b p, we obtain that b p, ŷ = x b p, y b p Θ Θ. Joining this with z = c pq 1, we have d pq 1, y z Ψ, which impies (4.19) since x z, dpq 1 = cpq 1, dpq 1 Ψ. Second, assume that x, y Pairs (N pq ) and z L \ S pq. We sti have {x, y} {0 L, 1 L } =, by the definition of Θ. Since x < z < 1 L impies x Dp q {dpq 6 }, the definition of Θ gives {x, y} Dp q {dpq 6 }. By (4.8) and the definition of Θ, we obtain x, y Θ. We have x z since x z. If we had z U p, then z Nbh 1 (D p U p ), x D q U q Nbh 1 (D q U q ), and the vaidity of (A16 ) in L woud impy x = z L x = 0 L, whence x = 0 L, contradicting {x, y} {0 L, 1 L } =. Hence, z / U p, and (4.13) yieds that x z, y z = x L z, y L z Θ Θ. This proves (4.19). Since Θ is an equivaence reation, (4.19) and its quasi-dua impy that Θ is a congruence on L. Since it is distinct from L, L satisfies (A11). Next, we prove the converse of (4.18). Assume that x, y Pairs (L ) such that γ ( x, y ) 1 H ; we want to show that con L (x, y) L. Since this is cear if x = y, we assume x y. First, if x, y L, then et r = γ( x, y ). Appying (C1) to γ and (A4) to L, we obtain con L (x, y) = con L (a r, b r ). Hence Θ, which we used in the previous paragraphs, coapses x, y, and con L (x, y) Θ L. Second, if {x, y} L =, then there exists a pair x, y Pairs ({a p } U p ) Pairs ({b q } D q ) Pairs (L) such that γ ( x, y ) = γ ( x, y ) and x, y is projective to x, y. For exampe, if x, y Pairs ({c pq 1 } Uq p), then we et x, y = x, y. Thus, since projective pairs generate the same congruence, this case reduces to the first case. Finay, L {x, y} = 1 is excuded by (4.17). Now, after verifying the converse of (4.18), we have proved that, for a x, y Pairs (L ), γ ( x, y ) = 1 H con L (x, y) = L. (4.20) Observe that γ is isotone in the sense that if w 1 w 2 w 3 w 4, then γ ( w 2, w 3 ) ν γ ( w 1, w 4 ). (4.21) This foows from (4.17), from the definition of γ, and from the fact that γ is isotone by (C1) and (C2). Next, to prove that γ satisfies (C1), assume that u 1, v 1 and u 2, v 2 beong to Pairs (L ) such that γ ( u 1, v 1 ) ν γ ( u 2, v 2 ). Let r i = γ ( u i, v i ), for i {1, 2}. We have to show con L (u 1, v 1 ) con L (u 2, v 2 ).

21 Vo. 00, XX Principa congruences of a countabe attice 21 By (4.20), we can assume that r 2 1 H. We aso have that r 1 1 H, since otherwise 1 H = r 1 ν r 2 and the satisfaction of (A2) for L woud give r 2 = r 1 = 1 H. Simiary, we can assume that r 1 and, consequenty, aso r 2 differ from 0 H, since otherwise u 1 = v 1, and so con L (u 1, v 1 ) = L woud ceary impy con L (u 1, v 1 ) con L (u 2, v 2 ). Thus, r 1, r 2 H 01 = (H ) 01. By the construction of L, u i, v i is projective to some u i, v i Pairs (L) such that γ ( u i, v i ) = γ ( u i, v i ). Projectivity impies con L (u i, v i ) = con L (u i, v i ). Therefore, we can assume that u 1, v 1, u 2, v 2 Pairs (L), because otherwise we coud work with u 1, v 1 and u 2, v 2. According to (4.3), we distinguish two cases. First, assume that r 1 ν r 2. Since γ extends γ, we have that γ( u 1, v 1 ) = γ ( u 1, v 1 ) = r 1 ν r 2 = γ ( u 2, v 2 ) = γ( u 2, v 2 ). Appying (C1) to γ, we obtain u 1, v 1 con L (u 1, v 1 ) con L (u 2, v 2 ). Using Lemma 4.5 (i) (iii) in L and then Lemma 4.5 (iii) (i) in L, we obtain that con L (u 1, v 1 ) con L (u 2, v 2 ). Second, assume that r 1 ν p and q ν r 2. Since γ ( a p, b p ) = γ( a p, b p ) = p and γ ( a q, b q ) = q by (A4), the argument of the previous paragraph yieds con L (u 1, v 1 ) con L (a p, b p ) and con L (a q, b q ) con L (u 2, v 2 ). Ceary (or appying Lemma 4.5 within S pq ), we have con L (a p, b p ) con L (a q, b q ). Hence, transitivity yieds con L (u 1, v 1 ) con L (u 2, v 2 ). Consequenty, γ satisfies (C1). Next, to prove that γ satisfies (C2), we assume that u 1, v 1, u 2, v 2 Pairs (L ) such that con L (u 1, v 1 ) con L (u 2, v 2 ). Our purpose is to show that γ ( u 1, v 1 ) ν γ ( u 2, v 2 ). We can assume that u 1 v 1 and, by (4.20), that con L (u 2, v 2 ) L. That is, {con L (u 1, v 1 ), con L (u 2, v 2 )} is disjoint from { L, L }. We obtain from (4.17) that none of u 1, v 1 and u 2, v 2 is a mixed pair. If u i, v i is a new pair, that is, if {u i, v i } Pairs (N pq ), then we can consider an od pair u i, v i such that γ ( u i, v i ) = γ ( u i, v i ) and so, since γ satisfies (C1), con L (u i, v i) = con L (u i, v i ). Hence, we can assume that u 1, v 1 and u 2, v 2 are od pairs, that is, they beong to Pairs (L). The starting assumption con L (u 1, v 1 ) con L (u 2, v 2 ) is witnessed by Lemma 4.5. Let x j, y ij, z ij L, for i {1,..., n} and j {0,..., n}, be eements that satisfy (4.14); see aso Figure 9. To ease our terminoogy, the ordered pairs y ij, z ij wi be caed witness pairs. Since con L (u 2, v 2 ) L, none of the witness pairs generate L. Thus, by (4.17) and (4.20), none of the witness pairs is mixed or 1 H -coored. (4.22) Take two consecutive witness pairs, y i,j 1, z i,j 1 and y ij, z ij. Here i, j {1,..., n}, and (4.14) says that y i,j 1, z i,j 1 is weaky perspective into y ij, z ij. We want to show that γ ( y i,j 1, z i,j 1 ) ν γ ( y ij, z ij ). (4.23)

22 22 G. Czédi Agebra univers. To make the notation easier, we et y 0 = y i,j 1, z 0 = z i,j 1, p 0 = y 0, z 0, y 1 = y ij, z 1 = z ij, and p 1 = y 1, z 1. With this notation, (4.23) turns into γ (p 0 ) ν γ (p 1 ); (4.24) this is what we have to prove now. We assume y 0 < z 0 since (4.24) triviay hods otherwise. Hence, y 1 < z 1 aso hods. By (4.21), we can aso assume perspectivity rather than weak perspectivity. That is, as depicted in Figure 10, either z 0 y 1, z 0 y 1 = y 0 and z 0 y 1 = z 1, (4.25) or y 0 z 1, y 0 z 1 = y 1 and y 0 z 1 = z 0 ; (4.26) By (4.22), there are three cases to consider. Figure 10. Perspectivities (4.25) and (4.26) Case 4.8. If both p 0 and p 1 are od, then con L (p 0 ) = con L (p 1 ). Appying (C2) for L, we concude the reation γ(p 0 ) ν γ(p 1 ). Thus, since γ extends γ, (4.24) hods for od witness pairs. Case 4.9. If p 0, p 1 Pairs (N pq ), that is, if both p 0 and p 1 are new, then there are ony few cases, and (4.24) foows in a straightforward way from our assumptions and the definition of γ. For exampe, et p 0, p 1 Pairs (Up), q and consider the od pairs (p 0 ) := (y 0 ), (z 0 ) and (p 1 ) := (y 1 ), (z 1 ). Since the maps in the second row of (4.8) are reciproca attice isomorphism between {a p } U p and {c pq 1 } Uq p, it foows that (p 0) and (p 1 ) are perspective. Hence, con L ((p 0 ) ) = con L ((p 1 ) ), and we obtain from (C2), appied for L, that γ((p 0 ) ) = ν γ((p 1 ) ). On the other hand, we have γ (p 0 ) = γ((p 0 ) ) and γ (p 1 ) = γ((p 1 ) ) by (4.8) and the definition of γ. Hence, (4.24) foows by transitivity. Case Assume that one of p 0 and p 1 is od and one is new. We caim γ (p 0 ) = ν γ (p 1 ), (4.27) which is a stronger statement than (4.24). Since the roe of p 0 and p 1 in (4.27) is symmetric, we can assume that p 0 is od and p 1 is new. By quasi-duaity, we aso assume (4.25). Since 0 L < y 0 < y 1, y 0 is an od eement, and y 1 is a new one, we have y 1, z 1 {c pq 1 } Uq p and z 0 U p. First, assume that y 1 = c pq 1 ; see Figure 11. We concude from (4.9) that y 0 a p. Since z 0 c pq 1, we have z 0 a p. Hence, ẑ 0 = z 0 b p z 1 by (4.11). Let u = z 0 b p ; we have y 0 u. Denote γ( b p, ẑ 0 ) and γ( u, z 0 ) by r and r, respectivey. Since b p, ẑ 0 and u, z 0 are perspective in L, (C2) yieds r = ν r. (4.28)

23 Vo. 00, XX Principa congruences of a countabe attice 23 Figure 11. Case 4.10 with y 1 = c pq 1 Since y 0 a p < b p and, by assumption (4.25), y 0 = z 0 c pq 1, it foows that y 0 = b p y 0 b p = b p z 0 c pq 1 b p = u a p. Denote γ( y 0, u ) by p. If u a p, then y 0, u is perspective to a p, b p in L since a p b p by (A3), so (C2) yieds p = ν p. Otherwise, if u a p, then we obtain from y 0 u z 0 a p z 0 c pq 1 = y 0, (A1), and (A2) that p = γ( u, u ) = 0 H ν p. Hence, p ν p, and even p = ν p if u a p. (4.29) As a subcase, assume that ẑ 0 = b p. (We woud obtain this situation from Figure 11 by coapsing each of the r -coored and r-coored intervas to its bottom.) We have that y 0 < z 0 = z 0 b p = u. This excudes u a p, because z 0 y 1 = c pq 1. Hence, p = ν p by (4.29). Since z 1 = z 0 y 1 = ẑ 0 c pq 1 = b p c pq, we concude 1 = dpq 1 γ (p 0 ) = γ( y 0, u ) = p = ν p = γ( a p, b p ) = γ ( c pq 1, dpq 1 ) = γ (p 1 ); that is, (4.27) hods. We are eft with the subcase ẑ 0 > b p. We obtain p ν r from (A7). Hence, using Lemma 4.6, we obtain γ( a p, ẑ 0 ) = ν r. Since z 1 = z 0 c pq 1 = ẑ 0 c pq 1 = (ẑ 0 ) + by (4.13) and (4.2), it foows that γ (p 1 ) = γ( a p, ẑ 0 ), and we obtain γ (p 1 ) = ν r. Since (4.29), p ν r, and (4.28) impy p ν r by transitivity, Lemma 4.6 yieds γ (p 0 ) = ν r. Therefore, (4.27) and (4.24) foow from (4.28). Second, assume that y 1 Up; q see Figure 12. Let r = γ (p 0 ) = γ(p 0 ), r = γ (p 1 ), and et u = ẑ 0 y 1 U p. Since z 1 = z 0 y 1 = ẑ 0 y 1 by (4.13), u, ẑ 0 is up-perspective to p 1. Since z 0 ẑ 0 and u ẑ 0, we have z 0 u ẑ 0. This, together with (4.11), z 0 z 0 u and z 0 u S pq, impies z 0 u = ẑ 0. Hence, p 0 is perspective to u, ẑ 0. Thus, denoting γ ( u, ẑ 0 ) = γ( u, ẑ 0 ) by r, the vaidity of (C2) in L gives r = ν r. On the other hand, the subattice U p Up q is isomorphic to U p C 2 by definitions, see aso (4.8). Therefore, since u, ẑ 0 is up-perspective to p 1, it is straightforward to see that u, ẑ 0 is perspective to (y 1 ), (z 1 ). By the definition of γ and