ON THE UBIQUITY OF HERRINGBONES IN FINITELY GENERATED LATTICES

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1 proceedings of the american mathematical society Volume 82, Number 3, July 1981 ON THE UBIQUITY OF HERRINGBONES IN FINITELY GENERATED LATTICES I. RIVAL, W. RUCKELSHAUSEN and b. sands Abstract. Every finitely generated infinite lattice of finite width contains a subset isomorphic to the "herringbone" or its dual. 1. Introduction. Aficionados of lattice diagrams have long noticed that a particular infinite lattice, which we illustrate in Figure 1, appears repeatedly in the diagrams of finitely generated infinite lattices. This lattice, the herringbone, can be seen within Rolfs diagrams of F(2 + 2) and F(l + 4) [3]; more explicitly in Wille [4] and numerous other papers, usually as an integral part of a counterexample; in unpublished researches of the third author; finally in recent results of Bauer and Poguntke [1] and Poguntke and Sands [2], where it assumes a leading role. Figure 1. The herringbone The main result of this paper may be considered as an explanation of the seemingly ubiquitous nature of the herringbone. Theorem. Every finitely generated infinite lattice of finite width contains a subset order-isomorphic to the herringbone or its dual. If L is a finitely generated infinite lattice of width three, Poguntke and Sands [2] have shown that L will contain a sublattice isomorphic to the herringbone or its dual (see Figure 2(a) for an example). On the other hand, no such strengthening of our theorem is possible for lattices of width five (see Figure 2(b)). Received by the editors November 6, 1979 and in revised form, April 16, AMS (MOS) subject classifications (1970). Primary 06B American Mathematical Society /81 / /S02.S0

2 336 I. RIVAL, W. RUCKELSHAUSEN AND B. SANDS (0) (b) Figure 2 2. Infinite descending chains. Descending chains isomorphic to tcd play an important role in this investigation. We first record, without proof, the following elementary observation. Lemma 1. Let P be an ordered set of finite width, and let C = {c \i = 1, 2,... } and D = {d \i = 1,2,...} be infinite subsets of P. (a) If cx > c2 > c3 >..., and, for each i, d c but there exists j > i such that d > Cj, then C and D contain subsequences C = {c > c 2 >... } and D' = {d > d >... } respectively such that, for eachj, c, { d but c, < d,. (b) If dx > d2 > d3 >..., and, for each i, c, < d but there exists j > i such that c ^ dj, then C and D contain subsequences C = {c > c, >... } and D' = {d > d >... } respectively such that, for eachj, c < d, but c, ^ d. The proof of the theorem rests in no little measure on the following simple, but important, result. Proposition 2. Let C = {c, > c2 >... } be a chain in a finitely generated lattice L. Then the subset Ac = {x G L\ for all i, c, ^ x} is infinite. Moreover, for each i there are x, y G Ac satisfying x V V, Ö Ac and c ^ x V V,- Proof. Let L be a finitely generated lattice and let C = {c \i = 1, 2,... } be a subset of L satisfying c, > c2 >... Let us suppose that there is an i such that, for all x, y G Ac, either c, <x\/yotx\/ye Ac. Then Ac u [c,), where [c ) = {x G L c, < x], is a sublattice of L. Moreover, for all7 > 1, Ac u [cj) is a sublattice. As Rj = L\ (Ac u [cj)) is nonempty, it follows that Rj must contain a generator gj of L, for each j > i. In particular, there is jx > i such that g > c}. For the generator gjt G RJt, gjt cjx, whence g, = gjt. Now there isj2 >jx such that gjx > ch and the generator gj2 G RJí satisfies gh ^ c^. Therefore, the generators g, gjx, and gj are all distinct. In this way we construct an infinite set { g, gjt, gj2,... } of generators of L, which is a contradiction. We conclude from this that, for each /", there is x, y, G Ac such that x V v, Ö Ac and c, f x V v,.

3 herringbones in finitely generated lattices 337 Finally, that Ac is infinite follows from Lemma 1. Indeed, from the conclusion above there are subsets X = {x \i = 1, 2,... } and Y = {y \i = 1, 2,... } of Ac, and a subsequence c\> c'2>... of C such that, for all i, x V V, > c' but x Vv, ^ c/-i- It follows that (x, V v, j' = 1, 2,... } and hence X U Y is infinite, whence ylc is infinite. For an ordered set P, let ßy, denote the set of all those subsets of P which are isomorphic to ad. In the case that P is a lattice of finite width, we can endow QP with a strict order. Let C, D G Qp. We write C < D if, for any infinite subset >' of D, there is an infinite subset C of C such that (Cl) For every c G C and for every d E D',c 1? d, and (C2) there is a map/of C to >' satisfying: c < c' if and only if c </(c/). Remarks. 1. An instance of this ordering is illustrated in Figure If C < Z> in ep, and C çc,i)'ç Z), and /: C -> D' are as above, then c < c' in C if and only if/(c) < f(c') in >'; that is,/is j/r/cf order-preserving. D'»D pf(c,) f(c2) f<c > yff(c4) I Figure 3 Lemma 3. Let L be a lattice of finite width. Then < is a strict order on QL. Proof. Let C, D E 6L. If D < C then there is an infinite subset D' of D such that for each d G D' and for each c E C, d 1f c. If also C <D then there is an infinite subset C of C and a map /of C' to D' satisfying c <fi c) for each c E C. As/(c) G D' this is a contradiction. Therefore, < is antisymmetric. Let L have width n. We now show that < is transitive on ßL. To this end let C, D, and be members of QL satisfying C < D and D < E. Let " be an infinite subset of E. Then there is an infinite subset D' of D such that (i) for each d E D' and each e E E',d % e, and (ii) there is a map g of D' to E' satisfying: d < d' if and only if / < g(d').

4 338 I. RIVAL, W. RUCKELSHAUSEN AND B. SANDS Likewise, there is an infinite subset C of C such that (i) for each c E C and each d E D', c "% d, and (ii) there is a map/of C to D' satisfying: c < c' if and only if c <f(c'). Let C = {cx > c2 >... }. Let us suppose that, for some i, c < g(f(c2a+i)). We shall show that {ci+j \/ f(c2n+ _j)\j = 0, 1, 2,..., n} is an (n + l)-element antichain in L, which is impossible since L has width n. Note that, for each j = 0,1,2,...,n, cl+j < ct < g(f(c2n+i)) < g(/(c2n+l_y)). Let 0 < k < k' < n, and set c = c2n+i_k and c' = c2n+j_k,. Observe that c < c'. Then ci+k V /(c) < g(/(c)). It follows that/(c') ^ c,+ft V/(c), for otherwise/(c') < g(f(c)) and so/(c') < /(c) which is a contradiction to c < c'. Therefore, ci+vvf(c')ici+kväc). Since ' < n, c' = c2n+,_a, < ci+k,, so e/+ä, V/(0 < f(ci+j. Thus, c,+jt $ ci+/t, V f(c') and c, +,V/(c)^c,+it,V/(c'). Therefore, ci+k V /(c) is noncomparable to ci+k,\j f(c') for all 0 < & < ' < «, which is impossible. We conclude that,/or a// /, <\*g(/(c2n+,)). (*) Finally, let C" = {c2nk+x\k = 0, 1, 2,... }. For every c E C" and for every e E E', c e (for otherwise,/(c) > e for some c G C" and some e G E', which is impossible since D < E). Now, let h = g f\c~. If c2nk+ x < c2nt.+, then Clnk+i <f(clnk'+\) <g{aclnk+\)) = K^nk+Ù- If c2n*+i < h(c2nk. + x) and c2rafe,+ 1 < c2njt+1 then k' > 1 + k, so c2na:,+ 1 < c2n+2^+1 and c2n^+1 < A(c2 +2rtt+1) which contradicts (*) with /' = 2nk + 1. This completes the proof. (b) Figure 4 If P is not a lattice then the relation < on GP need not be transitive, even if P has width two (see Figure 4(a)). In fact, it is not hard to show that, for an ordered set P, < is a strict order on QP if and only if P contains no subset isomorphic to the

5 HERRINGBONES IN FINITELY GENERATED LATTICES 339 ordered set of Figure 4(a). If P is a lattice with arbitrarily large finite antichains, then again < on QP need not be transitive (see Figure 4(b)). Lemma 4. Let P be an ordered set, let C, D E QP, and let C, D' be infinite subsets of C, D, respectively. If C < D then C < D'. \J Lemma 5. Let L be a lattice affinité width. Then QL has finite length. Proof. We show by induction on n that, for any chain C, < C2 < < Cn of members of GL and for any selection x G C, i = 1, 2,..., n, there is a system ax, a2,..., a of distinct representatives of C C2,..., C such that a < x for each i = 1,2,..., n and {a a2,..., an} is an «-element antichain. As C, < C for each i = 1, 2,..., n 1, there exist chains C[ and maps/ of C/ to C satisfying (Cl) and (C2). From Lemma 4C'X<C'2<_For each i = 1, 2,..., n - 1, set x = inf{jc sup C,'}. According to the induction hypothesis, there is, for each /= 1, 2,...,«1, an element a G C such that a < x' and such that {a,, a2,..., a _,} is an (n l)-element antichain. Finally, let an be an element of C satisfying an < inf{/(a,) i = 1, 2,...,«1}. Then {ax, a2,..., a ) is an n- element antichain. This shows that QL has length at most n 1 if L has width n. U &L may have infinite chains even if L contains no infinite antichains (see Figure 5). Figure 5 Notice that, if L is an infinite lattice of finite width, Lemma 5 guarantees that QL will contain minimal elements. Lemma 6. Let C be a minimal element of QL and let D be an infinite subset of C. Then D is also a minimal element of QL (although noncomparable with C). Proof. By way of contradiction, assume that E E QL is such that E < D. We shall prove that E < C, which is impossible. Let E' Q E and/: E' -* D be as in the definition of <. Let C be an infinite subset of C, define c, = max C, e, = max{e G E'\f(e) < cx), and inductively define, for each i > I, ct = max{c G C' c ^ e _x}, e, = max{<? G E'\f(e) < c,}. (These definitions are possible since C and E' are isomorphic to u>d.) It is now easy to see that the set E" = [ex, e2,... } and the map g: E" -* C defined by g(e ) = c for each i, demonstrate that E < C.

6 340 I. RIVAL, W. RUCKELSHAUSEN AND B. SANDS 3. Proof of the theorem. Let L be an infinite, finitely generated lattice of finite width. We may suppose that QL = 0. By Lemma 5, we may choose a minimal member C = {cx > c2 > c3 >... } of QL. In view of Proposition 2, for each c, G C there are elements x,y in Ac such that z, = x V v, ^ c, and z Ö Ac; that is, z, > Cj for some y > i. By Lemma 1, there is an infinite subset C = {c > c >... } of C and a corresponding infinite chain Z = {z > z, >... } such that, for each j, z( ^ c, and z, > &. From Lemma 6, C is minimal in C ; observing also that Ac = Ac., we see that C may be taken to be C, that is, we may assume that Z = {z, > z2 >... } and z, c,, z, > c1+, for each i. (This tactic of identifying an infinite set with an infinite subset possessing certain desirable properties we call, naturally enough, relabeling^) Next, since z1+, < z, = x V v,> we may assume that, for each i, x $ zi+, and so x ^ c1+2. An application of Lemma 1 followed by a relabeling yields xx > x2>... Let X = {xx, x2,... }. Recall that x G /4C, and thus for ally, jc, ^ ç,. It follows that, if for each c, there were Xj such that x, < c,-, we would have X < C, a contradiction; thus there exists A: such that, for all n > k, x *{; ck. By relabeling, we may assume that for all n, xn ^ c,. It is now evident that the subset C u Z u X is isomorphic to the ordered set illustrated in Figure 6. Selecting the subset consisting of {c \i even} u Z u [x \i even} produces the required herringbone. Figure 6 References 1. H. Bauer and W. Poguntke, Lattices of width three: an example and a theorem. Contributions of General Algebra, (Proc. Klagenfurt Conf., 1978), Verlag Johannes Heyn, Klagenfurt, 1979, pp W. Poguntke and B. Sands, On finitely generated lattices of finite width, Cañad. J. Math, (to appear). 3. H. L. Rolf, The free lattice generated by a set of chains, Pacific J. Math. 8 (1958), R. Wille, Jeder endlich erzeugte, modulare Verband endlicher Weite ist endlich, Mat. Casopis Sloven. Akad. Vied. 24 (1974), Department of Mathematics & Statistics, University of Calgary, Calgary T2N 1N4, Alberta, Canada