A Semilattice on the Set of Permutations on an Infinite Set

A SemiIattice on the Set of Permutations on an Infinite Set

By I. G. ROSENBERG of MontrQal

(Eingegangen am 25. 7. 1972)

1. Introduction

LEHMANN [a] has defined a partial order on the set S of all permutations on a finite set E (see also [5 ] ) . GUILBAUD and ROSENSTIEHL [7] and YANA- GIMOTO and OKAMOTO [6] have characterized i t in terms of partial orderings of E. Their result can be summed up as follows:

Let E = (0, 1, . . ., n - l}. Let S be the set of all n ! permutations of E. For any p E 8 set

( 1 )

and P = (rp 1 p E 81. It is shown in [6] that s E P if and only if s & 0, s partial order, and

(2) (e, z) E s, e < cz < z + (e, a) E s or (0, z) E s,

where o is the natural order on E. Further it is proved that @: p -+ rp is a bijection between S and P. Now P is naturally ordered by the set- theoretical inclusion of relations. This partial order is proved to be a lattice, whose structure can be transfered to S (using @-I).

The purpose of this paper is to extend these interesting results to the set S of all permutations on an infinite set E. In this case the original definition by means of transpositions, used in [a], is not sufficient ; therefore we will start directly from partial orderings. In order to get results com- parable with those obtained in the finite case, we assume that E is well- ordered and the type of E is an initial ordinal w . For simplicity we set E = {a 1 a < w>. The natural order on E will be denoted by 0; however, we will often simply write a b instead of (a , b) E 0. For any p E S we define rp by (1). The set P = {rp I p E S } is characterized by the condition (2) plus an ascending chain condition and a certain restriction on cardinalities of intervals in the complement of s in 0. The set (P, z) is proved to be an upper complete semilattice in which every closed interval is a complete lattice.

rp = {(a, B> E E 2 I a 5 B, P- (a) 5 P-I ( B ) }

192 Rosenberg, A Semilattice on the Set of Permutations on an Infinite Set

For any ordinal jb we set I il 1 = I ( a I a < A} 1 . We set8

i = { ( a , a ) j a E E } . A relation means always a binary relation on E. For any relation s w0 set s+ = s U i, s- = s \ i , S = (0 \ s)+, and 8-1 = ( (a , p) I (B, a) E s}. For any a , B E E, (a , B) E s the interval [a, B], in s is defined by

[a, PI, = (Y I ( a , Y) E 8, (Y, P ) E 4; for a E E set (a) , = ( y j (a , y ) E s-} . The relations is restricted if j (a)J < ! co 1 for every a E E. The relation s is NOETHERIAN if there is no sequence {tv I v < coo} in E such that

(t”, E V + A E 8- for every v < wo (i.e. if i t satisfies the ascending chain condition).

has a similar meaning. An order means a partial order on E. If T is a system of sets, U T denotes the union of the sets in T ; n T

2. The set P

In this section the following characterization of P is given:

Proposition 1. The set P is the set of all orders s & o such that S i s a

The mapping @: p ---f rp i s a bijection between X and P . Corollary 1. Let o = { s G o I s and S orders}. If w E P and s 2 w, then

Proof of the Corollary From s 2 w we get S 5 8. Because ZZ, is a restrict-

Remark 1. It is no difficult to prove that o is the set of all relations s

Corollary 2. ([6] 2 . 2 ) If E is finite, then P = o. Proof of the Corollary: It follows from Corollary I and i E P .

Remark 2. It is an interesting consequence of Corollary 2 that 1 o j = I E 1 ! for finite E. Note that i 6 P for infinite E while always o E P .

Proof of the proposition: Let Q be the set of all relations satisfying the conditions of the proposition.

I ) We will prove first that P & Q . Let p E 8. I n order to prove that rp and Y p are orders it suffices to check their transitivity. Suppose (a , B) € rp

and (b, y ) E r p . Then clearly a 5 B 5 y and p - l ( a ) 5 p-’(B) 5 p - f ( y ) proving ( a , y ) E rT,. Similarly, ( a , /3) E FP and (/I, y ) E F p implies a I: B 5 y

restricted NoETHERIan order.

s E P i f and only i f s E 0.

ed NOETHERian order, so is 8.

satisfying ( 2 ) .

Rosenberg, A Semilattice on the Set of Permutations on an Infinite Set 193

and p-1 (a ) 2 p-1 (8) 2 p - l ( y ) which again gives ( a , y ) E P p . Suppose that Y p is not NOETHERian. Then there exist 5” E E (Y < wo) such that,

p-1 ( t o ) > 21-1 (ti) > * * * . But this is clearly impossible in the well-ordered set E. Finally for any u E E we have

(ahp = (8 I a < 8, P-l(.) > P-l (8)) G (8 I P-I(a) > p-I (B)) .

Since p is a permutation and w an initial ordinal we have

I (8 I p - l ( a ) > P-I(B)) I = I b I P - I b ) > y l I < I w I and this proves that P p is restricted.

relation s on E set

(3)

2 ) We will prove that Q & P and that @: p --+ rp is injective. For any

s” = s u s-1.

In the following lemmas i t will be shown that €or s E Q tho relation s” is a well-ordering of E with order type w. Then it will be proved tha t the (unique) order isomorphism p : ( E , 0) --+ ( E , s”) satisfies rp = s.

Lemma 1. If s E 0, then s is a linear order. P r o o f . The relation s” is antisymmetric: Let a < 8 < co, (a , 8) E s”,

and (p, a ) E s”. Then ( a , 8) € s and (/?, a ) E 3-1, i.e. ( a , 8) E s and ( a , 8) E S, a contradiction. We will prove that $is transitive. Let ( a , 8) E s” and (8, y ) E s”. If‘ both (a , 8) and (8, y ) belong to s or both belong to 3-1, then (N, 8) E s“ from the transitivity of s and S-1. Thus we must consider two cases:

1) Let ( a , 8) E s and (8, y ) E 5-f. Consider first the case a 5 y < 8. Then, by Remark 1, (u , 8) E s and ( y , 8) E S imply ( a , y ) E s; hence ( a , y ) E s & s“, as required. If y < a < 8, by a similar argument we get ( y , u) E 6 and therefore again ( a , y ) E S-1

2 ) The case (u, 8) E S-’ and (8, y ) E s can be reduced to the case 1 by simply interchanging the roles of s and 3-1. Hence s” is an order.

JVe will prove that s” is a linear order. Assume that a < < co, ( a , 8) B s” and (p , a ) CE s” From (u, 8) B s” wet get ( a , 8) 6 s and from (8, a ) B s” we get ( x , B ) B S. Hence we have the contradiction (N, 8) B s U S = 0.

Lemma 2. If s E 0 and S i s NOETHERian, then s” is a well-ordering of E. Proof . Let 5, E E (Y < wo) satisfy (E,+i, 5,) E s“. We shall show first

that there is a sequence YO < v 1 < * - * < wo such that E,, < EVi < - . . Set Mv = { p 1 p > Y and E , > tV) and I = {Y 1 Y < mo}. Assume that all Ny are finite. Then we can choose the sequence {tn I A < mo} as follows: zL is chosen to be the least element of the non-empty set I \ Uv< M,. However, the elements t,, form a descending sequence E,, > EZi > * - -; and this con-

s”.

13 Math. Nachr. 1074, Bd. 60, €1. i -6


tradiction shows that there exists Y < coo with infinite J!Iv. Let v0 be the least ordinal with this property. Applying the same argument to the sequence EQ (p E MVo) we get v1 such that iMuo n Mvt is infinite. Here v l E L%!uo implies f,, < E,, . Continuing in this way we get the required sequence.

However this is impossible for the NOETHERian order S and therefore s” is a well-ordering.

Lemma 3. I f s E o und S is restricted and NOETHERiun, then the order type of s” i s w .

Proof. Let t be the order type of 3. Since both s” and o are well-ordering of E and w is an initial ordinal, we have w 5 t. Assume that w < t. Write E = (5, I a < t> where (t,, tp) E s” e a 5 j3 < t, and let y = 5,. Then (E , , y ) E s“ = s U 5-1 if and only if u < w . Therefore we have

From Eva+, > E v a , (5uj.+i, Eua) E 4 and (3) weget (Eva, Eva+,) E S for all A < (‘10.

I a < m l = G u I (L 7) E s> u it, 1 (7, E A E s> c - { A I1 5 7 ) u ( y ) e -

Because w is an initial ordinal, the first set has cardinality less than I (II 1 . By assumption the second set has also cardinality less than 1 w /. But this yields the contradiction 1 { E , I a < w } I < 1 w I and therefore w = t.

Lemma 4. Let s E Q and p E S. Then p is a n order-isomorphism f r o m ( E ; o) onto ( E ; s”) i f and only i f s = rp.

Proof. Let p be an order isomorphism: ( E ; 0) + ( E ; s”). Let (a , j3) E s. Then (g, j3) E s” and therefore p 1 ( a ) 5 p-1 (j3). This and ( a , j3) E s 5 o proves(a, j3)E rp. Conversely, let ( y , 6 ) E rp . Thenp-1 ( y ) 5 p - i ( 6 ) . Applying the isomorphism p we get ( y , 6 ) E s”. From rp o we get

( y , 6 ) E s” n o = s .

This shows that rp = s.

( p ( a ) , p(j3)) E rp Conversely suppose that s = rp and let a < j3. I f p ( a ) < p(j3) then

&as required. If p ( a ) > p ( / ? ) , then again

(P ( (XI, P (8)) E (0 \ rJ1 s 7p. Consequently p is an order isomorphism (E , 0) 4 (E, s”), as was to be shown.

Proof of the proposition: Let @: X -, P be defined by n -+ r, (n E 8). Let s E Q. It is well-known (e. g. [ 2 ] 111. tj 3, Lemma 2) that there exists precisely one order-isomorphism p : ( E ; 0) + ( E , 8) between the two well- orderings of the same type. Now from this and Lemmas 3 and 4 i t follows

Rosenberg, A Semilattice on the Set of Permutations on an Infinite Set 195

s = 'rp. Thus Q & P. We have proved in part 1 that P Q. Hence Q = P and @ is surjective. The injectivity of @ follows from the uniqueness of p . Thus @ is a bijection and the proof is complete.

3. The semilattice ( P ; g)

The set P is naturally ordered by inclusion. We will now study the poset (P; 5) and prove that it is a complete upper semilattice. We first introduce the following notion. For any relation k the transitive and reflexive hull of k is the relation k* defined as follows: (u, /?j E k* e there exists v < oo and So, . . . , S, such that a = So, B = S, and (d,, 6 , + , ) E k for all 1 = 0, 1, . . . , v - 1. We will need the following:

Lemma 5. Let m G o be a n order. T h e n %* is the least element in (0; !Z) of the set {sE o I s 2 %}.

Proof. We will prove first that %* E 0. For this it suffices to prove that l = o \ 6%" is transitive. Let (cr, /?) E 1, (8, y ) E 1, and assume (u, y } E %*. Then there exist 0 < v < wo and S o , . . ., 6, such that M = do, y = 6, and ( S , , S,+, ) E % for all L = 0, 1, . . ., Y - 1. From a < /? < y it follows that there is 0 5 x < v such that Sx 5 /? < 1 3 ~ + ~ . The assumption (Sx, /?) E +fi leads to the contradiction (a , /?) E fi*. Similarly (j3, Sx+, ) gives the contradiction (/?, y> € %*. Thus (Sx , /?> E m and (j3, a,+,) € m in contradiction to (Sx, 6,+,) E f i . Hence (a , y ) E o \ fi* and fi* E 0. Moreover, %* 2 % and for any s E 0, s 2 rF, we have s = s* 2 fi* proving that %* is the least element of {s E o I s 2 &}.

N o w we can easily prove:

Proposition 2. Let 0 + T P. T h e n sup T exists and is equal to (U T)*.

Proof. Let m = n (S 1 s E T} = (0 \ U T)+. Then by Lemma 5, the order j = +fi* = (U T)* is the least element of {s E o I s 2 U T } . Let t E T . Then 7- = o \ (0 \ m)* & o \ (0 \ m) = m & f ; hence j 2 t and j E P by virtue of Corollary 1; and this proves j = sup T.

The infimum may not exist.

Proposition 3. Let T & P and let k = o \ (1 T. If inf T exists, then inf T = (0 \ k*)+. iWoreover, the following conditions are equivalent:

(i) T has a lower bound, (ii) inf T exists,

(iii) k i s a n NoETHERian relation such that k* i s restricted, (iv) (o \ k*)T E P.

13'

196 Roscnberg, A Semilattice on the Set of Permutations on a n Infinite Set

Proof . Observing that I = n T is an order, by virtue of Lemma 5 we obtain that k* is the least element of { s E o [ s 2 k}. Then (0 \ k*)- is the greatest element of { S E o I s k}. Hence, byproposition 1, (iv) =+ (ii). Clearly (ii) + (i). Next (i) + (iii) follows as in the proof of Corollary 1 . Finally (iii) + (iv) follows from Proposition 1.

Corollary 3. Let T & P and let I TI < Icc) I . Then inf T exists i f und only if k = o \ n T is NOETHERian. Inpurticular, for x , y E P the infinimztrn x A y exists i f and only i f z = 5 U is NOETHERIAN. If x A y exists, it is equal to (0 \ z* )+ .

Proof. Lst k be NOETHERian and let M E E . It suffices to prove that I (a)k* 1 < [ o I. Since I T I < I cu 1, it results from k = U {i 1 t E T } that I ( a ) k l < 1 4 . Now,if I c u I > ~ o ,

I b ) k * I 5 ’ I + I (E lk2 1 + - - *

< I I + 1 I + = Kol(0 I = I w l .

Consider the case I cc) 1 = KO. Then I (j3)k j < is NOETHERian, we get [ ( M ) ~ , < Ho (e.g. [I] Ch. 3, Th. 2 . ) .

for each /3 < oo. Because k

Corollary 4. The poset ( P ; 2 ) i s u complete upper semilattice in which every interval i s a complete luttice.

4. Intervals in (P; 2)

We will fix our attention on lattice-theoretical properties of the intervals in (P ; E) and will prove that, in general, most of the usual properties are not satisfied. We will use the following order isoniorphism. Let mi (i = 1, 2 ) be initial ordinals, 0 < w 1 < cop and let Ei = {a 1 M < cui} with the natural order oi. Let Pi be the corresponding sets. Further let A be an order isomorphism (E i ; oi) ---+ (E2; 02) such that

B = A (El ) = { y I /3’ 5 y < B” } .

v 2 {(y , 6) I y < j?’ 5 6 < m } u {(A, p) I 2 < B” 5 p < OJ}

Let u E Pi and v E P2 satisfy

and A(u) = v n B2 (where A(u) denotes the image of u in A ) . Finally for every s E [zc, 09 set y ( s ) = v U A ( s ) . We will need the following:

Lemma 6. The mapping y is an order isomorphism of th,e intervul [u, 0J n (Pi; E) onto the interval [v, v u A (of)] in (P.; E).

R.osenberg, A Semilattice on the Set of Permutations on an Infinite Set 197

Proof. We wish to show first that w = v UA (s) E P, for every s E [u, oil. Because w 2 v E P,, by Corollary 1 it suffices to prove that w E 02 . The mapping A is an order isomorphism (Ei; oi) -+ (E2; 02); hence s & o1 yields y (s) = w U A (s) S o2 U o2 = o2 and we must only check the transitivity of w and 6. Assume (a , 8) E w and (,8, y ) E w. If both (a, /I) and (8, y ) belong to v or both belong to A ( s ) , then (a , y ) E w = v U A(s ) by the transitivity of v andA(s). Thus consider the case (a , ,8) E A ( s ) and (P, y ) E w. Then a E B and (a, 7) E w & w. The same argument applies in the remaining case (M, /I) E v, (8, y ) E A ( s ) whence w is transitive. We shall prove now that 22, = o2 \ w is transitive. Let ( a , ,8) E 6- and (/I, y ) E 6-. Then clearly (a, 8) B A ( s ) and (B, y ) B A (s f . Suppose ( a , y ) E A ( s ) . Then a, y E B and B convex implies ,8E B. This means (A- l (a ) , A-l(p)) E 3 and ( A - J ( b ) , A - l ( y ) ) E 3 . However, by virtue of s E PI, we get

(A-Ib), A-’(y)) E 3 ,

which contradicts (a , y ) E A(s) . Thus we have proved (a , y ) A(s ) . Next (tc, ,8) E 22, and (/I, y ) E 22, yield ( a , 8) E B and (,8, y ) E B. Using the assumption v E Pa we get (a , y ) E E , whence (a , y ) E [02 \ A (s)] n [02 \ v] = 8- as de- sired. Thus we have proved that w € P2.

It is immediate that y preserves & and therefore i t remains to prove that y is a bijection from [u, oil onto [w, w U A(ol)]. Choose any

z E rw, v UA(OJI.

It is easy to see that A(.) = v n B? z n B2 & A(oi). If we set

y = A - ~ ( z n B2)

we get t c y & oi. Moreover, x n B2 and

(02 n ~ 2 ) \ (2 n ~ 2 ) = (02\ 2) n ~2

are orders on B because x E P,. Further z BJ is NOETEERian and restricted since x has this property. But this means that y E PI, whence y is surjective. For injectivity, making use of the assumption, for any y E [ZL, oi] w e h a v e B z n y ( y ) = [ B 2 n v ] u [ B 2 n A ( y ) ] = A ( u ) UA(y)=A(y) .Thus y (y i) = y ( y,) implies A (yl) = A (y2), this gives yl = y2 ; and the lemma is established.

In (P; &) x < y designates x c y and x c u & y =+ ZL = y.

Proposition 4. Let M + 2 < 03 and

q = { ( a , a + I), (a , M + 2) , ( a + 1, a + 2 ) ) .

v 2 { ( y , @ I y < 5 6 < O3) U ( Y , P ) I y < a + 2 S P < w>, If w E P, q E 1? and


then [w, v U c1-J does not satisfy

(4) and is not relatively complemented.

Proof. Let El = (0, 1, 2 } , Ez = E , @: i + cc + i (i E Ei), 4c = i (iden- tity on Ei). The diagram for the lattice (Pi; &) [6] is given in Figure 1 .

x > x A Y , Y > x A Y =sx V Y > X

8+ Fig. 1.

Since this lattice does not satisfy (4) and is not relatively complemented, it suffices to apply Lemma 6.

Corollary 5. Let v satisfy the conditions of the proposition. Then the lattice [v, 03 i s not semi-modular and i s not relatively complemented. I n par- ticular, if 2 < w < oo then ( P ; s) i s not semi-modular and not relatively complemented.

Corollary 5 suggests the following problem : Problem 1. Determine all intervals in (P; &) that satisfy (4), are semi-

modular, distributive, or Booman algebras. Note that for o = 3 every proper interval is a chain and therefore

distributive. If o = 4 there are six intervals in ( P , 2 ) that are four-element BoOLEan algebras. The intervals that satisfy (4) are all distributive.

I f cr) < oo then S is clearly the complement of s in (P; z). Since the lattice is not relatively complemented we have

Problem 2. Determine all intervals in (P; g ) which are complemented (relatively complemented).

Similarly one could study maximal antichains in (P; 5) (see e.g. [3]).

Remark. In general the partial order induced on S by (P; s) (via 0 - 1 )

is incompatible with the multiplication (i. e. composition of permutations) in S. A counter-example can be found already for w = 3 (take the permutations (0, 1, 2 ) and (0, 1 ) and multiply them by (0, 2) ) .

Rosenberg, A Semilattice on the Set of Permutations on a n Infinite Set 1%

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[I] C. BEROE, Theorie des graphes et ses applications. Paris 1958. [2] G. BIRKHOFF, Lattice Theory. Amer. Math. SOC. Colloq. Publ., vol. 25, revised edition

dnier. Math. Soc., iYew York, N. Y. (1948). [3] D. KLEITMAN, M. EDELBERG and D. LUBELL, Maximal sized antichains in partial

orders. Discrete Mathematics 1, 47-53 (1971). [4] E. L. LEHMANN, Some Concepts of Dependence. Ann. Math. Statist. 37, 1137-1153

(1966). [5] I. R. SAVAGE, Contributions t o the theory of rank order Statistics. Applications of

Lattice Theory. Rev. Internat. Statist. Inst. 32, 52-64 (1964). 161 T. YANAGIMOTO and M. OKAMOTO, Partial ordering8 of Permutations and Monotonicity

of a Rank Correlation Statistic. Annals of the Institute of Statistical Studies 21,

[7] G. TH. GUILBAUD and P. ROSENSTIEHL, Analyse algebrique d'un scrutin, Ordres totaus finis (Travaux du seminaire sur les ordres totaux finis, Aix-en-Provence, juillet 1907), Editions Gauthier-Villars Paris 1971, 71-100.

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A Semilattice on the Set of Permutations on an Infinite Set