On relative ranks of the semigroup of orientation-preserving transformations on infinite chain with restricted range
aa r X i v : . [ m a t h . R A ] F e b On relative ranks of the semigroup oforientation-preserving transformations on infinitechain with restricted range
Ilinka Dimitrova ∗ Faculty of Mathematics and Natural ScienceSouth-West University ”Neofit Rilski”2700 Blagoevgrad, Bulgariae-mail: ilinka [email protected]
J¨org Koppitz
Institute of Mathematics and InformaticsBulgarian Academy of Sciences1113 Sofia, Bulgariae-mail: [email protected] of MathematicsPotsdam UniversityPotsdam, 14469, [email protected]
February 8, 2021
Abstract
Let X be an infinite linearly ordered set and let Y be a nonemptysubset of X . We calculate the relative rank of the semigroup OP ( X, Y )of all orientation-preserving transformations on X with restricted range Y modulo the semigroup O ( X, Y ) of all order-preserving transforma-tions on X with restricted range Y . For Y = X , we characterize therelative generating sets of minimal size. ∗ The author gratefully acknowledges support of DAAD, within the funding programme”Research Stays for University Academics and Scientists, 2019” number 57442043. ey words: transformation semigroups on infinite chain, restricted range,order-preserving transformations, orientation-preserving transformations, rel-ative rank.2010 Mathematics Subject Classification: 20M20 Let S be a semigroup. The rank of S (denoted by rank S ) is defined tobe the minimal number of elements of a generating set of S . The ranks ofvarious well known semigroups have been calculated [10, 11, 13]. For a set A ⊆ S , the relative rank of S modulo A , denoted by rank( S : A ), is theminimal cardinality of a set B ⊆ S such that A ∪ B generates S . It followsimmediately from the definition that rank( S : ∅ ) = rank S , rank( S : S ) = 0,rank( S : A ) = rank( S : h A i ) and rank( S : A ) = 0 if and only if A is agenerating set for S . The relative rank of a semigroup modulo a suitable setwas first introduced by Ruˇskuc [16] in order to describe the generating setsof semigroups with infinite rank.A total (i.e. linear) order < on a set X is said to be dense if, for all x, y ∈ X with x < y , there is a z ∈ X such that x < z < y . Let X be adensely infinite linearly ordered set. We denote by T ( X ) the monoid of allfull transformations of X (under composition). In [14], Howie, Ruˇskuc, andHiggins considered the relative ranks of the monoid T ( X ), where X is aninfinite set, modulo some distinguished subsets of T ( X ). They showed thatrank( T ( X ) : S ( X )) = 2, rank( T ( X ) : E ( X )) = 2 and rank( T ( X ) : J ) = 0,where S ( X ) is the symmetric group on X , E ( X ) is the set of all idempotenttransformations on X , and J is the top J -class of T ( X ), i.e. J = { α ∈T ( X ) : | Xα | = | X |} .A function f : A → X from a subchain A of X into X is said to be order-preserving if x ≤ y implies xf ≤ yf , for all x, y ∈ A . Notice that,given two subchains A and B of X and an order-isomorphism (i.e. an order-preserving bijection) f : A → B , then the inverse function f − : B → A is also an order-isomorphism. In this case, the subchains A and B arecalled order-isomorphic . We denote by O ( X ) the submonoid of T ( X ) of allorder-preserving transformations of X . The relative rank of T ( X ) modulo O ( X ) was considered by Higgins, Mitchell, and Ruˇskuc [12]. They showedthat rank( T ( X ) : O ( X )) = 1, when X is an arbitrary countable chain oran arbitrary well-ordered set, while rank( T ( R ) : O ( R )) is uncountable, byconsidering the usual order of the set R of real numbers. In [2], Dimitrova,2ernandes, and Koppitz studied the relative rank of the semigroup O ( X )modulo the top J -class of O ( X ), for an infinite countable chain X .A generalization of the concept of an order-preserving transformationis the concept of an orientation-preserving transformation, which was in-troduced in 1998 by McAlister [15] and, independently, one year later byCatarino and Higgins [1], but only for finite chains. In [8], Fernandes, Jesus,and Singha introduced the concept of an orientation-preserving transforma-tion on an infinite chain. It generalizes the concept for a finite chain. Definition 1 [8] Let α ∈ T ( X ). We say that α is an orientation-preserving transformation if there exists a non-empty subset X of X such that:(1) α is order-preserving both on X and on X = X \ X ;(2) for all a ∈ X and b ∈ X , we have a < b and aα ≥ bα .In the present paper, we will keep this meaning of X and X for anorientation-preserving transformation if it is clear from the context. Wecall to such a subset X an ideal of α . Notice that X is an order ideal of X . Recall that a subset I of X is called an order ideal of X if x ≤ a implies x ∈ I , for all x ∈ X and all a ∈ I .Denote by OP ( X ) the subset of T ( X ) of all orientation-preserving trans-formations. In [8], Fernandes, Jesus, and Singha proved that OP ( X ) is amonoid. Moreover, they proved that if α ∈ OP ( X ) is a non-constant trans-formation then α admits a unique ideal. Clearly, O ( X ) ⊆ OP ( X ) and wehave α ∈ O ( X ) if and only if α ∈ OP ( X ) and α admits X as an ideal. In[5], Dimitrova and Koppitz determined the relative rank of the semigroup OP ( X ) modulo O ( X ) for some infinite chains X .Given a nonempty subset Y of X , we denote by T ( X, Y ) the subsemi-group { α ∈ T ( X ) : im( α ) ⊆ Y } of T ( X ) of all elements with range (im-age) in Y . The semigroup T ( X, Y ) was introduced and studied in 1975by Symons (see [17]), and it is called semigroup of transformations withrestricted range. Recently, for a finite set X the rank of T ( X, Y ) was com-puted by Fernandes and Sanwong [9].In this paper, for the set X and a nonempty subset Y of X , we considerthe order-preserving and the orientation-preserving counterparts of the semi-group T ( X, Y ), namely the semigroup O ( X, Y ) = T ( X, Y ) ∩ O ( X ) = { α ∈O ( X ) : im( α ) ⊆ Y } and the semigroup OP ( X, Y ) = T ( X, Y ) ∩ OP ( X ) = { α ∈ OP ( X ) : im( α ) ⊆ Y } .For a finite chain X the ranks of the semigroups O ( X, Y ) and OP ( X, Y )were determined by Fernandes, Honyam, Quinteiro, and Singha [6, 7]. In[18], Tinpun and Koppitz studied the relative rank of T ( X, Y ) modulo O ( X, Y ). In [3], Dimitrova, Koppitz, and Tinpun studied rank properties3f the semigroup OP ( X, Y ). In [4], Dimitrova and Koppitz determined therelative rank of T ( X, Y ) modulo OP ( X, Y ) and the relative rank of thesemigroup OP ( X, Y ) modulo O ( X, Y ).In [19], Tinpun and Koppitz considered generating sets of infinite fulltransformation semigroups with restricted range. In this paper, we deter-mine the relative rank of OP ( X, Y ) modulo O ( X, Y ) for a densely infinitelinearly ordered set X and a nonempty proper subset Y of X . We considertwo cases for the set Y : 1) Y is a convex subset of X ; 2) Y is not a convexsubset of X but Y contains an open convex subset of X . For Y = X , wecharacterize the relative generating sets of minimal size.We begin by recalling some notations and definitions that will be used inthe paper. For any transformation α ∈ T ( X, Y ), we denote by ker α , dom α ,and im α the kernel, the domain, and the image (range) of α , respectively.The inverse of α is denoted by α − . For a subset A ⊆ T ( X, Y ), we denoteby h A i the subsemigroup of T ( X, Y ) generated by A . For a subset C ⊆ X ,we denote by α | C the restriction of α to C and by id C the identity mappingon C . A subset C of X is called a convex subset of X if z ∈ X and x < z < y imply z ∈ C , for all x, y ∈ C . Let C, D be convex subsets of X . We willwrite C < D (respectively, C ≤ D ) if c < d (respectively, c ≤ d ) for all c ∈ C and all d ∈ D . If C = { c } or D = { d } , we write c < D , C < d (respectively, c ≤ D , C ≤ d ) instead of { c } < D or C < { d } (respectively, { c } ≤ D or C ≤ { d } ). Note that C < ∅ and ∅ < D .For A < B ⊆ X and a, b ∈ X with a < b , a < B , and A < b , we put(
A, B ) = { x ∈ X : A < x < B } , for a convex subset of X which has no minimum and no maximum;[ a, B ) = { x ∈ X : a ≤ x < B } , for a convex subset of X which has a minimum a but no maximum;( A, b ] = { x ∈ X : A < x ≤ b } , for a convex subset of X which has no minimum but a maximum b ;[ a, b ] = { x ∈ X : a ≤ x ≤ b } , for a convex subset of X which has a minimum a and a maximum b .A convex subset of X without minimum and maximum will be called open convex subset of X . 4otice that if A = ∅ or B = ∅ then we have( A, ∅ ) = { x ∈ X : A < x } , ( ∅ , B ) = { x ∈ X : x < B } , [ a, ∅ ) = { x ∈ X : a ≤ x } , and ( ∅ , b ] = { x ∈ X : x ≤ b } . Proposition 2 [8] Let α ∈ OP ( X ) with ideal X . If X α ∩ X α = ∅ then X α ∩ X α = { c } , for some c ∈ X . Moreover, in this case, X α has aminimum, X α has a maximum, and both of these elements coincide with c . In the present paper, we consider densely infinite linearly ordered setswith the following both properties: (a) if X = X ′ ∪ X ′′ with X ′ < X ′′ and X ′ , X ′′ = ∅ is a decomposition of X then X ′ has a maximum or X ′′ has a minimum; (b) any two open convex subsets of X are order-isomorphic. Lemma 3
Let α ∈ OP ( X, Y ) \ O ( X, Y ) with ideal X and let c ∈ X bethe maximum of X (respectively the minimum of X ) then cα ∈ Y is themaximum of im α (respectively cα ∈ Y is the minimum of im α ). Proof.
Let c ∈ X be the maximum of X and let y ∈ im α ⊆ Y . Thenthere is x ∈ X such that xα = y . If x ∈ X then xα ≤ cα , since c is themaximum of X and α | X is order-preserving. If x ∈ X then xα ≤ zα forall z ∈ X , since X α ≤ X α ( α is orientation-preserving). Therefore, weobtain y ≤ cα , i.e. cα is the maximum of im α .Now, let c ∈ X be the minimum of X . Dually, we obtain that cα is theminimum of im α .From Lemma 3, we obtain: Corollary 4
Let Y ⊂ X has no minimum and no maximum. Then for all α ∈ OP ( X, Y ) \ O ( X, Y ) there exists l ∈ Y with { y ∈ Y : y < l } ∩ im α = ∅ or there exists h ∈ Y with { y ∈ Y : y > h } ∩ im α = ∅ . OP ( X ) modulo O ( X ) of minimal size In [5], Dimitrova and Koppitz determined the relative rank of the semigroup OP ( X ) modulo O ( X ) for certain infinite chains X . It remains a character-ization of the relative generating sets of OP ( X ) modulo O ( X ) of minimal5ize. This will be the main purpose of this section. In particular, this resultwill be used in the next section. Recall, Dimitrova and Koppitz have alreadyproved: Proposition 5 [5]1) Let the set X have a minimum a and a maximum b , and let c ∈ ( a, b )and d ∈ ( c, b ), i.e. a < c < d < b . Then OP ( X ) = hO ( X ) , γ i , where xγ = d, x = axµ , x ∈ ( a, c ) a, x = cxµ , x ∈ ( c, b ) c, x = b with the order-isomorphisms µ : ( a, c ) → ( d, b ) and µ : ( c, b ) → ( a, c ).2) Let the set X have a minimum a but no maximum and let c ∈ ( a, ∅ ).Then OP ( X ) = hO ( X ) , γ i , where xγ = c, x = axν, x ∈ ( a, c ) a, x = cxν − , x ∈ ( c, ∅ )with the order-isomorphism ν : ( a, c ) → ( c, ∅ ).3) Let the set X have no minimum but a maximum b and let c ∈ ( ∅ , b )and l ∈ ( ∅ , c ), i.e. l < c < b . Then OP ( X ) = hO ( X ) , γ i , where xγ = xτ , x ∈ ( ∅ , c ) l, x = cxτ , x ∈ ( c, b ) c, x = b with the order-isomorphisms τ : ( ∅ , c ) → ( c, b ) and τ : ( c, b ) → ( l, c ).In all the cases, rank( OP ( X ) : O ( X )) = 1 and { γ } is a relative gen-erating set of OP ( X ) modulo O ( X ). So, we have to indicate all singletonsets which are relative generating sets of OP ( X ) modulo O ( X ). First, weconsider the case X has a minimum and a maximum.6 roposition 6 Let X has a minimum a and a maximum b , i.e. X = [ a, b ].Let ϕ ∈ OP ( X ) \ O ( X ) with the ideal X . Then hO ( X ) , ϕ i = OP ( X ) ifand only if, for i ∈ { , } , there is a set Y i ⊆ X i which is order-isomorphicto an open convex subset of X such that ϕ | Y i is an order-isomorphism. Proof.
Suppose that for i ∈ { , } , there is a set Y i ⊆ X i which isorder-isomorphic to an open convex subset of X such that ϕ | Y i is an order-isomorphism. Let c, d ∈ X such that a < c < d < b . Then there are sets U ⊆ Y , order-isomorphic to [ a, c ), and U ⊆ Y , order-isomorphic to [ c, b ],with respect to the order-isomorphisms µ and µ , respectively. Then wedefine b ϕ : X → X by x b ϕ = (cid:26) xµ , x ∈ [ a, c ) xµ , x ∈ [ c, b ] . Clearly, b ϕ ∈ O ( X ). Further, there are order-isomorphisms µ : ( a, c ) µ ϕ → ( d, b ) µ : ( c, b ) µ ϕ → ( a, c ) . We define e ϕ : X → X by x e ϕ = a, x ∈ [ a, cµ ϕ ] xµ , x ∈ ( c, b ) µ ϕc, x ∈ [ bµ ϕ, aµ ϕ ) d, x = aµ ϕxµ , x ∈ ( a, c ) µ ϕb, x ∈ [ cµ ϕ, b ] . Note that ( c, b ) µ ϕ < ( a, c ) µ ϕ , since ϕ ∈ OP ( X ) \ O ( X ), ( a, c ) µ ⊆ X ,and ( c, b ) µ ⊆ X . Since µ , µ , µ , µ as well as ϕ | Y i ( i = 1 ,
2) are order-isomorphisms andmin X = a ≤ cµ ϕ < ( c, b ) µ ϕ < [ bµ ϕ, aµ ϕ ) < aµ ϕ < ( a, c ) µ ϕ < cµ ϕ ≤ b = max X,a < ( a, c ) = ( c, b ) µ ϕµ < c < d < ( d, b ) = ( a, c ) µ ϕµ < b, we have that e ϕ ∈ O ( X ).Further, we have( a, c ) b ϕϕ e ϕ = ( a, c ) µ ϕ e ϕ = ( a, c ) µ ϕµ and ( c, b ) b ϕϕ e ϕ = ( c, b ) µ ϕ e ϕ = ( c, b ) µ ϕµ . b ϕϕ e ϕ | ( a,c ) : ( a, c ) → ( d, b )and b ϕϕ e ϕ | ( c,b ) : ( c, b ) → ( a, c )are order-isomorphisms. Moreover, we can calculate that a b ϕϕ e ϕ = d , c b ϕϕ e ϕ = a , and b b ϕϕ e ϕ = c . Then b ϕϕ e ϕ is the transformation γ from Proposition 5case 1), which is used as relative generating set for OP ( X ) modulo O ( X ).Since b ϕϕ e ϕ ∈ hO ( X ) , ϕ i , we can conclude that hO ( X ) , ϕ i = OP ( X ).Suppose now that hO ( X ) , ϕ i = OP ( X ). Assume that there is i ∈ { , } such that there is no subset Z of X i which is order-isomorphic to an openconvex subset of X such that ϕ | Z is an order-isomorphism. Let p, q ∈ X with a < p < q < b . Further, let σ : [ a, p ) → [ q, b ) and let σ : [ p, b ] → [ a, p ]be order-isomorphisms. Then let α : X → X be defined by xα = (cid:26) xσ x ∈ [ a, p ) xσ x ∈ [ p, b ] . Clearly, α ∈ OP ( X ) \ O ( X ) and there are α , . . . , α n ∈ O ( X ) ∪ { ϕ } suchthat α = α · · · α n . Moreover, we put α = id X ∈ O ( X ).Assume that | X i ∩ im( α · · · α j − ) | < ℵ for all j ∈ { , . . . , n } with α j = ϕ . Then we obtain that the ideal of α = α · · · α n is finite or only finiteelements do not belong to the ideal. This is a contradiction, since the ideal of α is the infinite set [ a, p ) and the infinite set [ p, b ] does not belong to the idealof α . Hence there is k ∈ { , . . . , n } such that | X i ∩ im( α · · · α k − ) | = ℵ and α k = ϕ . Then there are x < x ∈ X i ∩ im( α · · · α k − ) such that α · · · α k − | e Z is order-preserving, where e Z = { x ( α · · · α k − ) − : x ∈ ( x , x ) ∩ im( α · · · α k − ) } . Since X i is convex, we have b Z = ( x , x ) ∩ im( α · · · α k − ) ⊆ X i . Moreover,since α = α · · · α n is injective, we obtain that α k | b Z is an order-isomorphism.On the other hand, e Z is a subset of X i and thus e Z is not order-isomorphic toany open convex subset of X . Hence, e Z is not a convex set, since b Z is order-isomorphic to e Z . Thus, there are x < x ∈ e Z and z ∈ ( x , x ) \ e Z . We have x < z < x and z belongs to the ideal of α · · · α k − if and only if e Z be-longs to it. This implies x ( α · · · α k − ) ≤ z ( α · · · α k − ) ≤ x ( α · · · α k − ),since α · · · α k − restricted to its ideal (and its complement, respectively) isorder-preserving. Clearly, z ( α · · · α k − ) ∈ im( α · · · α k − ). Moreover, thereare e x , e x ∈ b Z such that x = e x ( α · · · α k − ) − and x = e x ( α · · · α k − ) − .8his shows that x < e x = x ( α · · · α k − ) < z ( α · · · α k − ) < x ( α · · · α k − ) = e x < x , i.e. z ( α · · · α k − ) ∈ b Z . Thus z ∈ e Z , a contradiction.Now, we consider the case X has a minimum but no maximum. Proposition 7
Let X has a minimum a but no maximum, i.e. X = [ a, ∅ ).Let ϕ ∈ OP ( X ) \ O ( X ) with the ideal X . Then hO ( X ) , ϕ i = OP ( X ) ifand only if, for i ∈ { , } , there is a set Y i ⊆ X i which is order-isomorphicto an open convex subset of X such that ϕ | Y i is an order-isomorphism and( Y ϕ, ∅ ) = ∅ . Proof.
Suppose that for i ∈ { , } , there is a set Y i ⊆ X i which isorder-isomorphic to an open convex subset of X such that ϕ | Y i is an order-isomorphism and ( Y ϕ, ∅ ) = ∅ . Let c ∈ ( a, ∅ ). Then there are sets U ⊆ Y with ( U ϕ, ∅ ) = ∅ , order-isomorphic to [ a, c ), and U ⊆ Y , order-isomorphicto [ c, ∅ ), with respect to the order-isomorphisms µ and µ , respectively.Then we define b ϕ : X → X by x b ϕ = (cid:26) xµ , x ∈ [ a, c ) xµ , x ∈ [ c, ∅ ) . Clearly, b ϕ ∈ O ( X ). Further, there are order-isomorphisms µ : ( a, c ) µ ϕ → ( c, ∅ ) µ : ( c, ∅ ) µ ϕ → ( a, c ) . We define e ϕ : X → X by x e ϕ = a, x ∈ [ a, cµ ϕ ] xµ , x ∈ ( c, ∅ ) µ ϕc, x ∈ (( c, ∅ ) µ ϕ, aµ ϕ ] xµ , x ∈ ( a, c ) µ ϕ. The transformation e ϕ is well defined since ( U ϕ, ∅ ) = ∅ and ( c, ∅ ) µ ϕ < ( a, c ) µ ϕ (since ϕ ∈ OP ( X ) \ O ( X )). Since µ , µ , µ , µ as well as ϕ | Y i ( i = 1 ,
2) are order-isomorphisms andmin X = a ≤ cµ ϕ < ( c, ∅ ) µ ϕ < (( c, ∅ ) µ ϕ, aµ ϕ ] < ( a, c ) µ ϕ = U ϕ,a < ( a, c ) = ( c, ∅ ) µ ϕµ < c < ( c, ∅ ) = ( a, c ) µ ϕµ , we have that e ϕ ∈ O ( X ). 9urther, we have( a, c ) b ϕϕ e ϕ = ( a, c ) µ ϕ e ϕ = ( a, c ) µ ϕµ and ( c, ∅ ) b ϕϕ e ϕ = ( c, ∅ ) µ ϕ e ϕ = ( c, ∅ ) µ ϕµ . This shows that b ϕϕ e ϕ | ( a,c ) : ( a, c ) → ( c, ∅ )and b ϕϕ e ϕ | ( c, ∅ ) : ( c, ∅ ) → ( a, c )are order-isomorphisms. Moreover, we can calculate that a b ϕϕ e ϕ = c and c b ϕϕ e ϕ = a . Then b ϕϕ e ϕ is the transformation γ from Proposition 5 case 2),which is used as relative generating set for OP ( X ) modulo O ( X ). Since b ϕϕ e ϕ ∈ hO ( X ) , ϕ i , we can conclude that hO ( X ) , ϕ i = OP ( X ).Suppose now that hO ( X ) , ϕ i = OP ( X ). Let α ∈ OP ( X ) \ O ( X ) withthe ideal X such that α | X is injective, X α is convex, and ( X α, ∅ ) = ∅ .Then there are α , . . . , α n ∈ O ( X ) ∪ { ϕ } such that α = α · · · α n . Since α / ∈ O ( X ), there is k ∈ { , . . . , n } with α k = ϕ . Without loss of generality,we can assume that k is the greatest r ∈ { , . . . , n } with α r = ϕ andim( α · · · α r − ) ∩ X = ∅ , where α = id X .We put Z = im( α · · · α k − ) ∩ X and observe that Z ( α k · · · α n ) ≥ (im( α · · · α k − ) \ Z )( α k · · · α n ). Let m ∈ Z ( α k · · · α n ). Since ( X α, ∅ ) = ∅ ,we can conclude that ( m, ∅ ) ⊆ im α , since X α is convex and ( X α, ∅ ) = ∅ . Further, let x ∈ ( m, ∅ ). Because of Z ( α k · · · α n ) ≥ (im( α · · · α k − ) \ Z )( α k · · · α n ), we obtain x ( α k · · · α n ) − ⊆ Z ⊆ X . This shows that Y = ( m, ∅ )( α k · · · α n ) − ⊆ X . Because of Y ⊆ im( α · · · α k − ) and since α | X is injective, we can concludethat ( α k · · · α n ) | Y is injective. This implies that α k | Y is injective. On theother hand Y is order-isomorphic to the open convex set ( m, ∅ ). Hence, ϕ | Y is an order-isomorphism.Assume that ( Y ϕ, ∅ ) = ∅ . Then there is u ∈ ( Y ϕ, ∅ ). Then k
Let X has a maximum b but no minimum, i.e. X = ( ∅ , b ].Let ϕ ∈ OP ( X ) \ O ( X ) with the ideal X . Then hO ( X ) , ϕ i = OP ( X ) ifand only if, for i ∈ { , } , there is a set Y i ⊆ X i which is order-isomorphicto an open convex subset of X such that ϕ | Y i is an order-isomorphism and( ∅ , Y ϕ ) = ∅ . OP ( X, Y ) modulo O ( X, Y ) In this section, we determine the relative rank of OP ( X, Y ) modulo O ( X, Y )for certain proper subsets Y of X . Let Y be a proper convex subset of X .Notice that, the convex subsets of Y are also convex subsets of X and thus,any two open convex subsets of Y are order-isomorphic. Proposition 9 If Y has no minimum and no maximum then rank( OP ( X, Y ) : O ( X, Y )) is infinite.
Proof.
Assume that there is a finite set G ⊆ OP ( X, Y ) \ O ( X, Y ) suchthat OP ( X, Y ) = hO ( X, Y ) , G i . From Corollary 4 and since G is a finiteset, it follows that there are a, b ∈ Y such that { y ∈ Y : y < a } ∩ im σ = ∅ or { y ∈ Y : y > b } ∩ im σ = ∅ for all σ ∈ G .Since Y is a proper convex subset of X , we have that there is x ∈ X with x < Y or Y < x . First, we assume that there is x ∈ X with x < Y .Let b < h ∈ Y and let α ∈ OP ( X, Y ) \ ( O ( X, Y ) ∪ G ) with im α = { y ∈ Y : y ≤ h } . Then there are α , . . . , α k ∈ O ( X, Y ) ∪ G such that α = α · · · α k , for a suitable k > α / ∈ O ( X, Y ) ∪ G ). Because11m α ⊆ im α k , we can conclude that α k / ∈ G , i.e. α k ∈ O ( X, Y ). Then for x < Y , it follows that xα k ≤ Y α k . Moreover, since k > α k − ⊆ Y ,we have that im( α k | Y ) ⊇ im α = { y ∈ Y : y ≤ h } . Therefore, we obtain xα k ≤ { y ∈ Y : y ≤ h } . Since Y has no minimum, we obtain xα k / ∈ Y , acontradiction.One can dually consider the case if there is x ∈ X with Y < x . Then for a > l ∈ Y and α ∈ OP ( X, Y ) \ ( O ( X, Y ) ∪ G ) with im α = { y ∈ Y : y ≥ l } ,one obtains also a contradiction.Further, we will consider two cases with finite relative rank. First, let X has no maximum and no minimum. Theorem 10
Let X has neither a maximum nor a minimum. If Y has aminimum or a maximum then rank( OP ( X, Y ) : O ( X, Y )) = 1.
Proof.
Suppose that Y has a maximum but no minimum.Let α ∈ OP ( X, Y ) \ O ( X, Y ) with the ideal X . Then X α ≤ X α . Wehave that X α has a maximum y or X α has a minimum y or we can extend X α < X α to a decomposition Y < Y with X α ⊆ Y and X α ⊆ Y .Then Y has a maximum y or Y has a minimum y . In particular, we have h ∈ Y with X α ≤ h ≤ X α .We consider now the set b X which is the set X equipped with an addi-tional maximum x max .Let β : b X → Y with xβ = xα for all x ∈ X and x max β = h . Clearly, β ∈ OP ( b X, Y ), β | X = α and α = id X β .Let Z be a convex subset of Y such that there is an order-isomorphism µ : Z → Y and there are y − , y + ∈ Y with y − < Z ≤ y + . Since Y has amaximum and µ is a bijection, we have that Z has also a maximum. Letmax Z = z max .Further, we take an order-isomorphism ν : b X → Z . In particular, ν ∈O ( b X, Y ). Note that id X ν ∈ O ( X, Y ). Let δ = ν − βµ − . It is easy to verifythat δ ∈ OP ( Z ) and β = νδµ . From Proposition 8, it follows that there are δ , . . . , δ k ∈ O ( Z ) ∪ { γ } such that δ = δ · · · δ k for a certain transformation γ ∈ OP ( Z ) \ O ( Z ).For i ∈ { , . . . , k } , we can extend δ i to a transformation ˜ δ i ∈ OP ( X, Y ).If δ i ∈ O ( Z ) then we construct ˜ δ i in the following way: x ˜ δ i = y − , x ∈ ( ∅ , Z ) xδ i , x ∈ Zy + , x ∈ ( Z, ∅ ) .
12t is easy to see that ˜ δ i ∈ O ( X, Y ), ˜ δ i | Z = δ i , i.e. δ i = id Z ˜ δ i . Notice that˜ δ i µ ∈ O ( X, Y ).If δ i = γ then we define ˜ δ i = ˜ γ : X → Y by x ˜ γ = (cid:26) xγ, x ∈ Z ( z max ) γ, otherwise.Clearly, ˜ γ ∈ OP ( X, Y ), ˜ γ | Z = γ , i.e. γ = id Z ˜ γ .Finally, we have α = id X β = id X νδµ = id X νδ · · · δ k µ = id X ν id Z ˜ δ id Z ˜ δ · · · id Z ˜ δ k µ. Since id X ν id Z = id X ν , id X ν ˜ δ id Z = id X ν ˜ δ , . . . , id X ν ˜ δ · · · ˜ δ k − id Z =id X ν ˜ δ · · · ˜ δ k − , we have α = id X ν ˜ δ · · · ˜ δ k µ . Moreover, we have id X ν ∈O ( X, Y ), ˜ δ i ∈ O ( X, Y ), if i ∈ { , . . . , k } with δ i ∈ O ( Z ), and ˜ δ i = ˜ γ if i ∈ { , . . . , k } with δ i = γ . Moreover, ˜ δ k µ ∈ O ( X, Y ) if δ k ∈ O ( Z )and ˜ δ k µ ∈ OP ( X, Y ) otherwise. Therefore, α = (id X ν )˜ δ · · · ˜ δ k − (˜ δ k µ ) ∈hO ( X, Y ) , ˜ γ, ˜ γµ i . We define the transformation η : X → Y by xη = y − , x ∈ ( ∅ , Y ) xµ − , x ∈ Yy + , x ∈ ( Y, ∅ ) . Clearly, η ∈ O ( X, Y ) and η | Y = µ − . Therefore, we obtain ˜ γ = ˜ γ id Z =˜ γµµ − = ˜ γµ ( η | Y ) = ˜ γµη ∈ hO ( X, Y ) , ˜ γµ i .This shows that { ˜ γµ } is a relative generating set for OP ( X, Y ) modulo O ( X, Y ) and thus rank( OP ( X, Y ) : O ( X, Y )) = 1.One can dually consider the case Y has a minimum but no maximumand show that rank( OP ( X, Y ) : O ( X, Y )) = 1.Suppose that Y has both a maximum and a minimum. Then we put Z = Y , y − = y min , y + = y max . Clearly µ = id Y , ˜ γµ = ˜ γ and α =(id X ν )˜ δ · · · ˜ δ k ∈ hO ( X, Y ) , ˜ γ i . Therefore, in this case { ˜ γ } is a relative gen-erating set for OP ( X, Y ) modulo O ( X, Y ) and rank( OP ( X, Y ) : O ( X, Y )) =1. Now, we consider the case X has a minimum or a maximum. Theorem 11
Let X has a minimum or a maximum. If Y has a minimumor a maximum then rank( OP ( X, Y ) : O ( X, Y )) = 1.13 roof.
Let X has a minimum a but no maximum. Let e X be a properconvex subset of X with a / ∈ e X and ( e X, ∅ ) = ∅ , which is order-isomorphicto X with the order-isomorphism µ : X → e X . Further, let b X = X \ { a } .Let α ∈ OP ( X, Y ) \ O ( X, Y ). Then we define β : b X → Y µ by xβ = (cid:26) xµ − αµ for x ∈ e Xaαµ otherwise.It is easy to see that β ∈ OP ( b X, Y µ ). We also see that β | e X = µ − αµ . Thisprovides α = µ ( β | e X ) µ − = µβµ − . Note that b X has neither a maximumnor a minimum and that Y µ has a minimum or a maximum. We can applyTheorem 10 and obtain that there are β , . . . , β n ∈ O ( b X, Y µ ) ∪ { γ } for acertain γ ∈ OP ( b X, Y µ ) \ O ( b X, Y µ ) such that β = β · · · β n . Further, wehave µ − µ = id e X . Since im β i ⊆ e X for 1 ≤ i ≤ n , we obtain β = β µ − µβ µ − µ · · · β n − µ − µβ n and thus α = µβµ − = µβ µ − µβ µ − µ · · · β n − µ − µβ n µ − . For i ∈ { , . . . , n } , we have im β i ⊆ Y µ and im( β i µ − ) ⊆ Y µµ − = Y id X = Y . Hence, µβ i µ − ∈ OP ( X, Y ). Further, if β i ∈ O ( b X, Y µ ) then it is easyto verify that µβ i µ − ∈ O ( X, Y ). Therefore, α ∈ hO ( X, Y ) , µγµ − i , i.e. { µγµ − } is a relative generating set of OP ( X, Y ) modulo O ( X, Y ).One can dually consider the case X has a maximum but no minimumand show that rank( OP ( X, Y ) : O ( X, Y )) = 1.Suppose that X has both a minimum a and a maximum b , i.e. X = [ a, b ].Then we take e X to be a proper convex subset of X with a, b / ∈ e X , which isorder-isomorphic to X with the order-isomorphism µ : X → e X . Moreover,we put b X = X \ { a, b } and β ∈ OP ( b X, Y µ ) with xβ = aαµ x < e Xxµ − αµ for x ∈ e Xbαµ x > e X. Using the same method as above, we obtain that α = µβµ − ∈ hO ( X, Y ) , µγµ − i and thus, rank( OP ( X, Y ) : O ( X, Y )) = 1.Finally, we generalize Theorem 11 for the case that Y has a minimum anda maximum. For this let X has a minimum or a maximum, i.e. X = [ a, ∅ )or X = ( ∅ , b ] or X = [ a, b ]. 14 roposition 12 Let X has a minimum or a maximum and let Y be a subsetof X with a minimum and a maximum such that X is order-isomorphic toa subset of Y . Then rank( OP ( X, Y ) : O ( X, Y )) = 1.
Proof.
Let e Y ⊆ Y such that X is order-isomorphic to e Y . By Theorem11, there is γ ∈ OP ( X, e Y ) such that OP ( X, e Y ) = hO ( X, e Y ) , γ i . Since OP ( X, e Y ) ⊆ OP ( X, Y ), we have γ ∈ OP ( X, Y ). We will show that OP ( X, Y ) ⊆ hO ( X, Y ) , γ i . For this let α ∈ OP ( X, Y ) with the ideal X .Then there are order-isomorphisms ϕ : X → e Y and ϕ : X → e Y such that X ϕ < X ϕ . We define a transformation µ : X → e Y by xµ = (cid:26) xϕ , x ∈ X xϕ , x ∈ X . Clearly, µ ∈ OP ( X, e Y ) with the same ideal X as α . If X has a minimumthen we put m = (min X ) α . Otherwise, we put m = (max X ) α . Thus, X α ≤ m ≤ X α . Now, we define a transformation e µ : X → Y by x e µ = y min , x < X ϕ xϕ − α, x ∈ X ϕ m, X ϕ < x < X ϕ xϕ − α, x ∈ X ϕ y max , x > X ϕ , where y min and y max is the minimum and the maximum, respectively, in Y .Clearly, e µ ∈ O ( X, Y ) since ϕ − as well as ϕ − are order-isomorphisms and y min ≤ X α ≤ m ≤ X α ≤ y max . Since µ ∈ OP ( X, e Y ) = hO ( X, e Y ) , γ i ⊆hO ( X, Y ) , γ i , we obtain µ e µ ∈ hO ( X, Y ) , γ i . It remains to show that α = µ e µ .For this let x ∈ X . Then xµ e µ = xϕ e µ = xϕ ϕ − α = x id X α = xα .Analogously, we obtain xµ e µ = xα , whenever x ∈ X .Therefore, α ∈ hO ( X, Y ) , γ i , i.e. { γ } is a relative generating set of OP ( X, Y ) modulo O ( X, Y ). Since O ( X, Y ) is a proper subsemigroup of OP ( X, Y ) we have rank( OP ( X, Y ) : O ( X, Y )) = 1.