Glasgow Math. J.45(2003) 557–566.2003 Glasgow Mathematical Journal Trust. DOI: 10.1017/S0017089503001460. Printed in the United Kingdom
GENERATING THE FULL TRANSFORMATION SEMIGROUP
USING ORDER PRESERVING MAPPINGS
P. M. HIGGINS
Department of Mathematics, University of Essex, Wivenhoe Park, Colchester, CO4 3SQ, United Kingdom e-mail: [email protected]
J. D. MITCHELL and N. RU ˇSKUC
Mathematical Institute, North Haugh, St Andrews, Fife, KY16 9SS, United Kingdom e-mail: [email protected]
(Received 8 October, 2002; accepted 2 February, 2003)
Abstract. For a linearly ordered set X we consider the relative rank of the semigroup of all order preserving mappingsOXonXmodulo the full transformation semigroupTX. In other words, we ask what is the smallest cardinality of a setAof mappings such thatOX∪A =TX. WhenXis countably infinite or well-ordered (of arbitrary cardinality) we show that this number is one, while whenX=⺢(the set of real numbers) it is uncountable.
2000Mathematics Subject Classification.20M20, 06A05.
1. Introduction. For a semigroup S, the ‘classical’ idea of rank is concerned with finding minimum size generating sets forS; see [6] or [10]. When working with a finitely generated semigroupS determining the rank of S, denoted rank(S), is a natural consideration. However, for an uncountable semigroupSthe rank ofSis|S|, and so the classical notion of rank provides us with no information. We introduce a different rank property which allows us to ‘measure’, from a certain perspective, a given semigroup with respect to some distinguished subsemigroups. For a semigroup
S, ifA⊆Sthen we call the minimum cardinality of a setBsuch that
A∪B =S,
therelative rank of S modulo A. Alternatively, we may refer to this cardinality as the
relative rank of A in S; which we denote by rank(S:A). This subject has been studied in the context of groups in [2] and [14]. In these papers, so called large subgroups of the symmetric groupSX, over an infinite setX, were considered. The relative rank of the full transformation semigroupTX, over an infinite setX, modulo various standard subsemigroups was first considered in [7]. It was shown in [11] that the relative rank ofTX moduloSX is two. In the same paper it was shown that the relative rank of the set of all idempotent maps onX in TX is, also, two. Sierpi ´nski [15] showed that any countable set of maps from X to X is contained in a 2-generated subsemigroup of
analogues of these results in the semigroup of all binary relations, the semigroup of all partial maps, and the symmetric inverse semigroup were proven in [8].
In this paper we consider the relative rank of the full transformation semigroup
TX, where (X,≤) is an infinite linearly ordered set, modulo the subsemigroupOXof all order preserving maps onX. Recall, that a mapα∈TXisorder preservingif
x≤yimpliesxα≤yα,
for allx,y∈X.
For finiteX (of sizen, say) the semigroup OX has been studied extensively. Its order is (2nn−−11), its rank, in the classical sense, isn, it is idempotent generated and its idempotent rank is 2n−2 (see [4] and [9]). Furthermore, it is easy to see that
rank(TX :OX)=2.
Indeed, it is well-known that it is possible to generateTX using elements ofSX and an arbitrary map α with the property that |im(α)| =n−1. If we choose α∈OX then the result follows from the observation thatOX∩SX= {1X}and the fact that
rank(SX)=2.
In [8] the case whereX=⺞(with the usual ordering) was considered, and it was shown that
rank(T⺞:O⺞)=1.
In this paper we build on the above example and show that
rank(TX:OX)=1,
whenXis an arbitrary countable linearly ordered set or an arbitrary well-ordered set (of any cardinality) but that rank(TX :OX) can be uncountable for some (uncountable, non-well-ordered) linearly ordered setsX.
2. Countable linearly ordered sets. In this section we prove the following result.
THEOREM2.1.Let X be a countable linearly ordered set. The relative rank ofTX moduloOXis one.
For the remainder of this section X will be a fixed countably infinite linearly ordered set. Forx,y∈Xwithx<ywe define
[x,y]= {z∈X:x≤z≤y}, (x,y)= {z∈X:x<z<y},
(x,y]= {z∈X:x<z≤y}, [x,y)= {z∈X:x≤z<y}.
We call these setsintervals,delimitedbyxandy. An elementd ∈X is calleddiscrete
if there existx,y∈Xwithd ∈(x,y) such that (x,d)=(d,y)= ∅. We call an element
the current definitions. To remedy this, ifx0∈Xis the smallest element ofXand there
existsy∈Xsuch thaty=x0 and (x0,y)= ∅then we shall callx0 discrete; otherwise
we callx0left isolated. Analogously, ifXhas a largest element then it is either discrete
or right isolated. Finally, an elementt∈X(which is neither the largest nor the smallest element ofX) is called alimit pointif (x,t)= ∅and (t,y)= ∅for everyx<tand for everyy>t.
We now start a sequence of lemmas, leading to the proof of Theorem 2.1. Throughout we take⺞= {1,2, . . .}.
LEMMA2.2. Let X be a countable linearly ordered set consisting entirely of limit
points, and let λbe any function from X to ⺞. Then there exists an order preserving injectionαfrom X to X such that|X\im(α)| = |X|(= |im(α)|)satisfying xαλ≥xλ, for all x∈X .
Proof.We start by finding an intervalI⊆X such that for everyx,y,z∈I, with
x<y, there existst∈(x,y) such thattλ≥zλ. LetJ be an arbitrary interval. One of the following alternatives holds:
(i) for every subinterval ofJ delimited byx,y∈J, withx<y, and for every
n∈⺞there existsz∈(x,y) such thatzλ >n; or
(ii) there exists a subintervalK⊆J and there existsn∈⺞such thatxλ≤n, for eachx∈K.
If condition (i) holds then the intervalJ has the required property and we letI =J. Assume that condition (ii) holds. If for allx,y∈K, withx<y, there existsz∈(x,y) such thatzλ=n, thenKsatisfies the necessary condition and we letI=K. Otherwise, there exists a subintervalK1⊆Ksuch thatxλ≤n−1, for allx∈K1. We considerK1
in the same way as we have just consideredK, so that if for allx,y∈K1, withx<y,
there existsz∈(x,y) such thatzλ=n−1 then we letI=K1. Otherwise, there exists
a subintervalK2 ⊆K1such thatxλ≤n−2, for allx∈K2. We repeat this process to
give the sequence of non-empty subintervals:
K=K0⊇K1 ⊇K2⊇K3⊇ · · ·,
wherexλ≤n−ifor everyx∈Ki. Note that this process always terminates withi= n−1 at latest.
Let (a,b)⊆Ibe an arbitrary interval. Note that sinceaandbare both limit points we have|(a,b)| = |X|. We defineα∈OX inductively as follows. First we defineαon the pointsaandb, so thatbα=bandaα=c, wherec∈(a,b) withcλ≥aλ. Such a pointcexists from the defining property ofI.
Next, we enumerate the elements of [a,b]:
b=e0,a=e1,e2,e3, . . . .
Fork∈⺞, our inductive hypothesis is that the elementse0α,e1α,e2α, . . . ,ekα∈[c,b]
are defined so thatα is injective, order preserving and satisfieseiαλ≥eiλ for every
i∈ {0,1, . . . ,k}. To defineek+1α, we find the largestei∈[a,b] withei<ek+1, where i≤k, and the smallestej∈[a,b] withek+1<ej, wherej≤k. Sincei,j≤kthe elements eiα andejα are already defined andeiα <ejα. From the definition ofI there exists
y∈(eiα,ejα) such thatyλ≥ek+1λand so we defineek+1α=y. Finally, forx<aand
forx>b we letxα=x. Note that sinceais a limit point and [a,c)∩im(α)= ∅we
We use the above lemma to prove a more general result for linearly ordered countably infinite sets which contain elements that are not limit points.
LEMMA2.3.Let X be a countable linearly ordered set. Then there exists an order
preserving injectionαfrom X to X such that|X\im(α)| = |X|(= |im(α)|).
Proof.There are two cases to consider.
Case 1. There is an infinite sequence of consecutive discrete points in X .It is clear that any such sequence is either strictly increasing or strictly decreasing. Without loss of generality, we assume that{y1<y2<· · ·} ⊆X is an increasing sequence of
consecutive discrete points. DefineαfromXtoXby
xα=
y2i x=yi (i∈⺞) x x∈ {y1,y2, . . .}.
The mapαis an order preserving injection andX\im(α)= {y1,y3, . . .}.
Case 2. No infinite sequence of consecutive discrete points exists.Without loss of generality assume thatXhas no smallest or largest point. (Indeed, ifXhad a smallest point, say, then it would start asx1<x2<· · ·<xk<l, where each xi is a discrete
point andlis a left isolated point. But then we can considerX\{x1, . . . ,xk,l}.)
Consider an arbitrary right isolated pointr. Then there existsx1such thatr<x1
and (r,x1)= ∅. Clearly, x1 is either a discrete point or a left isolated point. If x1
is discrete then there exists x2 >x1 such that (x1,x2)= ∅. By assumption we can
continue this for only finitely many steps to obtain a finite sequencer,x1,x2, . . . ,xk,l
where eachxiis a discrete point andlis left isolated. Thus for every right isolatedr∈X
there exists a corresponding left isolatedl∈Xsuch that the interval (r,l) is finite. Let ρbe the equivalence relation with equivalence classes{t}, wheretis a limit point, and
{r,x1,x2, . . . ,xk,l}wherexiis a discrete point for eachi∈ {1,2, . . . ,k},lis left isolated
andris right isolated. LetX=X/ρ, the quotient ofXbyρ. The order onXinduces a (linear) order onX:
x/ρ ≤y/ρif and only ifx/ρ =y/ρorx<yfor allx∈x/ρ, y∈y/ρ.
We claim that every pointx/ρ ofX is a limit point. We have two cases to consider, whenx/ρ= {x}and whenx/ρ = {r,x1,x2, . . . ,xk,l}, for somek≥0.
In the first case, we have thatxis a limit point inX. Lety∈Xwithy/ρ <x/ρ. We show that (y/ρ,x/ρ)= ∅. From the definition of the order onX we havey<x
and so|(y,x)| = |X|. It follows that there existsz∈(y,x) such thatz=xandz∈y/ρ, sincey/ρ is finite. Sincex/ρ is a singleton this implies thatz/ρ∈(y/ρ,x/ρ) and so
(y/ρ,x/ρ)= ∅, as required. An analogous argument shows that for anyz∈X with
x/ρ <z/ρwe have (x/ρ,z/ρ)= ∅. It follows thatx/ρis a limit point.
In the second case, note that for anyy,z∈Xsuch thaty/ρ <x/ρ <z/ρwe have
y<r<l<z. Sinceris a right isolated point it follows that (y,r) is infinite. Again, since
y/ρandr/ρare finite there existst∈(y,r) such thatt∈y/ρandt∈r/ρ. This implies thatt/ρ∈(y/ρ,r/ρ)=(y/ρ,x/ρ) and so (y/ρ,x/ρ)= ∅, as required. An analogous argument shows that (x/ρ,z/ρ)= ∅. It follows thatx/ρis a limit point.
We now label each element ofXaccording to the size of its class, so that we may apply Lemma 2.2. More precisely, we defineλ:X→⺞by
By Lemma 2.2 there exists an order preserving bijection ¯αfromXtoXsatisfying
(x/ρ) ¯αλ≥(x/ρ)λ,
such thatX\im( ¯α) is infinite. We shall now ‘lift’ the function ¯αto a functionα:X→X
as follows:
xα=
y ifxis a limit point and (x/ρ) ¯α=y/ρ= {y}whereyis a limit point,
r ifxis a limit point and (x/ρ) ¯α= {r,x1,x2, . . . ,xk,l}, xi ifx=xiin{r=x0,x1, . . . ,xk,xk+1=l} ∈Xand
(x/ρ) ¯α= {r=x0,x1, . . . ,xs,xs+1=l}fors≥k.
In the final case, since ¯αsatisfies (x/ρ) ¯αλ= |(x/ρ) ¯α| ≥ |x/ρ| =(x/ρ)λ, it is clear that,
under ¯α, the image ofx/ρ = {r=x0,x1, . . . ,xk,xk+1=l}must be a set with at least k+2 elements. Note that,xα∈(x/ρ) ¯αfor everyx∈X.
For arbitraryx,y∈X, withx<y, we show thatxα <yαand henceαis order preserving and injective. There are two cases to consider. Firstly, ifx/ρ=y/ρ then (x/ρ) ¯α <(y/ρ) ¯α, since ¯αis order preserving and injective. It follows from the definition of the order on X that z<t, for all z∈(x/ρ) ¯α and for all t∈(y/ρ) ¯α, and hence
xα <yα. Secondly, if
x/ρ = {r=x0,x1, . . . ,xk,xk+1=l} =y/ρ
thenx=xiandy=xjfori<j. By definition we have
(x/ρ) ¯α= {r=x0,x1, . . . ,xs,xs+1=l} =(y/ρ) ¯α,
wheres≥kandxα=xi<xj=yα, as required. Note that im(α)⊆im( ¯α), and since
X\im( ¯α) is infinite, it follows thatX\im(α) is infinite.
The next two lemmas allow us to use methods similar to those in the proof of [8, Example 1.6] to encode an arbitrary map into an order preserving map.
LEMMA 2.4.Let Y be a countably infinite linearly ordered set. Then there exists
Z⊆Y such that either Z∼=⺪+or Z∼=⺪−
Proof.There are two cases to consider.
Case 1. All the points in Y are discrete points. We construct Z as follows. Let
z1 ∈Ybe arbitrary. Then at least one of the sets{y∈Y:y>z1}or{y∈Y:y<z1}
is infinite. Without loss of generality we assume that
|{y∈Y:y>z1}| = |Y|.
We may choose z2,z3, . . . so that (z1,z2)= ∅, (z2,z3)= ∅, etc. Then Z=
{z1,z2,z3, . . .} ∼=⺪+.
Case 2. There exists a left isolated point, a right isolated point or a limit point in Y .Without loss of generality, we assume that there is a right isolated pointr∈Y. Let
z1 <rbe arbitrary, sinceris right isolated we have (z1,r)= ∅. Hence we may choose z2 ∈(z1,r). Continue to choosez3∈(z2,r),z4∈(z3,r), etc. ThenZ= {z1,z2,z3, . . .} ∼=
LEMMA2.5.Let X be a countable linearly ordered infinite set. Let Z⊆X be such that Z∼=⺪+or Z∼=⺪−, and letαbe an order preserving map from Z to Z. Then there existsβ ∈OX such thatβZ=α.
Proof.We assume, without loss of generality, thatZ= {z1<z2<· · ·} ∼=⺪+ and
letαbe an order preserving map fromZtoZ. We defineβ ∈TXby
xβ=
x ifx<z1, ziα ifx∈[zi,zi+1),
x ifx>zi, for everyi∈⺞.
It is easy to verify thatβ is order preserving andβZ =α.
We are now in a position to prove the main result of this section.
Proof of Theorem2.1. LetY ⊆X such that|Y| = |X\Y| = |X|and let β be an order preserving bijection fromXtoX\Y; these exist by Lemma 2.3. Assume, without loss of generality, that there existsZ⊆Ysuch thatZ∼=⺪+and letZ= {z1<z2<· · ·},
as described in Lemma 2.4. SinceX\Y andZhave the same cardinality there exists a bijectionfromX\Y toZ. Note thatβis a bijection fromX toZ. Letδ be any mapping fromZtoXsuch that
zpj kδ =zk
−1β−1,
wherepkis thekth prime andj∈⺞arbitrary. Let∈TXbe any mapping such that
x=
x ifx∈X\Y xδ ifx∈Z
arbitrary ifx∈Y\Z
Letα∈TX be arbitrary. We show thatαcan be generated usingand elements of OX. We define γ from Z to Z inductively. First we define γ on z1, so that if z1−1β−1αβ=zkthen z1γ =zpk. For t>1 we assume thatγ is defined and order preserving onz1, . . . ,zt−1. To defineztγ we first let
M=max{i:zi∈ {z1, . . . ,zt−1}γ}.
Sinceβis a bijection fromXtoZthere existsx∈Xsuch that
zt−1β−1=x
and there existszs∈Zsuch that (xα)β=zs. We choosej∈⺞such thatpjs>Mand
we defineztγ =zpjs. Our choice ofjensures thatγ is order preserving. Letη∈TX be an extension ofγto an element ofOX, as described in Lemma 2.5.
We claim that
α=βη.
Letx∈X be arbitrary. Sinceβis a bijection (from X to Z) there exists a unique elementzt ∈Zsuch thatxβ=xβ=zt. Analogously, there exists a unique element zs∈Zsuch thatxαβ=xαβ=zs. Hence
xβη=xβη=ztη=ztγ =zpj
s=zpjsδ=zs
−1β−1=xα.
3. Well-ordered sets. In this section we extend the result of the previous section to well-ordered sets of arbitrary cardinality. Recall that an ordered set (X,≤) is well-ordered if every subset ofX contains a least element. We start by introducing some standard results concerning well-ordered sets which we shall use later. For more details see [5], [12] or [13].
For an arbitraryx∈Xwe call the sets(x)= {y∈X:y<x}theinitial segmentof
x. It is well-known that for any two well-ordered setsXandYeitherXis isomorphic to
Y,Xis isomorphic to an initial segment ofYorYis isomorphic to an initial segment ofX; see for example [12, Theorem 1]. This induces a natural (well) ordering on the class of all well-ordered sets, so that
X≤Yif and only ifX∼=YorXis isomorphic to an initial segment ofY. (1)
A natural reformulation of this result relates a well-ordered set to its subsets.
PROPOSITION3.6.Each subset of a well-ordered set X is either isomorphic to X or to
an initial segment of X .
Another useful and natural consequence of the order on the class of all well-ordered sets is:
PROPOSITION3.7.No well-ordered set is isomorphic to an initial segment of itself.
For a proof see [12, Lemma 2.2].
In light of the previous section a natural question to ask is whether, or not, there exists an uncountable setXfor which the relative rank ofTXmoduloOXis countable? The next result answers this question in the affirmative.
LEMMA3.8.Let X be an arbitrary infinite set and letdenote the least well-ordered
set of cardinality|X|. Then the relative rank ofTmoduloOis one.
Proof.Letx(x∈) denote subsets ofsuch that|x| = ||andx∩y= ∅
wheneverx=y. Since eachxhas cardinality|X|andis the smallest well-ordered
set of cardinality|X|, it follows from Proposition 3.6 and Proposition 3.7 thatx∼=,
for allx∈. Fix a mappingµ∈Tsuch that
yµ=y (y∈).
For an arbitrary α∈T, we define a map β ∈O by transfinite induction as follows. If xα=y then we define xβ =z, wherez∈y andz>tβ for every t<x.
Such an elementzexists sincey(∼=) is not isomorphic to an initial segment of,
by Proposition 3.7. For an arbitraryx∈, ifxα=ythen we have
xβµ=zµ=y=xα,
and soα∈ O, µ. It follows thatT= O, µand in particular rank(T:O)=1,
as required.
We use this lemma to prove the main result of this section:
THEOREM3.9.Let X be an arbitrary well-ordered set. The relative rank ofTXmodulo OXis one.
the latter case the result follows by Lemma 3.8. In the former case, letY denote the initial segment ofXisomorphic toT and letZ=X\Y. By the same argument as in the proof of Lemma 3.8, we may find pairwise disjoint setsY1,Y2,Y3⊆Ysuch that Y=Y1∪Y2∪Y3 andYi∼=Y, for eachi. By Lemma 3.8 there exists a map∈TY2 such thatOY2, =TY2. We define a map σ :TY2−→TX such that forx∈X and ρ∈TY2
(x)(ρσ)=
x x∈Z,
yρ x∈Y, wherey=min{z∈Y2:z≥x}.
In the second case, note that such an elementyalways exists sinceY2 is well-ordered
and is not a subset of an initial segment ofY. It is easy to see thatOY2σ ⊆OX. Since
Y∼=Y1 andy≤z, for everyy∈Y and for everyz∈Z, we may define an injection
β∈OXfromXtoY1∪Zsuch thatYβ=Y1andZβ=Z. SinceY1∪Zhas the same
cardinality asY2we may find a bijectionδfromY1∪ZtoY2. We let ¯µbe any order
preserving bijection fromY2toY3and defineµ∈TXby
xµ=
x x>y, for ally∈Y2; yµ¯ y=min{z∈Y2:z≥x}.
In fact,µis order preserving. In order to see this, it is enough to note that{x:x> yfor ally∈Y2} =ZsinceY2, being isomorphic to Y, is not contained in an initial
segment ofY. From the definition it is obvious thatµrestricted to eitherZorY=X\Z
is order preserving. In addition, Yµ=Y2µ=Y3⊆Y, Zµ=Z and y<z for all y∈Y,z∈Z.
Next, we define a mapγ :Y3 −→Xby
yγ =yµ¯−1δ−1β−1.
Note that the domains of the mapsδ, andγ are disjoint and that their union isX
itself. Hence we may define a mapη∈TXby
xη=
xδ x∈Y1∪Z x x∈Y2 xγ x∈Y3.
We claim thatηtogether withOXgeneratesTX. Letα∈TXbe arbitrary. First note that for anyθ∈TY2 there existsρ ∈ OY2σ, η ⊆ OX, η(⊆TX) such that ρY2=θ. This follows from (ζ σ)Y2=ζ for allζ ∈TY2,ηY2=andOY2, =TY2.
In particular, sinceβδis a bijection fromXtoY2, we may find a mapρ∈ OX, η
such thatρY2=δ−
1β−1αβδ. Then, for an arbitraryx∈X, withxβδ=y∈Y
2we have
xβηρµη=(xβδ)ρµη=yρµη=yδ−1β−1αβδµη=(xα)βδµη
=(xαβδµ)γ =xαβδµµ¯−1δ−1β−1=xα,
and soα∈ OX, ηandTX = OX, η.
LEMMA 4.10. Let X be an infinite linearly ordered set such that there exists a subset Y ⊆X with |Y| = |X| and where any order preserving map from Y to Y can be extended to an order preserving map from X to X . Thenrank(TY :OY)≤2implies
rank(TX:OX)≤2.
Proof.By assumption, there exist, δ∈TY such thatOY, , δ =TY, and for everyα∈OY there existsη∈OXsuch thatηY =α. Let, δ∈TX be any mappings
such thatY= andδY =δ. Letβ :X →Y be any bijection and letγ ∈TX be
arbitrary. We show that it is possible to generate γ using elements ofOX and four other mappings. SinceOY, , δ =TY we see that for any mapα∈TY there exists µ∈ OX, , δsuch thatµY =α. In particular, there existsµ∈ OX, , δsuch that
µY=β−1γβ.
Letνbe any extension ofβ−1to an element ofTX. For an arbitraryx∈Xifxβ =y
then
xβµν=yµν =yβ−1γβν =yβ−1γ =xγ.
We have shown thatγ ∈ OX, δ, , β, νand soTX= OX, δ, , β, ν. It follows from
[8, Corollary 1.2] that rank(TX:OX)≤2.
COROLLARY4.11.If X is an arbitrary linearly ordered set, such that there exists a
well-ordered subset Y ⊆X , with|Y| = |X|, thenrank(TX :OX)≤2.
Proof.Letαbe an order preserving map fromY toY. Thenα∈TXdefined by
xα=
x x>yfor ally∈Y yα y=min{z∈Y:z≥x},
is an order preserving map fromXtoX. The result follows by Theorem 3.9 and Lemma
4.10.
Having found the relative rank of T⺞ modulo O⺞ and the relative rank ofT⺡ moduloO⺡a natural question to ask is: what is the relative rank ofT⺢moduloO⺢?
EXAMPLE4.12. The relative rank ofT⺢moduloO⺢is uncountable. We show this by proving that the cardinality of the semigroup of all order preserving mappings on the reals⺢is 2ℵ0<22ℵ0 = |T⺢|. For an arbitraryα∈O⺢we show thatαis discontinuous at only countably many points in⺢. LetD= {x:αis discontinuous atx}. Forx∈D, letax=supt<x{tα}and letbx=inft>x{tα}. Next, defineβ:D−→G, whereGis the
family of all open subsets of⺢, by:
xβ =(ax,bx).
Observe that the family{xβ:x∈D}consists of non-empty pairwise disjoint open sets, hence
|{xβ:x∈D}| ≤ ℵ0.
The mapβis injective and so
Next, we show thatαis almost determined by its rational points. Forx∈⺢\⺡
definesx=supq∈⺡{qα:q<x}andtx=infq∈⺡{qα:q>x}and observe that xα∈[sx,tx],
sinceαis an order preserving map. Sinceαis discontinuous at only countably many points there are only countably many intervals [sx,tx] which are not singletons. It follows that there are only 2ℵ0maps inO
⺢.
The cardinality of the setPof all order preserving mappings⺡→⺢is 2ℵ0. Since every elementαofO⺢is almost determined byα⺡∈Pit follows that|O⺢| =2ℵ0too.
We conclude the paper with the following two questions:
OPENPROBLEM4.13.Is it true that ifrank(TX:OX)≤2for a linearly ordered set X then there exists Y⊆X such that|X| = |Y|and Y , or YR(the set Y with the order reversed), is well-ordered?
OPEN PROBLEM4.14. Does there exist an infinite linearly ordered set X such that rank(TX :OX)=2?
ACKNOWLEDGEMENTS. The authors are grateful to an anonymous referee for his/her comments. The second and third authors would also like to thank Dr Lars Olsen for helpful discussions. The third author acknowledges partial support from INTAS 99/1224.
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