ENDOMORPHISMS OF PREORDERS, GRAPHS, AND TOLERANCES
J. D. MITCHELL, M. MORAYNE, Y. P ´ERESSE, AND M. QUICK
ABSTRACT. LetΩΩbe the semigroup of all mappings of a countably infinite set
Ω. IfUandV are subsemigroups ofΩΩ, then we writeU ≈V if there exists a
finite subsetF ofΩΩsuch that the subsemigroup generated byUandF equals
that generated byV andF. The relative rank ofUinΩΩis the least cardinality
of a subsetAofΩΩsuch that the union ofUandAgeneratesΩΩ. In this paper
we study the notions of relative rank and the equivalence≈for semigroups of endomorphisms of binary relations onΩ.
The semigroups of endomorphisms of preorders, bipartite graphs, and toler-ances onΩare shown to lie in two equivalence classes under≈. Moreover such semigroups have relative rank0,1,2, ordinΩΩwheredis the minimum
cardinal-ity of a dominating family forNN. We give examples of preorders, bipartite graphs,
and tolerances onΩwhere the relative ranks of their endomorphism semigroups inΩΩare0,1,2, andd.
We show that the endomorphism semigroups of graphs, in general, fall into at least four classes under≈and that there exist graphs where the relative rank of the endomorphism semigroup is2ℵ0.
1. INTRODUCTION
1.1. Background and Preliminaries. Bergman and Shelah [2] introduced the fol-lowing preorder (i.e. reflexive and transitive binary relation) on the subsets of the symmetric groupSym(Ω)on a countably infinite setΩ. IfGandH are subsets of Sym(Ω), thenG4Hif there exists a finite subsetFofSym(Ω)such thatGis con-tained in the subgroup generated byH∪F. Galvin [6] proved that every countable set of permutations onΩis contained in a2-generated subgroup ofSym(Ω). Hence if there exists a countable subsetF such thatGis contained in the subgroup gen-erated byH∪F, thenG4H. The preorder4gives rise to an equivalence relation ≈on the subsets ofSym(Ω)defined byG≈H wheneverG4H andH 4G. In [2] it was shown that the subgroups ofSym(Ω)that are closed in the topology of pointwise convergence fall into four classes with respect to≈. Furthermore, the partial order on these four equivalence classes induced by4is a total order.
The situation for the semigroupΩΩof all mappings fromΩtoΩ(the semigroup theoretic analogue ofSym(Ω)) is somewhat different. Of course, it is straightfor-ward to give a definition of4forΩΩ: ifU, V are subsets ofΩΩ, thenU
4 V if there exists a finite subsetF ofΩΩsuch thatU is contained in the subsemigroup generated byV ∪F. Throughout the remainder of the paper we will denote the subsemigroup generated by a subsetU ofΩΩbyhU i. Analogous to the theorem
1991Mathematics Subject Classification. 20M20, 08A35.
Key words and phrases. full transformation semigroup, subsemigroups closed in the function topol-ogy, equivalence relation on subsemigroups, relative rank, endomorphism of binary relations.
of Galvin mentioned above, a classical theorem of Sierpi ´nski [12] states that every countable set of mappings on Ωis contained in a2-generated subsemigroup of ΩΩ. Hence ifU, V ⊆ΩΩsuch thatU ⊆ hV, F ifor some countableF ⊆ΩΩ, then U 4V.
Mesyan [11] proved an analogue of Bergman and Shelah’s theorem for a re-stricted collection of closed (again in the topology of pointwise convergence) sub-semigroups ofΩΩ. Namely for subsemigroups U with the following properties (throughout this article we will write functions on the right of their arguments and compose them from left to right):
• ifΣ⊆Ωis finite, thenU ≈ {f ∈U : σf =σfor allσ∈Σ}; and
• the set of functions inUthat are injective on a cofinite subset ofΩare dense inU.
Letting Ω = {α1, α2, . . .} andN = {1,2, . . .}, Mesyan showed that such
sub-semigroups must be equivalent under≈to one of the following semigroups: (i) the trivial semigroup{1Ω};
(ii) S1,α =
f ∈ΩΩ : αf ∈ {α
1, α}for allα∈Ω ; (iii) S2=
f ∈ΩΩ : {α2
n−1f, α2nf} ⊆ {α2n−1, α2n}for alln∈N ;
(iv) S≤ =f ∈ΩΩ : αnf ∈ {α1, α2, . . . , αn}for alln∈N ;
(v) the full transformation semigroupΩΩ.
It was also shown that ifF={f ∈ΩΩ : |Ωf|<ℵ0}, then {1Ω} ≺F≺S1,α ≺S2≺S≤ ≺ΩΩ
where ≺denotes4but not ≈. Mesyan also proved that4 contains an infinite chain and at least two incomparable elements. However, there is no complete characterisation of the closed subsemigroups of ΩΩ with respect to 4. It is not even known how many equivalence classes there are on subsets ofΩΩunder≈.
In this paper rather than considering all closed subsemigroups ofΩΩwe will consider subsemigroups arising as the endomorphism semigroups of preorders, graphs and tolerances (reflexive and symmetric binary relations). In the main the-orems of this paper, we will prove that if S is the endomorphism semigroup of a preorder, bipartite graph, or tolerance on Ω, then eitherS ≈ ΩΩ or S ≈ S
≤.
WhetherS≈ΩΩorS≈S≤depends on certain simple structural properties of the
underlying relation; further details can be found in Section 1.3.
The notion of≈among subsets ofΩΩis related to that of relative rank. The
relative rankof a subsetUofΩΩis defined to be the least cardinality of a setAsuch thathU, Ai= ΩΩand is denoted byrank(ΩΩ:U). Relative ranks of subsets ofΩΩ have been previously studied, for example, see [4], [5], or [9].
Using Sierpi ´nski’s Theorem [12] it is straightforward to prove thatrank(ΩΩ:U) is0,1,2or uncountable for anyU ⊆ΩΩ. Moreover, it follows immediately from the definitions thatrank(ΩΩ : U) = 0,1,2 if and only ifU ≈ ΩΩ. On the other hand, ifU, V ≤ ΩΩwithU 4 V andrank(ΩΩ : U) > ℵ0, then rank(ΩΩ : U) ≥ rank(ΩΩ:V).
Assuming the Continuum Hypothesis holds the relative rank of anyU inΩΩ is0,1,2, or2ℵ0. However, if the Continuum Hypothesis is not assumed, then it is
We will prove that if U and V are semigroups of endomorphisms of a pre-order, bipartite graph, or tolerance, whererank(ΩΩ:U),rank(ΩΩ:V)>ℵ0, then rank(ΩΩ : U) = rank(ΩΩ : V). We require the following well-known notion to define the cardinal equalling any such relative ranks.
IfΩis well-ordered by≤, then a functionf ∈ ΩΩis said todominateg ∈ΩΩif αf ≥αgfor allα∈Ω. The study of the notion of dominance and related ideas gave rise to the following cardinal number, introduced by van Douwen. Adominating family forΩΩis a subsetFofΩΩsuch that for allf ∈ΩΩthere existsg∈ Fwhereg dominatesf. Of course, whether a subset is a dominating family forΩΩdepends on the well-ordering ofΩ, but the least cardinality of a dominating family does not depend on the well-ordering. Thus we can define (without ambiguity) the cardinaldto be the least cardinality of a dominating family forΩΩ. The following relations are not hard to obtain:ℵ1≤d≤2ℵ0. If the Continuum Hypothesis holds,
thend= 2ℵ0. However, without the Continuum Hypothesis, it is consistent with
the usual axioms of set theory (ZFC) thatd=ℵ1<2ℵ0 =ℵ2orℵ1<d= 2ℵ0 =ℵ2, see [1].
1.2. Definitions and notation. As usual a binary relation R on a set Ωis just a subset ofΩ×Ω. LetΩandΛbe sets, andRandSbe binary relations onΩandΛ, respectively. Then ahomomorphismfrom(Ω, R)to(Λ, S)is a functionf : Ω−→Λ such that(αf, βf) ∈ S for all(α, β) ∈ R. A homomorphism is an isomorphism
if it is bijective and its inverse is also a homomorphism. An endomorphismis a homomorphism from(Ω, R)to(Ω, R). Anautomorphismis an isomorphism from (Ω, R)to(Ω, R). We denote the semigroup of endomorphisms on(Ω, R)under composition of mappings byEnd(Ω, R). LetR⊆Ω×Ωand letΛ⊆Ω. We define thesubrelationofRinduced byΛto beR∩Λ×Λ.
Awalkfromα∈Ωtoβ ∈Ωin(Ω, R)is a sequence of elements ofΩ α=γ0, γ1, γ2, . . . , γn=β
such that(γi, γi+1) ∈ Ror(γi+1, γi) ∈ Rfor alli. We will say that such a walk
has lengthn. Two points areconnectedif there exists a walk from one to the other. Being connected is an equivalence relation onΩand the equivalence classes are called thecomponentsof(Ω, R). We will say that(Ω, R)isconnectedif it only has one component. IfRis a binary relation onΩ, then apathin(Ω, R)is a walk in which all points are distinct.
The degree of an element α ∈ Ωis the size of the set {β ∈ Ω : (α, β) ∈ Ror(β, α) ∈ R}. We say that(Ω, R)islocally finiteif all the elements ofΩhave finite degree.
Apreorder is a reflexive and transitive binary relation. Apartial orderis a pre-order that is also anti-symmetric. A set with a partial pre-order is called a partially ordered setorposet. AgraphG= (Ω, E)is a setΩtogether with a binary relationE that is symmetric and irreflexive. IfGis a graph, then for the sake of consistency with the literature, we will call the elements ofΩtheverticesofG, the elements ofE theedgesofG, and a subrelation induced by a set will be referred to as the
In what follows we will always, unless stated otherwise, assume thatΩis the countably infinite set{α1, α2, . . .}and we will always assume thatN={1,2, . . .}.
1.3. Overview. LetRbe a preorder, bipartite graph, or tolerance. Then the main theorems of this paper can be summarized as follows:
• ifRhas finitely many components and is locally finite, thenEnd(Ω, R)≈ S≤andrank(ΩΩ: End(Ω, R)) =d;
• ifRhas infinitely many components or is not locally finite, thenEnd(Ω, R)≈ ΩΩandrank(ΩΩ: End(Ω, R))∈ {0,1,2};
see Theorems 2.3, 3.1, 4.4, 4.5, and 5.1.
The picture is more complicated for arbitrary non-bipartite graphs. In particu-lar, there exist examples of graphsGwhere:
• Ghas infinitely many components,End(G)≈ {1Ω}orEnd(G) ≈S2, and rank(ΩΩ: End(G)) = 2ℵ0;
• Ghas infinitely many components,End(G)≈S≤, andrank(ΩΩ: End(G)) = d;
• Gis connected and locally finite,End(G)≈ {1Ω}, andrank(ΩΩ: End(G)) = 2ℵ0;
• Gis connected and not locally finite,End(G)≈S≤, andrank(ΩΩ: End(G)) = d;
see Examples 6.1, 6.2, and 6.3.
The following weaker version of the theorems regarding bipartite graphs hold for an arbitrary graphG:
• ifGhas finitely many components and is locally finite, thenEnd(G)4S≤;
• if all the components of G are finite, then one of the following holds: End(G)≈ {1Ω},S1,α4End(G)4S≤, orEnd(G)≈ΩΩ;
see Theorems 2.4 and 4.3.
2. UNCOUNTABLE RANKS AND BINARY RELATIONS
The following theorem connects the notions of relative rank, domination, and the preorder4. We require the following notion for a subsetFofΩΩ. We say that F is analmost disjoint familyif for allf, g ∈F there are only finitely manyα∈Ω such thatαf =αg. It is reasonably straightforward to show that there exists an almost disjoint familyF inΩΩwith|F|= 2ℵ0; see, for example, [10, Theorem 1.3].
Theorem 2.1. LetUbe a subset ofΩΩ. IfU ≈S≤, thenrank(ΩΩ:U) =d.
On the other hand, ifU 4S2, thenrank(ΩΩ:U) = 2ℵ0.
Proof. For a proof of the fact thatrank(ΩΩ:S≤) =dsee [5, Lemma 3.5].
We will show thatrank(ΩΩ : S
2) = 2ℵ0. LetA be a subset ofΩΩ such that hS2, Ai= ΩΩ. Seeking a contradiction assume that|A|<2ℵ0. Let(a1, a2, . . . , am)
be anm-tuple of elements ofA. Then define
The semigroupΩΩcan be given as the union of the setsB(a1,a2,...,am)over all finite
tuples of elements ofA.
LetF ⊆ΩΩbe a family of almost disjoint functions of size2ℵ0. IfB
(a1,a2,...,am)∩
Fwere finite for all(a1, a2, . . . , am), then|F| ≤min{ℵ0,|A|}. But|F|= 2ℵ0and so
there exists a tuple(b1, b2, . . . , bn)of elements fromAsuch thatB(b1,b2,...,bn)∩Fis
infinite. Define
Cα={αh : h∈B(b1,b2,...,bn)}.
Then |Cα| ≤ 2n+1 for all α ∈ Ω by the definition of S2. Let N = 2n+1 and
f1, f2, . . . , fN+1 be distinct elements ofB(b1,b2,...,bn)∩F. Then, sinceF is a
fam-ily of almost disjoint functions, there existsβ ∈ Ωsuch thatβf1, βf2, . . . , βfN+1 are distinct. But|Cβ| ≤N, a contradiction.
It is straightforward to classify those binary relations whose endomorphism semigroups equalΩΩ. The proof follows immediately from the definitions and is omitted.
Lemma 2.2. LetΩbe an infinite set and letRbe a binary relation onΩ. Then the relative rank ofEnd(Ω, R)inΩΩis0if and only ifRis one of∅,Ω×Ω, or∆Ω={(α, α) :α∈Ω}.
In light of Lemma 2.2 we will assume throughout thatRis a non-empty proper subset ofΩ×Ωnot equal to∆Ω={(α, α) : α∈Ω}.
Theorem 2.3. LetRbe a reflexive binary relation onΩsuch that(Ω, R)has infinitely many components. Thenrank(ΩΩ: End(Ω, R))≤1and soEnd(Ω, R)≈ΩΩ.
Proof. Recall thatΩ ={α1, α2, . . .}. Let the components of(Ω, R)beL1, L2, . . .and letγi∈Libe fixed for alli. Defineg∈ΩΩbyαig=γi.
Letf ∈ΩΩbe arbitrary. Let
b
f ∈ΩΩmap all points inL
itoαif fori= 1,2, . . ..
SinceRis reflexive,fb∈End(Ω, R). Then for allαi∈Ωwe haveαigfb=γifb=αif.
Thusf ∈ hEnd(Ω, R), gi. Sincef was chosen arbitrarily we conclude thatΩΩ= hEnd(Ω, R), giand hencerank(ΩΩ: End(Ω, R))≤1.
Theorem 2.4. LetRbe a binary relation onΩsuch that(Ω, R)has finitely many compo-nents and is locally finite. ThenEnd(Ω, R)4S≤and hencerank(ΩΩ: End(Ω, R))≥d.
We require the following result to prove Theorem 2.4. Letd: Ω×Ω−→Rbe a
metric onΩ. A functionf ∈ΩΩisLipschitzif there exists a constantC ∈
Nsuch
thatd(αf, βf)≤Cd(α, β)for allα, β∈Ω. We may also say thatf isLipschitz with constantC. Denote the semigroup of all Lipschitz functions onΩbyLΩ.
Proposition 2.5. [5, Theorem 3.1]LetΩbe a countably infinite set and letdbe a metric onΩthat is unbounded on every infinite subset ofΩ. ThenLΩ 4 S≤ and rank(ΩΩ : LΩ)≥d.
Proof of Theorem2.4. LetL1, L2, . . . , Ln be the components of R. To show that
End(Ω, R)4S≤we define a metric onΩand prove thatEnd(Ω, R)⊆LΩ.
LetdLi :Li×Li−→N∪{0}be defined so thatdLi(α, β)is the minimal length of
i. We will now extend the metricsdLi to a metricdon the entire setΩ. Letγi ∈Li
be fixed. Then definedby
d(α, β) =
(
dLi(α, β) ifα, β∈Li
dLi(α, γi) +dLj(γj, β) + 1 ifα∈Liandβ ∈Ljwherei6=j.
It can easily be seen thatdis indeed a metric onΩand that it is unbounded above on every infinite subset.
We will now show that all functions inEnd(Ω, R)are Lipschitz with respect to d. Letf ∈End(Ω, R)be arbitrary and letM = max{d(γi, γjf) : 1 ≤i, j ≤n}.
Ifαandβare in the same componentLi, thenαf, βf ∈Ljfor somej and since
f ∈End(Ω, R)we have that
d(αf, βf) =dLj(αf, βf)≤dLi(α, β) =d(α, β).
Next, ifα∈Li,β ∈Ljwithi6=j, andαf∈Lk,βf ∈Ll, then
d(αf, βf)≤d(αf, γif) +d(γif, γk) +d(γk, γl) +d(γl, γjf) +d(γjf, βf)
≤dLk(αf, γif) +M + 1 +M +dLl(γjf, βf)
≤dLi(α, γi) +M + 1 +M+dLj(γj, β)
=d(α, β) + 2M
≤d(α, β) + 2M d(α, β) = (2M+ 1)d(α, β).
Thusf is Lipschitz with constant2M + 1. Therefore it follows from Theorem 2.5
thatEnd(Ω, R)4S≤.
3. PREORDERS
In this section we completely classify the endomorphisms of preordersvonΩ with respect to4. Since preorders are reflexive, the case where(Ω,v)has infinitely many components follows directly from Theorem 2.3. That is, ifvis a preorder onΩsuch that(Ω,v)has infinitely many components, thenEnd(Ω,v)≈ΩΩand rank(ΩΩ: End(Ω,v))≤1.
The case wherevis a partial order was considered in [9]. It was shown that the endomorphisms of a poset(Ω,v)have finite relative rank inΩΩprecisely when (Ω,v)is locally finite or (Ω,v) has infinitely many components. Here we will show that this classification extends to preorders and show that the only infinite value that can arise forrank(ΩΩ: End(Ω,v))isd.
Theorem 3.1. Letvbe a preorder onΩsuch that(Ω,v)has finitely many components.
(i) If(Ω,v)is locally finite, thenEnd(Ω,v)≈S≤andrank(ΩΩ: End(Ω,v)) =d.
(ii) If(Ω,v)is not locally finite, thenEnd(Ω,v)≈ΩΩandrank(ΩΩ: End(Ω,v)) ≤2.
It is natural to ask if the bound given in Theorem 3.1(ii) is the best possible. The answer is yes: two examples of connected posets withrank(ΩΩ : End(Ω,v)) = 1 and2, respectively, were given in [9].
Lemma 3.2. LetRbe a binary relation onΩ, letg ∈End(Ω, R)be any endomorphism with infinite image, letR0be the subrelation ofRinduced byim(g), and letSbe any rela-tion onΩsuch that(im(g), R0)is isomorphic to(Ω, S). ThenEnd(Ω, R)<End(Ω, S).
Proof. Let Ψ : (im(g), R0) −→ (Ω, S)be an isomorphism. Then gΨ ∈ ΩΩ is a surjective homomorphism from(Ω, R)to(Ω, S).
Letg∈ΩΩbe any function such thatαg∈α(gΨ)−1={β∈Ω : βgΨ =α}for allα∈Ω. ThenggΨ = 1Ωwhere1Ωdenotes the identity map onΩ. Likewise, if Ψ∗∈ΩΩis an extension ofΨ, thenΨ−1Ψ∗= 1Ω.
Letf ∈End(Ω, S)be arbitrary. ThengΨfΨ−1∈End(Ω, R). Thus f =ggΨfΨ−1Ψ∗∈ hEnd(Ω, R), g,Ψ∗i.
Sincef was arbitrary,End(Ω, S)⊆ hEnd(Ω, R), g,Ψ∗i.
Lemma 3.3. LetΩ ={α1, α2, . . .}and let
(i) R={(αi, αi+1),(αi+1, αi) : i∈N};
(ii) S={(α2i−1, α2i),(α2i+1, α2i) : i∈N}.
Then(Ω, R)is a graph withEnd(Ω, R)<S≤, and(Ω, S)is a poset withEnd(Ω, S)<
S≤.
Proof. It suffices to show thatEnd(Ω, R)∩End(Ω, S) < S≤ . Let g ∈ ΩΩbe
de-fined byαng =αn(n−1)+1for alln∈ Nand leth∈ΩΩbe any function such that
(α2n−1)h=αnfor everyn∈N.
Letf ∈S≤be arbitrary. We will define a functionfb∈End(Ω, R)∩End(Ω, S)in
two steps so thatfcan be written as a product off , gb , andh. The first step is to let b
f be defined on the elements of the formαn(n−1)+1by (αn(n−1)+1)fb=α2k−1 wheneverαnf =αk.
The second step is to definefbon all the elementsαmwith indices in the range
n(n−1) + 2ton(n+ 1). Ifαnf =αk andαn+1f =αl, thenk≤nandl ≤n+ 1
sincef ∈S≤. It follows that the length of the path on(Ω, R)fromα2k−1toα2l−1 is an even number not greater than2n. Hence there exists a walk
β0=α2k−1, β1, . . . , β2n=α2l−1 of length2n. The definition offbis completed by setting
(αn(n−1)+1+i)fb=βi
for alli∈ {1,2, . . . ,2n−1}. By construction,fbis an endomorphism of(Ω, R).
We will now show thatfbis also an element ofEnd(Ω, S). By construction,
{α1, α3, α5, . . .}fb⊆ {α1, α3, α5, . . .}and{α2, α4, α6, . . .}fb⊆ {α2, α4, α6, . . .}.
Letα, β∈Ωwith(α, β)∈S. Thenα=α2i−1andβ =α2iorα2i−2for somei∈N.
Sinceαandβ are adjacent in(Ω, R)their imagesαfbandβfbare also adjacent in
(Ω, R). Thus either(αf , βb fb) ∈ S or(βf , αb fb) ∈ S. In fact, (αf , βb fb) ∈ S since
To conclude the proof, letαi∈Ωbe arbitrary and letαj =αif. Then
αigf hb = (αi(i−1)+1)f hb = (α2j−1)h=αj=αif.
ThusS≤⊆ hEnd(Ω, R)∩End(Ω, S), g, hiand soEnd(Ω, R)∩End(Ω, S)<S≤.
Lemma 3.4(K ¨onig’s Lemma). LetGbe an infinite connected locally finite graph. Then there exists an infinite path inG, that is, a sequence of distinct verticesβ1, β2, . . .such thatβiandβi+1are adjacent for alli.
For a proof see [3, Lemma 19.2.1].
The following lemma is an analogue of K ¨onig’s Lemma for arbitrary binary re-lations. It is also slightly stronger, in so far as when it is applied to graphs the subgraph induced byβ1, β2, . . . from Lemma 3.4 is isomorphic to the graph de-fined in Lemma 3.3(i).
Lemma 3.5. Let Ωbe countably infinite and let R ⊆ Ω×Ω be such that (Ω, R) is connected and locally finite. Then there exists a sequenceγ1, γ2, . . .of distinct elements of Ωsuch that, fori6=j,γiRγjorγjRγiif and only ifiandjare consecutive integers.
Proof. Let E be the symmetric closure ofR\∆Ω. ThenG = (Ω, E)is a graph. Hence by Lemma 3.4 there exist a infinite pathβ1, β2, . . .inG. Butβiis adjacent to
βi+1inGif and only if(βi, βi+1)or(βi+1, βi)∈R.
Letγ1=β1. Assume thatγi−1has been defined for somei >1. Then define ni= max{n∈N : (γi−1, βn)or(βn, γi−1)∈R}
and setγi=βni. The numberniexists since(Ω, R)is locally finite. The sequence
γ1, γ2, . . .obtained in this way has the required property.
Proof of Theorem3.1. (i). As(Ω,v)is locally finite, it follows immediately from Theorem 2.4 thatEnd(Ω,v)4S≤.
To prove thatEnd(Ω,v) <S≤, we show that there existsg ∈ End(Ω,v)such
that the preorder induced by the image ofgis isomorphic to that given in Lemma 3.3(ii). This allows us to apply Lemma 3.2 to conclude the proof.
Since(Ω,v)has finitely many components there is at least one infinite com-ponent. By Lemma 3.5 that component contains a sequence of distinct elements γ1, γ2, . . .such thatγivγjorγj vγiif and only ifiandjare consecutive integers.
Letγn be arbitrary. Ifγn v γn+1, thenγn+1 w γn+2 as otherwiseγn v γn+2 by transitivity ofv, a contradiction. Likewise, ifγn w γn+1, then γn+1 v γn+2.
Assume without loss of generality thatγ1 v γ2. We conclude that the subposet induced by{γ1, γ2, . . .}is isomorphic to that defined in Lemma 3.3(ii).
Next, we specifyg ∈End(Ω,v)with image equal to{γ1, γ2, . . .}by defining it on the components of(Ω,v). LetKbe any component of(Ω,v). Then sincevis transitive and(Ω,v)is locally finite, it follows that there existsβ1 ∈Ksuch that for allβ ∈ Kwithβ vβ1we have thatβ wβ1. Note that, in some sense,β1is a minimal element ofK.
and
L2i+1={β ∈K : there existsδ∈L2iwithβvδ} \(L1∪ · · · ∪L2i).
Of course, since(Ω,v)is locally finite,Liis finite for alli∈N. AsKis connected,
every element inKlies in someLi. Also, ifKis infinite, thenLiis non-empty for
alli.
So, ifgK :K−→Ωis defined so thatαgK =γifor allα∈Li, then by
construc-tiongK is a homomorphism from(K,v)to the preorder induced by{γ1, γ2, . . .}.
Letg: Ω−→ {γ1, γ2, . . .}be the union of the functionsgKover all the components
Kof(Ω,v). Theng∈End(Ω,v)and, as(Ω,v)has at least one infinite component, gis surjective.
IfR is the preorder induced byγ1, γ2, . . ., then, by Lemma 3.2, End(Ω,v) < End(Ω, R). Moreover, by Lemma 3.3, it follows thatEnd(Ω, R)<S≤and the proof
of this case is concluded.
(ii).Recall that in this case we assume that(Ω,v)is not locally finite. Ifα, β∈Ω such that α v β and β v α, then we will writeα ≡ β. If all the equivalence classes of ≡are finite, then there are infinitely many such classes and they can be given asE1, E2, . . .. Letβn ∈ En be fixed for everyn ∈ Nand letg ∈ ΩΩbe
defined byαg=βnfor allα∈Enand for alln. It is straightforward to verify that
g ∈ End(Ω,v). Furthermore the preorder induced by the image ofg is a partial order which is not locally finite. In [9] it was shown that the set of endomorphisms of a non-locally finite poset is always equivalent under≈toΩΩ. Thus by Lemma 3.2 we have that End(Ω,v) ≈ ΩΩ and so, as mentioned in the introduction, it follows by Sierpi ´nski [12] thatrank(ΩΩ: End(Ω,v))≤2.
Next, we assume that there exists an infinite equivalence classEof≡. Letk : Ω−→Ebe any bijection and letk∗ ∈ΩΩbe any extension ofk−1. Letf ∈ΩΩbe arbitrary and definefb∈ΩΩby
αfb= (
αk−1f k ifα∈E α ifα∈Ω\E.
Thenfb ∈ End(Ω,v)since f fixes Ω\E pointwise and maps elements ofE to
elements ofE. Furthermore, ifα∈Ω, then
αkf kb ∗=αk(k−1f k)k∗=αf.
ThusΩΩ=hEnd(Ω,v), k, k∗iand soEnd(Ω,v)≈ΩΩ. In fact,k∈End(Ω,v)and soΩΩ=hEnd(Ω,v), k∗iandrank(ΩΩ: End(Ω,v)) = 1.
4. GRAPHS
In this section we consider semigroups of endomorphisms of graphs. These semigroups fall into more equivalence classes under≈than endomorphisms of preorders and we do not achieve a full classification in this case.
Lemma 4.1. IfGcontains a subgraph isomorphic to the complete graphKΩonΩ, then
Proof. LetHdenote the subgraph ofGisomorphic toKΩ, letH1, H2, . . .be infinite sets partitioning the vertices of H, and let g ∈ ΩΩ be a function that maps all elements ofHitoαifori= 1,2, . . .. Note thatg6∈End(G).
Pick an arbitraryf ∈ΩΩ. Let
b
f be an injection such thatαifb∈ Hj whenever
αif =αj. Since im(fb)⊆Hall image points are adjacent and sofb∈End(G). Now
αif gb =αj=αif for allαi∈Ω. HenceΩΩ=hEnd(G), gi.
LetGbe a graph and defineK(G)to be the set of components. IfL, M ∈K(G), then we will writeLM whenever there exists a homomorphism fromLtoM. Denote byLthe set{M ∈K(G) : LM}.
Theorem 4.2. LetGbe a graph such that for infinitely many componentsLofGthe set
Lis infinite. Thenrank(ΩΩ: End(G))≤2.
Proof. LetL1, L2, . . .be the components ofGwithLi infinite for alli∈N.
First, let
{A(i,1), A(i,2), . . .} ⊆Li
such that{A(i,1), A(i,2), . . .} ∩ {A(j,1), A(j,2), . . .}=∅fori6=j.
LetΩ = {α1, α2, . . .}, letg ∈ ΩΩ be any function withαig ∈ Li, leth ∈ ΩΩ
be any function such thatαh =αj for allα∈ A(i,j), and letf ∈ΩΩbe arbitrary.
SinceA(i,k) ∈ Li for alli, k, there exists a homomorphism fromLi toA(i,k). A
function that is a homomorphism on all the components ofGis an endomorphism ofG. So there existsfb∈End(G)such thatLifb⊆A(i,k)wheneverαk =αif. Let
αi∈Ωbe arbitrary and letαk =αif. Thenαig∈Liand so(αig)fb∈A(i,k). Hence
αigf hb =αk =αif. Sof =gf hb andΩΩ=hEnd(G), g, hi.
In Theorem 2.3, we prove that endomorphisms of reflexive relations with infin-itely many components have relative rank at most1inΩΩ. However for graphs the analogous statement is not true. Examples of graphsGandH satisfying the hy-pothesis of Theorem 4.2 whererank(ΩΩ: End(G)) = 1andrank(ΩΩ: End(H)) = 2 can be found in Example 6.4 and Proposition 7.8, respectively.
We use a result from Mesyan [11] to show that the converse of Theorem 4.2 holds in the case that all the components ofGare finite.
Theorem 4.3. LetGbe a countably infinite graph such that every component of Gis finite. Then the following are equivalent:
(i) Lis finite for all but finitely many componentsLofG;
(ii) rank(ΩΩ: End(G))>2; (iii) rank(ΩΩ: End(G))≥d;
(iv) S1,α 4End(G)4S≤orEnd(G)≈ {1Ω}.
Proof. By Theorem 2.1 it follows that (iv) implies (iii). Also (iii) implies (ii) imme-diately. Theorem 4.2 tells us that (ii) implies (i).
It was shown in [11, Section 7] thatE(ρ) 4 S≤ for such a preorderρ. It follows
from the definition ofE(ρ)thatEnd(G)⊆E(ρ)and thusEnd(G)4S≤.
It remains to prove that eitherEnd(G)<S1,αorEnd(G)≈ {1G}. There are two
possibilities. Suppose that, for all but finitely many componentsL, the only homo-morphism fromLintoGis the identity map. It follows thatEnd(G)is countable since all the components ofGare finite. ThusEnd(G)≈ {1G}as the equivalence
class of{1G}, consists of all countable subsets ofΩΩ.
On the other hand, suppose there exist infinitely many componentsL1, L2, . . . ofGand non-identity homomorphismsgi:Li −→Gfor alli∈N. We will define
an infinite subset{δ1, δ2, . . .}of the union ofL1, L2, . . .such that (a) ifδiandδjare in the same component, theni=j;
(b) ifδi∈Lj, thenδigj 6∈ {δ1, δ2, . . .}for alli∈N.
Sincegiis not the identity onLi, for alli∈Nthere existsγi∈Lisuch thatγigi6=γi.
There are two cases to consider. If there existsj ∈Nsuch that
A={γi : γigi=γj}
is infinite, thenAsatisfies conditions (a) and (b) above.
Otherwise, we define{δ1, δ2, . . .}recursively as follows. Letδ1 =γ1. Assume thatδ1, δ2, . . . , δn−1∈ {γ1, γ2, . . .}have already been defined and set
Bn={γi : γigi∈ {δ1, δ2, . . . , δn−1} }.
Since by assumption{γi : γigi=δj}is finite for allj∈ {1, . . . , n−1},Bnis finite.
Hence we may chooseδnto be any element of
{γ1, γ2, . . .} \(Bn∪ {δ1, δ2, . . . , δn−1}).
It follows, by construction, that{δ1, δ2, . . .}satisfies (a) and (b).
Leth : Ω −→ {δ1, δ2, . . .}be the map defined byαih = δi and letk ∈ ΩΩ be
defined by
αk=
(
αi ifα=δifor somei
α1 ifα6∈ {δ1, δ2, . . .}.
Letf ∈S1,αbe arbitrary. Then definefb∈ΩΩas follows. Letα∈Ωand letLj
be the component ofGcontainingα. Ifδi∈Ljfor somei∈Nandαif =α1, then
we define
αfb=αgj.
Otherwise defineαfb=α. Sincefbis a homomorphism on each component,fb∈
End(G).
Letαi ∈ Ωbe arbitrary. Then eitherαif = α1 ori > 1andαif = αi. In the
former case, ifδi∈Lj, then
αihf kb =δif kb =δigjk=α1=αif
asδigj 6∈ {δ1, δ2, . . .}.
In the latter case,
αihf kb =δif kb =δik=αi=αif.
In Example 6.3 we give an instance of a graph Gwith infinitely many com-ponents, all of which are finite, and where End(G) ≈ S≤. In Example 6.2 we
show that there exists a graph with infinitely many components and whereS1,α≺
End(G) ≈ S2 ≺ S≤. It is not known if there exists a graph Gsuch thatS1,α ≈
End(G).
IfGis a graph with finitely many components andGis locally finite, then it fol-lows immediately from Theorem 2.4 thatEnd(G)4S≤andrank(ΩΩ: End(G))≥ d. The converse of this statement does not hold and Example 6.1 is a counterexam-ple. This contrasts with the analogous situation for preorders described in Theo-rem 3.1. In Lemma 3.3 and Example 6.2 we give examples of graphsGandHwith finitely many components and whereEnd(G)≈S≤andEnd(H)≈ {1Ω} ≺S≤.
Note that in the proofs of Theorems 4.2 and 4.3 neither symmetry nor irreflex-ivity is used and that these theorems generalise to arbitrary binary relations with infinitely many components. We chose not to phrase these results in the most general way since the only other kinds of relations considered in this paper are preorders and tolerances for which the much stronger Theorem 2.3 holds.
We have not succeeded in proving any general theorem relating to graphs with finitely many components that are not locally finite. However, we will show that there exist such graphs where the relative rank of their endomorphisms inΩΩis any of1,2,d, or2ℵ0. Moreover, if we restrict our attention to the class of bipartite
graphs, then we again obtain a complete classification.
Theorem 4.4. LetGbe a graph with infinitely many bipartite components. Thenrank(ΩΩ: End(G)) = 1and soEnd(G)≈ΩΩ.
Proof. There are two cases to consider.
Case 1: there exist infinitely many singleton components{β1},{β2}, . . .in G. Let g ∈ΩΩbe defined byα
ig =βifor alli ∈N. Iff ∈ΩΩis arbitrary, then definefb
byβifb=αif for alliandαfb=αfor allα6=βi for anyi. Thenfb∈End(G)and
αigfb=βifb=αif. HencehEnd(G), gi= ΩΩ.
Case 2:there exist infinitely many bipartite componentsL1, L2, . . .inGwith at least
two vertices.Letγn∈Lnbe fixed for alln∈Nand let
I={i∈N : αi6∈Ljfor allj∈N}.
Then, by definition,γm6=αnfor allm∈Nand for alln∈I. AlsoN\Iis infinite
as clearly there are infinitely verticesαiinL1∪L2∪ · · ·. It follows that there exists
an injectiveg∈ΩΩsuch thatγ
ig=αifor alli∈Iand where(Ω\ {γi:i∈I})g⊆
{γi:i∈N\I}. Henceg2is an injection andim(g2)⊆ {γi:i∈N\I}.
Let Li and Lj be arbitrary and letα ∈ Li and β ∈ Lj. SinceLi and Lj are
bipartite and contain at least two vertices, there exists a homomorphismφα,β :
Li−→Ljsuch thatαφα,β =β.
Letf ∈ ΩΩbe arbitrary. We require two endomorphismsf1b andf2b ofGthat
together withgwill generatef.
We definefb1on an arbitrary componentLas follows. Either there existi ∈ I, j ∈ N, andα∈ Ωsuch thatαf =αi,L =Lj, andαg2 =γj, or not. Ifi, j, andα
exist, then define
for allβ ∈L. Otherwise, we defineβf1b =βfor allβ ∈L. In particular, ifαf =αi
for somei6∈I, thenf1b fixesαg2.Sincef1b is a homomorphism on every component
ofG, it is an element ofEnd(G).
We definef2b on an arbitrary componentL ofGas follows. As above, either
there existi ∈N\I,j ∈N, andα∈Ωsuch thatαf =αi,L=Lj, andαg3=γj,
or not. Ifi, j,andαexist, then, sincei6∈I, there existsk∈Nsuch thatαi∈Lk. It
follows thatφγj,αiis well-defined and so we define
βf2b =βφγj,αi
for allβ ∈L. Otherwise, we defineβfb2 = βfor allβ ∈ L. In particular, ifi ∈ I, then, from the definition ofI,αi6∈Ljfor allj∈Nand sof2b fixesαi.Again since
b
f2is a homomorphism on all the components ofG, it follows thatfb2∈End(G). We will now show thatg2f1gb f2b =f. Letα∈Ωbe arbitrary. Thenαf =αifor
somei∈N. Ifi∈Iandαg2=γj for somej, then
αg2f1gb f2b =γjf1gb f2b =γjφγj,γigf2b =γigf2b =αif2b =αi=αf.
Ifi6∈Iandαg3=γ
kfor somek, then
(αg2)fb1gfb2=αg3fb2=γkfb2=γkφγk,αi =αi=αf.
ThusΩΩ=hEnd(G), giandrank(ΩΩ: End(G)) = 1.
Theorem 4.5. LetGbe a bipartite graph with finitely many components. Then either:
(i) Gis locally finite,End(G)≈S≤, andrank(ΩΩ: End(G)) =d; or
(ii) Gis not locally finite,End(G)≈ΩΩ, andrank(ΩΩ: End(G))≤2.
Before we prove Theorem 4.5 we require the following lemma.
Lemma 4.6. Let Gbe the graph with edges(α1, αi)for all i > 1 (see Figure 1 for a diagram). Thenrank(ΩΩ: End(G)) = 1.
u
α1
T
T
T
T
T
l l
l l
l l ,
, , , , , u
α2 uα3 uα4 uα5 uα6 . . .
FIGURE1. The graph from Lemma 4.6
Proof. Note that iff : Ω−→Ωsuch thatα1f =α1andαif 6=α1for alli >1, then
f ∈End(G). Letg, h∈End(G)be defined by
αig= (
αi i= 1
αi+1 i >1
αih= (
αi 1≤i≤2
αi−1 i >2.
Lett∈ΩΩbe a transposition withα
Letf be an arbitrary element ofΩΩ. Define the functionfbbyα1fb= α1 and
αi+1fb = αk+1 wheneverαif = αk. Then fb ∈ End(G)by our earlier remark.
Furthermore, for an arbitrary vertexαi∈Ωwithαif =αkwe have that
αigtf thb =αi+1f thb =αk+1th=αk=αif
and sohEnd(G), ti= ΩΩ.
Proof of Theorem4.5. LetGbe a bipartite graph with finitely many components L1, L2, . . . , Ln.
(i). IfGis locally finite, then by Theorem 2.4 we have thatEnd(G) 4S≤. We
will show thatEnd(G) <S≤. By Lemma 3.5, there exists a sequenceγ1, γ2, . . .of
vertices that induce a subgraphHofGisomorphic to the graph defined in Lemma 3.3(i).
Letδi ∈Libe fixed. Form= 0,1,2, . . .define
Lmi +1={α∈Li : the shortest path fromαtoδihas lengthm}.
Letg∈ΩΩmap every point inLm
1 ∪Lm2 ∪. . .∪Lmn toγm. SinceGis locally finite
and at least oneLi is infinite, it follows that g is surjective. If(α, β) ∈ E, then,
sinceGis bipartite,α∈Lmj andβ ∈Lmj+1 orβ ∈Lmj andα∈ Lmj +1for somej andm. Hence(αg, βg) = (γm, γm+1) ∈ E or(αg, βg) = (γm+1, γm) ∈ E. Thus
g∈End(G). So, by Lemma 3.2 and Lemma 3.3 it follows thatEnd(G)<S≤.
(ii). SinceGis bipartite we may partition Ωinto setsR and B such that the edges ofGonly join vertices inRto vertices inB. SinceGis not locally finite it has a vertex of infinite degree. Without loss of generality we assume thatα1 ∈R and thatα1has infinite degree.
Letgbe any function such thatαg=α1for allα∈Rand Bg⊆ {β ∈Ω : (α1, β)∈E} ⊆B
with |Bg| = ℵ0. Theng is an endomorphism of Gand the image of g induces a graph isomorphic to that defined in Lemma 4.6. So, by Lemmas 3.2 and 4.6 it follows thatEnd(G)≈ΩΩ.
Lemma 4.6 provides an example of a graphGsatisfying the hypothesis of The-orem 4.5(ii) and whererank(ΩΩ: End(G)) = 1. In Section 6 we give an example of such a bipartite graphHwithrank(ΩΩ: End(H)) = 2.
5. TOLERANCES
We completely classify the semigroups of endomorphisms of tolerancesRon Ωaccording to4. If(Ω, R)has infinitely many components, then it follows from Theorem 2.3 thatrank(ΩΩ: End(Ω, R)) = 1.
Theorem 5.1. LetRbe a tolerance onΩsuch that(Ω, R)has finitely many components. Then either:
(i) (Ω, R)is locally finite,End(Ω, R)≈S≤, andrank(ΩΩ: End(Ω, R)) =d; or
(ii) (Ω, R)is not locally finite,End(Ω, R)≈ΩΩ, andrank(ΩΩ: End(Ω, R))≤2.
Proof. Recall thatRis a symmetric and reflexive relation, and letL1, L2, . . . , Lnbe
the components of(Ω, R).
(i). By Theorem 2.4, it follows thatEnd(Ω, R) 4 S≤. We must prove that
End(Ω, R) < S≤. Then, by Lemma 3.5, there exists Γ = {γ1, γ2, . . .} such that,
fori6=j,(γi, γj)∈Rif and only if{i, j}={k, k+ 1}for somek∈N.
LetLm
i be the sets andg∈ΩΩbe the function defined in the proof of Theorem
4.5(i). If(α, β) ∈ R, then eitherα, β ∈Lm
j orα ∈ Lmj andβ ∈ L m+1
j for somej
andm. In the first case,(αg, βg) = (γm, γm)∈Rand in the second case(αg, βg) =
(γm, γm+1)∈R. Henceg∈End(Ω, R).
Let R0 be the subrelation of R induced by Γ. Then by Lemma 3.2 we have thatEnd(Ω, R) < End(Ω, S)where (Ω, S)is isomorphic to(Γ, R0). Now, (Ω, S\ ∆Ω)is a graph isomorphic to that defined in Lemma 3.3(i). Thus, by Lemma 3.3, End(Ω, S\∆Ω)<S≤. AsEnd(Ω, S)⊇End(Ω, S\∆Ω), it follows thatEnd(Ω, R)<
End(Ω, S)<End(Ω, S\∆Ω)<S≤.
(ii). There exists an element ofΩwith infinite degree. Assume without loss of generality thatα1has infinite degree, that is,A={β ∈Ω : (α1, β)∈R}is infinite. It is a straightforward consequence of Ramsey’s Theorem [3, Theorem 10.6.1], ap-plied to(Ω, R\∆Ω), that the subrelation induced byAcontains an infinite subset Bsuch that(B×B)∩R=B×Bor∆B.
Note that(Ω, R\∆Ω) is a graph andEnd(Ω, R\∆Ω) ⊆ End(Ω, R). If (B× B)∩R = B ×B, then, by Lemma 4.1, rank(ΩΩ : End(Ω, R\∆Ω)) = 1and so rank(ΩΩ: End(Ω, R)) = 1.
If(B×B)∩R = ∆B, then defineg ∈ΩΩbyαg=αfor allα∈ Band define
αg = α1 for allα∈ Ω\B. SinceRis reflexive and(α1, β) ∈ Rfor allβ ∈ B, it follows thatg ∈ End(Ω, R). Therefore by an argument analogous to that in the previous paragraph, by Lemmas 3.2 and 4.6,rank(ΩΩ: End(Ω, R))≤2.
IfG= (Ω, E)is the graph in Lemma 4.6, then(Ω, E∪∆Ω)is a tolerance where rank(ΩΩ : End(Ω, E ∪∆
Ω)) = 1. In Section 8 we construct a tolerance with rank(ΩΩ: End(Ω, R)) = 2.
6. EXAMPLESI
The following example shows that, in general, the converse of Theorem 2.4 is not true.
Example 6.1. Let Gdenote the graph with edges (α1, αi) and (αi, αi+1) for all
i∈N(for a diagram see Figure 2). ThenGis not locally finite. However, we will
show thatEnd(G)4S≤and thusrank(ΩΩ: End(G))≥d. u
α1
T
T
T
T
T
l l
l l
l l ,
, , , , , u
α2 uα3 uα4 uα5 uα6 . . .
FIGURE2. The graph from Example 6.1
Let F = {f ∈ End(G) : α1f = α1 } and U = End(G)\ F. If H is the graph obtained fromGby deleting all the edges incident toα1, thenF⊆End(H). But End(H) ≈ S≤ by Theorem 4.5 and so F 4 S≤. In Section 1.1 we defined F={f ∈ΩΩ : |Ωf|<ℵ0}and we noted that Mesyan [11] proved thatF≺S
≤.
SinceU 4F, this implies thatU ≺S≤. It follows thatEnd(G) =U∪F4S≤.
In fact, an argument analogous to that used in the proof of Lemma 3.3 shows thatEnd(G)≈S≤and sorank(ΩΩ: End(G)) =d.
Example 6.2. A graph G is calledrigid if End(G) = {1Ω}. It follows from [8, Theorem 3] that there exists a locally finite countably infinite rigid graphH with infinitely many components.
We will construct a graphGfrom the components ofHsuch thatEnd(G)≈S2. LetL1, L2, . . .be distinct components ofH. Then defineGto have components M1, M2, . . .andN1, N2, . . .such thatMi 6=Nj andMi,Ni, andLiare isomorphic
for alli, j ∈N. The only homomorphisms between components ofGare the
iso-morphismshi :Mi −→ Ni. Thus for allα∈Ωthe set{αf : f ∈End(G)}has
two elements:αandαhiifα∈Mifor someiorαandαh−i1ifα∈Nifor somei.
IfΩis enumerated in such a way that{α2i−1, α2i}={α2if : f ∈End(G)}for all
i∈N, then clearlyEnd(G)≤S2and, in particular,End(G)4S2.
To show thatEnd(G)<S2=
f ∈ΩΩ:{α
2i−1f, α2if} ⊆ {α2i−1, α2i}for alli∈ N , letm1∈M1, m2∈M2, . . .be fixed, letg: Ω−→ {mi, mihi : i∈N}be defined
byα2i−1g =miandα2ig =mihi ∈Ni, and lethbe any mapping extendingg−1
to an element of ΩΩ. Iff ∈ S2 is arbitrary, then there exists
b
f ∈ End(G)such thatmifb= (α2i−1f)g ∈ {mi, mihi} andmihifb= (α2if)g ∈ {mi, mihi}. Hence
α2i−1gf hb =mif hb =α2i−1f gh=α2i−1f and, likewise,α2igf hb =α2if. Therefore
End(G)<S2, as required.
LetAut(G)denote the group of automorphisms from a graphGtoG. Acycle
of lengthnis a graphGwith verticesβ1, β2, . . . , βn and with edges(β1, βn)and
(βi, βi+1)for1≤i≤n−1.
We will show that End(G) ≈ Aut(G) ≈ S≤ and so rank(ΩΩ : End(G)) =
rank(ΩΩ : Aut(G)) = d. It is well-known (and not difficult to verify) that the image of any element inO2i+1 under an endomorphism ofGlies inO2j+1 with j ≤i; for a proof see [7, Corollary 1.4]. In other words,|O2i+1| ≤ifor alli∈N. It
follows, by Theorem 4.3, thatEnd(G)4S≤.
Letω(i,1), ω(i,2), . . . , ω(i,2i+ 1)be the vertices ofO2i+1. Then defineg, h∈ΩΩ
byαig=ω(i,1)and(ω(i, j))h=αjfor alli, j∈N.
Letf ∈S≤be arbitrary and lett:N−→Nbe the map such thatαif =αitfor all
i∈N. Note thatit≤i <2i+1for alliand so the vertexω(i, it)exists for alli. Now, for alli∈Nthere exists an automorphism ofO2i+1mappingω(i,1)toω(i, it). Let
b
f ∈ΩΩbe the union of these automorphisms. By definition,fb∈Aut(G)and
αigf hb = (ω(i,1))f hb = (ω(i, it))h=αit=αif.
ThusS≤⊆ hAut(G), g, hiand our claim follows.
Example 6.4. Ann-cliqueof a graphGis a subgraph ofGisomorphic to the com-plete graphKnwithnvertices. LetGbe a graph with only finite components and
letGhave arbitrarily largen-cliques. We will show thatrank(ΩΩ: End(G)) = 1. LetL1, L2, . . .be the components ofG. Then there exist infinitely many disjoint setsL0,L1,L2, . . .of components such that for allk∈N∪ {0}, the setLk contains
a component with ann-clique for alln∈N.
LetM1, M2, . . .be distinct elements ofL0whereMicontains a clique of size at
least|Li|for alli. Then definegto be any injective endomorphism so thatLigis
contained inMifor alli. Leth∈ΩΩbe any function which, forj≥1, maps every
vertex lying in a component belonging toLjtoαjand which maps the vertexαig
(belonging to one of the components inL0) into one of the components inLi.
Letf ∈ ΩΩbe arbitrary. Then let
b
f be any endomorphism ofGsuch that: if αj=αif, thenLfbequals the set of vertices of an|L|-clique in some component in Ljfor allL∈Liandαfb=αfor allαbelonging to a component inL0. Note that
sinceΩ ={α1, α2, . . .},iandjin the preceding definition are strictly greater than 0.
Ifαi ∈Ωis arbitrary, thenαighlies in a component inLi,i >0. Thus(αigh)fb
lies in a component inLj whereαj = αif andj > 0. So(αighfb)h =αj = αif.
Hencef =ghf hb andΩΩ=hEnd(G), hi.
The purpose of the next two examples is to show that Theorems 3.1 and 5.1 do not generalise to arbitrary reflexive binary relations.
Example 6.5. We construct a relationR onΩsuch that(Ω, R)is connected, not locally finite, andEnd(Ω, R) 4S≤. LetG = (Ω, E)be a connected, locally finite
graph, letB={(β0, γ) : γ∈Ω}for a fixedβ0∈Ω, and letR=E∪B∪∆Ω. The relationRwas constructed so that it is reflexive and(Ω, R)is not locally finite.
Let α, β ∈ Ω such that α, β are adjacent inG and let f ∈ End(Ω, R). Then (αf, βf) ∈Rand(βf, αf)∈R. Henceαf =βforαf andβf are adjacent inG. We conclude thatEnd(Ω, R) ⊆ End(Ω, E∪∆Ω) ≈ S≤ by Theorem 5.1(i) and so
u u u u u
u u u u u
u u u u @ @ @ @ @ @ @ @ @ @ @ @ - -- -
6 ? 6 ? 6
@ @ R @ @ R @ @ R @ @ R α1 α2 α3
α4 α5
α6 α7
α8 α9 α10 β1 β2 β3 β4 . . .
FIGURE 3. The binary relation from Example 6.6. The relations (α, α)for allα∈Ωare not shown.
Example 6.6. LetΩ ={α1, α2, . . .} ∪ {β1, β2, . . .}and define the following relation RonΩ. Let(α, α)∈Rfor allα∈Ωand let
(αi, αi+1),(α2i+2, α2i−1),(α2i−1, βi),(βi, α2i+2)∈R
for alli ∈ N. A diagram of(Ω, R)can be found in Figure 3. The relationR is
reflexive and(Ω, R)is connected and locally finite. We will prove thatEnd(Ω, R)4
F≺S≤.
Letf ∈End(Ω, R), letAi={α2i−1, α2i, α2i+1, α2i+2}, and letBi={α2i−1, α2i+2, βi}
for alli∈N. We start by proving that for alli∈None of the following holds:Aif
is a singleton,Aif = Aj, orAif =Bj for somej ∈N. We will also show that if
Aif =Aj, then
(1) βif =βjand(α2i−1f, α2if, α2i+1f, α2i+2f) = (α2j−1, α2j, α2j+1, α2j+2).
Sincef is a homomorphism,Aif = {γ1, . . . , γk}where1 ≤k ≤ 4and for all
1 ≤ j ≤ k−1we have that(γk, γ1),(γj, γj+1) ∈ R. The only subsets ofΩthat satisfy this condition are singletons,Aj, orBjfor somej ∈N. ThusAifis either a
singleton,Aif =Aj, orAif =Bjfor somej∈N.
In the case that,Aif =Aj, sincefis an endomorphism, we have that
(α2i+2f, α2i−1f),(α2i−1f, βif),(βif, α2i+2f)∈R.
The onlyγ, δ∈Ajwith(γ, δ)∈Rsuch that there existsλ∈Ωwith(δ, λ),(λ, γ)∈
Rareα2j−1andα2j+2. It follows thatβif =βjand(α2i−1f, α2if, α2i+1f, α2i+2f) =
(α2j−1, α2j, α2j+1, α2j+2).
We will now prove that there are only countably many elements ofEnd(Ω, R) with infinite image. Note that the only element ofΩnot in anyBjisα2. There are
3 cases to consider.
Case 1: A1f = Aj for some j ∈ N. In this case, from (1), β1f = βj and
(α1f, α2f, α3f, α4f) = (α2j−1, α2j, α2j+1, α2j+2). Since α3f andα4f are distinct,
A2f is not a singleton. Also if α3f ∈ Bi and α4f ∈ Bk, then i 6= k and so
A2f 6= Bi for alli ∈ N. HenceA2f = Ak for somek ∈ N. It follows from (1)
thatα3f =α2(j+1)−1andα4f =α2(j+1). ThusA2f =Aj+1and so again, from (1), α5f =α2(j+1)+1,α6f =α2(j+1)+2andβ2f =βj+1.
Repeating this process it follows thatαif = α2(j−1)+i and βif = β(j−1)+i for
alli ∈ N. In particular, there are only countably many endomorphismsf with
Case 2:A1f ⊆Bjfor somej∈N.In this case,
α3f, α4f ∈Bj ={α2j−1, α2j+2, βj}.
Since(α3f, α4f) 6= (α2k−1, α2k)for allk∈ N, it follows by (1) thatA2f 6=Ak for
allk∈N. Thus eitherA2f =BjorA2fis a single element ofBjand in either case
A2f ⊆Bj.
Repeating this argument, we conclude thatωf ∈ Bj for all ω ∈ Ωand f has
finite image.
Case 3: A1f ={α2}. In particular,α3f =α4f and so|A2f| < 4. Thus by (1) A2f 6=Ak for allk ∈N. Furthermore,α2 6∈Bkfor allk∈Nand soA2f 6=Bk for
allk∈N. ThusA2f ={α2}. Repeating this argument it follows thatim(f) ={α2}.
Since there are only countably many endomorphisms of (Ω, R) with infinite image we conclude thatEnd(Ω, R)4F ≺S≤. Note that, on the other hand, it is
possible to show that|End(Ω, R)|= 2ℵ0 and soEnd(Ω, R) {1Ω}.
7. EXAMPLESII –GRAPHS WITH RANK2
In this section we construct two examples of graphsG, one connected and one with infinitely many components, such thatrank(ΩΩ: End(G)) = 2.
Lemma 7.1. LetU be a subsemigroup ofΩΩsuch thatf ∈Uis injective if and only iff
is surjective. Thenrank(ΩΩ:U)≥2.
Proof. Letg∈ΩΩbe arbitrary. Seeking a contradiction assume thathU, gi= ΩΩ. Let h ∈ ΩΩ be injective but not surjective and let k ∈ ΩΩbe surjective but not injective. Then there existh1, h2, . . . , hm, k1, k2, . . . , kn ∈ U ∪ {g}such thath =
h1h2· · ·hmandk=k1k2· · ·kn.
Let
M = min{i : h1h2· · ·hiis not surjective}
and
N = max{i : kiki+1· · ·knis not injective}.
ThenhM is injective, ashis injective, and sohM = g. On the other hand,kN is
surjective, askis surjective, and sokN =g. But thengis injective and not injective,
a contradiction.
An example of a connected but not locally finite poset (Ω,v)where the only injective or surjective endomorphism is the identity is given in [9, Section 6]. It follows from Theorem 3.1 and Lemma 7.1 thatrank(ΩΩ: End(Ω,v)) = 2. We will use this poset to define a bipartite graph with the same property. The poset(Ω,v) is described as follows.
LetA = {ai : i ∈ N}be a countably infinite set. LetE denote the set of all
finite subsetsEofAsuch that|E| ≥2and wherean∈Eimplies that|E| ≤n+ 1.
Thus any set inE containinga1 has cardinality2, any set inE containinga2has cardinality 2 or 3, any set in E containing a3 has cardinality 2, 3 or 4, etc. We enumerate the elements ofEasA1, A2, . . .Now, we assign in a one-to-one way a new elementbE, not inA, to everyE inE. LetB = {bE : E ∈ E }. Also, let
s c0 s c2 s c4 s c6 s c8 s c10 s c1 s c3 s c5 s c7 s c9 s c11 A A A A A A A A A A A A @ @ @ @ @ @ Q Q Q Q Q Q Q Q Q H H H H H H H H H H H H a a a a a a a a a a a a a a a . . .
FIGURE4. The posetvrestricted toC
s J J J J J J J J J J
bAi
Ai
s s s c2si
s
c2i+2
s
s s s s
c2i+1
s s . . . .
FIGURE5. A portion of the poset(Ω,v).
We define the partial order von the elements of Ω = A∪B ∪C by: a v bEfor alla∈E;c2i+1 vc0for alli ≥0;xvc2for allx∈ {c1, c3, c5};c2i−1vc2i,
c2i+1vc2i for alli≥2; andc2i+1 vbAi for alli≥0. See Figures 4 and 5 for two
diagrams of portions of(Ω,v).
Theorem 7.2. Letvbe the partial order defined above and letf ∈End(Ω,v)be injective or surjective. Thenf is the identity mapping onΩ.
For a proof see [9, Theorem 6.7].
We construct a graphG= (Ω, E)from the poset(Ω,v)by letting (α, β),(β, α)∈Ewheneverα6=βandαvβ.
LetP =A∪ {c2i+1 : i∈N∪ {0} }andQ=B∪ {c2i : i∈N∪ {0} }. Note that if
α, β∈Ωwithα6=βandαvβ, thenα∈Pandβ ∈Q. Note that every edge inG connects a vertex inPto one inQand soGis bipartite.
Lemma 7.3. Let f ∈ End(G). If there exists α ∈ P such that αf ∈ P, then f ∈ End(Ω,v). Likewise, if there existsα∈Qsuch thatαf∈Q, thenf ∈End(Ω,v).
Using Lemma 7.3 we prove that the graph obtained from(Ω,v)has no non-identity injective or surjective endomorphisms. To do so, we will make use of the following notion.
IfRis a binary relation onΩ,α∈Ωandn∈N, then let
B(α, n) ={β∈Ω : there exists a path of length at mostnfromαtoβ}. The proof of the following lemma is straightforward and omitted.
Lemma 7.4. Let R ⊆ Ω×Ωand letα ∈ Ω. If B(α, n) = Ωfor somen ∈ N and
f ∈End(Ω, R)is surjective, thenB(αf, n) = Ω.
Theorem 7.5. LetGbe the graph defined above and letf ∈ End(Ω, G)be injective or surjective. Thenf is the identity mapping onΩ.
Proof. Letg ∈ End(G)be injective. Note that all vertices ofA ⊆ P have infinite degree butc0is the only vertex ofQwith infinite degree. Since injective endomor-phisms map vertices of infinite degree to vertices of infinite degree, it follows that ag ∈Qfor at most onea∈A. In particular, there existsa∈Asuch thataf ∈P and so, by Lemma 7.3,g ∈ End(Ω,v). By Theorem 7.2 this implies thatgis the identity onΩ.
Leth∈End(G)be surjective. We will show thatc0h=c0. From the definition ofGwe have thatB(c0,1) ={c0} ∪ {c2i+1 : i∈N∪ {0} }and thusB(c0,2) =B∪C andB(c0,3) = Ω. We will prove thatB(α,3)6= Ωfor allα6=c0.
Ifai∈A, thenB(ai,3)∩ {c2k+1 : k∈N∪ {0} }={c2j+1 : ai∈Aj} 6={c2k+1 : k∈N∪ {0} }. IfbE∈B, thenB(bE,3)∩B ={bF ∈B : E∩F 6=∅ } 6=B. Ifi≥0,
thenB(c2i+1,3)∩A={aj∈A : aj∈Ai} 6=A. Finally, ifi≥1, thenB(c2i,3)∩A
is finite.
Thusc0is the unique vertexαofGsuch thatB(α,3) = Ω. It follows by Lemma 7.4 thatc0h=c0. Thush∈End(Ω,v)by Lemma 7.3 and hencehis the identity on Ωby Theorem 7.2.
Corollary 7.6. LetGbe the graph obtained from(Ω,v). Thenrank(ΩΩ: End(G)) = 2.
Proof. Since G is bipartite and not locally finite, by Theorem 4.5(ii), rank(ΩΩ : End(G)) ≤ 2. On the other hand,Ghas no non-identity injective or surjective endomorphisms by Theorem 7.5. Thusrank(ΩΩ: End(G))≥2by Lemma 7.1.
The following example shows that there are graphsGwith infinitely many com-ponents andrank(ΩΩ : End(G)) = 2. We require the following notion. A graph is acoreif every endomorphism is an automorphism. IfGis a graph where every component is a core and no two components are isomorphic, then the preorder defined in Section 4 is a partial order on the set of components ofG.
Theorem 7.7. [7, Theorem 3.3]LetP be a countable poset. Then there exists a graph
G where every component is a finite core and the set of components ofG under is isomorphic toP.
Example 7.8. LetGbe a graph with finite components the distinct coresL1, L2, . . . and M1, M2, . . .such that there exists a homomorphism fromLi −→ Mj for all
Now rank(ΩΩ : End(G)) ≤ 2 by Theorem 4.3. Furthermore, every injective endomorphism ofGmust fix each component setwise. It follows that every injec-tive endomorphism is surjecinjec-tive. Likewise all surjecinjec-tive endomorphisms are also injective. So, using Lemma 7.1, we conclude thatrank(ΩΩ: End(G)) = 2.
8. EXAMPLESIII – ATOLERANCE WITH RANK2
LetΩ =A∪BwhereAandBare the sets defined in Section 7 and letvbe the partial order defined in Section 7 restricted toA∪B.
Lemma 8.1. Letvbe the partial order defined above and letf ∈End(Ω,v)be surjective. Thenf is the identity mapping onΩ.
For a proof see [9, Lemma 6.5].
We define a toleranceRbased onvby letting(α, β),(β, α)∈Rwheneverα=β orαvβ.
The following lemma is routine and the proof omitted.
Lemma 8.2. Iff ∈End(Ω, R)such thatAf⊆A, thenf ∈End(Ω,v).
Next, we prove that(Ω, R)has no non-identity surjective endomorphisms. Lemma 8.3. LetRbe the tolerance defined above and letf ∈ End(Ω, R)be surjective. Thenf is the identity mapping onΩ.
Proof. Letai ∈A. For anyaj ∈Athere existsbE ∈B such thatai, aj ∈E. Hence
B(ai,2) ⊇Aand soB(ai,3) = Ω. On the other hand, ifbE ∈B is arbitrary, then
B(bE,3)∩B = B(bE,2)∩B ={bF ∈ B : E∩F 6= ∅ } 6= B by construction.
ThusB(bE,3)6= Ω. Letf ∈End(Ω, R)be surjective. It follows by Lemma 7.4 that
Af ⊆A. Hencef ∈End(Ω,v)by Lemma 8.2 and thusf is the identity onΩby
Lemma 8.1.
Although(Ω, R)has no non-identity surjective endomorphisms, it does have injective endomorphisms that are not surjective. So, in order to apply Lemma 7.1, we will define a new toleranceR∗ on a setΣbased on(Ω, R)such that f ∈ End(Σ, R∗)is injective if and only iffis surjective.
Let{c(i, j) : i, j∈N}be a set of new points with no elements inAandB, let
Bbe as above, leta∗i ={c(i,1), c(i,2), . . . , c(i, i+ 2)}, letC=a∗1∪a∗2∪ · · · and let Σ = B∪C. Then defineR∗ to be the symmetric and reflexive closure of the set containing:
(i) (c(i, j), c(i, j+ 1))for allj∈ {1, . . . , i+ 1}and(c(i, i+ 2), c(i,1))for alli; (ii) (bE, c)for allc∈ai∗and for allisuch thatai∈E.
Note thata∗i is a cycle of lengthi+ 2for alli.
Theorem 8.4. Let (Σ, R∗) be the tolerance defined above. Then f ∈ End(Σ, R∗) is injective if and only iff is surjective.
Proof. Letc(i, j) ∈CandbE ∈ B. Then, by a similar argument to the one in the
proof of Lemma 8.3,B(c(i, j),3) = ΣandB(bE,3) 6= Σ. Letf ∈ End(Σ, R∗)be
f is a homomorphism, for anyi ∈Nwe have thata∗if ⊆a∗j for somej ∈N. We
may thus definefb∈ΩΩ(recall thatΩ =A∪B) by
αfb=
aj ifα=aianda∗if ⊆a∗j
aj ifα∈Bandαf ∈a∗j
αf ifα∈Bandαf ∈B.
Thenfbis surjective sincef is surjective. Moreover, ifai ∈ E, then(ai, bE) ∈ R
and so(c(i, j), bE)∈R∗for allj. Hence(c(i, j)f, bEf)∈ R∗for allj. Ifa∗if ⊆a∗j
andbEf ∈C, thenbEf ∈a∗j and so(aif , bb Efb) = (aj, aj)∈R. Otherwise,bEf =
bF ∈Bfor someF ∈ Eand soaj∈F. Hence(aif , bb Efb) = (aj, bF)∈R. Therefore b
f ∈End(Ω, R)and it follows thatfbis the identity by Lemma 8.3. Thereforebf =b
for allb∈Band the componentsa∗jare fixed setwise byf. Since everya∗jis finite andf is surjective, it follows thatf ∈Aut(Σ, R∗).
Letf ∈ End(Σ, R∗)be injective. Since every element inChas infinite degree and every element in B has finite degree, it follows that Cf ⊆ C. Hence for all i ∈ Nwe have that a∗if ⊆ a∗j for some j ∈ N. But since f is injective and |a∗
if|=|a∗i| =i+ 2it follows thatj ≥i. On the other hand, there does not exist
an injective homomorphism from the cyclea∗i to any cyclea∗jwherej > i. Hence i=jand sof ∈Aut(Σ, R∗). Corollary 8.5. Let(Σ, R∗)be the tolerance defined above. Thenrank(ΣΣ: End(Σ, R∗)) =
2.
Proof. By Theorem 5.1,rank(ΣΣ : End(Σ, R∗))≤2. By Theorem 8.4 and Lemma
7.1,rank(ΣΣ: End(Σ, R∗))≥2.
REFERENCES
[1] T. Bartoszy ´nski, H. Judah, and S. Shelah, The Cicho ´n diagram,J. Symbolic Logic58(1993) 401–423. [2] G. M. Bergman and S. Shelah, Closed Subgroups of the infinite symmetric group,Algebra
Univer-salis55(2006) 137–173.
[3] P. J. Cameron,Combinatorics: topics, techniques, algorithms. Cambridge University Press, Cam-bridge, 1994.
[4] J. Cicho ´n, J. D. Mitchell and M. Morayne, Generating continuous mappings with Lipschitz map-pings,Trans. Amer. Math. Soc.359(2007) 2059–2074.
[5] J. Cicho ´n, J. D. Mitchell, M. Morayne and Y. P´eresse, Relative ranks of Lipschitz mappings on countable discrete metric spaces, submitted.
[6] F. Galvin, Generating countable sets of permutations,J. London Math. Soc.(2)51(1995) 230–242. [7] P. Hell, J. Neˇsetˇril,Graphs and Homomorphisms, Springer Verlag, New York, Heidelberg, Berlin. [8] P. Hell and J. Neˇsetˇril, Groups and monoids of regular graphs (and of graphs with bounded
de-grees),Canad. J. Math.25(1973) 239–251.
[9] P. M. Higgins, J. D. Mitchell, M. Morayne and N. Ruˇskuc, Rank properties of endomorphisms of infinite partially ordered sets, Bull. London Math. Soc.38(2006) 177–191.
[10] K. Kunen,Set Theory, North Holland, Studies in Logic, Amsterdam, New York, Oxford, Tokyo, 1980.
[11] Z. Mesyan, Generating self-map monoids of infinite sets,Semigroup Forum75(2007) 649–676. [12] W. Sierpi ´nski, Sur les suites infinies de fonctions d´efinies dans les ensembles quelconques,Fund.
Math.24(1935) 209–212.
e-mail: [email protected]
J. D. Mitchell, Y. P´eresse, and M. R. Quick: Mathematics Institute, University of St Andrews, North Haugh, St Andrews, Fife, KY16 9SS, Scotland.
