Reconstructing the topology on monoids and polymorphism clones of reducts of the rationals
aa r X i v : . [ m a t h . L O ] M a r Reconstructing the topology on monoids and polymorphismclones of reducts of the rationals
John K. Truss, University of Leeds, andEdith Vargas-Garc´ıa, ITAM, Mexico.
Abstract
We extend results from an earlier paper giving reconstruction results for the endomorphismmonoid of the rational numbers under the strict and reflexive relations to the first order reductsof the rationals and the corresponding polymorphism clones. We also give some similar resultsabout the coloured rationals.
In [3] we showed how to reconstruct the topology on the endomorphism monoid of the set of rationalnumbers, under the strict and reflexive relations < and ≤ and the polymorphism clone Pol( Q , ≤ ).This is a so-called ‘automatic-homeomorphicity’ result, meaning that isomorphisms of a certain formmust necessarily also be homeomorphisms. In this paper we extend these results to the reducts ofthe rationals described by Cameron in [6], which are the betweenness relation, circular order, andseparation relation on Q . (The other reduct, namely the trivial structure, is already treated in[4].) We also consider the case of the coloured version of the rationals Q C for a set of colours with2 ≤ | C | ≤ ℵ , here just dealing with the analogues of the reducts of Q rather than all reducts(which according to [8] may be quite complicated). In most cases the earlier results can be invokedfairly directly, or else suitably adapted. En route we need to verify the small index property for theautomorphism group of the circular rational order ( Q , circ ), and the corresponding results in thecoloured case, which may or may not have been remarked before (or are ‘folklore’).By saying that a transformation monoid M on a countable set Ω has automatic homeomorphicity is meant that any isomorphism from M to a closed submonoid of the full transformation monoid T r (Ω ′ ) on a countable set Ω ′ is necessarily a homeomorphism. Here the topology on M is givenby taking as basic open sets B BC = { f ∈ M : f B = C } where B and C are finite subsets of Ω.The motivation for studying such properties is that in the case where M is a (closed) subgroup ofthe symmetric group on Ω, M has automatic homeomorphicity if and only if it has the small indexproperty SIP, which says that any subgroup of M of index < ℵ contains the pointwise stabilizerof a finite set. Thus automatic homeomorphicity represents a statement which one can attempt toestablish even when the statement of SIP makes no sense (in the monoid case, where we do not haveLagrange’s Theorem).The key steps presented in [3] were as follows. For M = Emb( Q , ≤ ), the monoid of order-preserving embeddings of Q , we established that any injective endomorphism of M which fixesall elements of G = Aut( Q , ≤ ), is the identity. Then invoking the fact that G has the small index Department of Pure Mathematics, University of Leeds, Leeds LS2 9JT, UK, and Departamento Acad´emico deMatem´aticas, ITAM. R´ıo Hondo No. 1, Col. Tizap´an San Angel, Del. Alvaro Obreg´on, C.P. 01080 Ciudad de M´exico,e-mails [email protected], [email protected]. The second author gratefully acknowledges the financial supportof Asociaci´on Mexicana de Cultura A.C. M is the closure of G under the above topology, and Lemma 12 from [4], automatichomeomorphicity of M follows at once. To prove automatic homeomorphicity of E = End( Q , ≤ ), inwhich G is not dense, a more detailed analysis of the way that E acts on Ω ′ was required; and thecorresponding proof for the polymorphism clone Pol( Q , ≤ ) was given.Here we follow a similar method for the following cases: ( Q , betw ), ( Q , circ ), and ( Q , sep ), whichare defined as follows: betw ( x, y, z ) if x ≤ y ≤ z or z ≤ y ≤ x , circ ( x, y, z ) if x ≤ y ≤ z or y ≤ z ≤ x or z ≤ x ≤ y , sep ( x, y, z, t ) if circ ( x, y, z ) ∧ circ ( x, t, y ) or circ ( x, z, y ) ∧ circ ( x, y, t ).We also treat the analogues for these of the C -coloured rationals Q C , where 2 ≤ | C | ≤ ℵ , whichis defined to be Q together with a colouring function F : Q → C such that for all c ∈ C , F − { c } is a dense subset of Q . This exists and is unique up to isomorphism (and is homogeneous and forthe case that C is finite, ℵ -categorical). The three structures that we treat for Q are the proper‘reducts’ of ( Q , ≤ ), as was shown in [6]. The analogous structures for the coloured case are indeedreducts of ( Q C , ≤ ); however, they are not all the reducts here, and the complete list of reducts of( Q C , ≤ ) is currently unknown; see [8] section 6 for a discussion of this case (for finite C ).All these structures come in reflexive versions, as we have stated them, and strict versions.However, in the present context it is good enough to distinguish these by means of monoids whichcapture the same ideas. Thus Emb( Q , betw ) for instance is the family of 1–1 maps of Q whichpreserve betw , and these are precisely the maps which preserve the corresponding ‘strict betweenness’ x < y < z ∨ z < y < x . As in [3], automatic homeomorphicity is established for the monoid ofembeddings (the endomorphisms preserving the strict relation), the monoid of all endomorphisms(preserving just the reflexive relation), and the polymorphism clone for the reflexive relation.Finally we give the result from section 6 of [3] which deduces automatic homeomorphicity forthe polymorphism clone generated by a monoid in the cases discussed here.Throughout we use G , M , E , P to stand for the automorphism group of the structure beingconsidered, or its monoid of embeddings, or endomorphisms, or the polymorphism clone of thereflexive relation, respectively.We outline the methods used in [3], and explain how they are adapted here. Of the threecases treated, that of the monoid of embeddings M is the easiest, in that we can concentrate onproving a combinatorial lemma, and then appeal to the results of [4] mentioned to complete theproof. The combinatorial lemma proved, which is attractive in its own right, is that any injectiveendomorphism ξ of M which fixes G pointwise is equal to the identity. That this statement isrelated to reconstruction matters is plausible, in that we are saying that somehow the monoid canbe ‘captured’ or ‘described’ from the group (and reconstruction from the group is just the smallindex property). The method for proving this lemma presented in [3] involved consideration of acertain family Γ of members of M , which intuitively are those whose image is as ‘spread out’ aspossible. It was shown that members of Γ are fixed by ξ , and the proof was completed by showingthat all members of M can be suitably expressed in terms of members of Γ. In fact, for ( Q , ≤ ) (andin this paper also for ( Q C , ≤ )) it was also necessary to consider variants of Γ, written Γ + , Γ − , andΓ ± , allowing for members of M with support bounded above or below or both. Here since we aredealing with suitable reducts, mainly based on the circular ordering of Q , only the analogue of Γis needed. For completeness we give the construction and arguments, even though these amount tostraightforward modifications of the ones given in [3].The harder part of the argument given in [3] was for the monoid of endomorphisms E of ( Q , ≤ ).The scenario we have to consider is that in which there is an isomorphism θ from E to a closedsubmonoid E ′ of the full transformation monoid T r (Ω) on a countable set Ω, and we have to showthat it is a homeomorphism. The first step was to show that the image of M under θ is a closedsubset of E ′ , which was Lemma 4.1 of [3]. From that we deduce by what has already been shownthat on M , θ is a homeomorphism. Next we consider the orbits of the action of G on Ω, andshow by appeal to the small index property that each of these can be identified with the family[ Q ] n of n -element subsets of Q for some n ≥
0, termed the ‘rank’ of the orbit. Thus we may writeΩ = S i ∈ I Ω i for some index set I , where Ω i = { a iB : B ∈ [ Q ] n i } where n i is the rank of the orbitΩ i , and furthermore the action is ‘natural’ in the sense that for each g ∈ G , θ ( g )( a iB ) = a igB . Next2t is shown that this naturality extends to the action of M , and even to members of E insofar asthey act injectively on the relevant set B . For this it is fairly easy to check directly that θ is ahomeomorphism.Here for completeness we go over the similar arguments in reasonable detail in the first casetreated, namely that of the betweenness relation on Q . All the other cases follow a similar pattern,so in these we omit most of the details, concentrating on the aspects which are genuinely different. Q We show that it is quite easy to ‘lift’ the automatic homeomorphicity results for ( Q , < ) and ( Q , ≤ )proved in [3] to the corresponding betweenness relation, essentially exploiting the fact that theoriginal group has index 2 in the bigger one. The following basic lemma is needed. Lemma 2.1. If B and B are finite subsets of Q , and G = Aut( Q , ≤ ) , then G B ∩ B = h G B , G B i (where these are the pointwise stabilizers).Proof. The fact that h G B , G B i ≤ G B ∩ B is immediate, so we concentrate on the reverse inclusion.Let g ∈ G B ∩ B , and we show by induction on n = |{ q ∈ B ∪ B : g ( q ) = q }| that g ∈ h G B , G B i .If n = 0 then already g ∈ G B . Otherwise choose the greatest q ∈ B ∪ B moved by g , andassume that g ( q ) > q (which may be arranged by passing to g − if necessary). Thus all members of( q, ∞ ) ∩ ( B ∪ B ) are fixed by g . Then there is h ∈ G which fixes all members of B ∪ B \ { q } andagrees with g on q . Thus h ∈ G B or G B and h − g ∈ G q . But h − g also fixes all members of B ∪ B which are fixed by g , so by induction hypothesis, lies in h G B , G B i . Hence also g ∈ h G B , G B i .From this lemma we can deduce that if H is a subgroup of G which contains G B for some finite B ⊆ Q , and B is of minimal size such that this is true, then H equals G B . To see this, take any h ∈ H . Then H = hHh − ≥ hG B h − = G hB , so by the lemma, H ≥ G B ∩ hB . By minimality of B , hB = B , so h ∈ G { B } = G B . In the other cases we consider the setwise and pointwise stabilizersmay not be equal, so there are more possibilities for H , but it is still always the case that if B is ofleast size such that H ≥ G B , then G B ≤ H ≤ G { B } (which we verify in the individual cases).In what follows we have occasion several times to use a key result from [4] (their ‘Lemma 12’),which we quote here: Lemma 2.2.
Let M be a closed submonoid of O (1) whose group of invertible elements G is densein M and has automatic homeomorphicity. Assume that the only injective endomorphism of M thatfixes every element of G is the identity function id M on M . Then M has automatic homeomorphicity. Theorem 2.3.
Emb( Q , betw ) has automatic homeomorphicity.Proof. Writing M and G for the monoid of embeddings of ( Q , betw ) to itself, and its automorphismgroup, respectively, we first note that G is dense in M . This is because any order-reversing memberof M is the composition of an order-preserving member of M (which can be arbitrarily well approx-imated by members of G ∩ Aut( Q , ≤ )) and the map i sending q to − q for all q which lies in G . Sothis means that we can appeal to Lemma 2.2 as before, and focus on consideration of an injectiveendomorphism ξ of M which fixes G pointwise.We recall the set Γ from [3], which comprises those order-preserving embeddings f of Q whoseimage arises from a copy of the 2-coloured rationals in which each point is replaced by an intervalisomorphic to Q and each red interval contains exactly one point of the image of f and each blueinterval is disjoint from the image of f . In the proof of Lemma 2.2 in [3] we defined S ( g ) = { ( α, β ) ∈ G : αg = gβ } for any order-preserving embedding g , where G is the subgroup of order-preservingpermutations. We use the same notation, even if g is order-reversing (though the group elementsconsidered at this point have to be order-preserving). Now let i be the involution above. Then g is order-preserving if and only if ig is order-reversing, and vice versa. We now find the connectionbetween S ( g ) and S ( ig ): S ( ig ) = { ( α, β ) ∈ G : αig = igβ } = { ( α, β ) ∈ G : iαig = gβ } = { ( iαi, β ) ∈ G : αg = gβ } (since as α ranges over G , so does iαi ) = { ( iαi, β )) : ( α, β ) ∈ S ( g ) } .3t was shown in [3] that for any order-preserving embedding g ∈ Γ, and rationals u and s , g ( u ) = s if and only if ∀ ( α, β ) ∈ S ( g )( β ( u ) = u → α ( s ) = s ). We can now deduce the sameequivalence in the case that g ′ is an order-reversing embedding, of the form g ′ = ig for g ∈ Γ. For g ′ ( u ) = s ⇔ g ( u ) = i ( s ) ⇔ ∀ ( α, β ) ∈ S ( g )( β ( u ) = u → α ( i ( s )) = i ( s )) ⇔ ∀ ( α, β ) ∈ S ( g ′ )( β ( u ) = u → iαi ( i ( s )) = i ( s )) ⇔ ∀ ( α, β ) ∈ S ( g ′ )( β ( u ) = u → α ( s ) = s ). The argument given towards theend of the proof of Lemma 2.2 in [3] shows that S ( ξ ( g )) = S ( g ), and we can therefore deduce that ξ ( g ) = g precisely as in the previous case. Since ξ fixes all members of Γ, by our earlier result, italso fixes the order-preserving members of M pointwise, and multiplying by i (which is known to befixed by ξ ), it follows that it also fixes all of M pointwise.In fact it is also necessary to consider members of associated classes Γ + , Γ − , and Γ ± —see [3] forthe precise definitions, but the details are similar and are omitted here.The remaining point to check to be able to appeal to Lemma 2.2, is to verify the small indexproperty for Aut( Q , betw ), but this follows easily from the fact that it holds for Aut( Q , ≤ ), andAut( Q , ≤ ) has index 2 in Aut( Q , betw ).Now we move on to consider the reflexive relation, in other words, the monoid E = End( Q , betw ).As in [3] we are given an isomorphism θ from E to a closed submonoid of T r (Ω) for some countableset Ω, and our task is to show that it is a homeomorphism. We may partition Ω into orbits under G . The identification of each orbit X with a set of the form [ Q ] n for some n is however not quiteso straightforward, and this is because although by the small index property, the stabilizer G x of amember x of X contains the pointwise stabilizer G B of a finite B ⊆ Q of minimal size, it may not equal it, since it may contain only order-preserving permutations, or alternatively order-preservingand reversing ones. The upshot of this is that X may either be identified with some [ Q ] n or the setof increasing or decreasing orderings of n -element subsets of Q under the natural action.The easiest method to reach this conclusion is to note that G x equals either the setwise orpointwise stabilizer of B in G , G { B } or G B . This may be deduced from the order-preserving caseby considering G x ∩ Aut( Q , ≤ ), which has index either 1 or 2 in G x . In each case, the orbit X containing x is equal to { θ ( g )( x ) : g ∈ G } , but the indexing to identify it with [ Q ] n or the set ofincreasing or decreasing orderings of n -element subsets of Q is a little different in the two cases.We first treat the case that G x = G { B } . For each g ∈ G , we write a gB = θ ( g )( x ). To justify thisnotation we note that for all g , g ∈ G , g B = g B ⇔ g − g ∈ G { B } ⇔ g − g ∈ G x ⇔ θ ( g )( x ) = θ ( g )( x ). This shows that the orbit X can be precisely identified with { a gB : g ∈ G } , and as G acts transitively on the family of n -element subsets of Q , also with [ Q ] n . More to the point, thisidentification also corresponds to the natural (left) action of G , since θ ( f )( a gB ) = θ ( f ) θ ( g )( x ) = θ ( f g )( x ) = a fgB .In the second case, G x = G B . This time we write a gB = θ ( g )( x ) for each g , where B is B together with its ordering (as a subset of Q ). Note that gB is ordered in the ‘correct’ way (as asubset of Q ) if g ∈ Aut( Q , ≤ ), and with the reverse ordering otherwise. This time the calculationjustifying the notation is as follows: g B = g B ⇔ g − g ∈ G B ⇔ g − g ∈ G x ⇔ θ ( g )( x ) = θ ( g )( x ).So there are two orbits of X under the action of Aut( Q , ≤ ), namely those in which the orderingof gB agrees with that of Q , or disagrees with it. Once more, the notation is respected by the leftaction of G since θ ( f )( a gB ) = θ ( f ) θ ( g )( x ) = θ ( f g )( x ) = a fgB .In the special case | B | = 0 or 1, G B = G { B } , and we choose the ‘first’ case. In each case we saythat X is an orbit of ‘rank’ n .This discussion enables us to write Ω = S i ∈ I Ω i where Ω i are the orbits, containing x i say, andassociated with each i ∈ I we have the rank n i of Ω i , and t i = 0 or 1, its ‘type’, being 0 if G x i = G { B } for some B , and 1 if G x i = G B for some B . We also write Ω i = { a iB : B ∈ [ Q ] n i } in the first case,and Ω i = { a iB : B ∈ [ Q ] n i , B the increasing or decreasing ordering of B } in the second.The first step in establishing automatic homeomorphicity is to use the result already proved for M . For that it is necessary to know that the image of M under θ is a closed submonoid of T r (Ω).This is provided by the following key result from [3] (Lemma 4.1):If θ : E → E ′ is an isomorphism where E ′ is a closed submonoid of the full transformationmonoid T r (Ω) on a countable set Ω, then the image M ′ of M is closed in T r (Ω) and the restriction4 ↾ M : M → M ′ is a homeomorphism.We do not need to reprove this here, since G also has automatic homeomorphicity and is densein M . Applying this here, we may appeal to automatic homeomorphicity of M to deduce that on M , θ is a homeomorphism. Lemma 2.4.
For each f ∈ M and i ∈ I , if t i = 0 then for each B ∈ [ Q ] n i , θ ( f )( a iB ) = a ifB , and if t i = 1 then for each B ∈ [ Q ] n i and increasing or decreasing ordering B of B , θ ( f )( a iB ) = a ifB .Proof. This is proved by a continuity argument. Since G is dense in M , there is a sequence ( g n ) ofmembers of G having f as limit. We treat just the case t i = 0 (and t i = 1 is done similarly). Let B = { q , . . . , q n i } and r k = f ( q k ). Then for each k f lies in the basic open set B qr = { h ∈ M : h ( q ) = r } , and as g n → f there is N k such that ( ∀ n ≥ N k ) g n ∈ B q k r k . Hence if n ≥ max ≤ k ≤ n i N k , g n ∈ T ≤ k ≤ n i B q k r k , so g n ( B ) = f ( B ).As remarked above, the restriction of θ to M is continuous. Hence θ ( g n ) → θ ( f ). Let θ ( f )( a iB ) = a jC . Thus θ ( f ) ∈ C ijBC . From θ ( g n ) → θ ( f ) it follows that ( ∃ N )( ∀ n ≥ N ) θ ( g n ) ∈ C ijBC . Hence forthis N , ( ∀ n ≥ N ) θ ( g n )( a iB ) = a jC . But we know that θ ( g n )( a iB ) = a ig n ( B ) as g n ∈ G . Hence for such n , j = i and g n ( B ) = C . Taking n ≥ N, max ≤ k ≤ n i N k , it follows that j = i and C = g n ( B ) = f ( B ).Thus θ ( f )( a iB ) = a if ( B ) as required.We extend this even to some actions of members of E . Lemma 2.5. If f ∈ E , i ∈ I , if t i = 0 , and a iB ∈ Ω i , where | f ( B ) | = n i = | B | , then θ ( f )( a iB ) = a if ( B ) , and if t i = 1 , and a iB ∈ Ω i , where B is the increasing or decreasing ordering of B , and | f ( B ) | = n i = | B | , then θ ( f )( a iB ) = a if ( B ) .Proof. It is easiest to appeal to the methods of [3] by composing, in the order-reversing case, withan order-reversing automorphism of Q which fixes B (setwise). First then suppose that f is order-preserving and surjective. Then there is an order-preserving h ∈ E such that for each q , f h ( q ) = q ,obtained by choosing h ( q ) ∈ f − ( q ), and such h is necessarily injective, so lies in M . Furthermore,if q ∈ B we let hf ( q ) = q (possible since f is 1–1 on B ). By appealing to Lemma 2.4, θ ( f )( a iB ) = θ ( f )( a ihfB ) = θ ( f ) θ ( h )( a ifB ) = θ ( f h )( a ifB ) = a ifB , in the case t i = 0, with a similar argument if t i = 1.Next if f is order-preserving but not necessarily surjective, as shown in [3] Lemma 3.3, we maywrite f = f f where f ∈ S and f ∈ M are order-preserving. Then | f f B | = | B | = | f B | .Hence by the surjective case just done, θ ( f )( a if B ) = a if f B = a ifB , so by Lemma 2.4 again, θ ( f )( a iB ) = θ ( f ) θ ( f )( a iB ) = θ ( f )( a if B ) = a ifB .Finally, suppose that f is order-reversing. Then f j is order-preserving where j is an (order-reversing) involution fixing B setwise, so θ ( f )( a iB ) = θ ( f j ) θ ( j )( a iB ) = θ ( f j ) a ijB = a ifj B = a ifB .If f ∈ E ‘collapses’ a set B , then we can certainly not deduce that θ ( f )( a iB ) = a jC for j = i ,since Ω i and Ω j will have different ranks. If B = ∅ , in [3] was shown that there is an idempotentorder-preserving endomorphism h whose image is B , from which it follows by Lemma 2.5 that θ ( h )( a iB ) = a iB , and this is also valid, even if the subscript is ordered. Lemma 2.6.
Let i ∈ I and B ∈ [ Q ] n i , B = ∅ . If t i = 0 and a iB ∈ Ω i , then there is an idempotentorder-preserving endomorphism h ∈ E having B as image such that θ ( h )( a iB ) = a iB , and if t i = 1 ,and a iB ∈ Ω i , where B is the increasing or decreasing ordering of B , then there is an idempotentorder-preserving endomorphism h ∈ E having B as image such that θ ( h )( a iB ) = a iB .Proof. The case when t i = 0 was considered in Lemma 4.4 of [3]. If t i = 1, then similarly to Lemma4.4 of [3] we obtain an idempotent endomorphism h ∈ E fixing all elements of B and satisfyingim( h ) = B by subdividing Q into | B | pairwise disjoint intervals, each containing a single member of B , and mapping the whole of each such interval to the member of B it contains. Since h ( B ) = B where B ∈ [ Q ] n i , we can apply Lemma 2.5 to get θ ( h )( a iB ) = a ih ( B ) = a iB .5or the proof of openness in the main theorem, we still need some information about C , namelythat it is contained in f ( B ). Lemma 2.7. If f ∈ E , i, j ∈ I , and B ∈ [ Q ] n i , C ∈ [ Q ] n j are such that θ ( f )( a iB ) = a jC (possiblywith orderings on the subscripts, depending on the values of t i , t j ), then C ⊆ f B .Proof. By Lemma 2.6, if B = ∅ there is an idempotent order-preserving endomorphism h whoseimage is B , from which it follows by Lemma 2.5 that θ ( h )( a iB ) = a iB , and this applies here too,even if the subscript is ordered. From this it follows that if f , f ∈ E are order-preserving, andthey agree on their actions on B , then θ ( f )( a iB ) = θ ( f )( a iB ) since f h = f h , so that θ ( f )( a iB ) = θ ( f )( a ihB ) = θ ( f ) θ ( h )( a iB ) = θ ( f h )( a iB ) = θ ( f h )( a iB ) = θ ( f )( a iB ). This even applies if B = ∅ ,since then by Lemma 2.5, θ ( f )( a iB ) and θ ( f )( a iB ) are both equal to a iB . If now C f ( B ), choose q ∈ C \ f ( B ) and let h ∈ G be order-preserving taking q to h ( q ) C but fixing all members of f ( B ).Then f and hf agree on B , so a jC = θ ( f )( a iB ) = θ ( hf )( a iB ) = θ ( h ) θ ( f )( a iB ) = θ ( h )( a jC ), contrary to hC = C , and giving the result.Finally the result for order-reversing f may be deduced by composing with an order-reversingmember of G { B } as in the proof of Lemma 2.5.Using the ideas from above, we can demonstrate automatic homeomorphicity of E = End( Q , betw ). Theorem 2.8. E = End( Q , betw ) has automatic homeomorphicity.Proof. The sub-basic open sets in E and E ′ are of the form B qr = { f ∈ E : f ( q ) = r } and C ijBC = { f ∈ E ′ : f ( a iB ) = a jC } or variants of these with orderings on the subscripts B, C , so to establishcontinuity we have to show that each θ − ( C ijBC ), θ − ( C ijBC ) etc is open in E . We concentrate onthe case of C ijBC , the others being easy modifications of the same argument. Let B = { q , . . . , q m } and if f ∈ θ − ( C ijBC ) is order-preserving let r k = f ( q k ), so that f ∈ T mk =1 B q k r k . We show thatthis open set is contained in θ − ( C ijBC ), which suffices. Let f ′ ∈ T mk =1 B q k r k . Then f and f ′ agreeon B . If f ′ is also order-preserving, by the proof of Lemma 2.7, θ ( f ′ )( a iB ) = θ ( f )( a iB ) = a jC , whichgives f ′ ∈ θ − ( C ijBC ). If f ′ is not order-preserving, then as f and f ′ agree on B , we must have m ≤
1, giving t i = 0. Let g ∈ G B be an involution. Then θ ( g )( a iB ) = a igB = a iB , and f and f ′ g are order-preserving, and agree on B , so by what we have just shown, θ ( f ′ )( a iB ) = θ ( f ′ g )( a iB ) = θ ( f ′ g ) θ ( g )( a iB ) = θ ( f ′ g )( a iB ) = θ ( f )( a iB ) = a jC , again giving f ′ ∈ θ − ( C ijBC ). If f is order-reversing,the result is established by composing with an order-reversing automorphism in G { B } as before.Next we have to show that θ is open, so we consider the image θ ( B qr ) of any sub-basic open setand show that this is open. Let θ ( f ) ∈ θ ( B qr ) where f ( q ) = r , and we find i, j ∈ I and B, C ⊂ Q sothat θ ( f ) ∈ C ijBC ⊆ θ ( B qr ) (possibly with B and/or C ordered), or in one case, B , B , C , C ⊂ Q so that θ ( f ) ∈ C ijB C ∩ C ijB C ⊆ θ ( B qr ).The first step is to show that | im( f ) | ≥ n i > i ∈ I . For if not, for every i ∈ I , n i > ⇒ | im( f ) | < n i . For any a iB in Ω i , let θ ( f )( a iB ) = a jC (where the subscripts may be ordered).By Lemma 2.7, C ⊆ f ( B ), so n j ≤ | im( f ) | , giving n j = 0 and C = ∅ . Let g ( q ) = q + 1, so g ∈ G .Then θ ( gf )( a iB ) = θ ( g ) θ ( f )( a iB ) = θ ( g )( a j ∅ ) = a j ∅ = θ ( f )( a iB ), so that θ ( gf ) = θ ( f ). By injectivityof θ , gf = f , contradiction.Consider the case t i = 1, and since | im( f ) | ≥ n i we may choose B and C both of size n i such that f ( B ) = C and q ∈ B . Then by Lemma 2.5, θ ( f )( a iB ) = a if ( B ) = a iC , showing that θ ( f ) ∈ C iiB C .Now consider any member of C iiB C . Since we are working in E ′ which is the image of E , this hasthe form θ ( h ) for some h ∈ E and θ ( h )( a iB ) = a iC . By Lemma 2.7, C ⊆ h ( B ), so as | B | = | C | , h ( B ) = C . Since f maps q in B to the corresponding entry of C , it follows that h does too, andhence h ( q ) = r , which shows that θ ( h ) ∈ θ ( B qr ) as required.Now look at the case t i = 0, and suppose first that for some i , n i < | im f | , and f is order-preserving. Choose B, C ∈ [ Q ] n i +1 such that f ( B ) = C and q ∈ B . Let B = { q , q , . . . , q n i +1 } inincreasing order, and similarly for C = { r , r , . . . , r n i +1 } , so that f ( q k ) = r k for each k . Let B = { q , q , . . . , q n i } , B = { q , q , . . . , q n i +1 } , and similarly for C , C . Then θ ( f ) ∈ C iiB C ∩ C iiB C .We show that C iiB C ∩ C iiB C ⊆ θ ( B qr ). For this, take any member θ ( h ) of C iiB C ∩ C iiB C .6hen by the above calculations, h ( B ) = C and h ( B ) = C . If h is order-reversing, the firstequation implies that h ( q ) = r n i − and the second that h ( q ) = r n i +1 . This contradiction showsthat actually h is order-preserving, and as before it follows that h ( q ) = r , and hence θ ( h ) ∈ θ ( B qr ).A similar proof applies if f is order-reversing.This reduces us to the case in which for every i , n i = | im( f ) | or 0, and we suppose f order-preserving, with a similar argument in the order-reversing case. Then if im( f ) = C , for any B ′ ∈ [ Q ] n i on which f is 1–1, θ ( f )( a iB ′ ) = a iC , and so if θ ( h ) ∈ C iiBC , for some such B , also θ ( h )( a iB ′ ) = a iC for every B ′ on which h is 1–1. If h is order-preserving, then we argue as be-fore, so suppose that h is order-reversing. We show that θ ( f ) = θ ( h ). Let B = { q , q , . . . , q n i } ,listed in increasing order, and let q ′ k , q ′′ k ∈ ( q k , q k +1 ) be such that f ( −∞ , q ′ ) = h ( q ′′ n i − , ∞ ) = { r } , f ( q ′ , q ′ ) = h ( q ′′ n i − , q ′′ n − ) = { r } , . . . , f ( q ′ n i − , ∞ ) = h ( −∞ , q ′′ ) = { r n i } . Let g ∈ G { B } beorder-reversing taking each q k to q n i +1 − k and q ′ k to q ′′ n i − k . Then by considering the action on eachinterval ( −∞ , q ′ ), ( q ′ , q ′ ) , . . . , ( q ′ n i − , ∞ ) we see that f = hg , and hence θ ( f )( a iB ) = θ ( hg )( a iB ) = θ ( h )( a igB ) = θ ( h )( a iB ). If n i = 0 then we get the same equality immediately from Lemma 2.5. Itfollows that θ ( f ) = θ ( h ), but as f is order-preserving and h is order-reversing, this is contrary toinjectivity of θ . Q The (strict) circular order on Q is a ternary relation which may be defined by circ ( x, y, z ) if x < y < z or y < z < x or z < x < y . In this section we demonstrate automatic homeomorphicity for itsmonoid of embeddings. We adapt the techniques from [3] section 2, already mentioned above whenconsidering the betweenness relation. There we defined families Γ, Γ + , Γ − , Γ ± of embeddings.For these we used the ‘2-coloured version of the rationals’ Q , which is taken to be the orderedrationals together with colouring by 2 colours such that each colour occurs densely. Analogously wemay form the C of ( Q , circ ), which is taken to be Q under the same (circular)relation, coloured by two colours, ‘red’ and ‘blue’, each of which occurs densely. This again existsand is unique up to isomorphism. Note that for any x, y ∈ Q , we may form the closed interval[ x, y ] = { z : circ ( x, z, y ) } , even if y < x (in which case it actually equals [ x, ∞ ) ∪ ( −∞ , y ] for ‘usual’intervals).This leads us to the analogue of the class Γ in this case (since there are no endpoints, Γ + , Γ − , Γ ± are not needed). For any embedding f of ( Q , circ ), we define ∼ by x ∼ y if [ x, y ] or [ y, x ] containsat most one point of the image of f . Each ∼ -class is then an interval containing at most one pointof im f ; if one point, then the interval is red ; if no point, then it is blue . Then Γ is taken to be theset of all members f of M all of whose ∼ -classes are non-empty open intervals, and the red and blueclasses form a copy of C .For any g ∈ M we let ∼ be the equivalence relation defined above, and we let P be the family ofall pairs ( a, b ) of finite partial automorphisms of Q satisfying the following properties:(1) a is colour-preserving, and ∼ -preserving (meaning that for x, y ∈ dom( a ), x ∼ y ⇔ a ( x ) ∼ a ( y )),(2) if x ∈ dom( a ) lies in a red interval containing a point y of im( g ), then y ∈ dom( a ),(3) if x ∈ im( a ) lies in a red interval containing a point y of im( g ), then y ∈ im( a ),(4) if x ∈ dom( b ), then g ( x ) ∈ dom( a ), and gb ( x ) = ag ( x ),(5) if x ∈ im( b ), then g ( x ) ∈ im( a ), and a − g ( x ) = gb − ( x ),(6) if x ∈ im( g ) ∩ dom( a ), then g − ( x ) ∈ dom( b ), and gbg − ( x ) = a ( x )(7) if x ∈ im( g ) ∩ im( a ), then g − ( x ) ∈ im( b ), and a − ( x ) ∈ im( g ). Moreover, b − g − ( x ) = g − a − ( x ). Lemma 3.1. If f ∈ Γ , then any ( a, b ) ∈ P can be extended to a pair of automorphisms ( α, β ) of ( Q , circ ) such that αf = f β .Proof. Let Q = S { A q : q ∈ C } where each A q is an open interval, circularly ordered by the naturalrelation determined from that on C , and where | A q ∩ im f | = 1 if q is red, and A q ∩ im f = ∅ if q isblue. Let a ( q ) = r if there is x ∈ A q ∩ dom( a ) such that a ( x ) ∈ A r . Then a is a finite colour and7circular-)order preserving partial automorphism of C . Extend a to an automorphism α of C , andlet α be an automorphism extending a preserving im( f ) such that for each q ∈ C , α ( A q ) = A α ( q ) .Let β = f − αf . Lemma 3.2.
Any injective endomorphism ξ of M which fixes G pointwise also fixes every memberof Γ .Proof. Let g ∈ Γ, and S ( g ) = { ( α, β ) ∈ G : αg = gβ } . Consider elements u and s of Q with s = g ( u ). We construct ( α, β ) such that α ( s ) = s and β ( u ) = u . We consider two cases:1. If s ∈ im( g ), then s and g ( u ) lie in different red intervals. Let A s be the red interval containing s . Since im( g ) ∼ = Q , there is t ∈ im( g ) such that circ ( g ( u ) , s, t ). Since g is circ -preserving, circ ( u, g − ( s ) , g − ( t )). Hence a = { ( g ( u ) , g ( u )) , ( s, t ) } and b = { ( u, u ) , (( g − ( s ) , g − ( t )) } arefinite partial automorphisms. We can verify that ( a, b ) ∈ P (as defined before the previouslemma).2. If s / ∈ im( g ), we consider two cases:(i) If s lies in a blue interval A q , we choose t = s in the same interval. Since A q ∼ = Q , a = { ( s, t ) , ( g ( u ) , g ( u )) } and b = { ( u, u ) } are finite partial automorphisms. Again ( a, b ) ∈ P .(ii) If s lies in a red interval A r , with r ∈ im( g ), we choose t = s in A r on the same sideof s (which also allows for the possibility that r = g ( u )), meaning that circ ( g ( u ) , r, s )and circ ( g ( u ) , t, s ), or g ( u ) = r and circ ( g ( u ) , t, s ), or circ ( g ( u ) , s, r ) and circ ( g ( u ) , s, t ).Then a = { ( g ( u ) , g ( u )) , ( r, r ) , ( s, t ) } and b = { ( u, u ) , ( g − ( r ) , g − ( r )) } are finite partialautomorphisms, and once more we can verify that ( a, b ) ∈ P .In each case we can extend ( a, b ) to ( α, β ) such that αg = gβ by appealing to Lemma 3.1, thus( α, β ) lies in S ( g ), and satisfies β ( u ) = u , α ( s ) = t = s . Now the element g ( u ) can be recoveredfrom S ( g ), namely as g ( u ) = s ⇐⇒ ∀ ( α, β ) ∈ S ( g ) ( β ( u ) = u → α ( s ) = s ) (1)For if g ( u ) = s and ( α, β ) ∈ S ( g ) with β ( u ) = u , then α ( s ) = α ( g ( u )) = gβ ( u ) = g ( u ) = s. Conversely, if g ( u ) = s , then by the above we can construct ( α, β ) ∈ S ( g ) such that β ( u ) = u and α ( s ) = s .Finally, from Equation (1) we obtain ξ ( g ) = g ,( u, s ) ∈ g ( ) ⇐⇒ ∀ ( α, β ) ∈ S ( g ) ( β ( u ) = u → α ( s ) = s ) ⇐⇒ ∀ ( α, β ) ∈ S ( ξ ( g )) ( β ( u ) = u → α ( s ) = s ) ⇐⇒ ( u, s ) ∈ ξ ( g )Now we consider how the members of Γ and M interact. If g ∈ Γ and f ∈ M , then any ∼ gf -classis a union of a convex family of ∼ g -classes. This is because im( gf ) ⊆ im( g ) and so x ∼ g y ⇒| [ x, y ] ∩ im( g ) | ≤ | [ y, x ] ∩ im( g ) | ≤ ⇒ | [ x, y ] ∩ im( gf ) | ≤ | [ y, x ] ∩ im( gf ) | ≤ ⇒ x ∼ gf y .Since all ∼ g -classes are isomorphic to Q , so are all the ∼ gf -classes. The family of red ∼ gf classesis ordered like Q , since it corresponds precisely to the image of gf , which is a copy of Q . And theblue ∼ gf classes occupy some cuts among the red ones. Two distinct blue ∼ gf classes must occupydistinct cuts, as if they had no red ∼ gf class between them, then by definition of ∼ , they’d have tobe in the same ∼ gf -class. This means that we may write Q as a disjoint union of sets A q for q lyingin some subset Q of C , where A q ∼ = Q and all the red members of C lie in Q . This describes thegeneral situation. Depending on the precise values of g and f , we may find that gf ∈ Γ or not. Wefirst see that if they both lie in Γ, then the product necessarily does too.
Lemma 3.3. If g and g lie in Γ then so does g g . roof. From the above remarks, we just need to see that between any two g g -red intervals thereis a g g -blue one, where this now means in the sense of the circular order. Let g g x and g g y liein distinct intervals. Since g ∈ Γ, there is a g -blue interval ( a, b ) ⊆ ( g x, g y ), and its endpoints a and b are irrationals which are limits of points of im( g ). Let a = sup g a n , b = inf g b n where( a n ) is an increasing sequence, and ( b n ) is a decreasing sequence. Clearly g ( a, b ) is disjoint fromim( g g ). It is contained in a g g -blue interval (which therefore lies (strictly) in between g g x and g g y ) because the only way in which it could lie in a g g -red interval ( c, d ) would be if there wasa single point g g ( z ) of im( g g ) lying in it; but then g g a n < g g z < g g b n for all n , giving g a n < g z < g b n so g z ∈ ( a, b ), contrary to ( a, b ) ∩ im( g ) = ∅ , so this cannot happen. (It ispossible that the g g -blue interval is larger than the convex hull of g ( a, b ), but this does not affectthe argument.)To conclude, note that by definition of ∼ , there cannot be consecutive blue intervals, or aconsecutive pair of red/blue intervals, and as the red intervals are ordered like Q there are no twoconsecutive red intervals either. From this it easily follows that the family of intervals is orderedlike C . Lemma 3.4.
For any f ∈ M , there is g ∈ Γ such that gf ∈ Γ .Proof. We start by taking any g ∈ Γ, and then we see that we can describe g f fairly well. Thenwe take another g ∈ Γ, which will be chosen so that g g f ∈ Γ. Appealing to Lemma 3.3, we maylet g = g g to conclude the proof.By the discussion above, there is a subset Q of C containing all the red points, such that Q = S q ∈ Q A q where the A q are copies of Q such that circ ( r, s, t ) in Q implies that the corresponding A r , A s , A t are circularly ordered in the same way (as sets) and if q ∈ Q is red, then A q is a g f -redinterval, and if it is blue, then A q is a g f -blue interval. Now we choose a countable dense set B of(blue) irrationals such that the family of sets A q for red q ∈ Q and the set of members of B whichare cuts of this family, together form a copy of C . Note that B will have a lot more members thanthese particular cuts, but these are the crucial ones which will ensure that our g g f lies in Γ. Notethat in addition Q ∪ B also forms a copy of C , and we use it to find g ∈ Γ. Now each g f -redinterval gives rise to a g g f -red interval. This is because it clearly still just has one point in theimage, and it doesn’t extend ‘any further’ because of the presence of the dense set B . The imagesof the members of B which were inserted densely between the sets A q for red q ∈ Q are g g f -blueintervals which enable us to see that the result is itself a copy of C . Corollary 3.5.
Any injective endomorphism ξ of M which fixes G pointwise also fixes every memberof M .Proof. Let f ∈ M . By Lemma 3.4, there is g ∈ Γ such that gf ∈ Γ. By Lemma 3.2, ξ fixes g and gf .Hence ξ ( g ) ξ ( f ) = ξ ( gf ) = gf = ξ ( g ) f . Since g is left cancellable, so is ξ ( g ), and hence ξ ( f ) = f . Lemma 3.6.
Aut ( Q , circ ) has the small index property.Proof. This follows easily from the observation that Aut( Q , < ) has countable index in Aut( Q , circ )(as follows by the orbit-stabilizer theorem) and the fact that Aut( Q , < ) has the small index property([3]). Theorem 3.7.
Emb( Q , circ ) has automatic homeomorphicity.Proof. This follows from Corollary 3.5 and Lemma 2.2, since G is dense in M , and G has the smallindex property, so that by [4] we know that G has automatic homeomorphicity.We now adapt the ideas of [3] and section 2 to demonstrate automatic homeomorphicity forEnd( Q , circ ). Once more by the small index property, if H is a subgroup of G of countable index,there is a minimal finite subset B of Q such that G B ≤ H . To show that H ≤ G { B } , we deducefrom Lemma 2.1 its analogue in the current situation. Lemma 3.8. If B and B are finite subsets of Q , and G = Aut( Q , circ ) , then G B ∩ B = h G B , G B i . roof. We exploit the fact that the stabilizer G a of any point a ∈ Q is isomorphic to Aut( Q , ≤ ),and can then deduce the result from Lemma 2.1. As before we just have to check that G B ∩ B ≤h G B , G B i . Pick a ∈ B ∩ B if this is non-empty. Then g ∈ ( G a ) B ∩ B ≤ h ( G a ) B , ( G a ) B i byLemma 2.1, ≤ h G B , G B i .If B ∩ B = ∅ , we start by writing an arbitrary f ∈ G as the product of two elements, eachhaving a fixed point. Take any a ∈ Q , and let b = f ( a ). There is h ∈ G taking a to b , andfixing some (rational) point of ( b, a ). Then h and h − f each has a fixed point (since h − f fixes a ).Given this observation, it suffices to show that any member of g of G having a fixed point lies in h G B , G B i . Let a be fixed by g . Running the same argument as in the first paragraph, we find that g ∈ G ( B ∪{ a } ) ∩ ( B ∪{ a } ) = ( G a ) B ∩ B ≤ h ( G a ) B , ( G a ) B i ≤ h G B , G B i .Now that we know that H ≤ G { B } , we need to consider what the options are for such H (insection 2 there were only two). This time, if B = { b , b , . . . , b n − } in cyclic order is non-empty (thatis, n ≥ | G { B } : G B | = | B | , since the cyclic ordering on B has to be preserved and each cyclicpermutation is possible. It easily follows that for some factor m of n , if s m ( b i ) = b i + m where thesubscripts are taken modulo n , then H = { g ∈ G : g acts on B as a power of s m } . Let us say that m is the ‘type’ of the orbit. Given this, we can just adapt the machinery from section 2. Namely, Ωmay be written as the union of G -orbits Ω i for i ∈ I , n i and m i are specified, and for each i , Ω i isa family of elements of the form a iB mi where B ∈ [ Q ] n i . Here, B m is the set of images of B underpowers of s m , so that the orderings of B which arise are in the correct anticlockwise cyclic order, andform an orbit under h s m i . This is all done so that the action of θ is compatible with this enumerationfor members of G . More precisely, we let Ω i = { a ihB mi : h ∈ G } , where a ihB mi = θ ( h ) a iB mi . Thepoint is that for g, h ∈ G , a igB mi = a ihB mi ⇔ θ ( h − g ) fixes a iB mi ⇔ h − g acts on B as a power of s m i .Given this background, our remaining task is to show how the machinery developed in theprevious section for the betweenness relation carries over to this setting. The analogue of Lemma2.4 holds here by a similar continuity argument, and the analogues of Lemma 2.5 and 2.7 also carryacross straightforwardly. Next we have the analogue of Lemma 2.6. Lemma 3.9.
Let i ∈ I and B ∈ [ Q ] n i where n i = 0 , and Ω i have type m i . Let a iB mi ∈ Ω i . Thenthere is an idempotent endomorphism h ∈ E having image B such that θ ( h ) fixes a iB mi .Proof. Subdivide the circularly ordered Q into n i pairwise disjoint intervals, each containing a singlemember of B . Then h ( B ) = B , so h also fixes B m i , so by the analogue of Lemma 2.5 for this case, θ ( h )( a iB mi ) = a ihB mi = a iB mi .The final result of this section is as follows. Theorem 3.10.
End( Q , circ ) has automatic homeomorphicity.Proof. Let E ′ be a closed submonoid of T r (Ω) where | Ω | = ℵ , and let θ be an isomorphism from E to E ′ , which we have to show is a homeomorphism. We decompose Ω into orbits Ω i as above, and thistime the sub-basic open sets in E and E ′ are of the form B qr = { f ∈ E : f ( q ) = r } and C ijB mi C mj = { f ∈ E ′ : f ( a iB mi ) = a jC mj } , so for continuity we have to show that each θ − ( C ijB mi C mj ) is open in E , and for openness that each θ ( B qr ) is open in E ′ .For openness of θ − ( C ijB mi C mj ), let B = { q , q , . . . , q n i } , where B is listed in increasing order.Let f ∈ θ − ( C ijB mi C mj ) and let f ( q k ) = r k (the r k need not be distinct). Thus f ∈ T n i k =1 B q k r k , andwe have to show that this set is contained in θ − ( C ijB mi C mj ). Let f ′ be any member of T n i k =1 B q k r k .Thus for each k , f ′ ( q k ) = r k , and hence f and f ′ agree on B . By Lemma 3.9 there is an idempotent h ∈ E with image B such that θ ( h ) fixes a iB mi . Then f ′ h = f h , and so θ ( f ′ )( a iB mi ) = θ ( f ′ h )( a iB mi ) = θ ( f h )( a iB mi ) = θ ( f )( a iB mi ) = a jC mj , and therefore f ′ ∈ θ − ( C ijB mi C mj ) as required.Next we show that θ ( B qr ) is open for any q, r . As in the proof of Theorem 2.8 we may find i suchthat n i ≤ | im( f ) | . Choose B and C of size n i with q ∈ B and such that f ( B ) = C . By the analogue10f Lemma 2.5, θ ( f )( a iB mi ) = a ifB mi = a iC mi , showing that θ ( f ) ∈ C iiB mi C mi . Let θ ( h ) lie in this set.As before, h ( B ) = C , and in fact h ( B m i ) = C m i . As before, the problem is that we do not knowthat h takes q to r . For this, we follow a similar strategy to that adopted in the proof of Theorem2.8. First if for some i , | im( f ) | > n i we find ‘overlapping’ cyclically ordered sequences of length n i ,and use the extra room thus created to recover sufficiently the structure, so that endomorphismslying in the intersection of two sets of the form C iiB mi C mi must take q to r . Finally, if n i = | im( f ) | or 0 for every i , and the endomorphism h which arises in the proof satisfies h ( q ) = r , we show that θ ( h ) = θ ( f ), contrary to the injectivity of θ . More precisely, let B = { q , q , . . . , q n i } in increasingorder, and C = { r , r , . . . , r n i } be enumerated so that f ( q k ) = r k for each k . Let h ( q k ) = r k + t foreach k , fixed t , where t = 0. Then there is g ∈ G taking q k to q k − t for each k (where all the sufficesare taken modulo n i ), and we find that hg ( q k ) = h ( q k − t ) = r k = f ( q k ), giving hg = f . Here thefact that h ( B m i ) = C m i ensures that m i divides t from which it follows that g fixes B m i . Therefore θ ( f )( a iB mi ) = θ ( hg )( a iB mi ) = θ ( h )( a igB mi ) = θ ( h )( a iB mi ) and so θ ( f ) = θ ( h ), as stated. Q Since Aut( Q , circ ) has index 2 in Aut( Q , sep ), we may use the ‘same’ method as in section 2 to deduceautomatic homeomorphicity for Emb( Q , sep ) from the corresponding result for Emb( Q , circ ). Themain step as usual is to consider an injective endomorphism ξ of Emb( Q , sep ) to itself, which fixes allgroup elements, and show that it must be the identity. We fix some involution i in Aut( Q , sep ), whichinterchanges the sets of orientation preserving and orientation reversing members of Emb( Q , sep ),and we consider the class Γ as in section 3. If for any g ∈ M we define S ( g ) to be { ( α, β ) ∈ G : αg = gβ } , where G = Aut( Q , circ ), the same calculations used in section 3 for members of Emb( Q , circ )apply to Emb( Q , sep ) to show that ξ fixes all members of Γ, and hence (since i is necessarily fixedby ξ ) also all members of M . This establishes the following result. Theorem 4.1.
Emb( Q , sep ) has automatic homeomorphicity. We now indicate how the methods of sections 2 and 3 are adapted in this case to yield proofsof automatic homeomorphicity of End( Q , sep ). Once more if H is a subgroup of G = Aut( Q , sep )of countable index, then there is a unique (finite, minimal) B ⊂ Q such that G B ≤ H ≤ G { B } .If the two stabilizers are equal, then as usual we can identify the orbit with [ Q ] n for n = | B | .Otherwise, | B | ≥ | G { B } : G B | = 2 n , and we have to consider rotations and reflections. If we let B = { b , b , . . . , b n − } in increasing order, then the value of H can be ‘captured’ a combination of thetypes from the two previous sections, that is, it will be a pair consisting of 1 or 0, to tell us whether H has an orientation-reversing member or not, and a factor m of n such that the orientation-preservingsubgroup of H acts on B as a power of s m defined at the end of section 3. In section 3 we wereessentially considering the action of a cyclic group, but here the corresponding action is dihedral.We omit the details, but state the main theorem which applies here, and which is proved by methodssimilar too those in sections 2 and 3 (with some adaptations).As usual, by the small index property, if H is a subgroup of G of countable index, there is aminimal finite subset B of Q such that G B ≤ H , and using a combination of the tricks from sections2 and 3, also H ≤ G { B } , so that B is uniquely determined. This time there are extra options forwhat H can be, obtained by reversing the orientation of B (so it is essentially the dihedral groupthat is now acting). If we write B = ( b , b , . . . , b n − ) in increasing order, we look at the family ofsequences that arise by applying members of H . Since H ≤ G { B } , all members of H preserve the set B , but may perform ‘rotations’ or ‘reflections’. As in section 3, we can capture the possibilitiesvia the set of images under H , which are indexed by subgroups of the dihedral group of order 2 n .We write the set as B H (where it isn’t really H which is relevant—rather its induced action on B ).Thus we may write Ω as the union of G -orbits Ω i for i ∈ I , n i and H i are specified, and for each i , Ω i is the family of elements of the form a iB Hi where B ∈ [ Q ] n i . The same lemmas as before arenow proved in this case. We obtain the analogue of Lemma 2.4 by means of a continuity argument,and this leads to the analogue of Lemma 2.5. The analogue of Lemma 2.6 is as follows.11 emma 4.2. Let i ∈ I and B be a non-empty member of [ Q ] n i . Then there is an idempotentseparation-preserving endomorphism h ∈ E having B as image such that θ ( h ) fixes a iB Hi . For this we can take h to be orientation-preserving, so use the same method as in Lemma 3.9.Lemma 2.7 carries over straightforwardly to the new situation, and the main result is as follows. Theorem 4.3.
End( Q , sep ) has automatic homeomorphicity.Proof. For this we use a combination of the methods of Theorems 2.8 and 3.10. In fact the groupis a degree 2 extension of that for the circular ordering, so this case bears the same relationship tothe circular ordering as does the betweenness relation to the linear ordering.To give a few details, once more, let E ′ be a closed submonoid of T r (Ω) where | Ω | = ℵ , andlet θ be an isomorphism from E to E ′ . Write Ω = S i ∈ I Ω i where Ω i = { a iB Hi : B ∈ [ Q ] n i } , H i a subgroup of the dihedral group of order 2 n i . The sub-basic open sets in E and E ′ are of theform B qr = { f ∈ E : f ( q ) = r } and C ijB Hi C Hj = { f ∈ E ′ : f ( a iB Hi ) = a jC Hj } . The proof that θ is continuous is as before. For openness we show that each θ ( B qr ) is open in E ′ . We find i such that n i ≤ | im( f ) | , and choose B and C of size n i with q ∈ B and such that f ( B ) = C . Asbefore, θ ( f )( a iB Hi ) = a ifB Hi = a iC Hi , showing that θ ( f ) ∈ C iiB Hi C Hi . Let θ ( h ) lie in this set. Then h ( B ) = C , and h ( B H i ) = C H i . If for some i , | im( f ) | > n i we argue as for the circular orderingcase. If however n i = | im( f ) | or 0 for every i , and h ( q ) = r , we show that θ ( h ) = θ ( f ), contraryto the injectivity of θ . Assume that f is orientation-preserving (with a similar argument in theorientation-reversing case). If B = { q , q , . . . , q n i } in increasing order, and C = { r , r , . . . , r n i } are enumerated so that f ( q k ) = r k for each k , and h is also orientation-preserving, then we usethe argument from Theorem 3.10, where if h ( q k ) = r k + t for each k , we use g ∈ G taking q k to q k − t for each k . If however h is orientation-reversing, it must take the form h ( q k ) = r t − k for somefixed t , and instead we find g ∈ G such that g ( q k ) = q t − k for each k . This gives hg = f . Since f ( B H i ) = C H i and h ( B H i ) = C H i , it follows that hg ( B H i ) = h ( B H i ), and as h is 1–1 on B , that g ( B H i ) = B H i . Therefore θ ( f )( a iB Hi ) = θ ( hg )( a iB Hi ) = θ ( h )( a igB Hi ) = θ ( h )( a iB Hi ), showing that θ ( f ) = θ ( h ), contrary to θ injective. Q for the reflexive case Our aim in this section is to carry across the results from [3] for the polymorphism clone of therational numbers under the reflexive ordering to the reducts discussed earlier in the paper, between-ness, circular order, and separation relations. For definitions of the relevant notions here we referthe reader to [4], but mention a few notations that are needed. Denoting by O A the collection of allfinitary operations f : A n → A ( n ≥
0) on a set A , a subset C ⊆ O A is called a (‘concrete’) clone on A if it is closed under the operations of composition when defined (that is, the ‘arities’ are correct) andit contains all ‘projections’. These are the maps π ( n ) i : A n → A given by π ( n ) i ( a , a , . . . , a n ) = a i ,where 1 ≤ i ≤ n . The collection of all polymorphisms of a relational structure always forms a clone,and clones arising in this way are precisely the ones that are topologically closed. Of central interesthere are the clones Pol( Q , betw ) of polymorphisms of ( Q , betw ) and Pol( Q , circ ) of polymorphismsof ( Q , circ ), which are the families of all n -ary functions on Q for n ≥ betw and circ , respectively. Spelling out precisely what this means, f : Q n → Q lies in Pol( Q , betw ) pro-vided that if ( a , a , . . . , a n ) , ( b , b , . . . , b n ) , ( c , c , . . . , c n ) ∈ Q n and betw ( a i , b i , c i ) for all i , then betw ( f ( a , a , . . . , a n ) , f ( b , b , . . . , b n ) , f ( c , c , . . . , c n )). Similarly, f ∈ Pol( Q , circ ) if circ ( a i , b i , c i )for all i , implies that circ ( f ( a , a , . . . , a n ) , f ( b , b , . . . , b n ) , f ( c , c , . . . , c n )).We also study the clone Pol( Q , sep ) of polymorphisms of ( Q , sep ), which is the family of all n -ary functions on Q for n ≥ sep . In other words f : Q n → Q lies in Pol( Q , sep )if ( a , a , . . . , a n ) , ( b , b , . . . , b n ) , ( c , c , . . . , c n ) , ( d , d , . . . , d n ) ∈ Q n and sep ( a i , b i , c i , d i ) for all i implies that sep ( f ( a , a , . . . , a n ) , f ( b , b , . . . , b n ) , f ( c , c , . . . , c n ) , f ( d , d , . . . , d n )).12here is a corresponding notion of ‘abstract clone’, which we do not require here. Let us notealso that the set O A of all finitary operations on A forms a clone, even a polymorphism clone (e.g., O A = Pol( A, =)). This is the analogue of Sym( A ) for the automorphism group and Tr( A ) for theendomorphism monoid. In each of these cases, betweenness, circular, and separation relations wewrite M , E , and G for the monoids of embeddings, endomorphisms, and the group of automorphisms,respectively, and P for the corresponding polymorphism clone.The set-up is as follows. An isomorphism θ is given from P to a closed subclone P ′ of the fullpolymorphism clone O Ω on a countable set Ω, and our task is to show that it is a homeomorphism.Relying on Proposition 27 of [4], when proving automatic homeomorphicity of the clone P ineach of the cases mentioned above, it will suffice to verify that any clone isomorphism between P and a closed clone on some countable set is continuous. Theorem 5.1.
Pol( Q , betw ) has automatic homeomorphicity, meaning that any isomorphism θ from P = Pol( Q , betw ) to a closed subclone P ′ of O Ω , for a countable set Ω , is a homeomorphism.Proof. Openness follows from Proposition 27 of [4]. To demonstrate that θ is continuous, we usethe machinery from section 2 to provide the assumptions of Lemma 5.1 of [3]. Note that these areproperties of the restriction θ ↾ E : P (1) → P ′ (1) , which is a monoid isomorphism between the unaryparts P (1) = E and E ′ := P ′ (1) (these are closed monoids because P and Tr( Q ), and P ′ and Tr(Ω)are closed sets). Namely, we have to verify that for every b ∈ Ω we can find an endomorphism h ∈ E with finite image such that θ ( h )( b ) = θ ↾ E ( h )( b ) = b . However, this is precisely the content ofLemma 2.6 applied to θ ↾ E .Similarly, using Proposition 27 of [4], Lemma 5.1 of [3] and Lemmas 3.9, and 4.2, respectively,one can prove following theorem. Theorem 5.2.
Pol( Q , circ ) , Pol( Q , sep ) have automatic homeomorphicity. In this section we show how to ‘lift’ the automatic homeomorphicity results for the polymorphismclones h End ( Q , < ) i and h End ( Q , ≤ ) i generated by End ( Q , < ) and End ( Q , ≤ ) respectively proved in[3] to the reducts discussed earlier in the paper, betweenness, circular order and separation relations. Theorem 6.1. h Emb ( Q , betw ) i , h Emb ( Q , circ ) i , and h Emb ( Q , sep ) i have automatic homeomor-phicity.Proof. The following paragraph is an almost verbatim copy of the proof of Lemma 6.5 of [3].We consider the case when M = Emb ( Q , betw ). Let θ : h M i → C be a clone isomorphismbetween h M i and another closed clone C on a countable set Ω. Since by Theorem 2.3, M hasautomatic homeomorphicity, and the unary part of C is closed as C (1) = C ∩ Tr(Ω) and both setsare closed, the restriction θ ↾ M : M → C (1) is a homeomorphism. By [3] Corollary 6.3 we concludethat θ is continuous. To see that it must be open too, we use Proposition 32 from [4], which holdsfor clone isomorphisms and is applicable here since Aut ( Q , betw ) acts transitively on Q and θ ↾ M is open. Similarly, we obtain automatic homeomorphicity for h M i , where M = Emb ( Q , circ ), fromTheorem 3.7 and Lemma 6.5 of [3], which is applicable here since Aut ( Q , circ ) acts also transitivelyon Q . Finally, we get automatic homeomorphicity for h Emb ( Q , sep ) i , from Theorem 4.1 and Lemma6.5 of [3].Finally, by appealing to Lemma 6.5 of [3] and Theorems 2.8, 3.10 and 4.3, respectively, we get: Theorem 6.2. h End ( Q , betw ) i , h End ( Q , circ ) i , and h End ( Q , sep ) i have automatic homeomorphic-ity. The coloured case
In this section we remark (without giving full details) how the results earlier in the paper can beeasily extended to coloured versions. That is, we start with the C -coloured version of the rationals Q C where 2 ≤ | C | ≤ ℵ , and form the corresponding reducts, namely the betweenness, circular,and separation relations, and obtain analogous automatic homeomorphicity results. Note that, asexplained in [8], these are not by any means all the non-trivial reducts, but since exactly what theseare is unknown, we just deal with the analogues of the ones for Q .We start by considering ( Q C , ≤ ) itself. The main ‘trick’ to deal with this case (and also thecircular ordering on Q C ) is to define the correct analogue of the classes Γ , Γ + , Γ − , Γ ± (for thecircular ordering just Γ). The main thing to suppose initially is that an injective endomorphism ξ : M → M is given which fixes all members of G , and we have to show that it is the identity.Note that exactly as in the monochromatic case, G is dense in M , so by Lemma 2.2 automatichomeomorphicity for M follows, since by [9] Theorem 4.5 we know that Aut( Q C , ≤ ) has the smallindex property,.The definition of Γ (etc) in this case is carried out as follows. Given a subset A of Q C isomorphicto Q C , we again define x ∼ y on Q C to mean that there is at most one point of A strictly between x and y . As before, this is an equivalence relation, and the equivalence classes are convex and intersect A in at most one point. This time their colours are however relevant. We choose one ‘extra’ colour c ∗ (i.e. some point not lying in C ), and we colour an equivalence class by the colour of its uniquemember of A , if any, and we colour it by c ∗ otherwise. we can now take Γ in this setting to be thefamily of all f ∈ M such that Q C may be written as the disjoint union S { A q : q ∈ Q C ∪{ c ∗ } } ofconvex subsets of Q C such that q < r ⇒ A q < A r , each A q is isomorphic to Q C , if q ∈ Q C ∪{ c ∗ } iscoloured by c ∈ C then A q has a single point of im f , which is coloured c , and if q is coloured c ∗ then A q is disjoint from im f . The definitions of Γ + , Γ − ,Γ ± are similar.The lemmas used in [3] to derive automatic homeomorphicity are now transcribed, with appro-priate modifications, to prove automatic homeomorphicity of Emb( Q C , ≤ ) and End( Q C , ≤ ).The passage from the ordered case to the betweenness relation, and from the circular relation tothe separation relation are performed as before, since the index of the smaller group in the larger isagain 2. The main technical lemmas from [3] now carry over to the new situation, with colours in C inserted at all the appropriate points, and the methods used earlier in this paper used where needed.The conclusion is that the following all have automatic homeomorphicity: Emb( Q C , ≤ ), End( Q C , ≤ ),Emb( Q C , betw ), End( Q C , betw ), Emb( Q C , circ ), End( Q C , circ ), Emb( Q C , sep ), End( Q C , sep ), theclones generated by all of these, and also Pol( Q C , ≤ ), Pol( Q C , betw ), Pol( Q C , circ ), and Pol( Q C , sep ).Briefly, Lemmas 2.1 and 3.8 are readily adapted to the coloured situation, so that for the orderedcase, any subgroup H of small index may be written as G B for some uniquely determined finite B ⊂ Q C . For the reducts, we can still find a unique finite B such that G B ≤ H ≤ G { B } , andwe have the same range of possibilities for H as for the monochromatic situation. That is, for thebetweenness relation, H = G B or G { B } , for the circular ordering, H acts on B as a power of somefixed circular map on B , and for the separation relation, it acts on B as a subgroup of a finitedihedral group. This means that the machinery developed earlier all carries through to the colouredcase. Note that there are more restrictions here. Thus for instance, for the betweenness relation,the possibility that H = G { B } can only arise if B is ‘symmetrically’ coloured, since otherwise, B will not be preserved setwise by any order-reversing automorphism. Similar remarks apply in theother cases. It is still true that all possibilities for H must lie in this list, which suffices to make thearguments go through. Conclusions and problems
In summary we have given some extensions of the automatic homeomorphicity results of [4] and[3], by not too complicated modifications of the arguments of the second paper. As demonstratedin [3] and here, the methods which apply to the ordered rationals and its reducts have a ratherdifferent flavour from those used in [4]. It is not entirely clear how all these cases can be extended orgeneralized. Obvious instances are the (many) remaining reducts of ( Q C , ≤ ) alluded to in the paper.Even to describe what these are may be complicated, as explained in [8]. A natural extension would14e to the case of (2-transitive) trees, originally described in [7], and whose reducts are discussed, inat least one case, in [5], and to more general classes of ℵ -categorical structures. Finally, we notethat our results for the reconstruction of the polymorphism clone apply just to the reflexive case,and even for the strict relation on the ordered rationals, this remained open, though an answer hasbeen given in [2] (see also the remark at the end of [4]). References [1] S. A. Adeleke and Peter M. Neumann, Relations related to betweenness: Their structure andautomorphisms, Memoirs of the American Mathematical Society 623 (1998).[2] Robert Barham, Automatic homeomorphicity of locally moving clones, arXiv:1512.00251, July2016.[3] Mike Behrisch, John K. Truss, and Edith Vargas-Garcia, Reconstructing the topology onmonoids and polymorphism clones of the rationals, Studia Logica 105 (2017), 65-91.[4] Manuel Bodirsky, Michael Pinsker, and Andr´as Pongr´acz, Reconstructing the topology of clones,Transactions of the American Mathematical Society 369 (2017), 3707-3740.[5] Manuel Bodirsky, David Bradley-Williams, Michael Pinsker, and Andr´as Pongr´acz, The uni-versal homogeneous binary tree, arXiv:1409.2170, November 2016.[6] Peter Cameron, Transitivity of permutation groups on unordered sets, Math. Zeit. 148 (1976),127-139.[7] Manfred Droste, Structure of partially ordered sets with transitive automorphism groups, Mem-oirs of the American Mathematical Society 334 (1985).[8] Markus Junker and Martin Ziegler, The 116 reducts of ( Q , <, a, <, a