aa r X i v : . [ m a t h . L O ] J a n A New Weak Choice Principle
Lorenz Halbeisen, Riccardo Plati, Salome SchumacherDepartment of Mathematics, ETH Z¨urich
Abstract
For every natural number n we introduce a new weak choice principle nRC fin : Given any infinite set x , there is an infinite subset y ⊆ x and aselection function f that chooses an n -element subset from every finite z ⊆ y containing at least n elements. By constructing new permutation models built on a set of atoms obtained asFra¨ıss´e limits, we will study the relation of nRC fin to the weak choice principlesRC m (that has already been studied in [3] and [6]): Given any infinite set x , there is an infinite subset y ⊆ x with a choicefunction f on the family of all m -element subsets of y . Moreover, we prove a stronger analogue of the results in [6] when we study therelation between nRC fin and kC − fin which is defined by: Given any infinite family F of finite sets of cardinality greater than k , there is an infinite subfamily A ⊆ F with a selection function f that chooses a k -element subset from each A ∈ A . key-words: weak forms of the Axiom of Choice, consistency results, Ramsey Choice,Fraenkel-Mostowski permutation models of ZFA+ ¬ AC, Pincus’ transfer theorems, par-tial n -selection for infinite families of finite sets In this paper we will use the following terminology: • By ω we denote the set of all natural numbers { , , , . . . } and fin( ω ) denotesthe set of finite subsets of ω . • Given a set x and a natural number n , [ x ] n is defined as the set of all the subsetsof x with cardinality n . Similarly, [ x ] >n is the set of all the finite subsets of x with cardinality greater than n . 1 Given a permutation model M and a statement φ , we will write M | = φ toindicate that φ holds in M . • BFM is the well known Basic Fraenkel Model.Furthermore, we shall use the following notation for weak choice principles: • RC n is the following axiom: given any infinite set x , there exists an infinite subset y ⊆ x with a choice function f : [ y ] n → y such that, for all z ∈ [ y ] n , f ( z ) ∈ z . • nRC fin is the following axiom: given any infinite set x , there exists an infinitesubset y ⊆ x with a selection function f : [ y ] >n → [ y ] n such that, for all z ∈ [ y ] >n , f ( z ) ⊆ z . • C n is the following axiom: any infinite family A of sets of cardinality n has achoice function f : A → S A such that, for all A ∈ A , f ( A ) ∈ A . • C − n is the following axiom: given any infinite family F of non-empty sets withcardinality n , there exists an infinite subfamily A ⊆ F which has a choice function f : A → S A such that, for all A ∈ A , f ( A ) ∈ A . • nC fin is the following axiom: any infinite family A of finite sets with cardinalitygreater than n has a selection function f : A → [ S A ] n such that, for all A ∈ A , f ( A ) ⊆ A . • nC − fin is the following axiom: given any infinite family F of finite sets with car-dinality greater than n , there exists an infinite subfamily A ⊆ F which has aselection function f : A → [ S A ] n such that, for all A ∈ A , f ( A ) ⊆ A . • ACF − is the following axiom: given any infinite family F of non-empty finite sets,there exists an infinite subfamily A ⊆ F which has a choice function f : A → S A such that, for all A ∈ A , f ( A ) ∈ A . Following the terminology used in [1], we introduce a new class of diminished choiceprinciples nRC fin and study its relation with the two classes RC n and nC − fin , which havein general inspired the new one; we will indeed obtain analogous results to [3] and[8]. The title of each section refers to the following diagram. The labels of the arrowsindicate in which section we analyze that specific implication. Here, n,k,m,j stand allfor natural numbers. 2RC fin kC − fin RC m C − jSec . . . . . To be more precise, we will prove the following results: • Relation between nRC fin and RC m : – For each n ∈ ω , RC n ; nRC fin in ZF+ ¬ AC. – For all k, n ∈ ω , nRC fin ⇒ RC kn+1 . – fin ⇒ RC n whenever n is odd and greater than 4. • Relation between nRC fin and kC − fin : – For each n ∈ ω , nC − fin ; nRC fin in ZF+ ¬ AC. – For all n ∈ { , , , } , nRC fin ⇒ nC − fin . – For all primes p and all k ∈ ω we have that p k RC fin ⇒ p k WOC − fin . • A relation between nRC fin and C − k : – fin ⇒ C − . • Relation between nRC fin and kRC fin : – For all k, n ∈ ω , nRC fin ⇒ knRC fin . – Let k, n ∈ ω with k > n . If k is not a multiple of n , then nRC fin ; kRC fin in ZF+ ¬ AC.
We will prove independence between choice principles in ZF via permutation models. Ina few words, we can say that a permutation model is built from a ground model, whichis a model of ZFA: a variation of ZF set theory in which the axiom of extensionalityis weakened in order to allow the existence of new objects (called atoms) containingno elements, but which are still distinct from the empty set. From this ground model(which satisfies AC), one can extract a submodel of ZFA in which AC fails. For detailsregarding this construction, see, for example, [2]. We will simply denote a permutationmodel by the structure of the set of atoms A , the normal ideal I on A and the group ofpermutations G : the normal filter on G will always be the one generated by I . Given apermutation model, we will get conclusion regarding ZF in the following way: Suppose3e manage to build a permutation model in which a certain choice principle Ax1 holdsand some other Ax2 fails. Using the results of [7], we can conclude that if Ax1 andAx2 both belong to a certain class of statements (in which case the statements aresaid to be injectively boundable), then there is a model of ZF in which Ax1 holds, Ax2fails, and both, Ax1 and Ax2, have the same meaning as in the permutation model,i.e., cardinalities and cofinalities remain unchanged between the two models. For thedefinition of injectively boundable, see [7] or [4]. Once done, it is not hard to see thatall the choice principles we will consider are injectively boundable: an injection of ω inan infinite set gives an infinite subset of which the power set admits a choice function. In this section we show that for any positive m ∈ ω , ( ∀ n ∈ ω RC n ) does not imply m RC fin . To this end, we use the model which in [8] is called V fin . The results containedin this section are not new and can be found stated in [4] and proved in [5].The model V fin is constructed from a countable set of atoms A partitioned in a wellordered family of blocks { B i : i ∈ ω } , such that for every i ∈ ω , B i has cardinality p i , where p i is the i -th prime number. For each i ∈ ω , fix a cyclic permutation ϕ i on B i that has no fixed points. The considered group of permutations G is given by allthe permutations ϕ on A that move only finitely many atoms and such that, for every i ∈ ω , ϕ restricted to B i equals some power of ϕ i . The corresponding normal filter isgenerated by the normal ideal of all finite subsets of A . Theorem 4.1.
We have that V fin | = ∀ n ∈ ω (C n ∧ ¬ nRC fin ) . Since evidently C n implies RC n , the theorem proves what we claimed in this section. In this section we briefly mention that any conjunction of mC − fin does not imply anynRC fin . Theorem 5.1.
We have that
BFM | = ∀ n ∈ ω (nC − fin ∧ ¬ nRC fin ) .Proof. It is known (see, e.g., [4]) that ACF − holds in BFM, and it is easy to see thatby n consecutive applications of ACF − one obtains nC − fin . The conclusion follows fromnoticing that any nRC fin fails on the set of atoms. This subsection starts with the very few cases in which we have a full positive answers.4 emma 6.1.
For n ∈ { , } we have nRC fin ⇒ nC − fin .Proof. We prove each case separately. Let n = 2 and A = { A j : j ∈ J } be an infinitefamily of pairwise disjoint finite sets (we can assume they are disjoint by replacingeach A i with the unique function A ∗ i : A i → { A i } ). Set x = S A and apply 2RC fin to get an infinite y ⊆ x and a function g : [ y ] > → [ y ] such that, for all Y ∈ [ y ] > , g ( Y ) ⊆ Y . Since every element of A is finite, there must be an infinite subset I of J such that, for all i ∈ I , A i ∩ y = ∅ . If for infinitely many i ∈ I , | A i ∩ y | = 2 the claim isobvious, and likewise if i ∈ I , | A i ∩ y | > i ∈ I , then we are doneby defining for each and every such i ∈ I the function f : A i g ( A i ∩ y ). If that isnot the case, apply a second time 2RC fin to S { A i : i ∈ I } \ y , to get another infinitesubset z ⊆ x with z ∩ y = ∅ . If, again, { i ∈ I : | A i ∩ z | ≥ } is finite, then we get that K = { i ∈ I : | A i ∩ y | = | A i ∩ z | = 1 } is infinite, together with the obvious function f : A k A k ∩ ( y ∪ z ), for all k ∈ K .For the other case we start similarly: let n = 3 and A = { A j : j ∈ J } be an infinitefamily of pairwise disjoint finite sets. Set x = S A and apply 3RC fin to get an infinite y ⊆ x and a function g : [ y ] > → [ y ] such that, for all Y ∈ [ y ] > , g ( Y ) ⊆ Y . Sinceevery element of A is finite, there must be an infinite subset I of J such that, for all i ∈ I , A i ∩ y = ∅ . If { i ∈ I : | A i ∩ y | = 1 or | A i ∩ y | ≥ } is infinite, with a perfectlyanalogous approach to the previous case we get the conclusion. Otherwise, for all butfinitely many i ∈ I , | A i ∩ y | = 2. At this point we use Montenegro’s result that RC implies C − , which in turns implies C − . Since, by Lemma [9.1], 3RC fin implies RC , byapplying C − to { A i : | A i ∩ y | = 2 } we get to a case which has already been solved,namely the one in which { i ∈ I : | A i ∩ y | = 1 } is infinite.In the proofs of the following two theorems we use the ideas Montenegro needed in[6] to show the implication RC = ⇒ C − . Theorem 6.2. fin ⇒ − fin .Proof. Let A = { A j : j ∈ J } be an infinite family of pairwise disjoint finite sets. Set x = S A and apply 4RC fin to get an infinite y ⊆ x and a function g : [ y ] > → [ y ] suchthat, for all Y ∈ [ y ] , g ( Y ) ⊆ Y . Since every element of A is finite, there must be aninfinite subset I of J such that, for all i ∈ I , A i ∩ y = ∅ . With perfectly analogousarguments as in the previous lemma, it is easy to see that the only difficult case iswhen, for all i ∈ I , | A i ∩ y | = 3. The following part of the proof shows that 4RC fin implies C − .For all i ∈ I , set B i = A i ∩ y . We define a directed graph G ⊆ I on I : let ( i, j ) bean edge if and only if B j * g ( B i ∪ B j ). The idea behind this definition is that every time( i, j ) is an edge, then the function g selects one element from B j whenever consideredtogether with B i (we choose the element in B i \ g ( B i ∪ B j ) if | B i ∪ g ( B i ∪ B j ) | = 2).We say that an i ∈ I has outdegree k whenever |{ j ∈ I : ( i, j ) ∈ G }| = k . Notice thatwe can assume that no i ∈ I has infinite outdegree, otherwise we could easily selectone element from infinitely many B i . Now we claim that for each k ∈ ω there are5nly finitely many i ∈ I such that i has outdegree k . To prove this, assume towards acontradiction that there exists some k ′ ∈ ω and e I ⊆ I such that | e I | = 2 k ′ + 3 and thatfor all i ∈ e I , i has outdegree k ′ . By construction, if n is the number of edges containedin e I , then (cid:0) | e I | (cid:1) ≤ n ≤ | e I | k ′ , from which follows k ′ + 1 ≤ k ′ , a contradiction. We haveobtained a well ordered partition of I into finite classes according to the outdegree ofevery i ∈ I . In symbols: I k = { i ∈ I : i has outdegree k } for every k ∈ ω. Applying 4RC fin to I , we extract at most 4 elements from each class, and for some1 ≤ m ≤
4, we get exactly m elements from infinitely many classes, so we can assumethat we get m elements from every class. Write f ( I k ) for the m extracted elementsfrom I k . We finish the proof by analyzing each of these cases separately.If m = 1, then, for all k ∈ ω , there is at least one edge between the element of f ( I k ) and the one of f ( I k +1 ), and this allows us to select one element from B k or B k +1 .If m = 2, we just consider, for all k ∈ ω , the edges (there must be at least one)between the two elements of f ( I k ), and conclude the proof as in the previous case.If m = 3, consider again, for all k ∈ ω , all the inner edges contained in f ( I k ) . Sinceeach i ∈ f ( I k ) can be chosen at most 2 times and there are at least 3 inner edges, weare always able to choose one element from some some B j with j ∈ f ( I k ).If m = 4, consider, for all k ∈ ω , g ( ∪ i ∈ f ( I k ) B i ). If it selects less than 4 elementsfrom f ( I k ), we are in one of the previous cases and if it selects exactly one elementfrom each B i , with i ∈ f ( I k ), we are also done. This concludes the proof.In the proof of the following theorem we will also use, without going into details,techniques and arguments which were carefully explained in the two preceding proofs. Theorem 6.3. fin ⇒ − fin .Proof. Let A = { A j : j ∈ J } be an infinite family of pairwise disjoint finite sets. Set x = S A and apply 6RC fin to get an infinite y ⊆ x and a function g : [ y ] > → [ y ] . Asusual, we can assume | A j ∩ y | < j ∈ J . Moreover, it is possible to assume | A j ∩ y | < j ∈ J , as well. To see it, take for instance the case in which | A i ∩ y | = 5 for all the infinitely many i ∈ I ⊆ J . As in the previous theorem, definean oriented graph G ⊆ I and let ( i, j ) be an edge if and only if A j * g ( A i ∪ A j ). Thisway, we obtain a well ordered partition of I into finite classes I k , for k ∈ ω , accordingto the outdegree of each i ∈ I . Apply 6RC fin to I and extract a finite set f ( I k ) ofat most 6 elements from each class I k . Then extract again at most 6 elements fromeach ∪ i ∈ f ( I k ) ( A i ∩ y ). The only case which is not solved by the last extraction is when | f ( I k ) | = 1 for all k ∈ ω , but this is easily handled as the case m = 1, at the end ofthe previous proof. The case when | A i ∩ y | = 4 for infinitely many i ∈ I ⊆ J can besolved in the same way.Now, given A = { A j : j ∈ J } and y ⊆ S A , let I = { i ∈ J : A i ∩ y = ∅} . Apply6RC fin to ∪ i ∈ I A i \ y to get an infinite z ⊆ S A\ y . Similarly, if K = { k ∈ I : A k ∩ z = ∅} ,6pply 6RC fin to ∪ k ∈ K A k \ ( y ∪ z ) to get an infinite w ⊆ S A \ ( y ∪ z ). A straightforwardanalysis shows that the only non trivial case is given, modulo symmetries, by the onein which | A j ∩ y | = 3 , | A j ∩ z | = | A j ∩ w | = 2 and | A j | = 7 , for all j ∈ J. Our goal is to select one element either from | A j ∩ y | or | A j ∩ z | , for infinitely many j ∈ J .In order to do this, we consider the family of edges E = { E j := ( A j ∩ y ) × ( A j ∩ z ) : j ∈ J } and the corresponding partitions F ja = { e ∈ ( A j ∩ y ) × ( A j ∩ z ) : e (1) = a } , a ∈ A j ∩ y,G jb = { e ∈ ( A j ∩ y ) × ( A j ∩ z ) : e (2) = b } , b ∈ A j ∩ z. Notice that for all j ∈ J , | F ja | = 2 and | G jb | = 3. It is easy to see that whenever weselect a proper subset of E j for some j ∈ J , we are able to select one element from A j ∩ y or from A j ∩ z . Also for this reason, when applying 6RC fin to E := S E , wecan assume that we get a selection function f on the set of all edges E . To simplifythe notation, let e f be defined as e f : [ E ] → E , e f : S S \ f ( S ). Now, for j ∈ J and b ∈ A j ∩ z , define the degreedeg( G jb ) = |{ F ia ∪ F ia ′ : i ∈ J ∧ a, a ′ ∈ ( A j ∩ z ) ∧ e f ( G jb ∪ F ia ∪ F ia ′ ) ∈ F ia ∪ F ia ′ }| . We can assume that every G jb has finite degree, since we would be otherwise ableto select a proper subset from infinitely many E j . In addition, assume that for some k ∈ ω there are infinitely many G jb with degree equal to k . Then order k + 1 distinct4-element sets of the form F ia ∪ F ia ′ for some i ∈ J and a, a ′ ∈ ( A j ∩ z ). For each G jb ,there must be a first of these k + 1 sets with the property that e f ( G jb ∪ F ia ∪ F ia ′ ) ∈ G jb ,but this fact allows us to select one edge from each G jb with degree equal to k . Thus,assume that for each k ∈ ω there are only finitely many G jb with degree k . This givesus a well ordered partition of { G jb : j ∈ J ∧ b ∈ A j ∩ z } into finite subclasses. Explicitlyinto the subclasses H k = { G jb : deg( G jb ) = k } . Apply one last time the function f to each S H k and notice that the only case in whichwe are not able to select a proper subset from infinitely many E j , is when, for all butfinitely many k ∈ ω , f ( S H k ) = E i for some i ∈ J . We conclude the proof by solvingthis last case. Suppose that for infinitely many k ∈ ω , given f ( S H k ) = E i , there isat least one G ib ⊆ E i such that for an l ∈ J and a k ′ ∈ ω , with f ( S H k ′ ) = E l , it ispossible to select an element from G ib by considering the setsel( G ib , l ) := { e f ( G ib ∪ F la ∪ F la ′ ) : | F la ∪ F la ′ | = 4 ∧ F la ∪ F la ′ ⊆ E l } . Then we can conclude by choosing, for each such k ∈ ω , the first k ′ ∈ ω with thementioned property. If that is not the case, fix k ∈ ω with f ( H k ) = E i such that forinfinitely many j ∈ J and k ∈ ω with f ( H k ) = E j we have that | sel( G ib , j ) | = 3 for both b ∈ A i ∩ z, b ∈ A i ∩ z and a ∈ A i ∩ y . This selects a proper subsetfrom infinitely many E j , namely that unique F ja ∪ F ja ′ ⊆ E j such that e f ( G ib ∪ F ja ∪ F ja ′ ) = ( a , b ) . We conclude the subsection with an example of how it is possible to obtain aweaker implication than nRC fin = ⇒ nC − fin for some infinite class of cases. p k WOC − fin isessentially the same axiom as p k C − fin . The only difference is that we require the familyof finite sets to be well-ordered. Theorem 6.4.
For all primes p and all natural numbers k , p k RC fin ⇒ p k WOC − fin .Proof. Let A = { A i : i ∈ ω } be a well-ordered family of finite sets such that | A i | > p k for all i ∈ ω . p k WOC − fin is basically obtained by repeated applications of p k RC fin to S A , together with the following two considerations: The first is that if a finite sum P a of divisors of p k is such that P a > p k , then it is possible to extract a subsum P a ′ such that P a ′ = p k . The second consideration, which allows us to conclude the proof,is the following: Given a well-ordered family B = { B i : i ∈ B } of finite sets of the samesize m ∤ p k , with p k RC fin we can extract a family of subsets B ′ = { B ′ i : i ∈ B ′ ⊆ B } such that for every i ∈ B ′ , ∅ ( B ′ i ( B i . To see this, it is enough to apply p k RC fin to S B and, if needed, to choose p k elements from the union the first l sets, where l is theleast natural number such that lm > p k , and repeat for every next block of l elementsof the family B . A partial negative answer is provided by the models V , introduced and used in [3],to which we refer for more detailed explanations. In general, the model V n has acountable set of atoms A partitioned in blocks A i = { a i , . . . , a in } , i ∈ Q , of size n whichare linearly ordered isomorphically to Q . The normal ideal is the one given by the finitesubsets and the permutation group G is the one generated by all those permutations ϕ i on A that act as the identity on A \ A i and as the cycle ( a i , . . . , a in ) on A i , for some i ∈ Q . V n is generalized to V n ,...,n l , which is built basically in the same way, but inwhich the set of atoms is partitioned in l distinct and disjoint Q -lines of blocks. Wehave the following result. Theorem 6.5.
Let l ∈ ω , p , . . . , p l be distinct primes and a , . . . , a l natural numbersgreater than . Then, we have that V p a ,...,p all | = nRC fin ⇐⇒ V p a ,...,p all | = nC − fin ⇐⇒ n is a multiple of l Y k =1 p a k k . roof. It suffices to prove the theorem for l = 1. The general case then follows from V p a ,...,p all | = nRC fin ⇐⇒ l ^ k =1 V p akk | = nRC fin , and V p a ,...,p all | = nC − fin ⇐⇒ l ^ k =1 V p akk | = nC − fin . In [8, Proposition 5.3, Lemma 5.4] it is shown that V p a | = nC − fin ⇐⇒ n is a multiple of p a . It remains to show that the same holds for nRC fin . We start with showing that p a RC fin holds. In order to do so, we use the construction from [3, Fact 4] , which we now brieflyrecall. To help the reader, we use the same notation. Let x be an infinite not well-orderable set with support E and z ∈ x an element with support E z which is notsupported by E . Let A r be a block of atoms included in E z but not in E . Then, if wedefine the set f as f = { ( ϕ ( z ) , ϕ ( A r )) : ϕ ∈ fix G ( E z \ A r ) } , the following statements hold: • f is supported by E z \ A r ; • f is a function with dom( f ) ⊆ x and ran( f ) = { A q : q ∈ I } for some possiblyunbounded interval I ⊆ Q ; • if y = dom( f ) and Y = { f − ( A q ) : q ∈ I } , then Y is a linearly orderable partitionof y ; • the elements of Y are finite sets all having the same cardinality, which has to bea divisor of p a ; • we can write Y = { U ϕ : ϕ ∈ fix G ( E z \ A r ) } , where for ϕ ∈ fix G ( E z \ A r ), U ϕ = { ηz : η ∈ fix G ( E z \ A r ) , ϕ − η ( A r ) = A r } . Consider now the orbits O s = { ϕ ( s ) : s ∈ [ y ] >p a , ϕ ∈ fix G ( E z \ A r ) } and write O = { O s : s ∈ [ y ] >p a } . The goal is to show that it is possible to choose for each O s asubset ˜ s ( s such that | ˜ s | = p a and if O s = O t with ϕ ( s ) = t , then ϕ (˜ s ) = ˜ t. Noticethat this is equivalent to requiring that every time ϕ ( s ) = s for some ϕ ∈ fix G ( E z \ A r ),then ϕ (˜ s ) = ˜ s. Now, fix an O s ∈ O . Notice that if s is a union s = S { U ϕ : ϕ ∈ P s } for some subset P s ⊆ fix G ( E z \ A r ) the conclusion is trivial. To deal with the othercases, once more we will fully rely on the fact that if a sum of divisors of p a is greaterthan p a , then there is a subsum equal to p a . Indeed, notice that for all a ∈ s , the9ardinality of { ϕ ( a ) : ϕ ∈ fix G ( E z \ A r ) , ϕ ( s ) = s } has to be a divisor of p a . Theconclusion is given by the last claim together with the fact that if ˜ s ⊆ s is a union oforbits in the form { ϕ ( a ) : ϕ ∈ fix G ( E z \ A r ) , ϕ ( s ) = s } , then ϕ ( s ) = s implies ϕ (˜ s ) = ˜ s .To finish the proof we have to show that nRC fin is false in V p a whenever n is nota multiple of p a . But this can easily be shown on the set of all atoms. Fix a positive integer n and let L n be the signature containing an ( m + n )-placerelation symbol Sel m for each m ∈ ω with m > n . Let T n be the L n -theory containingthe following axiom schema: For each m ∈ ω with m > n , we have Sel m ( x , . . . , x m , x ′ , . . . , x ′ n ) if and only if the following holds: • V ≤ i For any positive integer n there exists a model F | = T n with domain ω such that: Given a non empty M ∈ C , F admits infinitely many submodels isomorphicto M . • Any isomorphism between two finite submodels of F can be extended to an auto-morphism of F .Proof. The construction of F is by induction on ω . Let F = ∅ . F is trivially a modelof T n and, for every element M of C with | M | ≤ F contains a submodel isomorphicto M . Let F s be a model of T n with a finite initial segment of ω as domain and suchthat for every M ∈ C with | M | ≤ s , F s contains a submodel isomorphic to M . Let • { A i : i ≤ p } be an enumeration of [ F s ] ≤ n , • { R k : k ≤ q } be an enumeration of all M ∈ C with 1 ≤ | M | ≤ s + 1, • { j l : l ≤ u } be an enumeration of all the embeddings j l : F s | A i ֒ −→ R k , where i ≤ p , k ≤ q and | R k | = | A i | + 1.For each l ≤ u , let a l ∈ ω be the least natural number such that a l / ∈ F s ∪ { a l ′ : l ′ < l } .The idea is to add a l to F s , extending F s | A i to a model F s | A i ∪ { a l } isomorphic to R k ,where j l : F s | A i ֒ −→ R k . Define F s +1 := F s ∪ { a l : l ≤ u } .In [2], F s +1 is made into a model of T n in a non-controlled way, while here we imposethe following: Let { x , . . . , x m ′ } be a subset of F s +1 from which we have not alreadychosen an n -element subset. Suppose m ′ > n and that i > j implies x i > x j (recall that F s +1 is a subset of ω ). Then we simply impose Sel( { x , . . . , x m ′ } ) = { x m ′ − n +1 , . . . , x m ′ } .The desired model is finally given by F = S s ∈ ω F s .We conclude by showing that every isomorphism between finite submodels can beextended to an automorphism of F . Let i : M → M be an isomorphism of T n -models. Let a be the least natural number in ω \ ( M ∪ M ). Then M ∪ M ∪ { a } iscontained in some F n and by construction we can find some a ′ ∈ ω such that F | M ∪{ a } is isomorphic to F | M ∪{ a ′ } . Extend i to i : M ∪ { a } → M ∪ { a ′ } by imposing i ( a ) = a ′ . Let a be the least integer in ω \ ( M ∪ M ∪ { a , a ′ } ) and repeat theprocess. The desired automorphism of F is i = S t ∈ ω i t . Definitions. Let us fix some notations and terminology. The elements of the model F above constructed will be the atoms of our permutation model. Since for each atom a there is a unique triple s, i, k such that F s | A i ∪ { a } is isomorphic to R k , each atom a corresponds to a unique embedding j a : F s | A i ֒ −→ R k . We shall call the domain ofthe embedding j a the ground of a . Furthermore, given two atoms a and b , we say that a < b in case a < ω b according to the natural ordering. Notice that this well orderingof the atoms does not exist in the permutation model.Let A be the domain of the model F of the theory T n . Then the permutation modelMOD n is built as follows: Consider the normal ideal given by all the finite subsets of A and the group of permutations G defined by π ∈ G ⇐⇒ ∀ X ∈ [ ω ] fin , π (Sel( X )) = Sel( πX ) . heorem 7.2. MOD n | = nRC fin .Proof. Firstly, notice that because for any m > n the function Sel selects an n -elementset from each m -element set of atoms, nRC fin holds in MOD n for any infinite set ofatoms. So, for an infinite set X in MOD n , it is enough to construct a bijection betweenan infinite set of atoms and a subset of X — the function Sel on the finite sets of atomswill then induce a selection function on the finite subsets of some infinite subset of X .Let X be an infinite set in MOD n with support S ′ . If X is well ordered, theconclusion is trivial, so let x ∈ X be an element not supported by S ′ and let S be asupport of x with S ′ ⊆ S . Let a ∈ S \ S ′ . If fix G ( S \ { a } ) ⊆ Sym G ( x ) then S \ { a } is a support of x , so by iterating the process finitely many times we can assume thatthere exists a permutation τ ∈ fix G ( S \ { a } ) such that τ ( x ) = x . Our conclusion willfollow by showing that there is a bijection between an infinite set of atoms and a subsetof X , namely between { π ( a ) : π ∈ fix G ( S \ { a } ) } and { π ( x ) : π ∈ fix G ( S \ { a } ) } .Suppose towards a contradiction that there are two permutations σ, σ ′ ∈ fix G ( S \{ a } ) such that σ ( x ) = σ ′ ( x ) but σ ( a ) = σ ′ ( a ). Then, by direct computation, thepermutation σ − σ ′ is such that σ − σ ′ ( a ) = a and σ − σ ′ ( x ) = x . Let b = σ − σ ′ ( a ).Then { b } ∪ ( S \ { a } ) is a support of x . By construction, the set { π ( a ) : π ∈ fix G ( { b } ∪ ( S \ { a } )) } is infinite, from which we deduce that also the set L = (cid:8) a ∈ A : ∃ π ∈ fix G ( S \ { a } ) such that π ( x ) = x and π ( a ) = a (cid:9) is infinite. Now, by assumption there is a permutation τ ∈ fix G ( S \ { a } ) such that τ ( x ) = x . Let y := τ ( x ). Then a standard argument shows that also R = (cid:8) a ∈ A : ∃ π ∈ fix G ( S \ { a } ) such that π ( x ) = y and π ( a ) = a (cid:9) must be infinite.First note that in L (and similarly also in R ) there are infinitely many elements withground S \ { a } . This is because ( S \ { a } ) ∪ { a } ⊆ F is a finite model of T n andin the construction of our permutation model we add infinitely many atoms a l (wherefrom outside, a l ∈ ω ), such that ( S \ { a } ) ∪ { a l } and ( S \ { a } ) ∪ { a } are isomorphicvia an isomorphism δ with δ | S \{ a } = id | S \{ a } and δ ( a l ) = a . We can extend δ to anautomorphism δ ∈ fix G ( S \ { a } ). By definition of L we have that a l ∈ L .Let r ∈ R and p, l ∈ L all having the same ground S \ { a } such that r ≥ p , l ≥ p andmin( { p, q, r } ) > max( S \ { a } ). We want to show that every map γ : ( S \ { a } ) ∪ { p } ∪ { l } → ( S \ { a } ) ∪ { p } ∪ { r } with γ | ( S \{ a } ) ∪{ p } = id ( S \{ a } ) ∪{ p } and γ ( l ) = r is an isomorphism of T n -models. Let X ⊆ ( S \ { a } ) ∪ { p } ∪ { l } . If { p, l } ∩ X = ∅ we have that γ (Sel( X )) = Sel( γ ( X )).If l ∈ X and p / ∈ X let π l , π r ∈ fix G ( S \ { a } ) with π l ( a ) = l and π r ( a ) = r .Then π r ◦ π − l | X = γ | X . So since π r ◦ π − l ∈ G we have γ (Sel( X )) = Sel( γ ( X )). In12he last case, when { p, l } ⊆ X , the selection function n biggest elements because ofthe particular care we took in the construction of the selection function on the set ofatoms and since p, r and l have ground S \ { a } . So we can extend γ to a function τ ′ ∈ fix G ( { p } ∪ ( S \ { a } )) with τ ′ ( l ) = r .Let π r ∈ fix G ( S \ { a } ) such that π r ( a ) = r and π r ( x ) = y . Let π l ∈ fix G ( S \ { a } )with π l ( a ) = l and π l ( x ) = x . Then we have that π − r ◦ τ ′ ◦ π l ( a ) = a which impliesthat π − r ◦ τ ′ ◦ π l ( x ) = x because the function fixes S . So τ ′ ( x ) = τ ′ ◦ π l ( x ) = π r ( x ) = y . (1)Now let π p ∈ fix G ( S \ { a } ) with π p ( a ) = p and π p ( x ) = x . Since S is a support of x , π p ( S ) = { p } ∪ ( S \ { a } ) is also a support of π p ( x ) = x . Therefore, τ ′ ( x ) = x . This is a contradiction to (1). So we showed that for all σ, σ ′ ∈ fix G ( S \ { a } ), σ ( x ) = σ ′ ( x ) implies σ ( a ) = σ ′ ( a ), from which we get the desired bijection.Due to the following theorem, the class of models MOD n will not tell us anythingabout the horizontal implications in the diagram. Theorem 7.3. For each n ∈ ω , MOD n | = ACF − .Proof. Fix n ∈ ω and let A = { A i : i ∈ I } be a family of finite sets. By applyingnRC fin to S A , it is enough to show that for all m ≤ n , C − m holds in MOD n . Fix m ∈ ω with m ≤ n and suppose A = { A i : i ∈ I } is a family of m -element sets, and let P be a support of A . If S A is well-orderable we are done, so let x ∈ A be an elementwhich is not supported by P , let S ′ be a support of x and a ∈ S ′ \ P an atom suchthat for some π ∈ fix G ( P ∪ ( S ′ \ { a } )), we have that π ( x ) = x , as in the previous proof.Set S = P ∪ ( S ′ \ { a } ) and X = { π ( x ) : π ∈ fix G ( S ) } . Now, we can replace A with { A i ∩ X : i ∈ I } since a choice function on this last set gives a choice function on theprevious A as well, and let us assume that A is family of m ′ -element sets for some m ′ ≤ m . As in the proof of Theorem 7.2 we can show that there is a bijection betweenthe infinite set X and the set of atoms Y := { π ( a ) | π ∈ fix G ( S ) } . So we can withoutloss of generality assume that A is a family of m ′ -element subsets of the atoms. Let A i ∈ A with A i ∩ S = ∅ , let a ∈ A i and let R ′ ⊆ A \ ( S ∪ A i ) be an ( n − r ∈ A \ ( S ∪ A i ∪ R ′ )such that ∀ a ∈ A i \ { a } (Sel( R ′ ∪ { r } ∪ { a } ) = R ′ ∪ { r } )and Sel( R ′ ∪ { r } ∪ { a } ) = R ′ ∪ { a } . R := R ′ ∪ { r } . Again by construction of the permutation model, we can findinfinitely many b ∈ A that behave the same way as a with respect to R ∪ S ∪ ( A i \{ a } ).In other words, if repl is the function that replaces a by b , i.e.repl : A → Ax a if x = b ; b if x = a ; x otherwise,we have that for all X ⊆ R ∪ S ∪ ( A \ { a i } )repl(Sel( X ∪ { a } )) = Sel(repl( X ∪ { a } )) . (2)Define γ : S ∪ R ∪ A i → S ∪ R ∪ ( A i \ { a } ) ∪ { b } by γ := repl | S ∪ R ∪ A i . With (2) we see that γ is an isomorphism of T n -models becausefor all X ⊆ R ∪ S ∪ A i γ (Sel( X )) = Sel( γ ( X )) . So we can extend γ to the whole model F . Since γ ∈ fix G ( S ∪ R ) , γ ( A i ) ∈ A . Sothere are infinitely many A j ∈ A such that there is exactly one element a ∈ A j with a ∈ Sel( R ∪ { a } ). Choose this element a . This gives a choice function with support R ∪ S .We just mention that fact that all of C n and nC fin for n ∈ ω are false in every MOD m :it is enough to consider the family of all set of atoms of correspondent cardinalities. In this subsection there is only to notice the straightforward: Lemma 8.1. For all k, n ∈ ω , nRC fin ⇒ knRC fin . Theorem 8.2. Let m, n ∈ ω with n > m . For every n which is not a multiple of m , MOD m = nRC fin .Proof. Consider the set of the atoms and suppose that there is an infinite subset A with a function f which selects n elements from every finite and large enough subsetof A . Let S be a support of f . Let M be any model of the theory T m with cardinality | M | = mk for k ∈ ω such that m ( k − < n < mk . Then it is possible to find an mk -element subset N = { x , . . . , x mk } ⊆ ω such that:14. N and M are isomorphic as models of T m ;2. Sel( Z ) can be fixed arbitrarily whenever Z ⊆ S ∪ N with | Z ∩ S | ≥ | Z ∩ N | ≥ { x im +1 , . . . , x mk } ) = { x im +1 , . . . , x ( i +1) m } holds for all i < k .Notice that that condition 3 is only a matter of reordering. Consider the followingpermutation of N , written as a finite product of finite cycles: e π = Y i As an immediate consequence of Lemma 8.1, we get the following. Lemma 9.1. For all k, n ∈ ω , nRC fin ⇒ RC kn +1 . It is interesting to notice that Lemma 9.1 and the next theorem are here provenusing qualitatively the same approach. Despite this fact, the forthcoming proof is morecomplex than the other. Theorem 9.2. fin ⇒ RC .Proof. Let A be an infinite set and apply 4RC fin to get an infinite subset B ⊆ A with afunction e f : [ B ] > → [ B ] . Let S be a 7-element subset of x . In this proof we are goingto consider all the possible ways in which the function e f can act on the subsets of S inorder to show that it is always possible to choose a particular element of S , and hence toverify RC . Though making use of symmetries in a few passages, it will substantially bea case-by-case analysis. Let S = { x, y, z, a, b, c, d } with e f ( S ) = { a, b, c, d } . To simplifythe notation, define the two functions • f : [ S ] > → P ( S ) given by f : T T \ e f ( T ); • g : [ S ] < → [ S ] < given by g : T f ( S \ T ).For simplicity we will write, for instance, g ( a ) instead of g ( { a } ). We can assume thatfor all l ∈ e f ( S ) = { a, b, c, d } we have that g ( l ) ∩ { x, y, z } = ∅ . Otherwise, thereis a natural way to choose an element from S . Now we build, step by step, all thepossibilities for { g ( a ) , g ( b ) , g ( c ) , g ( d ) } which do not allow us to immediately choose anelement from S .1. By symmetry, we can fix g ( d ) = f ( x, y, z, a, b, c ) = { a, b } .2. There are now only two non-equivalent cases:(a) g ( d ) = { a, b } and g ( c ) = { a, b } ;(b) g ( d ) = { a, b } and g ( c ) = { a, d } , which is equivalent to the third possiblechoice g ( c ) = { b, d } .3. The two cases branch now in five:(a) g ( d ) = { a, b } , g ( c ) = { a, b } and g ( b ) = { a, c } . This is symmetric to whichis symmetric to g ( d ) = { a, b } , g ( c ) = { a, b } and g ( b ) = { a, d } .(b) g ( d ) = { a, b } , g ( c ) = { a, b } and g ( b ) = { c, d } .(c) g ( d ) = { a, b } , g ( c ) = { a, d } and g ( b ) = { a, c } .17d) g ( d ) = { a, b } , g ( c ) = { a, d } and g ( b ) = { c, d } .(e) g ( d ) = { a, b } , g ( c ) = { a, d } and g ( b ) = { a, d } .Notice how the option 3.c can be ignored, since it allows us to choose a in S inde-pendently from g ( a ). With similar arguments we can show that the only four non-symmetric choices for g ( a ) in which we cannot immediately choose an element from S are:1. g ( d ) = { a, b } , g ( c ) = { a, b } , g ( b ) = { a, c } and g ( a ) = { b, d } ;2. g ( d ) = { a, b } , g ( c ) = { a, b } , g ( b ) = { c, d } and g ( a ) = { c, d } ;3. g ( d ) = { a, b } , g ( c ) = { a, d } , g ( b ) = { c, d } and g ( a ) = { b, c } ;4. g ( d ) = { a, b } , g ( c ) = { a, d } , g ( b ) = { a, d } and g ( a ) = { c, d } .For each of the above cases we can check that the only permutations on { a, b, c, d } thatpreserve g are given by1. ( a, b )( c, d );2. ( a, b ), ( c, d ), ( a, b )( c, d ), ( a, c )( b, d ), ( a, d )( b, c );3. ( a, c )( b, d );4. ( a, d )( b, c ).In each of these cases, it is possible to select a particular double transposition (incase 2, pick ( a, b )( c, d )). Last, consider how g acts on the six distinct pairs includedin { a, b, c, d } . A double transposition selects exactly two of these pairs: for instance( a, b )( c, d ) selects { a, b } and { c, d } . We conclude the proof by considering the uniquelydetermined g ( g ( a, b ) ∪ g ( c, d )). Corollary 9.3. fin implies RC n whenever n is odd and greater than .Proof. We have that either n = 1 + 4 k or n = 3 + 4 k for a k ∈ ω . The first case followsdirectly by Lemma 9.1. In the second case let x be an infinite set and apply 4RC fin toget an infinite subset y ⊆ x with a selection function f : [ y ] > → [ y ] . Let z ⊆ y bean n -element subset. Apply f exactly k times to find a 3-element subset z of z . Then | z ∪ f ( z ) | = 7 and we can use Theorem 9.2. Theorem 9.4. Let m, n ∈ ω \ { } . Then MOD m = RC n whenever for some prime p divisor of m , n p ) . roof. Let n, m ∈ ω \ { } and let p be a prime divisor of m such that n p ).Consider the set of the atoms and suppose that there is an infinite subset A with afunction f which selects an element from every n -element subset of A . Let S be asupport of f . Let M be any T m -model with cardinality | M | = n and write n = pk + r for unique k, r ∈ ω , with 1 < r < p . Then it is possible to find an n -element subset N = { x , . . . , x n } of C such that:1. N and M are isomorphic as models of T ;2. Sel( Z ) can be arbitrarily fixed whenever Z ⊆ S ∪ N with | Z ∩ S | ≥ | Z ∩ N | ≥ { x im +1 , . . . , x n } ) = { x im +1 , . . . , x ( i +1) m } holds for all i < k .Notice that that condition 3 is only a matter of reordering. Consider the followingpermutation of N , written as finite product of finite cycles. e π = ( x pk +1 , x pk +2 , . . . , x n ) k − Y i =0 ( x pi +1 , x pi +2 , . . . , x p ( i +1) )Our conclusion will follow by showing that there is a model M of T m , a correspondingsubset N ⊆ C and a permutation π ∈ fix G ( S ) such that π acts on N exactly as e π acts on M . Notice that every cycle in the definition of e π is non trivial if and only if r = 1. First we want to find a T m -model M = { x , . . . , x n } such that M and e πM areisomorphic as T m -models. Naturally we first impose condition 3, namely that for all i < k Sel( { x im +1 , x im +2 , . . . , x n } ) = { x im +1 , x im +2 , . . . , x ( i +1) m } . The main idea of the proof is the following: Let L be a subset of M with | L | > m , L = { x im +1 , x im +2 , . . . , x n } for every i < k . Consider the orbit { e π l L : l ∈ ω } . Now wewill choose an m -element subset L ′ ⊆ L and define Sel( L ) := L ′ . Extend this choice tothe whole orbit by defining Sel( e π l L ) = e π l (Sel( L )) . The choice of Sel( L ) has to be suitable in the sense that e π j L = L must imply e π j (Sel( L )) = Sel( L ). • First of all assume that for some I ⊆ k , | L ∩ ( [ i ∈ I { x pi +1 , . . . , x p ( i +1) } ) | ∈ { m, m − | L ∩ { x pk +1 , . . . , x n }|} . Then a suitable choice for Sel( L ) is given by either L ∩ ( S i ∈ I { x pi +1 , . . . , x p ( i +1) } )or by L ∩ ( S i ∈ I { x pi +1 , . . . , x p ( i +1) } ∪ { x pk +1 , . . . , x n } ).19 Otherwise, let J ⊆ k be the set of indices j such that 0 < | L ∩{ x jp +1 , . . . , x ( j +1) p }| 6 = p . Moreover, replace J by J ∪ { k } if | L ∩ { x kp +1 , . . . , x n }| either is 1 or doesnot divide r . For the sake of notation, let us write { x kp +1 , . . . , x ( k +1) p } in-stead of { x kp +1 , . . . , x n } . If | L \ S j ∈ J { x jp +1 , . . . , x ( j +1) p }| ≤ m , then we claimthat a suitable choice for Sel( L ) is given by any m -subset of L which includes L \ S j ∈ J { x jp +1 , . . . , x ( j +1) p } . The claim follows by the fact that, given a set { y , . . . , y p ′ } for some prime p ′ ∈ ω , if τ is the permutation ( y , . . . , y p ′ ) andsome power τ a fixes a proper subset H ( { y , . . . , y p ′ } , then τ a is the identity on { y , . . . , y p ′ } .Note that we covered every possible case. Indeed, if we are not in the last case, thenfor some k ′ , r ′ ∈ ω with r ′ ≤ r it is true that m < k ′ p + r ′ . Then, since r < p and p | m ,we are actually in the first case.Now we can show that S is not a support of the selection function f we choseat the beginning of the proof. Let M be the T m -model we constructed above thatsatisfies e πM = M . Let N ⊆ ω be a T m -model that is isomorphic to M and satisfiesconditions 1,2 and 3. With the proof above and condition 2 we can choose N suchthat π (Sel( L )) = Sel( π ( L )) for all L ⊆ N ∪ S . So π can be extended to a function π ∈ fix G ( S ) on the whole model F . Note that for all n -element subsets of N we havethat π ( N ) = N . So S is indeed not a support of the selection function f . This is acontradiction.With the same arguments it is possible to emulate the previous result in the fol-lowing way. Theorem 9.5. Let m ∈ ω be greater than . Then for all < n < m , MOD m = RC n .Proof. Exactly as in the previous theorem: just consider the permutation e π = ( x , . . . , x n )and impose that Sel( L ) ⊃ L ∩ N whenever L ∩ N = ∅ , with L ⊆ S ∪ N .As an immediate consequence of the last results, we get the following Corollary: Corollary 9.6. Let k ∈ ω \ { } , { p , . . . , p k } be distinct prime numbers and n = Q ki =1 p i . Then nRC fin ⇒ RC m if and only if m ≡ n ) . 10 Open Questions • For n ∈ { , , , } we have that nRC fin ⇒ nC − fin . Does this implication hold for n = 5? Or more generally: For which n ∈ ω does this implication hold? • Write a natural number as unique product of powers of primes n = Q ki =1 p m i i . Isit the case that nRC fin ⇒ RC m if and only if m > n and m ≡ Q ki =1 p i )?20 eferences [1] O. De La Cruz, C. A. Di Prisco, Weak Forms of the Axiom of Choice and Partitionsof Infinite Sets. In: Set Theory (Di Prisco C.A., Larson J.A., Bagaria J., andMathias A.R.D., eds.), Springer, Dordrecht (1998).[2] L. Halbeisen, Combinatorial Set Theory: With a Gentle Introduction toForcing , (revised and extended second edition), Springer Monographs in Mathe-matics, Springer, London, 2017.[3] L. Halbeisen and E. Tachtsis, On Ramsey Choice and Partial Choice for infinitefamilies of n-element sets . Archive for Mathematical Logic , 59 (2020), 583–606.[4] P. Howard and J. E. Rubin, Consequences of the Axiom of Choice ,[Mathematical Surveys and Monographs, vol. 59], American Mathematical Society,Providence, RI, 1998.[5] A. L´evy, Axioms of multiple choice , Fundamenta Mathematicae , 50 (1962),475–485.[6] C. H. Montenegro, Weak versions of the axiom of choice for families of finite sets ,in Models, algebras, and proofs , Selected papers of the X Latin Americansymposium on mathematical logic held in Bogot´a, Colombia, June 24–29, 1995(X. Caicedo and C. Montenegro, eds.), [Lecture Notes in Pure and Applied Math-ematics, vol. 203], Marcel Dekker, New York · Basel, 1999, pp. 57–60.[7] D. Pincus,