Four Cardinals and Their Relations in ZF

. For a set M , ﬁn( M ) denotes the set of all ﬁnite subsets of M , M 2 denotes the Cartesian product M × M , [ M ] 2 denotes the set of all 2-element subsets of M , and seq 1-1 ( M ) denotes the set of all ﬁnite sequences without repetition which can be formed with elements of M . Furthermore, for a set S , let | S | denote the cardinality of S . Under the assumption that the four cardinalities | [ M ] 2 | , | M 2 | , | ﬁn( M ) | , | seq 1-1 ( M ) | are pairwise distinct and pairwise comparable in ZF , there are six possible linear orderings between these four cardinalities. We show that at least ﬁve of the six possible linear orderings are consistent with ZF .


Introduction
Furthermore, for a set A, let |A| denote the cardinality of A. We write |A| = |B|, if there exists a bijection between A and B, and we write |A| ≤ |B|, if there exists a bijection between A and a subset B ′ ⊆ B (i.e., |A| ≤ |B| if and only if there exists an injection from A into B). Finally, we write |A| < |B| if |A| ≤ |B| and |A| = |B|. By the Cantor-Bernstein Theorem, which is provable in ZF only (i.e., without using the Axiom of Choice), we get that |A| ≤ |B| and |A| ≥ |B| implies |A| = |B|. It is natural to ask whether some of these equalities can be proved also in ZF, i.e., without the aid of AC. Surprisingly, this is not the case. In [1], a permutation model was constructed in which for an infinite cardinal m we have seq(m) < fin(m) (see [1,Thm. 2] or [4,Prp. 7.17]). As a consequence we obtain that the existence of an infinite cardinal m such that seq 1-1 (m) < fin(m) is consistent with ZF. This consistency result was modified to the existence of an infinite cardinal m for which m 2 < [m] 2 (see [4,Prp. 7.18]), and later, it was strengthened to the existence of an infinite cardinal m for which seq(m) < [m] 2 (see [3] or [ Consistency results as well as ZF-results concerning the relations between these cardinals with other cardinals can be found, for example, in [7,8] or [2]. Concerning the four cardinalities [m] 2 , m 2 , fin(m), and seq 1-1 (m), a question which arises naturally is whether for some infinite cardinal m, fin(m) < m 2 is consistent with ZF (see [5, Related Result 20, p. 133]). Moreover, assuming that m is infinite and the four cardinalities [m] 2 , m 2 , fin(m), seq 1-1 (m) are pairwise distinct and pairwise comparable in ZF, one may ask which linear orderings on these four cardinalities are consistent with ZF.
Since for all cardinals m, we cannot have [m] 2 > fin(m) or m 2 > seq 1-1 (m), there are only the following six linear orderings on these four cardinalities which might be consistent with ZF (where for two cardinals n 1 and n 2 , n 1 −→ n 2 means n 1 < n 2 ). Diagram Z Below we show that each of the five diagrams N, Z, N, C, Z is consistent with ZF.

Permutation Models
In order to show, for example, that for some infinite cardinals m and n, m < n is consistent with ZF, by the Jech-Sochor Embedding Theorem (see, for example, [6,Thm. 6.1] or [5,Thm. 17.2]), it is enough to construct a permutation model in which this statement holds. The underlying idea of permutation models, which will be models of set theory with atoms (ZFA), is the fact that a model V |= ZFA does not distinguish between the atoms, where atoms are objects which do not have any elements but which are distinct from the empty set. The theory ZFA is essentially the same as that of ZF (except for the definition of ordinals, where we have to require that an ordinal does not have atoms among its elements). Let A be a set. Then by transfinite recursion on the ordinals α ∈ Ω we can define the α-power P α (A) of A and P ∞ (A) = α∈Ω P α (A). Like for the cumulative hierarchy of sets in ZF, one can show that if M is a model of ZFA and A is the set of atoms of M, then M = P ∞ (A). The class M 0 := P ∞ (∅) is a model of ZF and is called the kernel. Notice that all ordinals belong to the kernel. By construction we obtain that every permutation of the set of atoms induces an automorphism of M, where the sets in the kernel are fixed.
Permutation models were first introduced by Adolf Fraenkel and, in a precise version (with supports), by Andrzej Mostowski. The version with filters, which we will follow below, is due to Ernst Specker (a detailed introduction to permutation models can be found, for example, in [5,Ch. 8] or [6]).
In order to construct a permutation model, we usually start with a set of atoms A and then define a group G of permutations or automorphisms of A.
The permutation models we construct below are of the following simple type: For each finite set E ∈ fin(A), let Fix G (E) := π ∈ G : ∀a ∈ E (πa = a) , and let F be the filter of subgroups of G generated by the subgroups {Fix G (E) : E ∈ fin(A)}. In other words, F is the set of all subgroups H ≤ G, such that there exists a finite set E ∈ fin(A), such that Fix G (E) ≤ H.
Then, a set x is symmetric if and only if there exists a set of atoms E x ∈ fin(A), such that We say that E x is a support of x. Finally, let V be the class of all hereditarily symmetric objects; then V is a transitive model of ZFA. We call V a permutation model. So, a set x belongs to the permutation model V (with respect to G and F ), if and only if x ⊆ V and x has a finite support E x ∈ fin(A). Because every a ∈ A is symmetric, we get that each atom a ∈ A belongs to V.

A Model for Diagram N
We first show that in every model for Diagram N, we have that the cardinality m is transfinite.
Proceeding this way, we finally have an injection from ω into A, which shows that ℵ 0 ≤ m. ⊣ Proof. By the assumption, there exists an injection h : ω → A, and for each i ∈ ω, let is a bijection between A and A \ C. Thus, it is enough to construct a finite-to-one function g : seq(A \ C) → fin(A). Let s = a 0 , . . . , a n−1 ∈ seq(A \ C) and let ran(s) := {a 0 , . . . , a n−1 }. The sequence s gives us in a natural way an enumeration of ran(s), and with respect to this enumeration we can encode the sequence s by a natural number i s ∈ ω. Now, let g(s) := ran(s) ∪ {x 2is }. Then, since there are just finitely many enumerations of ran(s), g is a finiteto-one function. ⊣ The following result is just a consequence of Proposition 2 and Lemma 1. We now introduce the technique we intend to use in order to build a permutation model from which it will follow that for some infinite cardinal m, the relation fin(m) < m 2 is consistent with ZF. Notice that this relation is the main feature of Diagram N and that this relation implies that ℵ 0 ≤ m. In the next section, we shall use a similar permutation model in order to show the consistency of Diagram Z with ZF.
Let K be the class of all the pairs (A, h) such that A is a (possibly empty) set and h is an injection h : fin(A) \ {∅} → A 2 . We will also refer to the elements of K as models. We define a partial ordering ≤ on K by stipulating When the functions involved are clear from the context, with a slight abuse of notation we will just write A ≤ B instead of (A, h) ≤ (B, f ) and A ∈ K instead of (A, h) ∈ K.
Before proceeding, we give two preliminary definitions. Given a model (M, f ) and a countable subset A ⊆ M , we define the closure cl(A, M ) as the smallest superset of A that is closed under f and pre-images with respect to the same function. Constructively, we can characterize cl(A, M ) as a countable union as follows: Define cl 0 = cl 0 (A, M ) := A and, for all i ∈ ω, in order to finally define cl(A, M ) := i∈ω cl i . Furthermore, we set a standardized way to extend a partial model (A, f ′ ), where f ′ is only a partial function, to an element of K: For what concerns the injection f ′ j+1 , we naturally require the inclusion f ′ j ⊆ f ′ j+1 , as well as We are now in the position of defining the plain extension of (A, f ′ ) as Given the previous definitions, we remark that given a model M ∈ K and a countable subset A ⊆ M , we have that cl(A, M ) ≤ M , which proves the following: Proposition 5 (CH). There is a model M * of cardinality c in K such that: • M * is ℵ 1 -homogeneous, i.e., if N 1 , N 2 ≤ M * are countable and π : N 1 → N 2 is an isomorphism then there exists an automorphism π * of M * such that π ⊆ π * .
• If N ≤ M * and A ⊆ M * are countable, then there is an automorphism π of M * that fixes N pointwise, such that π(A) \ N is disjoint from A.
Proof. We construct the model M * by induction on ω 1 , where we assume that ω 1 = c. Let M 0 = ∅. When M α is already defined for some α ∈ ω 1 , we can define The construction of M α+1 , starting from M α , consists of a disjoint union of two differently built sets of models. First, for each element N ∈ C α , let S N be a system of representatives for the strong isomorphism classes of all the models M ∈ K such that N ≤ M with M countable.
Here, by strong we mean that, for two models M 1 and M 2 with N ≤ M 1 , M 2 , it is not enough to be isomorphic in order to belong to the same class, but we require that there exists an isomorphism between M 1 and M 2 that fixes N pointwise, which we can express by saying that M 1 is isomorphic to M 2 over N . We first extend M α by the set where " " indicates that we have a disjoint union, and now we define M α+1 as the plain extension of M α ⊔ M ′ α . Finally, for non-empty limit ordinals δ define M δ = ∪ α∈δ M α , and let It remains to show that the model M * has the required properties: First we notice that M * has cardinality |M * | = c, as required, and since, by construction, In order to show that M * is ℵ 1 -homogeneous, we make use of a back-andforth argument. Let N 1 , N 2 ≤ M * be countable models and π : N 1 → N 2 an isomorphism. Let {x α : α ∈ ω 1 } be an enumeration of the elements of M * and let I 0 := N 1 . If x δ 1 is the first element (with respect to this enumeration) in M * \I 0 , then, by Fact 4, there exists a countable model for the same reason as above we can find countable models J 1 , I 1 such that I ′ 1 ≤ I 1 ≤ M * and J ′ 1 ≤ J 1 ≤ M * , together with x γ 1 ∈ J 1 and the fact that there exists an isomorphism π 1 : I 1 → J 1 with π ′ 1 ⊆ π 1 . Proceed inductively with x δ α+1 being the first element in M * \ I α and find countable models and π α ⊆ π ′ α+1 . As in the second part of the base step, let x γ α+1 be the first element in M * \ J ′ α+1 and find countable models with an isomorphism π α+1 : I α+1 → J α+1 such that x γ α+1 ∈ J α+1 and π ′ α+1 ⊆ π α+1 . We naturally take the union at limit stages and finally obtain π * = ∪ α∈ω 1 π α , which is the required automorphism of M * .
To show the last property of the theorem, let N ≤ M * and A ⊆ M * be both countable. Since the cofinality of ω 1 is greater than ω, we can find by construction both a countable model , where A is a set and we have three injections f : and h : seq 1-1 (A) → fin(A), which will be used below to show the consistency of Diagram Z with ZF.
Given Proposition 5, we consider the permutation model V N that arises naturally by considering the elements of the ℵ 1 -universal and ℵ 1 -homogeneous model M * as the set of atoms and its automorphisms Aut(M * ) as the group G of permutations. In particular, each permutation in G preserves the injection h : fin(M * ) \ {∅} → M 2 * that the model (M * , h) comes with. We are now ready to prove the following result.

Proof. The existence of an injection
* given by the specific permutation model, together with the fact that h cannot be surjective, which will be shown later. So, we only need to prove that in V N , there is no reverse injection from M 2 * into fin(M * ), and that there are no injections from In order to show that there is neither an injection from M 2 * into fin(M * ), nor an injection from Let S be a finite support of both functions f 1 and f 2 (if they exist). In other words, S ∈ fin(M * ) and for each automorphism π ∈ Fix G (S) we have π(f 1 ) = f 1 and π(f 2 ) = f 2 , respectively. Let N 1 be a countable model in K with S ⊆ N 1 ≤ M * . Let (N 2 , g) be a countable model in K such that (N 1 , h| N 1 ) ≤ (N 2 , g), constructed as follows: The domain of N 2 is the disjoint union Furthermore, we define the injection g : fin(N 2 ) \ {∅} → N 2 2 such that g ⊇ h| N 1 and for E ∈ fin(N 2 ) \ fin(N 1 ) we define g(E) = e 1 , e 2 such that g is injective and satisfies the following conditions (recall that since N 2 is countable, also fin(N 2 ) is countable): • If E ∩ {x, y, z} = ∅ then e 1 , e 2 = a n , a m for some n, m ∈ ω.
• If |E ∩ {x, y, z}| = 1, then e 1 , e 2 = u, a k for some k ∈ ω, where u is the unique element in E ∩ {x, y, z}.
Notice that there are automorphisms of (N 2 , g) that just permute x, y, z and fix all other elements of N 2 pointwise. By construction of M * , we find a model N ′ 2 ∈ K such that N 1 ≤ N ′ 2 ≤ M * and N ′ 2 is isomorphic to N 2 over N 1 . For this reason we can refer to N 2 as a legit submodel of M * that extends N 1 in the way we described. Let us now consider f 1 ( x, y ), where we assumed in V N the existence of an injection f 1 : N 2 , then we can apply the third property of Proposition 5 with respect to f 1 ( x, y ) and N 1 and N 2 respectively, which gives us a contradiction. If {x, y} ⊆ f 1 ( x, y ) or {x, y} ∩ f 1 ( x, y ) = ∅, we could swap x and y while fixing every other element of N 2 pointwise and get f 1 ( x, y ) = f 1 ( y, x ), which would imply that f 1 is not injective. So, assume that |{x, y} ∩ f 1 ( x, y )| = 1 and without loss of generality assume that {x, , y ), we similarly obtain a contradiction by swapping z and x, while if z / ∈ f 1 ( x, y ) we get a contradiction by swapping z and y. This shows that For what concerns f 2 , let us consider the set S consisting of sequences without repetition of {x, y, z} of length 2 or 3. Notice that |S | = 12. Now, for each element s ∈ S , if f 2 (s) = a, b , then a and b are such that a = b and a, b ∈ {x, y, z} 2 -notice that otherwise, for example, if {a, b} ∩ {x, y} = ∅, then we can swap x and y and hence move s without moving a, b , which is not consistent with S being a support of f 2 . We get the conclusion by noticing that, because of this restriction, there are only six possible images of elements of S , which implies that f 2 cannot be an injection.
It remains to show that in V N there are no injections from fin(M * ) into [M * ] 2 . For this, assume towards a contradiction that there exists such a function f 3 in V N and assume that S is a finite support of f 3 . Then, let N 1 be a countable model in K with S ⊆ N 1 ≤ M * . We will construct a countable model (N 2 , g) ∈ K satisfying (N 1 , h| N 1 ) ≤ (N 2 , g) ≤ M * with a finite subset u ∈ fin(N 2 \ N 1 ) such that, for all x, y ∈ N 2 2 \ N 2 1 , one of the following holds: • there exists an automorphism π of N 2 over N 1 with π(u) = u and π{x, y} = {x, y}.
Let now G 2 0 be an extension of G * 0 by adding a copy of G 1 0 \ N 1 , where the "copy function" is denoted by τ 0 . Notice that at this stage, together with an extension of h 1 0 defined as Notice that if a ∈ G 1 0 \ N 1 , then τ 0 (a) ∈ G 2 0 \ G * 0 . The construction carried out so far is actually the first of countably many analogous extension steps we will consequently apply in order to consider the union of all the progressive extensions. That is, assume that for some i ∈ ω we have already defined G 2 i and h 2 i . Define G 1 i+1 = G 2 i and, for each non-empty finite set E ∈ fin(G 1 i+1 ) which is not in the domain of h 2 i , consider a pair of new elements {x E , y E } and define Let now G 2 i+1 be an extension of G * i+1 by adding a copy of G 1 i+1 \ N 1 , where the "copy function" is now τ i+1 . More formally, Notice that every automorphism of (G 1 i+1 , h 2 i ) can be extended to an automorphism of (G consists of pairs {x, y}, each of which corresponds to a unique nonempty finite subset of G 1 i+1 and moves according to this finite subset. Finally, we conclude that every automorphism of (G * i+1 , h 1 i+1 ) can be extended to an automorphism of (G 2 i+1 , h 2 i+1 ), which follows from the following two facts: G 2 i+1 \ G * i+1 consists of a copy through τ i+1 of G 1 i+1 \N 1 , and therefore supports the same automorphisms. Furthermore, every automorphism of (G * i+1 , h 1 i+1 ) is an extension of some automorphism of (G 1 i+1 , h 2 i ), which follows from the fact that for all E ∈ fin(G We claim that (N 2 , g) satisfies the required properties. Indeed, if x, y ∈ N 2 2 \ N 2 1 and there is some finite set E ∈ fin(N 2 ) \ {∅} with g(E) = h(E) = x, y , then by construction of g we necessarily have x, y ∈ (N 2 \ N 1 ) 2 and the following fact: either there exists some index n ∈ ω such that x, y ∈ (G * n \ G 1 n ) 2 , or there are indices n, k ∈ ω with k > n such that for some ordered pair x ′ , y ′ ∈ (G * n \ G 1 n ) 2 we have x, y = τ k (x ′ ), τ k (y ′ ) . Each of the two conditions implies that there exists an automorphism π of N 2 over N 1 acting as follows: π x, y = y, x and πu = π{a 0 , b 0 , c 0 } = τ n (a 0 ), τ n (b 0 ), τ n (c 0 ) , which in particular means π{x, y} = {x, y} and πu = u, as desired. We can finally consider the image f 3 (u) = {x, y}: If {x, y} N 2 or {x, y} ⊆ N 1 , then we can apply the third property of Proposition 5 with respect to {x, y} and N 1 and N 2 respectively, which gives us a contradiction. Thus {x, y} ⊆ N 2 and {x, y} N 1 , and if there exists some finite set E ∈ fin(N 2 ) \ ∅ with g(E) = h(E) = x, y , then by the reasoning above we find that some automorphism of N 2 over N 1 does not preserve f 3 , a contradiction. In every other case, we consider cl(N 1 ∪ {x, y}, M * ) and notice that, since for no E ∈ fin(N 2 ) \ ∅ we have h(E) = x, y or h(E) = y, x , then we claim that u cannot be a subset of cl(N 1 ∪ {x, y}, M * ), which allows us to fix cl(N 1 ∪ {x, y}, M * ) pointwise, while not preserving u, a contradiction as well. In order to prove the claim, let U 0 := u and for i ∈ ω, let U i+1 := U i ∪ {τ i (a) : a ∈ U i } and define U := i∈ω U i . Furthermore, we define a rank-function rk : is consistent with ZF.

A Model for Diagram Z
We are now going to set an analogue framework to the one for Diagram N, just with the definitions adapted, in order to show the consistency of Diagram Z. In fact, as mentioned above, we can state the same proposition, guaranteeing the existence of a suitable ℵ 1 -universal and ℵ 1 -homogeneous model.
Let K be the class of all the quadruples (A, f, g, h) such that A is a (possibly empty) set and f, g, h are the following three injections: where the function h satisfies h(∅) = ∅. As before, we define a partial ordering ≤ on K by stipulating (A, f 1 , g 1 , h 1 ) ≤ (B, f 2 , g 2 , h 2 ) if and only if Proposition 7 (CH). There is a model M * of cardinality c in K such that: • M * is ℵ 1 -homogeneous, i.e., if N 1 , N 2 ≤ M * are countable and π : N 1 → N 2 is an isomorphism then there exists an automorphism π * of M * such that π ⊆ π * .
• If N ≤ M * and A ⊆ M * are countable, then there is an automorphism π of M * over N such that π(A) \ N is disjoint from A.
Proof. The proof is essentially the same as the one of Proposition 5.
in order to finally define cl(A, M ) := i∈ω cl i . Furthermore, we set a standardized way to extend a partial model (A, f ′ , g ′ , h ′ ), where f ′ , g ′ , h ′ are only partial functions, to an element of K: For what concerns the injections We are now in the position of defining the plain extension of (A, f ′ , g ′ , h ′ ) as and we can finally prove, in three analogous steps, that neither of the three injections of the model (M * , f, g, h) admits a reverse injection.

Assume there is an injection
To see this, notice first that since N 1 ∪ {a, b} ⊆ N 2 , we build the closure cl N 1 ∪ {a, b}, M * within the plain extension N 2 , and recall that for If there are no x ′ , y ′ such that f ( x ′ , y ′ ) = {a, b}, then, since {a, b} ⊆ N 2 and {a, b} N 1 , {a, b} cannot be a superset of any ran g({u, v}) for some u, v, or of any h( x 1 , . . . , x n ) for some x 1 , . . . , x n , so we have that {x, y} cl(N 1 ∪ {a, b}, M * ), which is a contradiction to ( * ). Now, since f ( x ′ , y ′ ) = {a, b}, by the construction of the plain extension N 2 we find an automorphism π of N 2 that fixes N 1 ∪ {x, y} pointwise and for which we have π(a) = b and π(b) = a. Hence, i(π{x, y}) = a, b = b, a = πi({x, y}), which is a contradiction.
Finally, assume there is an injection i : fin(M * ) → seq 1−1 (M * ) with finite support S. Let N 1 ∈ K be a countable model such that N 1 ≤ M * and S ⊆ N 1 . Let {x, y, z} ∈ [M * ] 3 be such that N 1 ∩ {x, y, z} = ∅, let M 0 = N 1 ⊔ {x, y, z}, and let N 2 be the plain extension of M 0 . Consider a j : j ∈ n = i({x, y, z}) for some n ∈ ω. It is easy to see that we must have {a j : j ∈ n}∩(N 2 \M 0 ) = ∅, and as before it must also hold {x, y, z} ⊆ cl N 1 ∪{a j : j ∈ n}, M * . In what follows, for a natural number n, we will refer to a cyclic permutation of order n by using the term "n-cycle". Our next step is to prove that a 3-cycle π applied to {x, y, z} cannot leave each element of {a j : j ∈ n} ∩ (N 2 \ M 0 ) unchanged, for at least one automorphism σ of N 2 extending π does not fix any element of N 2 \ M 0 , as the following argument shows: Assume there is such an automorphism σ fixing a first appearing element c ∈ M n \M n−1 in the construction of the plain extension of M 0 . Notice that σ moves every unordered pair, ordered pair and injective sequence with non-empty intersection with {x, y, z}. Now, by looking at the possible ways that c could have appeared in N 2 , one sees that if c is fixed, then some pair p ∈ [M n−1 ] 2 such that c ∈ ran(g(p)) also satisfies σ(p) = p. Now, since c was the first appearing element being fixed, we can say that for p = {a, b} ∈ [M k ] 2 , for which p = {a, b} / ∈ [M k−1 ] 2 holds, it is true that a, b are the first appearing elements being swapped by σ. Similarly as before, this implies that there are two pairs p ′ , p ′′ ∈ [M k−1 ] 2 such that σ(p) = p ′ and σ(p ′ ) = p, which in turn requires either the existence of two ordered pairs in M 2 k−1 swapped by σ, either contradicting the fact that a and b were the first appearing swapped elements, or implying the existence of a four-element set {s, r, t, p} ⊆ N 2 \ M 0 on which σ acts as a four-cycle, but we can extend π to at least one σ such that σ 3 = Id N 2 , which does not allow four-cycles, and this concludes the proof. : : is consistent with ZF.

A Model for Diagram N
We show that Diagram N holds in the model constructed in [3] (see also [5, p. 209 ff]), where m is the cardinality of the set of atoms of that model. The atoms of the permutation model V N for Diagram N are constructed as follows: (α) Let A 0 be an arbitrary infinite set.
(β) G 0 is the group of all permutations of A 0 .
Proof. In [3] it is shown that By extending E 1 if necessary, we may assume that if (n + 1, a 0 , . . . , a l−1 , ε) ∈ E 1 , then also a 0 , . . . , a l−1 belong to E 1 as well as the atom (n + 1, a 0 , . . . , a l−1 , 1 − ε). We say that a finite subset of A satisfying this condition is closed.
[m] 2 = fin(m): We show that there is no injection g 2 : fin(A) → [A] 2 . Assume towards a contradiction that there is such an injection with closed finite support E 2 . For a large enough number k ∈ ω we choose again a k-element set X ⊆ A 0 \ E 2 such that | fin (X) | > [E 2 ∪ X] 2 and such that we can find a subset S ⊆ X with P := g 2 (S) \ (E 2 ∪ X) = ∅ and |S| ≥ 2. If |P | = 1 it is clear that we can find a permutation π ∈ G which fixes E 2 pointwise and for which π(S) = S and π(g 2 (S)) = g 2 (S). Likewise, we find a contradiction also if |P | = 2 and P is not in the form {(l, p, 0), (l, p, 1)} for some l ∈ ω \ {0} and p ∈ l j=0 A j l−1 , so let us assume that P is indeed in that form. Consider the extension P ′ of P to the smallest closed superset P ⊆ P ′ . If we can find x ∈ S \ P ′ then G contains a permutation π with π(x) / ∈ S and π(P ) = P , which is a contradiction, so S \ P ′ must be empty. We notice that if a ∈ A is an atom, Y is the smallest closed set of atoms containing a and b ∈ A 0 ∩ Y , then for all permutations π ∈ G we have that π(b) = b implies π(a) / ∈ Y . We can now conclude since every element of S is in the closure of P and we can find a permutation π ∈ G which fixes pointwise E 2 with π(S) = S but for which S is not fixed pointwise. ⊣ So, the permutation model V N witnesses the following Consistency Result 3. The existence of an infinite cardinal m satisfying is consistent with ZF.

A Model for Diagram C
We show that Diagram C holds in a permutation model V C which is similar to the Second Fraenkel Model, where m is the cardinality of the set of atoms of V C .
The permutation model V C is constructed as follows (see also [5, p. 197]): The set of atoms of the model V C consists of countably many mutually disjoint, cyclically ordered 3-element sets. More formally, A = n∈ω P n , where P n = {a n , b n , c n } (for n ∈ ω), and the cyclic ordering on P n is illustrated by the following figure: a n b n c n On each triple P n , we define the cyclic distance between two elements by stipulating cyc (a n , b n ) = cyc (b n , c n ) = cyc (c n , a n ) = 1 and cyc (a n , c n ) = cyc (b n , a n ) = cyc (c n , b n ) = 2 .
Let G be the group of those permutations of A which preserve the triples P n (i.e., πP n = P n for π ∈ G and n ∈ ω) and their cyclic ordering. The sets in V C are subsets of V C with finite support.
[m] 2 ≤ m 2 : We define an injective function f 1 : [A] 2 → A 2 . Let {x, y} ∈ [A] 2 and m, n ∈ ω be such that x ∈ P m and y ∈ P n . Without loss of generality we may assume that m ≤ n. If m < n, then f 1 ({x, y}) := x, y , and if m = n, then f 1 ({x, y}) := z, z where z := P m \{x, y}.
It is easy to see that f 1 is an injective function, and since f 1 has empty support, f 1 belongs to V C . . , x k−1 ∈ seq 1-1 (A) be a non-empty sequence without repetition of length k. Let j : k → ω be such that for each i ∈ k, x i ∈ P j(i) . Let E 0 := ∅, and by induction, Now, we define f 3 (s) := E k ∪ P qs . It is easy to see that f 3 is an injective function from seq 1-1 (A) into fin(A), and since f 3 has empty support, f 3 belongs to V C .
[m] 2 = m 2 : It is enough to show that there is no injection from A 2 into [A] 2 . Assume towards a contradiction that there exists an injection g 1 : Let E 1 ⊆ E for some non-empty set E ∈ fin(A) such that P n ⊆ E whenever E ∩ P n = ∅. Then E is also a support of g 1 . Now |[E] 2 | < |E 2 |, which implies that there exists a pair x, y ∈ E 2 such that g 1 ( x, y ) / ∈ [E] 2 . So, there exists a π ∈ Fix G (E) such that πg 1 ( x, y ) = g 1 ( x, y ), but π x, y = x, y , which contradicts the fact that E is a support of g 1 . m 2 = seq 1-1 (m): It is enough to show that there is no injection from seq 1-1 (A) into A 2 . Assume towards a contradiction that there exists an injection g 2 : seq 1-1 (A) → A 2 with finite support E 2 . Let E 2 ⊆ E for some non-empty set E ∈ fin(A) such that P n ⊆ E whenever E ∩ P n = ∅. Then E is also a support of g 2 . Now |E 2 | < | seq 1-1 (E)|, and by similar arguments as above, we obtain a contradiction. seq 1-1 (m) = fin(m): It is enough to show that there is no injection from fin(A) into seq 1-1 (A). Assume towards a contradiction that there is an injection g 3 : fin(A) → seq 1-1 (A) with finite support E 3 , where we can assume that for all n ∈ ω, P n ⊆ E 3 whenever E 3 ∩ P n = ∅.
Since E 3 is finite, there exists an n ∈ ω such that g 3 (P n ) / ∈ seq 1-1 (E 3 ). Let a ∈ A be the first element of the sequence g 3 (P n ) which does not belong to E 3 . Then we find a π ∈ Fix G (E 3 ) such that πa = a (which implies πg 3 (P n ) = g 3 (P n )) but πP n = P n , which contradicts the fact that E 3 is a support of g 3 .  Assume that for some n ∈ ω we have already constructed an (n + 1)-element set E n := E i : i ≤ n of pairwise disjoint non-empty finite subsets of A. Let E n+1 := i,j≤n x : ∃a ∈ E i ∃b ∈ E j ∃y f ( a, b ) = {x, y} \ i≤n E i , and let E n+1 := E n ∪ {E n+1 }. Notice that for k := | i≤n E i |, we have which implies that E n+1 = ∅. Proceeding this way, E n : n ∈ ω is a countably infinite set of pairwise disjoint non-empty finite subsets of A.
Now, we apply the function g. For every n ∈ ω, let S n := g(E n ). Furthermore, let S 0 := S 0 , and in general, for n ∈ ω let S n+1 := S n ⌢ S n+1 . In this way, we obtain an infinite sequence S ∞ of elements of A. Since g is injective and the non-empty finite sets E n are pairwise disjoint, the sequence S ∞ must contain infinitely many pairwise distinct elements of A. Now, let h be the enumeration of these pairwise distinct elements in the order they appear in S ∞ . Then h : ω → A is an injection.