Partitioning an Infinite Set

Partitioning an Infinite Set

Carlos G. Gonz´ alez

Universidade Federal de Uberlˆ andia gonzalez@defil.ufu.br;gonzalcg@gmail.com

September 6, 2010

Abstract

We study the relative interdependence in set theory without the Axiom of Choice of properties such as: “a set can be partitioned into ninfinite sets”, “a set can be partitioned into infinitely many infinite sets”, “a set can be partitioned into countably many sets”, “a set is Dedekind infinite”, and so on. We solve as well a problem of Degen concerning notions of infinity. Finally, we present some results about partitions of Cartesian products.

In Tarski [1924] is posed the question whether an infinite set can be partitioned into two infinite sets, andLindenbaum Mostowski [1938] an- nounced the independence of the statement “every infinite set is representable as a union of two infinite and mutually disjoint sets” in set theory with atoms orUrelemente. Furthermore,Mostowski[1938] gives a detailed proof of the non-equivalence of the properties “xis infinite” with “xis Dedekind infinite”, and of “x is Dedekind infinite” with “P(x) is Dedekind infinite”. Naturally, these questions make sense only in the absence of the Axiom of Choice (AC).

An important paper dealing with these questions isLevy[1958], where some results are shown for Zermelo–Fraenkel set theory allowing atoms, ZFA.

These results were transferred to ZF by Jech Sochor [1966a,b] and Pin- cus [1971,72]. The purpose of this paper is to prove some results about related questions. In particular we analyze the properties “a set x can be partitioned inton infinite sets”, forn∈ω, n6= 0, “a setxcan be partitioned

into infinitely many infinite sets” and “a set xcan be partitioned into count- ably many sets”. We also study the relationship of these properties with others, well know, ones, e. g. “x is T-infinite” (see below), “x is Dedekind infinite”, and some other statements related with them. Some of our results solve a problem posed by Degen.

We also present some results about partitions of Cartesian products and Dedekind cardinals.

1 Definitions and first results

We first define the notions we shall deal with. Some of these notions are found inTarski[1924] and others inLevy [1958]. In this paper “countable”

means “countably infinite”.

Definition 1 :

1. A set x is called infinite if there exists no bijection of x into a natural number n.

2. A set xis calledD–infinite if there exists a bijection of xonto a proper subset. By a well-known theorem, a set is D–infinite iff it has a count- able subset.

3. A set x is called P–infinite if there exists a bijection of P(x) onto a proper subset.

4. A set x is called T–infinite if there exists a infinite⊆–chain in P(x).

5. A set x is called ℵ₀–partible if there exists a countable partition of x.

6. A setxis called∞–partible if there exists a partition ofxinto infinitely many sets, each of which is infinite.

7. For n ∈ ω, n 6= 0, a set x is called n–partible if it can be partitioned into n infinite sets.

8. A set x is called partible if it is 2–partible.

9. A set x is called <ω–partible if x is n–partible for every n∈ω, n6= 0.

Remark 1 :

(i) If x is D–infinite, then x is P–infinite.

(ii) If x is ℵ₀–partible, then x is P–infinite.

(iii) If x is P–infinite, then x is ℵ₀–partible.

(iv) If x is ℵ₀–partible, then x is T–infinite.

(v) If x is ℵ₀–partible, then x is ∞–partible.

(vi) If x is ∞–partible, then x is <ω–partible.

(vii) For n ∈ω, n6= 0, if x is <ω–partible, then x is n–partible.

(viii) If x is n–partible and n > m, m6= 0, then x is m–partible.

Note that (i), (ii), (iv), (vi), (vii) and (viii) are obvious. The property (iii) was first proved by Kuratowski (see Tarski[1924], p. 94–95); an alternative proof is given in Gonzalez[1995]. (See also Hickmann [1973], p. 329 and the review by J. E. Rubin.) In Gonzalez [1995] we remark that if x is ℵ₀–partible, then there exists a countable partition P of x, such that every member of P is infinite, and thus we have (v). Furthermore, the following results can be found in Levy [1958]:

Theorem 1 (Levy) Ifxis infinite and can be ordered, thenxis T–infinite.

Theorem 2 (Levy) If x is T–infinite, then x is partible.

Theorem 2 can be generalized, but first we need a result about orders.

Definition 2 An ordered set hA, <i is called order–partible if there exists an x∈ A such that both the sets {y ∈A: y < x} and {y ∈A: y > x} are infinite.

Lemma 1 Let hA, <i be a linear order and A infinite. Then, hA, <i is order–partible or A is countable.

Proof. Assume hA, <iis uncountable and not order-partible. Clearly, if b⊆A is finite, thenArb has either a first or a last element. Inductively, we define a sequence{a_n}_n∈ω by: a_n+1 is the first element ofAr{a_i: i≤n} if it exists, otherwise let it be the last (anda₀ the first or last element of A).

We claim that this is a sequence of successive elements, in the sense that for no a and n, a_n < a < a_n+1 or a_n+1 < a < a_n. Otherwise, let k be the least n such that there exists an a such that, for example, a_n < a < a_n+1 holds. By construction, this can happen only when A_k = Ar{a_i ∈ A : i ≤ k} has no first element and hence a_k+1 must be the last element of A_k. Furthermore, for n > k, A_n has no first element and a_n+1 is the last element of A_n. Hence {a_i: i > k} is an interval of order type ω^∗. As A is uncountable,B =Ar{a_n}_n∈ω 6=∅and any b∈B partitionshA, <iinto two infinite intervals, contradicting our hypothesis that A is not order–partible.

Indeed, the set {a ∈A:a < b} is infinite since B has no first element, and {a∈A:a > b} is infinite sinceb < a_n forn > k. Hence the sequence{a_n}_n∈ω consists of successive elements, and is an initial interval of order type ω or a final interval of order type ω^∗.

We proceed analogously in defining the sequence{b_n}_n∈ω inB =Ar{a_n: n ∈ω}, since A is uncountable. AsB satisfies the hypothesis of the lemma, we have that the sequence {b_n}n∈ω is an initial interval of order type ω or a final interval of order typeω^∗. HenceA is order–partible, a contradiction.

Now we will prove a generalization of the result of Levy (Theorem 2):

Lemma 2 If x is T–infinite, then it is <ω–partible.

Proof. Let x be T–infinite and let n∈ω. For simplicity, given a set A⊆P(x), let<A be the restriction of the inclusion toA: p <A q ↔ p⊂q∧ p∈A ∧ q ∈A. Sincex is T–infinite, there exists an infinite y⊂P(x), such thathy, <_yiis a linear order. Ifyis countable, the lemma is immediate, since

“z is countable” implies “z isℵ0–partible” for everyz. Thus we can suppose that y is uncountable. Then, by Lemma 1, hy, <_yi can be partitioned into two infinite linear orders. We repeat this argumentntimes in order to obtain a partition of y inton sets.

From Lemma 1 and Theorem 1 we obtain:

Corollary 1 If x is infinite and can be ordered, then x is <ω–partible.

Finally, we remark:

Remark 2 If x is infinite, then x×x is ∞-partible.

For y ∈ x define py = {hy, zi: z ∈ x}. Then P = {p_y: y ∈ x} is the required partition.

2 Independence results

In this section we prove some independence results concerning the properties of Definition 1. First, let A be the basic Fraenkel model and A its set of atoms (see Jech [1973], p. 48). The following are well-known results:

Lemma 3 (Lindenbaum–Mostowski) The set A is not partible in A.

Theorem 3 (Jech–Sochor) The statement “there exists a set which is not partible” is not provable in ZF.

Now, we will show that “xis T–infinite” does not imply “xis∞–partible”.

For this, let Mbe the ordered Mostowski model (Jech [1973], p. 49–50) let A be its set of atoms, and<the order on A isomorphic to the natural order of the rationals. Then we have the following:

Lemma 4 The set of atoms of M is not ∞–partible in M.

Proof. For this, suppose without loss of generality that B ={A_n: n ∈ ω}is a partition ofAinMwhere everyAnis infinite and letE ={e1, . . . , en} a finite support of B. The elements of E induce a partition of ArE into open intervals: (−∞, e_i1),(e_i1, e_i2), . . . ,(e_in,+∞), with e_i < e_i+1. At least one of these intervals must have infinitely many elements of two members of the partition, say (e_im, e_im+1) contains infinitely many elements of A_j, A_k

(j 6= k). Exchanging, if necessary, A_j and A_k we may find a, b ∈ A_j and c ∈ A_k with a < b < c and a, b, c,∈ (e_im, e_im+1). To get a contradiction, we claim that there exists a permutation π ∈ G such that π ∈ fix(E) and π(B)6=B. (G is the permutation group definingM.) Notice first that (a, b) and (a, c) are order isomorphic and so are (b, e_im+1) and (c, e_im+1) (they are countable linearly densely ordered intervals without first or last element).

Now let π be such that:

(i) π is a order isomorphism of (a, b) onto (a, c), and of (b, e_im+1) onto (c, e_im+1),

(ii) π(a) = a and π(b) =c, and

(iii) π(x) =x for the remaining elements of A.

Since π is order preserving we have π ∈ G. Obviously, π ∈ fix(E). Fur- thermore, we have that π(A_j) 6∈ B, which implies π[B] 6= B and hence π(B)6=B.

Since every set can be ordered in M, from Theorem 1 and Lemma 4 we have immediately the following:

Lemma 5 The property “x is T–infinite” does not imply the property “x is

∞–partible” in set theory with atoms.

Using the technique due to Jech, Sochor and Pincus, a symmetric model can be constructed from the Mostowski ordered model where the Ordering Principle and the Prime Ideal Theorem (PI) for Boolean algebras hold (see Jech [1973], p. 113). Thus, from Lemma 5 we obtain:

Theorem 4 The property “x is T–infinite” does not imply the property

“x is ∞–partible” in ZF plus PI.

The following are immediate consequences of Theorem 4:

Corollary 2 The property “x is <ω–partible” does not imply the property

“x is ∞–partible” in ZF plus PI.

Corollary 3 The property “x is T–infinite” does not imply the property “x is ℵ₀–partible” in ZF plus PI.

To prove the independence of “x is T–infinite” from “x is ∞–partible”

in ZF, we need some preliminary results. As above, A denotes the basic Fraenkel model (see Jech [1973], p. 48) and A its set of atoms.

Definition 3 For a set x, let P^<ω(x) denote the set of finite subsets of x.

Definition 4 For n∈ω, let B⁽ⁿ⁾ ={x∈B: |x|=n}.

Lemma 6 No infinite subset B of P^<ω(A) can be linearly ordered in A.

Proof. Assume otherwise and let < be a linear order on B. We pro- ceed by cases. First, suppose that ∀n ∈ ω, B⁽ⁿ⁾ is finite. Hence there exist infinitely many n such that B⁽ⁿ⁾ 6=∅. Note that the cardinal of ^SB⁽ⁿ⁾ is unbounded for n ∈ω, in the sense that∀i∈ω ∃n∈ω ^SB⁽ⁿ⁾> i. Indeed, if n > i with B⁽ⁿ⁾ 6=∅, we have ^SB⁽ⁿ⁾> i. Now we define:

C₀ =^SB⁽¹⁾; and C_n+1 =^SB⁽ⁿ⁺¹⁾rC_n.

Obviously, the C_n’s are pairwise disjoint, and a simple argument shows that the set {n: C_n 6= ∅} is infinite. Hence A is partible, in contradiction with Lemma 3.

Now, suppose that there exists n ∈ω, such that B⁽ⁿ⁾ is infinite, and fix such an n. Let S be a finite support of ^DB⁽ⁿ⁾, <B⁽ⁿ⁾

E. For s ⊆S we define B⁽ⁿ⁾_s={x∈B⁽ⁿ⁾: x∩S =s} and let P ={B⁽ⁿ⁾_s: s ∈S}. Note thatP is finite and includes a partition of B⁽ⁿ⁾. Thus, there exists an t⊆S such that B⁽ⁿ⁾_tis infinite and hencex*Sforx∈B⁽ⁿ⁾_t. Then we fixx, y ∈B⁽ⁿ⁾_t, x6=y and define:

y⁰ =yrS ={a₀, . . . , a_m}; and

x⁰ =xrS ={b₀, . . . , b_m}.

(Recall that x∩S =y∩S =t, x, y *S and |x|=|y| =n.) Now we define a permutation π:

π(a_i) =b_i; π(b_i) =a_i; and

π(z) =z if z 6=a_i and z 6=b_i.

where 0 ≤ i ≤ m. Obviously, π(S) = S. Furthermore, note that π(x) = y and π(y) =x. Hence π does not preserve <, a contradiction.

From Lemma 6 we obtain:

Lemma 7 No infinite subset B of P(A) can be linearly ordered in A.

Proof. Assume otherwise. SinceA is not partible,B can be partitioned into B^f and B^c, where:

B^f ={x∈B: x is finite}

B^c ={x∈B: (Arx) is finite}

By Lemma 6 B^f must be finite and thus B^c is infinite. Then we define:

B^−c={Arx: x∈B^c}

Clearly,B^−c can be linearly ordered, in contradiction with Lemma 6.

With Dom(s) and Cod(s) we abreviate the domain and codomain of a relation s. Furthermore, for s ⊆ A × A and a ∈ Dom(s) let s[a] = Cods{a}

={b: ha, bi ∈s}. A set x is calledcofinite if Arx is finite.

Theorem 5 A×A is not T–infinite in A.

Proof. To get a contradiction, let C be an infinite chain in P(A×A).

Let D be the diagonal relation on A: D = {ha, ai: a ∈ A}. Furthermore, for s∈C, let s⁰ =srD, and C⁰ ={s⁰: s∈C}.

We claim thatC⁰ is infinite. Otherwise, there exists an infinitex⊆Csuch that r, s∈x → r⁰ =s⁰. We fix such an x. For s∈C we defines_D =s∩D and x_D ={s_D: s∈x}. Note that x_D is infinite, since r 6=s → r_D 6=s_D for r, s∈x. Sincer_D 6=s_D → ^{S S}r_D 6=^{S S}s_D and ^{S S}r_D ⊆A, for r, s∈C, we obtain an infinite, linearly ordered family of subsets ofA, in contradiction with Lemma 7. Furthermore, notice thatC⁰ is linearly ordered by ⊆.

For s∈C⁰, let s−=ⁿha, bi ∈s: s[a]6=Ar{a}^o and ˜s={r ∈C⁰: r−= s−}. We claim that each ˜s is finite. Otherwise, fix an s with ˜s infinite. Let f(r) = ⁿa ∈ A: r[a] = Ar{a}^o. We will see that r 6= t → f(r) 6= f(t) for r, t ∈ s. Suppose˜ r 6= t and, without loss of generality, suppose that there exist a, b ∈ A, a 6= b, such that ha, bi ∈ r and ha, bi 6∈ t. Since ha, bi 6∈t, we have b 6∈t[a], and a 6∈f(t). Furthermore, ha, bi 6∈t−, and then ha, bi 6∈r−. Therefore, r[a] =Ar{a}, then a∈f(r), andf(r)6=f(t). Thus we have an infinite, linearly ordered family of subsets of A, in contradiction with Lemma 7. Since each ˜s is finite, there exists infinitely many ˜s with s ∈ C⁰. Furthermore, since each ˜s is linearly ordered by ⊆, it has first element, which is called ˜s¹. Notice that (˜s¹)− = s−, for s ∈ C⁰. Now we define C⁰⁰ = {s−: s ∈ C⁰}, and let r− < s− ↔ r˜¹ ⊂ ˜s¹, for r, s ∈ C⁰. Since ˜r¹ 6= ˜s¹ → (˜r¹)− 6= (˜s¹)−, straightforward verification shows that <

is well-defined and that it is a linear order.

For r∈C⁰⁰, we define:

ha, bi ∈r_f ↔















ha, bi ∈r and r[a] is finite, or

ha, bi 6∈r, r[a] is infinite and a6=b.

Let S = {r_f : r ∈ C⁰⁰}. For r, s ∈ C⁰⁰ we define r_f <⁰ s_f ↔ r < s. To see that <⁰ is well-defined, that <⁰ is a linear order, and that S is infinite, it suffices to show that r₋6=s₋ → r_−f 6=s_−f, forr, s∈C⁰. Letr, s∈C⁰ and supposer−6=s−. Letaandbsuch thatha, bi ∈s−andha, bi 6∈r− (otherwise exchange r with s). Notice thata 6=b. We proceed by analyzing two cases.

Case 1: r⊆s. We consider two subcases.

Subcase 1.a: s−[a] is finite and hence ha, bi ∈ s−f. Furthermore, since ha, bi ∈s−, s[a] is finite, and ha, bi 6∈r−f, i. e. r−f 6=s−f.

Subcase 1.b: s−[a] is infinite and hence ha, bi 6∈ s−f. First, note that there exists c 6= a with ha, ci 6∈ s−, and hence ha, ci ∈ s−f. Furthermore, since ha, bi ∈ s₋, we have that s[a] 6= Ar{a}, and hence ha, ci 6∈ s. As a conse- quence, we obtain ha, ci 6∈ r and ha, ci 6∈ r−. On the one hand, if r−[a] is finite, we have that ha, ci 6∈ r−f. On the other hand, if r−[a] is infinite, we have that ha, bi ∈r_−f, and hence r_−f 6=s_−f.

Case 2: s ⊆ r. First, note that ha, bi ∈ s− implies ha, bi ∈ s, and hence we have that ha, bi ∈ r. Since ha, bi 6∈ r−, we have r[a] = Ar{a}. Hence, r−[a] =∅, and a 6∈ Dom(r−f). On the one hand, if s−[a] is finite, we have that ha, bi ∈ s−f. On the other hand, if s−[a] is infinite, then there exists c∈A with ha, ci 6∈ s−, and thus ha, ci ∈ s−f. Since a6∈ Dom(r−f), we have that neither ha, bi, nor ha, cibelongs to r−f. Hence we have r−f 6=s−f.

We claim that there exists an infinite T ⊆S such that ∀r, s∈T Cod(r) = Cod(s). First, we define for s ∈ S: ¯s = {r: Cod(r) = Cod(s)}. We will see that there exists an s such that ¯s is infinite. Otherwise, {¯s: s ∈ S}

is infinite. Furthermore, since ¯s is linearly ordered by <⁰ for each s ∈ S,

¯

s has a first element, which we denote by ¯s¹. Now, for r, s ∈ S, we define Cod(r) <⁰⁰ Cod(s) ↔ r¯¹ <⁰ s¯¹. Clearly, <⁰⁰ is well-defined. But <⁰⁰ is a linear order, in contradiction with Lemma 7. Therefore, ¯sis infinite for some s∈S, and it is the required T.

Let s ∈ T. We claim that Cod(s) is infinite. Otherwise, let Cod(s) = {a0, . . . , am}. For each aj, 0 ≤ j ≤ m, and r ∈ T, we define raj = {a ∈ A: ha, a_ji ∈ r}. Note that, for each r, t ∈ T we have r 6=t → ∃j ≤m t_aj 6= r_aj. Indeed, if r 6= t, then there exists a and a_j such that ha, a_ji ∈ r and ha, aji 6∈ t (otherwise exchange r with t). As a consequence, we have that {r_aj: r ∈ T ∧ 0 ≤j ≤ m} is infinite, and hence there exists a p≤m such that {r_ap: r ∈T} is infinite, in contradiction with Lemma 7.

Lets∈T. Note that, for eacha∈Dom(T),s[a] is finite, by construction of S and because every infinite subset of A is cofinite. Let Y be a finite support of s. Since Cod(s) is infinite, there exists b ∈ Cod(s) such that b 6∈ Y, and hence there exists a∈A such that ha, bi ∈ s. Since s[a] and Y are finite, there exists c∈A such that ha, ci 6∈s and c6∈ Y. Then we define a permutation π:

π(b) = c; π(c) = b; and

π(z) = z if z 6=b, c.

Clearly, π(Y) = Y. Furthermore, we have that π(a) = a, since a 6= b, c.

Hence π(ha, bi) =ha, ci, andπ(s)6=s. Therefore, T =∅, a contradiction.

From Remark 2 and Theorem 5 we obtain the following:

Lemma 8 The property “x is∞–partible” does not imply the property “xis T–infinite” in set theory with atoms.

Once more, using the technique of Jech and Sochor, we obtain:

Theorem 6 The property “x is ∞–partible” does not imply the property

“x is T–infinite” in ZF.

The following are immediate corollaries of Theorem 6:

Corollary 4 The property “xis∞–partible” does not imply the property “x is ℵ0–partible” in ZF.

Corollary 5 The propertyx is<ω–partible” does not imply the property “x is T–infinite” in ZF.

In the sequel, we will prove a result about the relationship between n–

partible and m–partible for m > n. First, we need a definition.

Definition 5 P¹0(x) =x,

P¹n+1(x) = ⁿ{y}: y∈P¹n(x)^o.

As above, A denotes the basic Fraenkel model and A its set of atoms.

Note that P¹n(A) is not partible inA for n∈ω. Then the following holds:

Lemma 9 For n, m ∈ ω, 0 < n < m, there exists in A a set which is n–partible but not m–partible.

Proof. Fix n and m. B = ^S_i<nP¹i(A) is n–partible. To get a contra- diction, suppose that B is m–partible in A and let {Pi: i < m} a partition of B where each P_i is infinite. For each j < m, only one P_i can include an infinite subset of P¹j(A), since P¹j(A) is not partible. Since n < m, at least one of thePi’s must be finite, a contradiction.

Applying the technique of Jech and Sochor, we obtain the following:

Theorem 7 For n, m∈ω, 0< n < m, the property “x is n–partible” does not imply “x is m–partible” in ZF.

For an alternative proof of the preceding results, seeDegen[1994], p. 114.

Figure 1 summarizes the results of the last two sections.

3 Notions of Infinity and Degen’s Main Open Problem

In this section we present some results about notions of infinity and solve a question posed in Degen [1994]. In that paper a general concept of “notion of infinity” is introduced, as follows:

Definition 6 A notion of infinity forZF is a formulaφ(x) with exactly the free variable xsuch that the following are theorems of ZF:

(i) ∀α≥ω φ(α);

(ii) ∀n < ω ¬φ(n);

(iii) ∀x∀y (|x|=|y| ∧ φ(x)⇒φ(y));

(iv) ∀x∀y (x⊆y ∧ φ(x)⇒φ(y)).

x is D–infinite

? 6

x is P–infinite ^- x isℵ₀–partible

@

@

@

@

@ R@

@

@

@

@ I

x is T–infinite ^-

x is∞–partible

@

@

@

@

@ R

@

@

@

@

@

I

x is<ω–partible

?_##6

...

?_##6

x isk+ 1–partible

?_##6

x isk–partible

?_##6

...

?_##6

x is 3–partible

?_##6

x is partible

?_##6

x is infinite

Figure 1: Properties of partition and infinity

Proposition 1 All the properties stated in Definition 1 are notions of infin- ity.

Definition 7 A set x is DO if there exists a linear dense order on x. A set x is IO ifx is infinite and there exists a linear order on x.

Proposition 2 The properties “x is IO” and “x is DO” are not notions of infinity.

Proof. (iv) fails in both cases. For IO, we consider the second Fraenkel modelB(seeJech[1973], p. 48). This model is constructed from a countable set of atoms A. The set A is partitioned in countably many disjoint pairs.

The set of all these pairs is called B and is in the model. B is countable in B, and hence linearly ordered inB, while A∪B cannot be linearly ordered.

For DO, we use the ordered Mostowski modelM(seeJech[1973], p. 49–50).

LetI be a proper closed interval of the set of atoms of M. Furthermore, let a 6∈ I and let B = I ∪ {a}. The set of atoms A of the ordered Mostowski model is DO. However, B is not DO (see Theorem 8 in Gonz´alez [1995]).

Definition 8 Two properties φ(x) and ψ(x) (i. e. formulas with only one free variable) are called mutually independent if neither ∀x (φ(x) → ψ(x)) nor ∀x(ψ(x)→φ(x)) are provable inZF.

Proposition 3 The following pairs of properties are mutually independent:

(i) x is T–infinite and x is ∞–partible.

(ii) x is IO and x is ∞–partible.

(iii) x is IO and x is ℵ₀–partible.

(iv) x is IO and x is D–infinite.

(v) x is DO and x is ∞–partible.

(vi) x is DO and x is ℵ0–partible.

(vii) x is DO and x is D–infinite.

Proof. For (i), use Theorems 4 and 6. For (ii)-(vii) observe, on the one hand, that the set of atoms of the ordered Mostowski model is IO and DO, but is neither∞–partible nor ℵ₀–partible nor D–infinite. On the other hand, Theorem 8.3 of Jech[1973], p. 123 shows that there exists a set with countable subset, which cannot be ordered. The transfer of this facts to ZF is immediate.

Degen [1994] p. 124 poses the following Main Open Problem: Are there two notions of infinity φ(x), ψ(x) and a model N of ZF such that there is a setu∈NwithN|= φ[u]∧ ¬ψ[u] and a setw∈NwithN|= ψ[w]∧ ¬φ[w]?

For this, we define a pasted model C constructed using the ideas of both the basic Fraenkel model and the ordered Mostowski model (seeJech[1973], p. 48–50). Generally, a pasted model is a model that uses the set of atoms and the permutations groups of two other models. The set of atoms of C is A∪B, withAand B countable and disjoint. Furthermore,B has an order<

isomorphic to the natural order of the rationals. The group Gused to define C consists of all the permutations π such that:

(i) π is a bijection of A onto A and of B ontoB; (ii) π preserves <.

C is the permutation model constructed from the group G and the normal filter Fgenerated by the ideal of finite subsets of A∪B (see Jech [1973], p.

46-47). With this notation, we have the following:

Theorem 8 (i) The set A×A is ∞–partible, but not T–infinite;

(ii) The set B is T–infinite but not ∞–partible.

Proof. Use the arguments of Remark 2, Lemma 4 and Theorem 5.

We transfer immediately this result by using the technique of Jech and Sochor (see Problem 1 inJech [1973], p. 94).

Theorem 9 (Solution to Degen’s Problem) There are incomparable ZF-notions of infinity.

It is possible to build pasted models also for the remaining pairs of notions in Proposition 3. For this, use the ordered Mostowski model pasted with the model in Theorem 8.3 of Jech[1973], p. 123: the set of atoms isA∪B with

|A|=ℵ₀, |B|=ℵ₁, A∩B =∅. Ahas a order<as in the ordered Mostowski model and B is partitioned into pairs P_α, with α <ℵ₁. We take the group of all permutations which preserve<, the pairsP_α and the partition{A, B}, and the filter generated by the ideal of all sets which contain finitely many elements of A and at most countably many ofB.

4 Partitions of Cartesian Products

In this section we present a few simple results concerning partitions of finite Cartesian products.

It is well known that if the Cartesian product x×xis countable, then the setxis countable as well. Furthermore, if the Cartesian product x×xhas a countable subset, then the set x also has a countable subset. However, it is clear from Remark 2 and Theorem 5, that the hypothesis “there exists a set xsuch that x×x is∞–partible andxis not partible” is consistent withZF.

The following result was stated without proof inMonro [1973], p. 415; for completeness’ sake we include a proof of it:

Proposition 4 If x×x is ℵ₀–partible, then x is ℵ₀–partible.

Let P = {x_n: n ∈ ω} be a countable partition of x×x. We proceed by cases. On the one hand, if there exists y ∈ x, such that y ∈ Dom(x_n) for infinitely many n ∈ω, we define z_n ={y⁰ ∈x: hy, y⁰i ∈x_n}, for n ∈ω.

Then, {z_n: n ∈ω ∧ z_n 6=∅} is the required partition. On the other hand, suppose that for each y∈ x there exists only finitely many n ∈ω such that y ∈ x_n. Then we define f : x −→ P^<ω(ω) by: n ∈ f(y) iff y ∈ x_n, for y ∈ x. Notice that Cod(f) is infinite. Furthermore, let g be a bijection of P^<ω(ω) onto ω, and let h be the composition of g and f: h = g ◦f. Note thath: x−→ω and thatCod(h) is infinite. Hence, there existsh⁰: x−→ω onto. Then, Q={h⁰⁻¹[{n}] : n∈ω} is the required partition.

In the ordered Mostowski model the Cartesian product A ×A of the set of atoms is T–infinite and ∞–partible. However, from Remark 2 and Proposition 4, we get that A×A is not ℵ0–partible, since A is not. Thus, we obtain:

Theorem 10 The property “xis T–infinite and ∞–partible” does not imply

“x is ℵ₀–partible” in ZF plus PI.

Definition 9 For a set x and n ∈ ω, n6= 0, let x_n be the n-fold Cartesian product of x:

n times

z }| {

x×x×. . .×x

Clearly, ifx_nis D–infinite, thenx_n+1is D–infinite, forn∈ω, n6= 0. Hence, if x is not D–infinite, no xn is D–infinite, for n ∈ ω, n 6= 0. Furthermore, the following holds:

Proposition 5 If x is D–finite and has more than one point, then there exists no bijection of x_n+1 onto x_n.

Proof. We show the contrapositive. Let f: x_n+1 −→x_n be a bijection, and let y ∈ x. Then we define g: x_n+1 −→ x_n+1 by: g(hy₁, . . . , y_n+1i) =

Df(hy₁, . . . , y_n+1i), y^E. The image of g is a proper subset of x_n+1, showing that x is D–infinite.

Thus, we have a very simple way of showing the existence of a countable set of Dedekind cardinals (i.e. cardinals of infinite but D-finite sets), with the aditional property that |x_n|<|x_n+1|.

5 References

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