Ultrafilters maximal for finite embeddability

In [1] the authors showed some basic properties of a pre-order that arose in combinatorial number theory, namely the finite embeddability between sets of natural numbers, and they presented its generalization to ultrafilters, which is related to the algebraical and topological structure of the Stone-\v{C}ech compactification of the discrete space of natural numbers. In this present paper we continue the study of these pre-orders. In particular, we prove that there exist ultrafilters maximal for finite embeddability, and we show that the set of such ultrafilters is the closure of the minimal bilateral ideal in the semigroup $(\bN,\oplus)$, namely $\overline{K(\bN,\oplus)}$. As a consequence, we easily derive many combinatorial properties of ultrafilters in $\overline{K(\bN,\oplus)}$. We also give an alternative proof of our main result based on nonstandard models of arithmetic.


Introduction
This paper is a planned continuation of the work [1] by Andreas Blass and Mauro Di Nasso. Both in [1] and in this present paper it is studied a notion that arose in combinatorial number theory (see [4] and [7], where this notion was implicitly used), the finite embeddability between sets of natural numbers: Definition 1.1. ( [4], §4) Let A, B be subsets of N. We say that A is finitely embeddable in B (notation: A ≤ f e B) if, for every finite subset F of A, there exists a natural number n such that n + F ⊆ B. We use the standard notation n + F = {n + a | a ∈ F }, and we use the standard convention that N = {0, 1, 2, ...}. In [1] the authors also studied the generalization of ≤ f e to ultrafilters: Definition 1.2. ( [1], §1) Let U, V be ultrafilters on N. We say that U is finitely embeddable in V (notation: U ≤ f e V) if for every set B ∈ V there is a set A ∈ U such that A ≤ f e B.
It can be easily proved (see [1], [6]) that both (P(N), ≤ f e ) and (βN, ≤ f e ) are preorders. We will write ≡ f e for the induced equivalence relations and, given A ⊆ N or U ∈ βN, we will denote by [A] f e and [U] f e the corresponding equivalence classes. Moreover we will still denote by ≤ f e the ordering induced to the quotient spaces. In [1] the authors studied the basic properties of ≤ f e ; in this present paper we study the existence of maximal and minimal elements. In Section 2 we study the existence of maximal and minimal elements in the preordered set (P(N), ≤ f e ). In Section 3 we turn our attention to (βN, ≤ f e ), and we prove the main Theorem of this paper, namely that the set of maximal ultrafilters in (βN, ≤ f e ) is K(βN, ⊕), the topological closure of the minimal bilater ideal in the semigroup (βN, ⊕). In Section 4 we use nonstandard analysis to directly prove that every ultrafilter in K(βN, ⊕) is maximal in (βN, ≤ f e ). We refer to [5] for all the notions about combinatorics and ultrafilters that we will use, to [2], §4.4 for the foundational aspects of nonstandard analysis and to [3] for all the nonstandard notions and definitions. Finally, we refer the interested reader to [6], Chapter 4 for other properties and characterizations of the finite embeddability.
2 Maximal and minimal elements in (P(N), ≤ f e ) Let n be a natural number. Throughout this section we will denote by P ≥n (N) the set similarly, we will denote by P ℵ0 (N) the set It is immediate to see that the relation ≤ f e on P(N) is not antysimmetric (e.g., if O is the set of odd numbers and E is the set of even numbers, we have O ≡ f e E), so to search for maximal and minimal sets we will actually work in (P(N)/≡ f e , ≤ f e ). The characterization of maximal sets is an immediate consequence of the following result: Proposition 2.1. Let A be a subset of N. The following three conditions are equivalent: The proof of Proposition 2.1 follows from the definitions in a straightforward manner. There are a few less trivial things that can be said about the existence of minimal elements in (P(N)/≡ f e , ≤ f e ). Two easy observations are that the empty set is the minimum in (P(N)/≡ f e , ≤ f e ) and that {0} is the minimum in (P(N) ≥1 /≡ f e , ≤ f e ). Moreover, if we identify each natural number n with the singleton {n}, it is immediate to see that (N, ≤) forms an initial segment of (P ≥1 (N), ≤ f e ). So the question becomes: given a natural number n, are there minimal elements in (P ≥n (N), ≤ f e )? Here we do not consider the equivalence classes w.r.t. ≤ f e because, given any two finite sets, it is immediate to show that A ≡ f e B ⇔ A = B.
The following property holds: if and only if 0 ∈ A and |A| = n.
Proof. If |A| > n then A is not minimal since, e.g., if B = A \ {max{A}} then B ∈ P ≥n (N) and In particular, we get that: The last question we answer is: what does it happen if we consider only the infinite subsets of N?
If both A S f e B and B S f e A we say that A, B are strongly mutually unembeddable (notation: A ≡ f e B).
Let us observe that, in the previous definition, we can equivalenty substitute the condition "|C| = 2" with "|C| ≥ 2". Proof. We have to construct A, B ⊆ X such that, for any C ⊆ A, D ⊆ B with |C| = |D| = 2, we have C f e B and D f e B. We construct A and B: let X = {x n | n ∈ N}, with x n < x n+1 for every n ∈ N. We set a 0 = x 0 , a 0 = x 1 and, recursively, we set Finally, we set A = {a n | n ∈ N} and B = {b n | n ∈ N}. Clearly A ∩ B = ∅, and both A, B are infinite subsets of X. Now we let a n1 < a n2 be any elements in A. Let us suppose that there are and let us assume that b n2 > a n2 (if the converse hold, we can just exchange the roles of a n1 , a n2 , b m1 , b m2 ). By construction, since b m2 > a n2 , we have Proof. Let us suppose that there is such a minimal element [X] f e . Let A, B ⊆ X be strongly mutually unembeddable. Since X is minimal and A, B are both finitely embeddable in X, by minimality we get that and B ≤ f e A, and this is absurd.

Preliminars
In this section we want to study (βN, ≤ f e ). An useful notion in this setting is that of cone of an ultrafilter: We recall two important properties that have been proved in [1] and in [6]: Proposition 3.2. The two following properties hold for every ultrafilters U, V ∈ βN: As already remarked in [1] and in [6], from Proposition (3.2) it follows directly that the pre-ordered set (βN, ≤ f e ) is upward directed. Moreover, from Proposition 3.2 it also follows that the relation ≤ f e is not antisymmetric: in fact, if R is a minimal right ideal in (βN, ⊕) and U ∈ R then To study more in detail the ordered set (βN/≡ f e , ≤ f e ) we start with the following easy considerations: 2. if U is the principal ultrafilter generated by n and V is the principal ultrafilter generated by m then U ≤ f e V if and only if n ≤ m.
We omit the proofs, which are straightforward. If we identify, as usual, each natural number n with the corresponding principal ultrafilter on n, the above proposition states that N is an initial segment of (βN, ≤ f e ). A first question that arises is if (βN, ≤ f e ) is, or if it is not, a total preorder, and the answer, as a consequence of Proposition 2.4, is no: There are nonprincipal ultrafilters U, V such that U is not finitely embeddable in V and V is not finitely embeddable in U.
Proof. Let A, B be strongly mutually unembeddable infinite sets (which existence is a consequence of Proposition 2.4). Let U, V be nonprincipal ultrafilters such that A ∈ U, B ∈ V and let us suppose that This is absurd, so U is not finitely embeddable in V. In the same way we can prove that V is not finitely embeddable in U.

Minimal Ultrafilters
By Proposition 3.3 it follows that N is the initial segment of (βN, ≤ f e ), so it is immediate to see that the principal ultrafilter on 0 is the minimum element. What does it happen if we restrict our attention to the nonprincipal ultrafilters?
The first property that we prove is a simple consequence of Proposition 2.4. In the following proposition, by Θ X we mean the clopen set {U ∈ βN | X ∈ U}. Proof. Let us suppose that such a minimum M exists, and let U ∈ Θ X be such that M = [U] f e . Let A, B ⊆ X be mutually unembeddable subsets of X and let V 1 , V 2 be nonprincipal ultrafilters such that A ∈ V 1 and B ∈ V 2 (in particular, V 1 , V 2 ∈ Θ X ). Since [U] f e is the minumum, there are C 1 , C 2 ∈ U such that C 1 ≤ f e A and C 2 ≤ f e B. So C 1 ∩ C 2 is finitely embeddable in A and in B, and this cannot happen: in fact, let c 1 < c 2 be any two elements in C 1 ∩ C 2 . Then there are n, m such that n + {c 1 , c 2 } = {a 1 , a 2 } ⊂ A and m + {c 1 , c 2 } = {b 1 , b 2 } ⊂ B, and this cannot happen, because in this case we In particular, by taking X = N, we prove that: We now want to prove one other property of (βN/≡ f e , ≤ f e ), namely that it contains many infinite descending chains. This property can be proved by applying the Lemmas 3.7 and 3.9: Lemma 3.7. Let m be a natural number, and let us suppose that U ≤ f e U ⊖m 1 . Then, for every A ∈ U, for every n ∈ N, there are elements x 1 , ..., x n , y 1 , ...y n ∈ A such that y 1 − x 1 = ... = y n − x n ≥ m.
Proof. We observe that, in general, a set A is in U if and only if the set Let A ∈ U. Let B ∈ U be finitely embeddable in A − m, and let C = A ∩ B. Let x 1 , ..., x n ∈ C. Since C ≤ f e A − m, there is a natural number a such that x 1 + a, ..., x n + a ∈ A − m. So, in particular, x 1 + a + m, ..., x n + a + m ∈ A. If we pose, for i ≤ n, y i = x i + a + m, we have the thesis.
From now on, given an ultrafilter U and a natural number m we say that the property P m (U) holds if for every A ∈ U, for every n ∈ N there are elements x 1 , ..., x n , y 1 , ...y n ∈ A such that y 1 − x 1 = ... = y n − x n ≥ m. Proof. Let [U] f e be minimal in ((βN\N)/≡ f e , ≤ f e ). We know that U ⊖m ≤ f e U for every m ∈ N (since U = (U ⊖ m) ⊕ m) so, since U is minimal, we have that U ⊖ m ≤ f e U. The thesis then follows by applying Lemma 3.7. Lemma 3.9. There are nonprincipal ultrafilters U such that P m (U) does not hold for any m ≥ 1.
In fact, let us suppose that a n2 − a n1 = a m2 − a m1 but n 2 < m 2 . Then m 1 ≤ m 2 − 1 and n 1 ≤ m 2 − 1, so a n2 − a n1 ≤ a m2−1 , while a m2 − a m1 ≥ a m2 − a m2−1 = a m2−1 + 1. So n 2 = m 2 , and then clearly we also have that n 1 = m 1 . Now let U be a nonprincipal ultrafilter with A ∈ U. Then P 1 (U) does not hold, so P m (U) does not hold for any m ≥ 1. Proof. Let B ∈ V, and let A ∈ U be such that A ≤ f e B, and let n ∈ N. Let x 1 , ..., x n , y 1 , ..., y n be elements in A such that y 1 − x 1 = ... = y n − x n ≥ m, and let k ∈ N be such that k + {x 1 , ..., x n , y 1 , ..., y n } ∈ B. Then, if for i = 1, ..., n we set z i = x i + k and w i = y i + k, we have that z 1 − w 1 = z 2 − w 2 = ... = z n − w n ≥ m, so P m (V) holds.
Proposition 3.11. If there is a natural number m ≥ 0 such that P m (U) does not hold then there are infinite descending chains in ((βN \ N) Proof. Let U be such that P m (U) does not hold. Then, by Lemma 3.7 and Lemma 3.10 we deduce that U f e U ⊖m f e U ⊖2m f e ... f e U ⊖km f e ..., There is a question that remains open: are there any minimal elements in ((βN \ N)/≡ f e , ≤ f e )? We do not know. We conjecture that the answer should be no for the following reason: due to a few similitarities of the finite embeddability with the Rudin-Keisler preorder, one might expect the selective ultrafilters, which are minimal in the Rudin-Keisler preorder, to be minimal also with respect to the finite embeddability. This is false: Proof. We prove that if U is selective then in U there is a set A = {a n | n ∈ N} such that, for every i 1 < i 2 , j 1 < j 2 , we have From this property it follows that P 1 (U) does not hold, so we deduce that [U] f e is not minimal by applying Corollary 3.8. The existence of the set A ∈ U can be proved as follows: let f : N → N be the function such that, for every a, b ∈ N, For every a ∈ N let Let B ∈ U be a selector for the partition N = a∈N X a (so that |B ∩ X a | = 1 for every a ∈ N). Let b 0 < b 1 < b 2 < ... be the elements of B. Clearly, f (b i ) = i for every index i ∈ N, so we have that In particular, for every i ∈ N we have that and we call A the one of them that belongs to U. Then, by construction, A has the desired property: in fact, let us suppose (for sake of simplicity) that A = B E . By contrast, let i 1 < i 2 , j 1 < j 2 be such that Let us suppose i 2 < j 2 . By construction, a 2j2 − a 2j1 ≥ 3 2j2−1 + 1, while a 2i2 − a 2i1 ≤ 3 2i2+1 − 1 and, if i 2 < j 2 , we have so a 2i2 − a 2i1 < a 2j2 − a 2j1 . Hence i 2 = j 2 , and consequently i 1 < j 1 .

Maximal Ultrafilters
In this section we want to prove the following claim: Claim: There is a greatest element in (βN/≡ f e , ≤ f e ).
To prove this claim we need two results. The first one, which we already mentioned in Section 3 and has already been proved in [1] and [6], is that (βN/≡ f e , ≤ f e ) is an upward directed ordered set. This property is important: if we prove that there are maximal elements in (βN/≡ f e , ≤ f e ), then we automatically know that there is a greatest element. In fact, let us suppose that So to prove our claim it is sufficient to show that there are maximal elements in (βN/≡ f e , ≤ f e ). As usual, the natural idea to study the existence of maximal elements in (βN ≡ f e , ≤ f e ) is to use Zorn's Lemma. We call ≤ f e -chains the chains in (βN, ≤ f e ) and ≤ f e -upper bounds their upper bounds. A technical lemma that we need is the following: Lemma 3.13. Let I be a totally ordered set. Then there is an ultrafilter V on I such that, for every element i ∈ I, the set Proof. We have just to observe that {G i } i∈I is a filter and to recall that every filter can be extended to an ultrafilter.
The key property of these ultrafilters is the following: Proposition 3.14. Let I be a totally ordered set and let V be given as in Lemma 3.13. Then for every A ∈ V, i ∈ I there exists j ∈ A such that i ≤ j.
We omit the proof, since it is straightforward.
In the next Theorem we use the notion of limit ultrafilter. We recall that, given an ordered set I, an ultrafilter V on I and a family U i of ultrafilters on N, the V-limit of the family U i | i ∈ I (denoted by V − lim i∈I U i ) is the ultrafilter such that, for every A ⊆ N, Theorem 3.15. Every ≤ f e -chain U i | i ∈ I has an ≤ f e -upper bound U.
Proof. Let V be an ultrafilter on I with the property expressed in Lemma 3.13. We say that the ultrafilter is an ≤ f e -upper bound for the ≤ f e -chain U i | i ∈ I . We have to prove that U i ≤ f e U for every index i; let A be an element of U. By definition, I A is a set in V so, by Proposition 3.14, there is an element j > i in I A . So A ∈ U j and, since U i ≤ f e U j , there exists an element B in U i with B ≤ f e A. So U i ≤ f e U, and the thesis is proved.
As an immediate consequence we have that: So (βN ≡ f e , ≤ f e ) is an upward directed ordered set with maximal elements. As we observed, this entails the following interesting result: Theorem 3.17. In (βN/≡ f e , ≤ f e ) there is a greatest element.
We denote this greatest element by M f e . By definition, for every ultrafilter U we have the following equivalences: [U] f e = M f e ⇔ U is maximal in (βN, ≤ f e ) ⇔ V ≤ f e U for every V ∈ βN.
In particular, we can characterize the set of maximal ultrafilters in terms of the cones generated by the elements in βN: Proof. We have just to observe that M ⊆ C(U) for every ultrafilter U and that, if U is a maximal ultrafilter, then C(U) = M f e . Now the question is: which are the maximal ultrafilters? Proof. This follows from this well-known characterization of K(βN, ⊕): an ultrafilter U is in K(βN, ⊕) if and only if every element A of U is piecewise syndetic (see, e.g., [5]).
As mentioned in the introduction, the notion of finite embeddability is related with some properties that arose in combinatorial number theory. A particularity of the maximal ultrafilters is that every set in a maximal ultrafilter satisfies many of these combinatorial properties: Definition 3.21. We say that a property P is ≤ f e -upward invariant if the following holds: for every A, B ⊆ N, if P (A) holds and A ≤ f e B then P (B) holds. We way that P is partition regular if the family S P = {A ⊆ N | P (A) holds} contains an ultrafilter (i.e., if for every finite partition N = A 1 ∪ ... ∪ A n there exists at least one index i ≤ n such that A i ∈ S P ). Proposition 3.22. Let P be a partition regular ≤ f e -upward invariant property of sets. Then for every maximal ultrafilter U, for every A ∈ U, P (A) holds.
Proof. Let P be given, let S P = {A ⊆ N | P (A) holds} and let V ⊆ S P (such an ultrafilter exists because P is partition regular). Let B ∈ U. Since U is maximal, V ≤ f e U. Let A ∈ V be such that A ≤ f e B. Since P is ≤ f e -upward invariant and P (A) holds, we obtain that P (B) holds, so we have the thesis.
E.g., as a consequence of Proposition 3.22 we can prove the following: Claim: B i ≤ f e A.
In fact, by construction α + B i ⊆ * A + i, so By Proposition 4.1, this entails that B i ≤ f e A, and this proves that V ≤ f e U for every ultrafilter V. Hence U is maximal.