Multiplicative finite embeddability
vs divisibility of ultrafilters
Boris Šobot
Department of Mathematics and Informatics,
Faculty of Sciences, University of Novi Sad,
Trg Dositeja Obradovića 4, 21000 Novi Sad, Serbia
e-mail<EMAIL_ADDRESS>ORCID: 0000-0002-4848-0678

###### Abstract

We continue the exploration of various aspects of divisibility of
ultrafilters, adding one more relation to the picture: multiplicative finite
embeddability. We show that it lies between divisibility relations $\mid_{M}$
and $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$. The set of its
minimal elements proves to be very rich, and the $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$-hierarchy is used to get a better intuition of this
richness. We find the place of the set of $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$-maximal ultrafilters among some known families of
ultrafilters. Finally, we introduce new notions of largeness of subsets of
${\mathbb{N}}$, and compare it to other such notions, important for infinite
combinatorics and topological dynamics.

2020 Mathematics Subject Classification: 54D35, 54D80, 11U99, 03E05, 03H15

Key words and phrases: Stone-Čech compactification, ultrafilter, divisibility,
finite embeddability

##  1 Introduction

As usual, by $\beta{\mathbb{N}}$ we denote the set of ultrafilters on the set
${\mathbb{N}}$ of natural numbers. For each $n\in{\mathbb{N}}$, the principal
ultrafilter $\\{A\subseteq{\mathbb{N}}:n\in A\\}$ is identified with $n$, so
$\beta{\mathbb{N}}$ is an extension of ${\mathbb{N}}$. The topology on
$\beta{\mathbb{N}}$ is generated by base sets $\overline{A}=\\{{\cal
F}\in\beta{\mathbb{N}}:A\in{\cal F}\\}$ for all $A\subseteq N$. All such sets
are clopen. Because of nonstandard methods used in previous papers on the
subject, we will avoid to denote $\overline{A}\setminus A$ by $A^{*}$ (so it
will not get confused with a nonstandard extension ${{}^{*}A}$).

Whenever $\star$ is an associative operation on ${\mathbb{N}}$, it can be
extended to an operation on $\beta{\mathbb{N}}$ (which we also denote by
$\star$):

${\cal F}\star{\cal
G}=\\{A\subseteq{\mathbb{N}}:\\{n\in{\mathbb{N}}:n^{-1}A\in{\cal G}\\}\in{\cal
F}\\},$

where $n^{-1}A=\\{m\in{\mathbb{N}}:n\star m\in A\\}$. Then
$(\beta{\mathbb{N}},\star)$ becomes a compact Hausdorff right-topological
semigroup. Various properties of this semigroup are described in [4]. We will
be interested in extensions of operations $+$ and $\cdot$ (the usual addition
and multiplication on ${\mathbb{N}}$). Instead of $n^{-1}A$ in the case of
multiplication we write $A/n=\\{\frac{a}{n}:a\in A\land n\mid a\\}$ and in the
additive case $A-n=\\{a-n:a\in A\land a>n\\}$. Both in $(\beta{\mathbb{N}},+)$
and $(\beta{\mathbb{N}},\cdot)$ the elements of ${\mathbb{N}}$ are
cancellative and commute with all other elements.

Let us fix some more notation. Caligraphic letters ${\cal F},{\cal G},\dots$
will mostly be reserved for ultrafilters, and small letters $x,y,\dots$ for
elements of ${\mathbb{N}}$. By $P$ we denote the set of prime numbers, and
ultrafilters ${\cal P}\in\overline{P}$ are called prime. Let $A$ be a subset
of ${\mathbb{N}}$. We denote $A^{c}={\mathbb{N}}\setminus A$,
$A\uparrow=\\{n\in{\mathbb{N}}:(\exists a\in A)a\mid n\\}$,
$A\downarrow=\\{n\in{\mathbb{N}}:(\exists a\in A)n\mid a\\}$, ${\cal
U}=\\{A\in P({\mathbb{N}})\setminus\\{\emptyset\\}:A=A\uparrow\\}$ and ${\cal
V}=\\{A\in
P({\mathbb{N}})\setminus\\{{\mathbb{N}}\\}:A=A\downarrow\\}=\\{A^{c}:A\in{\cal
U}\\}$. If $\kappa$ is a cardinal and $n\in{\mathbb{N}}$, then
$[A]^{\kappa}=\\{X\subseteq A:|X|=\kappa\\}$, $[A]^{<\kappa}=\\{X\subseteq
A:|X|<\kappa\\}$ and $A^{n}=\\{a^{n}:a\in A\\}$. For ${\cal
F}\in\beta{\mathbb{N}}$, ${\cal F}^{n}$ is the ultrafilter generated by
$\\{A^{n}:A\in{\cal F}\\}$. For subsets $X$ and $Y$ of $\beta{\mathbb{N}}$,
$X\cdot Y=\\{{\cal F}\cdot{\cal G}:{\cal F}\in X,{\cal G}\in Y\\}$. If one of
the sets $X$ or $Y$, say $X$, has only one element ${\cal F}$, we will write
${\cal F}\cdot Y$ instead of $\\{{\cal F}\\}Y$; for example if
$n\in{\mathbb{N}}$, then $n{\mathbb{N}}$ is the set of natural numbers
divisible by $n$. We remind the reader that, for any semigroup
$({\mathbb{N}},*)$, $(\beta{\mathbb{N}},*)$ has the smallest ideal, usually
denoted by $K(\beta{\mathbb{N}},*)$.

There are several ways to extend the divisibility relation $\mid$ from
${\mathbb{N}}$ to $\beta{\mathbb{N}}$. Some of them were considered in [9]:

-${\cal F}\mid_{L}{\cal G}$ if ${\cal G}={\cal H}\cdot{\cal F}$ for some ${\cal H}\in\beta{\mathbb{N}}$;

-${\cal F}\mid_{R}{\cal G}$ if ${\cal G}={\cal F}\cdot{\cal H}$ for some ${\cal H}\in\beta{\mathbb{N}}$;

-${\cal F}\mid_{M}{\cal G}$ if ${\cal G}={\cal H}_{1}\cdot{\cal F}\cdot{\cal H}_{2}$ for some ${\cal H}_{1},{\cal H}_{2}\in\beta{\mathbb{N}}$;

-${\cal F}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal G}$ if, for every $A\in{\cal F}$ holds $A\uparrow\in{\cal G}$.

The definition of $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ was
inspired by extension of functions $f:{\mathbb{N}}\rightarrow{\mathbb{N}}$ to
$\beta{\mathbb{N}}$, given by: $\mathaccent 869{f}({\cal F})$ is the
ultrafilter generated by $\\{f[A]:A\in{\cal F}\\}$. For example, for the
function $pow_{n}(x)=x^{n}$ and ${\cal F}\in\beta{\mathbb{N}}$, we have
$\mathaccent 1373{pow_{n}}({\cal F})={\cal F}^{n}$. The equivalent conditions
for ${\cal F}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal G}$,
given by

${\cal F}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal
G}\Leftrightarrow{\cal F}\cap{\cal U}\subseteq{\cal G}\Leftrightarrow{\cal
G}\cap{\cal V}\subseteq{\cal F},$

proved to be very useful. In this paper we will include another relation
$\leq_{fe}$ into consideration. Strictly speaking, it was not originally
defined as a divisibility relation but, as we will see in Theorem 2.8, it is
closely connected to the relations above.

Each of the mentioned relations is a quasiorder (reflexive and transitive), so
by cutting with a respective equivalence relation we get orders on sets of
equivalence classes. For example, we define ${\cal F}=_{\sim}{\cal G}$ if
${\cal F}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal G}$ and
${\cal G}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal F}$, so we
observe $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ as an order on
equivalence classes $[{\cal F}]_{\sim}$ of $\beta{\mathbb{N}}/=_{\sim}$. The
notation for other relations is analogous.

An ultrafilter ${\cal F}$ is divisible by $n\in{\mathbb{N}}$ (in any of the
considered relations) if and only if $n{\mathbb{N}}\in{\cal F}$ ([9], Lemma
5.1). Hence we can write just $n\mid{\cal F}$, not emphasizing the particular
divisibility relation.

In [11] it was shown that the ultrafilters in the “lower” part of the $\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-hierarchy can be organized
into $\omega$-many levels $\overline{L_{n}}$, where $L_{n}=\\{p_{1}p_{2}\dots
p_{n}:p_{1},p_{2},\dots,p_{n}\in P\\}$ is the set of numbers with exactly $n$
prime (not necessarily distinct) factors. In particular $L_{1}=P$, so
$\overline{L_{1}}$ is the set of all prime ultrafilters. Each proper divisor
of any ${\cal F}\in\overline{L_{n}}$ must belong to $\overline{L_{m}}$ for
some $m<n$. The next result is Theorem 5.5 from [11], with somewhat simplified
notation. We remark that it will be used only once in this paper, in Theorem
3.6.

###### Proposition 1.1

Let ${\cal A}_{n_{1},n_{2},\dots,n_{m}}^{{\cal P}_{1}^{k_{1}},{\cal
P}_{2}^{k_{2}},\dots,{\cal P}_{m}^{k_{m}}}$ be the family of all ultrafilters
containing

$F_{n_{1},n_{2},\dots,n_{m}}^{{\cal P}_{1}^{k_{1}},{\cal
P}_{2}^{k_{2}},\dots,{\cal
P}_{m}^{k_{m}}}=\\{(A_{1}^{k_{1}})^{(n_{1})}(A_{2}^{k_{2}})^{(n_{2})}\dots(A_{m}^{k_{m}})^{(n_{m})}:$
$A_{i}\in{\cal P}_{i}\land A_{i}\subseteq P\land(A_{i}=A_{j}\mbox{ if }{\cal
P}_{i}={\cal P}_{j}\mbox{ and }A_{i}\cap A_{j}=\emptyset\mbox{
otherwise})\\},$

where ${\cal P}_{1},{\cal P}_{2},\dots,{\cal P}_{m}$ are prime ultrafilters,
$n_{1},n_{2},\dots,n_{m},k_{1},k_{2},\dots,k_{m}\in{\mathbb{N}}$ and

$\displaystyle(A_{1}^{k_{1}})^{(n_{1})}(A_{2}^{k_{2}})^{(n_{2})}\dots(A_{m}^{k_{m}})^{(n_{m})}$
$\displaystyle=$
$\displaystyle\left\\{\prod_{i=1}^{m}\prod_{j=1}^{n_{i}}a_{i,j}^{k_{i}}:a_{i,j}\in
A_{i}\mbox{ for all }i,j\;\land\mbox{ all }a_{i,j}\mbox{ are
distinct.}\right\\}$

Then the level $\overline{L_{n}}$ is the disjoint union of all families ${\cal
A}_{n_{1},n_{2},\dots,n_{m}}^{{\cal P}_{1}^{k_{1}},{\cal
P}_{2}^{k_{2}},\dots,{\cal P}_{m}^{k_{m}}}$, for
$n_{1}k_{1}+n_{2}k_{2}+\dots+n_{m}k_{m}=n$.

For example, $\overline{L_{2}}$ consists of: (1) ${\cal P}^{2}$ for ${\cal
P}\in\overline{P}$; (2) ultrafilters containing some $F_{2}^{\cal P}$, and
hence having only one prime $\hskip 2.84526pt\mathaccent 869{\mid}\hskip
2.84526pt$-divisor ${\cal P}$ and (3) ultrafilters containing $F_{1,1}^{{\cal
P},{\cal Q}}$ for some distinct primes ${\cal P}$ and ${\cal Q}$. Of course,
the latter two kinds are only possible for nonprincipal ${\cal P}$ and ${\cal
Q}$. More on this hierarchy can be found in [11].

Ultrafilters not divisible by any $n\in{\mathbb{N}}\setminus\\{1\\}$ are
called ${\mathbb{N}}$-free. Sets $A\subseteq{\mathbb{N}}$ belonging to some
${\mathbb{N}}$-free ultrafilter are also called ${\mathbb{N}}$-free. Let us
denote by ${\cal U}_{N}$ the family of all ${\mathbb{N}}$-free sets from
${\cal U}$, and ${\cal V}_{N}=\\{A^{c}:A\in{\cal U}_{N}\\}$.

We recall several important notions of largeness for subsets of
${\mathbb{N}}$. With some of them we also include equivalent conditions;
proofs of these equivalences can be found in [1] and [4]. Since we want to
work with two versions (multiplicative and additive) for some of the
definitions, we will add the prefix A- when dealing with the $+$-version. If
$\langle x_{n}:n\in{\mathbb{N}}\rangle$ is a sequence of elements of
${\mathbb{N}}$, we denote $FS(\langle x_{n}\rangle)=\\{\sum_{n\in
F}x_{n}:F\in[{\mathbb{N}}]^{<\aleph_{0}}\\}$.

A set $A\subseteq{\mathbb{N}}$ is:

-A-IP if there is a sequence $\langle x_{n}:n\in{\mathbb{N}}\rangle$ in ${\mathbb{N}}$ such that $FS(\langle x_{n}\rangle)\subseteq A$, if and only if it is a member of some $+$-idempotent;

-A-central if it is a member of some minimal $+$-idempotent (idempotent belonging to $K(\beta{\mathbb{N}},+)$);

-A-IP* if it intersects every A-IP set, if and only if it is a member of every $+$-idempotent;

-A-central* if it intersects every A-central set, if and only if it is a member of every minimal $+$-idempotent;

-A-thick if it contains, for every $n\in{\mathbb{N}}$, a set $m+\\{1,2,\dots,n\\}=\\{m+k:1\leq k\leq n\\}$ for some $m\in{\mathbb{N}}$;

-A-piecewise syndetic (short: A-pcws) if there is a finite $F\subseteq{\mathbb{N}}$ such that the family $\\{(\bigcup_{t\in F}(A-t))-n:n\in{\mathbb{N}}\\}$ has the finite intersection property, if and only if it is a member of some ultrafilter in $K(\beta{\mathbb{N}},+)$;

-A-J set if for any finite set $F$ of functions from ${\mathbb{N}}$ to ${\mathbb{N}}$ there are $a\in{\mathbb{N}}$ and $H\in[{\mathbb{N}}]^{<\aleph_{0}}$ such that $a+\sum_{i\in H}f(i)\in A$ for all $f\in F$.

The corresponding multiplicative versions of these properties will be denoted
with the prefix M-. For example, $A\subseteq{\mathbb{N}}$ is M-IP if there is
a sequence $\langle x_{n}:n\in{\mathbb{N}}\rangle$ in ${\mathbb{N}}$ such that
$FP(\langle x_{n}\rangle):=\\{\prod_{n\in
F}x_{n}:F\in[{\mathbb{N}}]^{<\aleph_{0}}\\}\subseteq A$, if and only if it is
a member of a $\cdot$-idempotent. We note also that sometimes only sets of the
form $FS(\langle x_{n}\rangle)$ are called A-IP sets (not their supersets),
but we find the definition used here more convenient.

###### Remark 1.2

In general it is common to define, for a property P, that
$A\subseteq{\mathbb{N}}$ satisfies P* if it intersects all sets with property
P. Note that, if P is upwards closed (that is, $A$ has P implies that every
superset of $A$ has P), then $A$ has P* if and only if ${\mathbb{N}}\setminus
A$ does not have P. It follows that, in this case, P and P* are dual: $A$ has
P if and only if it intersects all sets with property P*. If we have another
property Q, clearly P$\Rightarrow$Q implies Q*$\Rightarrow$P*. Hence, for
upwards closed properties the reverse implication also holds.

The paper has six sections. Section 2 introduces multiplicative finite
embeddability $\leq_{fe}$ and proves some of its basic properties. In Section
3 we explore $\leq_{fe}$-minimal ultrafilters. Using the connection between
$\leq_{fe}$ and $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$, we
show that they are spread all over the lower half, but also present in the
upper half of the $\hskip 2.84526pt\mathaccent 869{\mid}\hskip
2.84526pt$-hierarchy. Section 4 deals with families of ultrafilters maximal
for some of the relations considered. Using these families, in Section 5 we
get new classes of large subsets of ${\mathbb{N}}$ and find their place among
some old notions of this kind. The closing section contains several problems
we were not able to resolve.

Throughout the paper we use many results from other sources. This is natural
since quite a few times we are able to draw conclusions simply by combining
these known results. Some of them we single out as separate propositions, but
doing so for all of them would make the paper undesirably long.

##  2 Multiplicative finite embeddability

The concept of finite embeddability (with respect to the addition operation)
was introduced (implicitly) by Ruzsa in [8] and made precise by Di Nasso in
[3]. It was futher investigated in [2] and [6]. The paper [7] generalized the
notion to $F$-finite embeddability with respect to an arbitrary set $F$ of
operations. We are interested in multiplicative finite embeddability, defined
with respect to the multiplication on ${\mathbb{N}}$. Many results on
$F$-finite embeddability apply to multiplicative finite embeddability, and
some more results on the additive version translate directly to the
multiplication version. It turns out that this relation is related to our
divisibility relations. For the sake of simplicity we keep the notation
$\leq_{fe}$.

###### Definition 2.1

For $A,B\in P({\mathbb{N}})$, $A$ is finitely embeddable in $B$
($A\leq_{fe}B$) if for every finite $F\subseteq A$ there is $k\in{\mathbb{N}}$
such that $kF\subseteq B$.

For ${\cal F},{\cal G}\in\beta{\mathbb{N}}$, ${\cal F}$ is finitely embeddable
in ${\cal G}$ (${\cal F}\leq_{fe}{\cal G}$) if for every $B\in{\cal G}$ there
is $A\in{\cal F}$ such that $A\leq_{fe}B$.

It is clear that the relations $\leq_{fe}$ (both on $P({\mathbb{N}})$ and
$\beta{\mathbb{N}}$) are reflexive and transitive. It is a special case of
[7], Proposition 2.6 that $A\subseteq{\mathbb{N}}$ is in the
$\leq_{fe}$-greatest class of sets if and only if it is M-thick; therefore
$\leq_{fe}$ for subsets of ${\mathbb{N}}$ is not antisymmetric. Neither is
$\leq_{fe}$ for ultrafilters, which will follow from Proposition 4.1.

Naturally we define ${\cal F}=_{fe}{\cal G}$ if ${\cal F}\leq_{fe}{\cal G}$
and ${\cal G}\leq_{fe}{\cal F}$ and think of $\leq_{fe}$ as an order on
equivalence classes $[{\cal F}]_{fe}$ of $=_{fe}$. Let us also denote ${\cal
F}<_{fe}{\cal G}$ if ${\cal F}\leq_{fe}{\cal G}$ but ${\cal F}\neq_{fe}{\cal
G}$.

The additive version of the following result is included in [2], Theorem 4
with several more equivalent conditions. We include a direct proof here.

###### Proposition 2.2

For $A,B\in P({\mathbb{N}})$, $A\leq_{fe}B$ if and only if there is ${\cal
H}\in\beta{\mathbb{N}}$ such that $\overline{A}\cdot{\cal
H}\subseteq\overline{B}$.

Proof. Firstly, $A\leq_{fe}B$ actually means that the family $\\{B/a:a\in
A\\}$ has the finite intersection property. Let ${\cal H}$ be an ultrafilter
containing this family. Then, for every ${\cal F}\in\overline{A}$,
$\\{n\in{\mathbb{N}}:B/n\in{\cal H}\\}\supseteq A\in{\cal F}$, so $B\in{\cal
F}\cdot{\cal H}$.

In the other direction, assume that $\overline{A}\cdot{\cal
H}\subseteq\overline{B}$, but $\\{B/a:a\in A\\}$ does not have the finite
intersection property. Then there is $a\in A$ such that $B/a\notin{\cal H}$,
so $a{\cal H}\notin\overline{B}$, a contradiction. $\Box$

###### Lemma 2.3

$\leq_{fe}$ coincides with each of the relations
$\mid_{R},\mid_{L},\mid_{M},\hskip 2.84526pt\mathaccent 869{\mid}\hskip
2.84526pt$ on ${\mathbb{N}}\times\beta{\mathbb{N}}$. Also ${\cal
G}\not\leq_{fe}m$ for all ${\cal G}\in\beta{\mathbb{N}}\setminus{\mathbb{N}}$,
$m\in{\mathbb{N}}$.

Proof. As we already noted, for $\rho\in\\{\mid_{R},\mid_{L},\mid_{M},\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt\\}$, $m\in{\mathbb{N}}$ and
${\cal F}\in\beta{\mathbb{N}}$, $m\rho{\cal F}$ is equivalent to
$m{\mathbb{N}}\in{\cal F}$. We show that the same holds for $\leq_{fe}$.

If ${\cal F}=n\in{\mathbb{N}}$, then $\\{n\\}\in{\cal F}$ so if there are a
set $A$ containing $\\{m\\}$ and $k\in{\mathbb{N}}$ such that
$kA\subseteq\\{n\\}$, it follows that $m\mid n$. On the other hand, if $m\mid
n$ (say $n=mk$) then every $B\in{\cal F}$ contains $n$, so if we take
$A=\\{m\\}$ then $kA=\\{n\\}\subseteq B$.

If ${\cal F}\in\beta{\mathbb{N}}\setminus{\mathbb{N}}$, assume first that
$m{\mathbb{N}}\in{\cal F}$. Then every $B\in{\cal F}$ contains an element $mk$
divisible by $m$, so there is $A=\\{m\\}$ such that $kA\subseteq B$. If we
assume that $m{\mathbb{N}}\notin{\cal F}$, then $(m{\mathbb{N}})^{c}\in{\cal
F}$. Now there can be no set $A$ containing $\\{m\\}$ such that
$A\leq_{fe}(m{\mathbb{N}})^{c}$: if $F=\\{m\\}$ then
$kF\subseteq(m{\mathbb{N}})^{c}$ does not hold for any $k\in{\mathbb{N}}$.

To prove the last statement, assume the opposite: that ${\cal G}\leq_{fe}m$.
For $B=\\{m\\}$ and any $A\in{\cal F}$, if $F\subseteq A$ is of cardinality at
least 2, then clearly $kF\subseteq B$ does not hold for any
$k\in{\mathbb{N}}$. $\Box$

Having in mind the lemma above, in the next few proofs it suffices to consider
only nonprincipal ultrafilters. The first one is a special case of [7],
Proposition 3.11, but we include a simple proof here.

###### Lemma 2.4

If ${\cal F},{\cal G}\in\beta{\mathbb{N}}$, then ${\cal F}\leq_{fe}{\cal
F}\cdot{\cal G}$ and ${\cal G}\leq_{fe}{\cal F}\cdot{\cal G}$.

Proof. Let $B\in{\cal F}\cdot{\cal G}$ be arbitrary. Then
$A:=\\{n\in{\mathbb{N}}:B/n\in{\cal G}\\}\in{\cal F}$. For every finite
$F\subseteq A$ the set $\bigcap_{n\in F}B/n$ is in ${\cal G}$, so it is
nonempty. For any $k\in\bigcap_{n\in F}B/n$ we have $kF\subseteq B$. Thus
${\cal F}\leq_{fe}{\cal F}\cdot{\cal G}$.

On the other hand, for any $n\in A$ we have $B/n\in{\cal G}$ and $n\cdot
B/n\subseteq B$, so in particular for every finite $F\subseteq B/n$ we have
$nF\subseteq B$. Hence $B/n\leq_{fe}B$ and ${\cal G}\leq_{fe}{\cal
F}\cdot{\cal G}$. $\Box$

###### Lemma 2.5

$\mid_{M}\subset\leq_{fe}$.

Proof. From Lemma 2.4 it follows directly that ${\cal F}\mid_{R}{\cal H}$
implies ${\cal F}\leq_{fe}{\cal H}$ and ${\cal G}\mid_{L}{\cal H}$ implies
${\cal G}\leq_{fe}{\cal H}$. But $\mid_{M}$ is the transitive closure of
$\mid_{L}\cup\mid_{R}$ and $\leq_{fe}$ is transitive, so
$\mid_{M}\subseteq\leq_{fe}$ as well.

We will see in Proposition 4.1 that the $\mid_{M}$-maximal class is
$K(\beta{\mathbb{N}},\cdot)$ and the $\leq_{fe}$-maximal class is
$\overline{K(\beta{\mathbb{N}},\cdot)}$, which strictly contains
$K(\beta{\mathbb{N}},\cdot)$. Therefore the inclusion is strict. $\Box$

Let us denote, for $m\in{\mathbb{N}}$ and $A,B\in P({\mathbb{N}})$, by
$A\leq_{me}B$ the statement: for every $F\in[A]^{m}$ there is
$k\in{\mathbb{N}}$ such that $kF\subseteq B$. Then, for ${\cal F},{\cal
G}\in\beta{\mathbb{N}}$, ${\cal F}\leq_{me}{\cal G}$ will mean: for every
$B\in{\cal G}$ there is $A\in{\cal F}$ such that $A\leq_{me}B$.

###### Lemma 2.6

For ${\cal F},{\cal G}\in\beta{\mathbb{N}}$, ${\cal F}\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal G}$ if and only if ${\cal
F}\leq_{1e}{\cal G}$.

Proof. If ${\cal F}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal
G}$ then for any $B\in{\cal G}$ we can take $A=B\downarrow\in{\cal F}$ and get
$A\leq_{1e}B$. On the other hand, if $B\in{\cal G}\cap{\cal V}$, $A\leq_{1e}B$
implies $A\subseteq B$, so $A\in{\cal F}$ implies $B\in{\cal F}$. $\Box$

###### Lemma 2.7

$\leq_{fe}\subset\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$.

Proof. $\leq_{fe}\subseteq\hskip 2.84526pt\mathaccent 869{\mid}\hskip
2.84526pt$ follows directly from the definition of $\leq_{fe}$ and the lemma
above. That the inclusion is strict will follow from the fact that there are
(many) $\leq_{fe}$-minimal ultrafilters that are not $\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal (prime), see Theorems
3.4 and 3.6. $\Box$

The results above show that $\leq_{fe}$ is related to the divisibility
relations as follows.

###### Theorem 2.8

$\begin{array}[]{c}\mid_{L}\\\ \\\
\mid_{R}\end{array}\begin{array}[]{c}\rotatebox[origin={c}]{-45.0}{$\subset$}\\\
\rotatebox[origin={c}]{45.0}{$\subset$}\\\
\end{array}\;\mid_{M}\;\subset\;\leq_{fe}\;\subset\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$

The strict inclusions of the diagram above not regarding $\leq_{fe}$ were
proved in [9]. Thus the following diagram is obtained, for any ${\cal
F}\in\beta{\mathbb{N}}$:

$\begin{array}[]{c}[{\cal F}]_{L}\\\ \\\ \;[{\cal F}]_{R}\\\
\end{array}\begin{array}[]{c}\rotatebox[origin={c}]{-45.0}{$\subseteq$}\\\
\rotatebox[origin={c}]{45.0}{$\subseteq$}\\\ \end{array}\;[{\cal
F}]_{M}\;\subseteq\;[{\cal F}]_{fe}\;\subseteq[{\cal F}]_{\sim}$

The inclusion between $[{\cal F}]_{M}$ and $[{\cal F}]_{fe}$ is not always
equality by Proposition 4.1. Neither is the inclusion between $[{\cal
F}]_{fe}$ and $[{\cal F}]_{\sim}$, as we will see in Theorem 4.5.

We have a few direct corollaries of Theorem 2.8. In a partial order with a
smallest element $1$, a set $X$ is called a strong antichain if, for every two
elements of $X$, $1$ is the only element below both of them.

###### Corollary 2.9

(a) For every ultrafilter ${\cal F}\in L$, its $=_{fe}$-equivalence class is a
singleton.

(b) $(\beta{\mathbb{N}}/=_{fe},\leq_{fe})$ has strong antichains of
cardinality $2^{{c}}$.

Proof. (a) By [11], Corollary 5.10, the equivalence classes of such
ultrafilters for $=_{\sim}$, and therefore for each of the relations $=_{L}$,
$=_{R}$, $=_{M}$ and $=_{fe}$, are singletons.

(b) Since $(\beta{\mathbb{N}}/=_{\sim},\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt)$ has strong antichains of cardinality $2^{{c}}$
(for example, the antichain of all prime ultrafilters), so does
$(\beta{\mathbb{N}}/=_{fe},\leq_{fe})$. $\Box$

For $p\in P$, $n\in{\mathbb{N}}$ and ${\cal F}\in\beta{\mathbb{N}}$, by
$p^{n}\parallel{\cal F}$ we denote that $p^{n}\mid{\cal F}$ but
$p^{n+1}\mathchar 13613\relax{\cal F}$. By Lemma 2.4, for any ${\cal
F}\in\beta{\mathbb{N}}$ and any $p\in P$, ${\cal F}\leq_{fe}p\cdot{\cal F}$.
It may happen that $p\cdot{\cal F}=_{fe}{\cal F}$, for example when ${\cal F}$
is $\leq_{fe}$-maximal. However, if $p^{n}\parallel{\cal F}$ then ${\cal
F}<_{fe}p\cdot{\cal F}$ (because $p^{n+1}\mid{\cal F}$, so even ${\cal
F}\neq_{\sim}p\cdot{\cal F}$) and it will turn out that in this case
$p\cdot{\cal F}$ is a direct $\leq_{fe}$-successor of ${\cal F}$.

(We know from Lemma 4.4. of [12] that $p\cdot{\cal F}$ is indeed a direct
$\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-successor of ${\cal
F}$. However, this does not directly imply the same for $\leq_{fe}$: an
ultrafilter ${\cal G}$ such that ${\cal F}<_{fe}{\cal G}<_{fe}p\cdot{\cal F}$
may be in the same $=_{\sim}$-equivalence class with ${\cal F}$ or
$p\cdot{\cal F}$.)

###### Theorem 2.10

Let $p^{n}\parallel{\cal F}$ for some $p\in P$ and some
$n\in{\mathbb{N}}\cup\\{0\\}$, and ${\cal H}=p\cdot{\cal F}$. Then there is no
ultrafilter ${\cal G}$ such that ${\cal F}<_{fe}{\cal G}<_{fe}{\cal H}$.

Proof. Assume the opposite. Clearly $p^{n}\mid{\cal G}$. We consider two
cases.

$1^{\circ}$ $p^{n+1}\mid{\cal G}$, i.e. $p^{n+1}{\mathbb{N}}\in{\cal G}$. Let
$B\in{\cal G}$ be arbitrary, and let $B^{\prime}=B\cap p^{n+1}{\mathbb{N}}$.
There is $A\in{\cal F}$ such that $A\leq_{fe}B^{\prime}$. Since
$p^{n+1}{\mathbb{N}}\notin{\cal F}$, we may assume that $A\cap
p^{n+1}{\mathbb{N}}=\emptyset$. But then, for any finite $F\subseteq A$, if
$kF\subseteq B^{\prime}$, then $k$ must be divisible by $p$. This means that
$pA\leq_{fe}B^{\prime}$, and thus $pA\leq_{fe}B$. We proved that ${\cal
H}\leq_{fe}{\cal G}$, a contradiction.

$2^{\circ}$ $p^{n+1}\mathchar 13613\relax{\cal G}$. Let $A_{0}\in{\cal F}$ be
arbitrary and $A:=A_{0}\cap p^{n}{\mathbb{N}}$. Then $A\in{\cal F}$ and
$pA\in{\cal H}$, so there is $B_{0}\in{\cal G}$ such that $B_{0}\leq pA$. Let
$B:=B_{0}\cap(p^{n}{\mathbb{N}}\setminus p^{n+1}{\mathbb{N}})$. We found
$B\in{\cal G}$ such that $B\leq_{fe}A_{0}$, proving that ${\cal
G}\leq_{fe}{\cal F}$, a contradiction again. $\Box$

We finish this section with some results on chains in the $\leq_{fe}$-orders.

###### Theorem 2.11

In $[{\mathbb{N}}]^{\aleph_{0}}$ there is a strictly $\leq_{fe}$-decreasing
chain $\langle A_{n}:n<\omega\rangle$ without a lower bound.

Proof. Let $\\{F_{m}:m<\omega\\}$ be an enumeration of $[{\mathbb{N}}]^{2}$.
We construct $A_{n}$ in such way that $F_{m}\not\leq_{fe}A_{n}$ for $m<n$. We
start with $A_{0}:={\mathbb{N}}$. Assume that $A_{n}$ is constructed and
define $A_{n+1}=\\{a_{m}:0<m<\omega\\}$ by recursion on $m$. Let $a_{0}$ and
$a_{1}$ be the smallest two elements of $A_{n}$. Given
$a_{0},a_{1},\dots,a_{m}$, for $a_{m+1}$ we choose the minimal $a\in A_{n}$
such that
$F_{0},F_{1},\dots,F_{n},\\{a_{0},a_{1}\\}\not\leq_{fe}\\{a_{1},a_{2},\dots,a_{m},a\\}$
(clearly, it suffices to take $a$ large enough). In the end, we get that
$A_{n+1}\subset A_{n}$ (so $A_{n+1}\leq_{fe}A_{n}$) but
$A_{n}\not\leq_{fe}A_{n+1}$. If $B\in[{\mathbb{N}}]^{\aleph_{0}}$ were a lower
bound for $\langle A_{n}:n<\omega\rangle$, for any $n$ such that
$F_{n}\in[B]^{2}$ we would have $F_{n}\leq_{fe}A_{n+1}$, a contradiction.
$\Box$

What about upper and lower bounds of chains in
$(\beta{\mathbb{N}}/=_{fe},\leq_{fe})$? Using limits by ultrafilters it is not
hard to construct an upper bound for any $\leq_{fe}$-chain $\langle{\cal
F}_{i}:i\in I\rangle$ in $\beta{\mathbb{N}}$. Namely, if ${\cal V}$ is any
ultrafilter on $I$ containing sets $\\{j\in I:j\geq i\\}$ for all $i\in I$,
then $\lim_{i\rightarrow{\cal V}}{\cal F}_{i}$ is the ultrafilter ${\cal F}$
defined by: $A\in{\cal F}$ if and only if $\\{i\in I:A\in{\cal
F}_{i}\\}\in{\cal V}$. $\lim_{i\rightarrow{\cal V}}{\cal F}_{i}$ is an upper
bound of the given chain; see Theorem 3.12 of [6] for details. In the proof of
the following result bounds are also obtained as limits by ultrafilters.

###### Proposition 2.12

([13], Lemma 4.1) (a) Every $\hskip 2.84526pt\mathaccent 869{\mid}\hskip
2.84526pt$-increasing chain $\langle[{\cal F}_{i}]_{\sim}:i\in I\rangle$ in
$\beta{\mathbb{N}}/=_{\sim}$ has the least upper bound $[{\cal G}]_{\sim}$. If
none of the ultrafilters ${\cal F}_{i}$ is $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$-divisible by some prime ultrafilter ${\cal P}$,
then neither is ${\cal G}$.

(b) Every $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-decreasing
chain $\langle[{\cal F}_{i}]_{\sim}:i\in I\rangle$ in
$\beta{\mathbb{N}}/=_{\sim}$ has the greatest lower bound $[{\cal G}]_{\sim}$.

In contrast with the situation for $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$-chains, for $\leq_{fe}$ we have the following
result.

###### Theorem 2.13

In $\beta{\mathbb{N}}/=_{fe}$ there is a strictly $\leq_{fe}$-increasing chain
without a smallest upper bound.

Proof. Let $\langle n_{i}:i<\omega\rangle$ be a sequence in ${\mathbb{N}}$
such that no two differences $n_{i}-n_{j}$ for $j<i<\omega$ are the same (it
is easily constructed by recursion on $i$, taking each $n_{i}$ large enough).
Let ${\cal V}$ be any nonprincipal ultrafilter containing $2{\mathbb{N}}$ and
${\cal W}$ any nonprincipal ultrafilter containing ${\mathbb{N}}\setminus
2{\mathbb{N}}$. Now choose any nonprincipal prime ultrafilter ${\cal P}$ and
define ${\cal P}^{(n)}=\underbrace{{\cal P}\cdot{\cal P}\cdot\dots\cdot{\cal
P}}_{n}$. Since ${\cal P}^{(n)}\in\overline{L_{n}}$, any upper bound for the
sequence $\langle{\cal P}^{(n_{i})}:i<\omega\rangle$ must be in
$\beta{\mathbb{N}}\setminus L$. We consider two such upper bounds: ${\cal
G}=\lim_{i\rightarrow{\cal V}}{\cal P}^{(n_{i})}$ and ${\cal
H}=\lim_{i\rightarrow{\cal W}}{\cal P}^{(n_{i})}$. We will show that no upper
bound for our sequence (in fact, no ultrafilter outside $L$) can be both below
${\cal G}$ and below ${\cal H}$.

Assume the opposite, that ${\cal F}\in\beta{\mathbb{N}}\setminus L$ is such
that ${\cal F}\leq_{fe}{\cal G}$ and ${\cal F}\leq_{fe}{\cal H}$. If we denote
$A_{0}=\bigcup_{i<\omega}L_{n_{2i}}$ and
$A_{1}=\bigcup_{i<\omega}L_{n_{2i+1}}$, then $A_{0}\in{\cal G}$ and
$A_{1}\in{\cal H}$. There are $C_{0},C_{1}\in{\cal F}$ such that
$C_{0}\leq_{fe}A_{0}$ and $C_{1}\leq_{fe}A_{1}$. For $C=C_{0}\cap C_{1}$ we
have $C\leq_{fe}A_{0}$ and $C\leq_{fe}A_{1}$. Since ${\cal F}\notin L$, $C$
must contain elements from different levels: $c_{i}\in C\cap L_{i}$ and
$c_{j}\in C\cap L_{j}$ (for some $i\neq j$). So there are
$k_{0},k_{1}\in{\mathbb{N}}$ such that $\\{k_{0}c_{i},k_{0}c_{j}\\}\subseteq
A_{0}$ and $\\{k_{1}c_{i},k_{1}c_{j}\\}\subseteq A_{1}$. However, if $k_{0}\in
L_{l_{0}}$ and $k_{1}\in L_{l_{1}}$, then $k_{0}c_{i}\in L_{l_{0}+i}\cap
A_{0}$, $k_{0}c_{j}\in L_{l_{0}+j}\cap A_{0}$, $k_{1}c_{i}\in L_{l_{1}+i}\cap
A_{1}$ and $k_{1}c_{j}\in L_{l_{1}+j}\cap A_{1}$. But
$(l_{0}+i)-(l_{0}+j)=(l_{1}+i)-(l_{1}+j)$, a contradiction with the choice of
levels $n_{i}$. $\Box$

##  3 Minimal ultrafilters

When speaking about minimal elements for divisibility relations, we exclude
$1$, since it would trivially be the smallest for each such order. Thus,
$\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal elements are
exactly the prime ultrafilters. The set of $\mid_{M}$-minimal elements is the
set of all ultrafilters in
$\beta{\mathbb{N}}\setminus((\beta{\mathbb{N}}\setminus\\{1\\})\cdot(\beta{\mathbb{N}}\setminus\\{1\\}))$.
The following result says that there are many more such ultrafilters beside
primes.

###### Proposition 3.1

([4], Theorem 6.35)
$(\beta{\mathbb{N}}\setminus{\mathbb{N}})\cdot(\beta{\mathbb{N}}\setminus{\mathbb{N}})$
is nowhere dense in $\beta{\mathbb{N}}\setminus{\mathbb{N}}$, i.e. for every
$A\in[{\mathbb{N}}]^{\aleph_{0}}$ there is $X\in[A]^{\aleph_{0}}$ such that
none of the elements of $\overline{X}$ can be represented as a product of two
elements from $\beta{\mathbb{N}}\setminus{\mathbb{N}}$.

Of course, for ${\cal F}\in\overline{X}$ the proposition above only rules out
$\mid_{M}$-divisors from $\beta{\mathbb{N}}\setminus{\mathbb{N}}$. We will see
in the proof of Theorem 3.6 how to find ultrafilters not divisible by elements
of ${\mathbb{N}}$ either.

It was left unresolved in [6] whether there are any minimal elements for
additive finite embeddability in $\beta{\mathbb{N}}\setminus{\mathbb{N}}$. For
multiplicative $\leq_{fe}$ we will show that there are many minimal elements
in $(\beta{\mathbb{N}}\setminus{\mathbb{N}},\leq_{fe})$.

###### Lemma 3.2

Every $\leq_{fe}$-minimal element is either a prime number $p\in P$ or an
${\mathbb{N}}$-free ultrafilter.

Proof. Assume the opposite, that a $\leq_{fe}$-minimal ultrafilter ${\cal F}$
is divisible by some $p\in P$, but not equal to $p$. Then $p\leq_{fe}{\cal
F}$. However, ${\cal F}\not\neq_{fe}p$ follows from Lemma 2.3: if ${\cal
F}\in{\mathbb{N}}$ it is a consequence of the fact that the restriction of
$\leq_{fe}$ to ${\mathbb{N}}^{2}$ is the ordinary divisibility, and otherwise
it follows from the fact that there are no ${\cal
F}\in\beta{\mathbb{N}}\setminus{\mathbb{N}}$ and $n\in{\mathbb{N}}$ such that
${\cal F}\leq_{fe}n$. $\Box$

However, not all ${\mathbb{N}}$-free ultrafilters are $\leq_{fe}$-minimal. For
example, let ${\cal P},{\cal Q}\in\overline{P}\setminus P$. Then ${\cal
P}\cdot{\cal Q}$ is ${\mathbb{N}}$-free, but it is not minimal: by Lemma 2.4
${\cal P}\leq_{fe}{\cal P}\cdot{\cal Q}$, and ${\cal P}\cdot{\cal
Q}\not\leq_{fe}{\cal P}$ (because ${\cal P}\cdot{\cal Q}\;\mathaccent
869{\mathchar 13613\relax}\;{\cal P}$).

###### Lemma 3.3

For every ultrafilter ${\cal F}$, ${\cal F}^{n}={\cal G}^{\prime}\cdot{\cal
H}^{\prime}$ holds if and only if ${\cal G}^{\prime}=m_{1}{\cal G}^{n}$,
${\cal H}^{\prime}=m_{2}{\cal H}^{n}$ and $m_{1}m_{2}=m^{n}$ for some
$m_{1},m_{2},m\in{\mathbb{N}}$ and ${\cal G},{\cal H}\in\beta{\mathbb{N}}$
such that ${\cal F}=m\cdot{\cal G}\cdot{\cal H}$.

Proof. First, let ${\cal F}_{1}={\cal G}\cdot{\cal H}$, ${\cal F}=m{\cal
F}_{1}$ and $m^{n}=m_{1}m_{2}\in{\mathbb{N}}^{n}$. For any $A\in{\cal F}_{1}$
we have $B:=\\{k\in{\mathbb{N}}:A/k\in{\cal H}\\}\in{\cal G}$. If
$s=k^{n}\in{\mathbb{N}}^{n}$, then $A^{n}/s=(A/k)^{n}$. Thus
$\\{s\in{\mathbb{N}}:A^{n}/s\in{\cal H}^{n}\\}\supseteq B^{n}\in{\cal G}^{n}$,
so $A^{n}\in{\cal G}^{n}\cdot{\cal H}^{n}$. This proves that ${\cal
F}_{1}^{n}={\cal G}^{n}\cdot{\cal H}^{n}$, so ${\cal F}^{n}=m^{n}{\cal
F}_{1}^{n}=m_{1}{\cal G}^{n}\cdot m_{2}{\cal H}^{n}$.

In the other direction, let ${\cal F}^{n}={\cal G}^{\prime}\cdot{\cal
H}^{\prime}$ for some ${\cal G}^{\prime},{\cal
H}^{\prime}\in\beta{\mathbb{N}}$. Let

$D:=\\{p_{1}^{s_{1}}p_{2}^{s_{2}}\dots
p_{r}^{s_{r}}:p_{1},p_{2},\dots,p_{r}\in P\mbox{ distinct }\land
0<s_{1},s_{2},\dots,s_{r}<n\\}\cup\\{1\\}.$

For any $d\in{\mathbb{N}}$, by $l(d)$ we denote the unique element od $D$ such
that $d\cdot l(d)\in{\mathbb{N}}^{n}$. Then
${\mathbb{N}}^{n}/d=l(d){\mathbb{N}}^{n}$. Hence

$C:=\\{d\in{\mathbb{N}}:{\mathbb{N}}^{n}/d\in{\cal
H}^{\prime}\\}=\\{d\in{\mathbb{N}}:l(d){\mathbb{N}}^{n}\in{\cal
H}^{\prime}\\}$

is nonempty only if $m_{2}{\mathbb{N}}^{n}\in{\cal H}^{\prime}$ for some
$m_{2}\in D$. But for distinct $d_{1},d_{2}\in D$, $d_{1}{\mathbb{N}}^{n}\cap
d_{2}{\mathbb{N}}^{n}=\emptyset$, so such an $m_{2}$ is unique. If
$m_{1}=l(m_{2})$ then $C=m_{1}{\mathbb{N}}^{n}\in{\cal G}^{\prime}$. Hence
${\cal G}^{\prime}=m_{1}{\cal G}^{n}$ and ${\cal H}^{\prime}=m_{2}{\cal
H}^{n}$ for some ${\cal G}$ and ${\cal H}$. But then $m_{1}m_{2}=m^{n}$ for
some $m\in{\mathbb{N}}$ and ${\cal F}^{n}=m^{n}\cdot{\cal G}^{n}\cdot{\cal
H}^{n}=(m\cdot{\cal G}\cdot{\cal H})^{n}$ (by the first direction). Since
$\mathaccent 1373{pow_{n}}$ is one-to-one, ${\cal F}=m\cdot{\cal G}\cdot{\cal
H}$. $\Box$

###### Theorem 3.4

For every $\leq_{fe}$-minimal ultrafilter ${\cal
F}\in\beta{\mathbb{N}}\setminus{\mathbb{N}}$ and every $n\in{\mathbb{N}}$, the
ultrafilter ${\cal F}^{n}$ is also $\leq_{fe}$-minimal.

Proof. Assume the opposite, that there is ${\cal G}\in\beta{\mathbb{N}}$ such
that ${\cal G}<_{fe}{\cal F}^{n}$. By Lemma 3.2 ${\cal F}$ must be
${\mathbb{N}}$-free, and hence so are ${\cal F}^{n}$ and ${\cal G}$. Let
$B\in{\cal F}$ be arbitrary. Then there is $A\in{\cal G}$ such that
$A\leq_{fe}B^{n}$. By Proposition 2.2 there is ${\cal H}\in\beta{\mathbb{N}}$
such that $\overline{A}\cdot{\cal H}\subseteq\overline{B^{n}}$. Then ${\cal
G}\cdot{\cal H}={\cal W}^{n}$ for some ${\cal W}\in\overline{B}$. By Lemma
3.3, since ${\cal G}$ is ${\mathbb{N}}$-free, we must have ${\cal G}={\cal
G}_{1}^{n}$ for some ultrafilter ${\cal G}_{1}$. But ${\cal
G}_{1}^{n}<_{fe}{\cal F}^{n}$ clearly implies ${\cal G}_{1}<_{fe}{\cal F}$, a
contradiction. $\Box$

In particular, all prime ultrafilters must be $\leq_{fe}$-minimal since they
are $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal; therefore
all the powers of ultrafilters from $\overline{P}\setminus P$ are also
$\leq_{fe}$-minimal.

The following well-known easy fact will be useful for constructing some
examples.

###### Proposition 3.5

Every sequence $\langle X_{n}:n<\omega\rangle$ of infinite subsets of
${\mathbb{N}}$ such that $X_{n+1}\subseteq X_{n}$ has a pseudointersection: an
infinite set $Y\subseteq{\mathbb{N}}$ such that $Y\setminus X_{n}$ is finite
for every $n<\omega$.

The next theorem shows that, in fact, there are many more ${\mathbb{N}}$-free
$\leq_{fe}$-minimal ultrafilters besides those from Theorem 3.4. We use
notation from Proposition 1.1.

###### Theorem 3.6

(a) Let $A_{1},A_{2},\dots,A_{m}$ be disjoint infinite subsets of $P$ and let
$k_{1},k_{2},\dots,k_{m},n_{1},n_{2},\dots,n_{m}\in{\mathbb{N}}$ be arbitrary.
Then, for some prime ultrafilters ${\cal P}_{1}\in\overline{A_{1}},{\cal
P}_{2}\in\overline{A_{2}},\dots,{\cal P}_{m}\in\overline{A_{m}}$, there is a
$\leq_{fe}$-minimal ultrafilter ${\cal F}\supseteq
F_{n_{1},n_{2},\dots,n_{m}}^{{\cal P}_{1}^{k_{1}},{\cal
P}_{2}^{k_{2}},\dots,{\cal P}_{m}^{k_{m}}}$.

(b) There is a $\leq_{fe}$-minimal ultrafilter in $\beta{\mathbb{N}}\setminus
L$.

Proof. (a) Let
$S=(A_{1}^{k_{1}})^{(n_{1})}(A_{2}^{k_{2}})^{(n_{2})}\dots(A_{m}^{k_{m}})^{(n_{m})}$,
and let $X_{n}=S\cap\bigcap_{1<l\leq n}(l{\mathbb{N}})^{c}$. By Proposition
3.5 there is an infinite set $Y\subseteq S$ containing, for any
$n\in{\mathbb{N}}\setminus\\{1\\}$, only finitely many elements divisible by
$n$. By Proposition 3.1 there is an infinite set $X\subseteq Y$ such that
$\overline{X}\cap(\beta{\mathbb{N}}\setminus{\mathbb{N}})\cdot(\beta{\mathbb{N}}\setminus{\mathbb{N}})=\emptyset$.
Let us show that any ultrafilter ${\cal F}\in\overline{X}\setminus X$ is
$\leq_{fe}$-minimal. Assume the opposite, that there is ${\cal G}<_{fe}{\cal
F}$. Since $Y\in{\cal F}$, ${\cal F}$ is ${\mathbb{N}}$-free, hence so is
${\cal G}$. There is $B\in{\cal G}$ such that $B\leq_{fe}X$, and $B$ must be
infinite. By Proposition 2.2 there is ${\cal H}\in\beta{\mathbb{N}}$ such that
$\overline{B}\cdot{\cal H}\subseteq\overline{X}$. Clearly ${\cal
H}\notin{\mathbb{N}}$, since otherwise ${\cal H}\cdot B\subseteq X\subseteq
Y$. But now, for any ${\cal W}\in\overline{B}\setminus B$, ${\cal W}\cdot{\cal
H}\in\overline{X}\cap(\beta{\mathbb{N}}\setminus{\mathbb{N}})\cdot(\beta{\mathbb{N}}\setminus{\mathbb{N}})$,
a contradiction.

(b) Let $X_{n}=L_{n}\uparrow\cap\bigcap_{1<l\leq n}(l{\mathbb{N}})^{c}$; by
Proposition 3.5 there is a pseudointersection $Y$ of all $X_{n}$. Any
ultrafilter ${\cal F}\in\overline{Y}\setminus Y$ is outside $L$ and
${\mathbb{N}}$-free. By Proposition 3.1 there is $X\subseteq Y$ such that all
${\cal F}\in\overline{X}\setminus X$ are $\mid_{M}$-minimal. As in (a) one
shows that any such ${\cal F}$ is also $\leq_{fe}$-minimal. $\Box$

Lemma 2.1 from [11] says that if, for some $A\in{\cal F}$, $f(n)\mid n$ for
all $n\in A$, then $\mathaccent 869{f}({\cal F})\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt{\cal F}$. Theorem 3.6 shows that no such result
holds for $\leq_{fe}$ (and consequently for $\mid_{M}$): if ${\cal
F}\in\overline{L_{n}}$ ($n>1$) is $\leq_{fe}$-minimal, any function such that
$f(n)\mid n$ and $f(n)\neq n$ for all $n\in A$ (for some $A\in{\cal F}$) would
map ${\cal F}$ to an ultrafilter ${\cal G}\in\overline{L_{m}}$ for some $m<n$,
but ${\cal G}\not\leq_{fe}{\cal F}$.

An ultrafilter ${\cal F}$ on ${\mathbb{N}}$ is selective (Ramsey) if, for
every family $\\{A_{n}:n<\omega\\}$ of disjoint subsets of ${\mathbb{N}}$ not
in ${\cal F}$, there is a $B\in{\cal F}$ such that $|B\cap A_{n}|\leq 1$ for
every $n\in\omega$. The existence of such ultrafilters can not be proved in
ZFC, but if they exist, then there are also ${\mathbb{N}}$-free selective
ultrafilters: if $X$ is a pseudointersection of sets $(n{\mathbb{N}})^{c}$ for
$n>1$, then any bijection $f:{\mathbb{N}}\rightarrow X$ maps any selective
ultrafilter to an ${\mathbb{N}}$-free selective ultrafilter.

###### Theorem 3.7

For any ${\mathbb{N}}$-free selective ultrafilter ${\cal F}$ there is no
${\cal G}\in L$ such that ${\cal G}<_{fe}{\cal F}$. In particular, if ${\cal
F}\in L$ then ${\cal F}$ is $\leq_{fe}$-minimal.

Proof. $1^{\circ}$ ${\cal F}\in\overline{L_{n}}$ for some $n\in\omega$. Assume
there is ${\cal G}<_{fe}{\cal F}$. ${\cal G}$ must belong to
$\overline{L_{m}}$ for some $m<n$. Let $k=n-m$ and let $T$ be the set of all
$k$-tuples $(i_{1},i_{2},\dots,i_{k})$ such that $1\leq
i_{1}<i_{2}<\dots<i_{k}\leq n$. Fix a $k$-tuple $(i_{1},i_{2},\dots,i_{k})\in
T$; for every $k$-tuple $p_{1}\leq p_{2}\leq\dots\leq p_{k}$ of primes, let

$\displaystyle A_{(i_{1},i_{2},\dots,i_{k})}^{(p_{1},p_{2},\dots,p_{k})}$
$\displaystyle=$ $\displaystyle\\{q_{1}q_{2}\dots
q_{n}:q_{1},q_{2},\dots,q_{n}\in P\land q_{1}\leq q_{2}\leq\dots\leq
q_{n}\land$ $\displaystyle q_{i_{j}}=p_{j}\mbox{ for }j=1,2,\dots,k\\}.$

None of the sets $A_{(i_{1},i_{2},\dots,i_{k})}^{(p_{1},p_{2},\dots,p_{k})}$
is in ${\cal F}$, because $p_{1}p_{2}\dots p_{k}{\mathbb{N}}\notin{\cal F}$.
Since they are disjoint (for different $(p_{1},p_{2},\dots,p_{k})$ and the
same $(i_{1},i_{2},\dots,i_{k})$) and ${\cal F}$ is selective, there is
$B_{(i_{1},i_{2},\dots,i_{k})}\in{\cal F}$ such that
$|B_{(i_{1},i_{2},\dots,i_{k})}\cap
A_{(i_{1},i_{2},\dots,i_{k})}^{(p_{1},p_{2},\dots,p_{k})}|\leq 1$ for all
$k$-tuples $p_{1}\leq p_{2}\leq\dots\leq p_{k}$. Hence

$B:=\bigcap\\{B_{(i_{1},i_{2},\dots,i_{k})}:(i_{1},i_{2},\dots,i_{k})\in
T\\}\in{\cal F}$

as well. Let $C\in{\cal G}$ be such that $C\leq_{fe}B$; we can assume that
$C\subseteq L_{m}$. However, if $r=|T|+1$, for any $F\in[C]^{r}$ there can be
no $s\in{\mathbb{N}}$ such that $sF\subseteq B$. Namely, if $s=p_{1}p_{2}\dots
p_{k}$ where $p_{1}\leq p_{2}\leq\dots\leq p_{k}$, then two of the elements of
$sF$ would belong to the same
$A_{(i_{1},i_{2},\dots,i_{k})}^{(p_{1},p_{2},\dots,p_{k})}$, but $B$
intersects each such set in at most one element; a contradiction.

$2^{\circ}$ ${\cal F}\notin L$. Since ${\cal F}$ is selective, there is
$B\in{\cal F}$ such that $|B\cap L_{n}|\leq 1$ for all $n$. Suppose there are
$m\in{\mathbb{N}}$ and ${\cal G}\in\overline{L_{m}}$ such that ${\cal
G}<_{fe}{\cal F}$. Let $C\in{\cal G}$ be such that $C\leq_{fe}B$; again we
assume that $C\subseteq L_{m}$. But for any $F\in[C]^{2}$ and any
$s\in{\mathbb{N}}$, the elements of $sF$ must belong to the same level
$L_{n}$, so $sF\subseteq B$ is not possible. $\Box$

##  4 Maximal ultrafilters

###### Proposition 4.1

(a) $K(\beta{\mathbb{N}},\cdot)$ is the set of $\mid_{M}$-maximal
ultrafilters.

(b) $\overline{K(\beta{\mathbb{N}},\cdot)}$ is the set of $\leq_{fe}$-maximal
ultrafilters.

(c) $K(\beta{\mathbb{N}},\cdot)$ is not topologically closed.

(a) is [10], Theorem 4.1(d), (b) is a special case of [7], Corollary 4.10 and
(c) is [4], Corollary 8.25.

Since ${\cal U}$ has the finite intersection property, there is a $\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-greatest class that we denote
$MAX$.

###### Proposition 4.2

([12], Lemma 4.6) For every ${\cal F}\in\beta{\mathbb{N}}$, ${\cal F}\in MAX$
if and only if $n\mid{\cal F}$ for all $n\in{\mathbb{N}}$.

The first two facts from the next lemma are a reformulation of Exercise 4.3.6
from [4].

###### Lemma 4.3

(a) $MAX$ is a topologically closed set.

(b) $MAX$ is a $\cdot$-ideal of $\beta{\mathbb{N}}$.

(c) $MAX$ is closed for $+$.

Proof. (a) $MAX=\bigcap_{A\in{\cal U}}\overline{A}$, an intersection of closed
sets.

(b) Let ${\cal F}\in MAX$ and ${\cal G}\in\beta{\mathbb{N}}$. Then ${\cal
F}\mid_{R}{\cal F}\cdot{\cal G}$, so by Theorem 2.8, ${\cal F}\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt{\cal F}\cdot{\cal G}$. Hence
${\cal F}\cdot{\cal G}\in MAX$ as well. That ${\cal G}\cdot{\cal F}\in MAX$ is
proved analogously.

(c) Let ${\cal F},{\cal G}\in MAX$. By Proposition 4.2, it suffices to show
that $m\mid{\cal F}+{\cal G}$ for all $m\in{\mathbb{N}}$. But ${\cal F}=m{\cal
F}_{1}$ and ${\cal G}=m{\cal G}_{1}$ for some ${\cal F}_{1},{\cal
G}_{1}\in\beta{\mathbb{N}}$, so ${\cal F}+{\cal G}=m({\cal F}_{1}+{\cal
G}_{1})$ (distributivity holds in $(\beta{\mathbb{N}},\cdot)$ for multiplying
by $m\in{\mathbb{N}}$). $\Box$

###### Example 4.4

$\beta{\mathbb{N}}\setminus MAX$ is not $\cdot$-closed; in other words we can
find ${\cal F},{\cal G}\notin MAX$ such that ${\cal F}\cdot{\cal G}\in MAX$.
Let $P=Q_{1}\cup Q_{2}$ be a partition of $P$ into two infinite sets,
$Q_{1}=\\{p_{n}:n\in{\mathbb{N}}\\}$ and $Q_{2}=\\{q_{n}:n\in{\mathbb{N}}\\}$.
Let $[{\cal F}]$ and $[{\cal G}]$ be the smallest upper $\hskip
2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-bounds of
$\langle\prod_{i=1}^{n}p_{i}^{n}:n<\omega\rangle$ and
$\langle\prod_{i=1}^{n}q_{i}^{n}:n<\omega\rangle$; here we use Proposition
2.12(a) and it is irrelevant which representatives of the
$=_{\sim}$-equivalence classes we take. Then ${\cal F},{\cal G}\notin MAX$
because $q_{1}\mathchar 13613\relax{\cal F}$ and $p_{1}\mathchar
13613\relax{\cal G}$. However, $p^{i}\mid{\cal F}\cdot{\cal G}$ for all $p\in
P$ and all $i\in{\mathbb{N}}$, so ${\cal F}\cdot{\cal G}\in MAX$ by
Proposition 4.2.

###### Lemma 4.5

$\overline{K(\beta{\mathbb{N}},\cdot)}\subseteq MAX$.

Proof. Lemma 4.3(b) implies that $K(\beta{\mathbb{N}},\cdot)\subseteq MAX$, so
Lemma 4.3(a) implies $\overline{K(\beta{\mathbb{N}},\cdot)}\subseteq MAX$.
$\Box$

The strict inclusion will be proved in Corollary 5.4.

An ultrafilter ${\cal F}\in\beta{\mathbb{N}}$ is called weakly summable if
every $A\in{\cal F}$ is an A-IP set. In [4], Theorem 12.17, it was shown that,
if we denote $\Gamma={\rm cl}\\{{\cal F}\in\beta{\mathbb{N}}:{\cal F}+{\cal
F}={\cal F}\\}$, then $\Gamma$ is exactly the set of weakly summable
ultrafilters. Thus every $+$-idempotent ultrafilter is weakly summable; the
reverse is false by [4], Theorem 8.24 or Theorem 12.22. The following
proposition is (a) [4], Theorem 5.20 or [1], Lemma 4.1, and (b) [4], Theorem
12.23.

###### Proposition 4.6

(a) $\Gamma$ is a left $\cdot$-ideal of $\beta{\mathbb{N}}$;

(b) $\Gamma$ is not $+$-closed.

###### Theorem 4.7

$\Gamma\subset MAX$.

Proof. By [4], Theorem 5.19.1 every $+$-idempotent is divisible by all
$n\in{\mathbb{N}}$, so Proposition 4.2 implies that the set of all
$+$-idempotents is included in $MAX$. By Lemma 4.3(a) MAX also contains its
closure $\Gamma$.

However, $\Gamma\neq MAX$. One way to see this is the fact that $MAX$ is
$+$-closed and $\Gamma$ is not. The other way is described in Example 4.8
below. $\Box$

###### Example 4.8

Let $\langle a_{n}:n\in{\mathbb{N}}\rangle$ be a sequence in ${\mathbb{N}}$
defined recursively: $a_{1}=1$ and, for all $n>1$, $n\mid a_{n}$ and
$a_{n}>\sum_{i=1}^{n-1}a_{i}$. Then the set $A=\\{a_{n}:n\in{\mathbb{N}}\\}$
intersects $m{\mathbb{N}}$ for every $m\in{\mathbb{N}}$. Thus ${\cal
U}\cup\\{A\\}$ has the finite intersection property, and there is ${\cal F}\in
MAX$ such that $A\in{\cal F}$. But $A$ is not an A-IP set, so there is no
$+$-idempotent containing $A$, and ${\cal F}\notin\Gamma$.

The set ${\mathbb{H}}=\bigcap_{n\in{\mathbb{N}}}\overline{2^{n}{\mathbb{N}}}$
plays an important role in the investigation of the algebraic structure of
$\beta{\mathbb{N}}$, see Chapter 6 of [4]. It is obvious that
$MAX\subset{\mathbb{H}}$, and by [4], Corollary 7.26, $MAX$ is algebraically
and topologically isomorphic to ${\mathbb{H}}$. The results above give us the
following image, showing the place of $MAX$ among these important families of
ultrafilters.

Recall that ${\cal U}_{N}$ is the family of all ${\mathbb{N}}$-free sets from
${\cal U}$, and ${\cal V}_{N}=\\{A^{c}:A\in{\cal U}_{N}\\}$. In [13] it was
shown that ${\cal
U}_{N}\cup\\{(n{\mathbb{N}})^{c}:n\in{\mathbb{N}}\setminus\\{1\\}\\}$ has the
finite intersection property, so there is a $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$-greatest ${\mathbb{N}}$-free class that we denote
$NMAX$. $A\subseteq{\mathbb{N}}$ is a strong antichain if every two elements
$m,n\in A$ are mutually prime.

###### Proposition 4.9

([13], Lemma 5.2) $S\subseteq{\mathbb{N}}$ is an ${\mathbb{N}}$-free set if
and only if it contains an infinite strong antichain.

###### Lemma 4.10

(a) $NMAX$ is a topologically closed set.

(b) $NMAX$ is a $\cdot$-ideal of $\beta{\mathbb{N}}_{F}:=\\{{\cal
G}\in\beta{\mathbb{N}}:{\cal G}\mbox{ is }{\mathbb{N}}$-free$\\}$.

Proof. (a) $NMAX=\bigcap_{A\in{\cal
U}_{N}}\overline{A}\cap\bigcap_{n\in{\mathbb{N}}\setminus\\{1\\}}\overline{(n{\mathbb{N}})^{c}}$,
an intersection of closed sets.

(b) Let ${\cal F}\in NMAX$ and ${\cal G}\in\beta{\mathbb{N}}_{F}$. Assume
$p\mid{\cal F}\cdot{\cal G}$ for some $p\in P$. Then $p\mid{\cal F}$ or
$p\mid{\cal G}$, a contradiction. So ${\cal F}\cdot{\cal G}$ is
${\mathbb{N}}$-free. But it is $\mid_{M}$-divisible by ${\cal F}$ and hence
$\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-divisible by all
${\mathbb{N}}$-free ultrafilters, so ${\cal F}\cdot{\cal G}\in NMAX$. $\Box$

$NMAX$ is clearly not an ideal in $\beta{\mathbb{N}}$, since for ${\cal F}\in
NMAX$ and $n\in{\mathbb{N}}\setminus\\{1\\}$ we have $n\cdot{\cal F}\notin
NMAX$. Also, it is not $+$-closed because a sum of two ultrafilters ${\cal
F},{\cal G}$ not divisible by 2 must be divisible by 2: if we denote by
$2{\mathbb{N}}+1$ the set of odd numbers, then
$\\{n\in{\mathbb{N}}:(2{\mathbb{N}}+1)-n\in{\cal G}\\}=2{\mathbb{N}}\in{\cal
F}$, so $2{\mathbb{N}}\in{\cal F}+{\cal G}$.

##  5 Large subsets of ${\mathbb{N}}$

To definitions of various largeness properties from Section 1 we now add a few
more, derived from our divisibility relation $\hskip 2.84526pt\mathaccent
869{\mid}\hskip 2.84526pt$. For each of them we also give some immediate
equivalents. A set $A\subseteq{\mathbb{N}}$ is:

-MAX-set if it a member of some ultrafilter in $MAX$, if and only if it is not contained in a set in ${\cal V}$, if and only if $A\downarrow={\mathbb{N}}$;

-NMAX-set if it is a member of an ultrafilter divisible by all ${\mathbb{N}}$-free ultrafilters, if and only if it is not contained in a set in ${\cal V}_{N}$;

-MAX*-set if it intersects every MAX-set, if and only if it is a member of every ultrafilter in $MAX$, if and only if it contains some set in ${\cal U}$;

-NMAX*-set if it intersects every NMAX-set, if and only if it is a member of every ultrafilter divisible by all ${\mathbb{N}}$-free ultrafilters, if and only if it contains some set in ${\cal U}_{N}$.

We will obtain the following situation:

Some of the implications are trivial: MAX$\Rightarrow$NMAX,
NMAX$\not\Rightarrow$MAX, NMAX* $\Rightarrow$MAX*,
A-IP*$\Rightarrow$A-central*$\Rightarrow$A-central$\Rightarrow$A-IP and
M-central*$\Rightarrow$M-central$\Rightarrow$M-IP. Proofs of most of the other
known implications can be found in [1]: A-central*$\Rightarrow$M-thick is
5.11, A-thick$\Rightarrow$A-central and M-thick$\Rightarrow$M-central are
5.10, A-central$\Rightarrow$A-pcws is 5.5, M-central$\Rightarrow$M-pcws is
5.7, M-central$\not\Rightarrow$A-IP is 5.16 and A-central$\not\Rightarrow$M-IP
is Proposition 5.14. A-central*$\not\Rightarrow$A-IP*,
M-central*$\Rightarrow$A-central and A-pcws$\Rightarrow$A-J (as well as
M-pcws$\Rightarrow$M-J) are, respectively, Theorem 16.8, Theorem 16.26.1 and
Theorem 14.8.3 from [4]. For sake of completeness, we include the proof of
A-IP$\not\Rightarrow$A-pcws in the theorem below.

Now let us consider the remaining implications, involving sets derived from
maximal divisibility classes.

###### Theorem 5.1

For subsets of ${\mathbb{N}}$ the following implications hold:

(a) A-IP$\Rightarrow$MAX but MAX$\not\Rightarrow$A-IP.

(b) M-IP$\not\Rightarrow$NMAX.

(c) M-J$\Rightarrow$MAX but MAX$\not\Rightarrow$M-J.

(d) A-pcws$\not\Rightarrow$MAX.

(e) A-pcws$\Rightarrow$NMAX.

(f) A-IP$\not\Rightarrow$A-pcws.

Proof. (a) Every A-IP set is contained in a $+$-idempotent, which is by Lemma
4.7 an element of $MAX$. On the other hand, in Example 4.8 we actually
constructed a set belonging to an ultrafilter in $MAX$ which is not an A-IP
set.

(b) Let $P=\\{p_{n}:n<\omega\\}\cup Q$ be a partition of the set $P$ of primes
into two infinite subsets. If $A=FP(\langle p_{n}\rangle)$, then $A$ is an
M-IP set. However, by Proposition 4.9 $Q\uparrow$ is a set in ${\cal U}_{N}$
and $A$ is contained in its complement, which is in ${\cal V}_{N}$.

(c) Assume the opposite, that there is a M-J-set $A$ such that $A\subseteq B$
for some $B\in{\cal V}$. Any number divisible by any element of the complement
$B^{c}$ also belongs to $B^{c}$. Hence, if we define a function
$f:{\mathbb{N}}\rightarrow B^{c}$ arbitrarily, $a\cdot\prod_{i\in H}f(i)\notin
B$ for all $a\in{\mathbb{N}}$, $H\in[{\mathbb{N}}]^{<\aleph_{0}}$, a
contradiction.

To prove that the reverse implication does not hold, let
$P=\\{p_{n}:n\in{\mathbb{N}}\\}$ be the increasing enumeration of the primes
and

$A=\\{p_{i_{1}}^{k}p_{i_{2}}^{k}\dots p_{i_{m}}^{k}:m,k\in{\mathbb{N}}\land
i_{1}<i_{2}<\dots<i_{m}\\}.$

Then $A\downarrow={\mathbb{N}}$, so $A$ is a MAX-set. To prove that it is not
M-J, let functions $f,g:{\mathbb{N}}\rightarrow{\mathbb{N}}$ be defined by
$f(n)=p_{2n}^{2}p_{2n+1}$ and $g(n)=p_{2n}p_{2n+1}^{2}$. Now, if
$H\in[{\mathbb{N}}]^{<\aleph_{0}}$, $a\in{\mathbb{N}}$, $x_{f}=a\prod_{i\in
H}f(i)$ and $x_{g}=a\prod_{i\in H}g(i)$, let $j=\min H$, and
$p_{2j}^{k}\parallel a$ and $p_{2j+1}^{l}\parallel a$ (i.e. $k,l$ are the
largest numbers for which these divisibilities hold). Then
$x_{f}=p_{2j}^{k+2}p_{2j+1}^{l+1}q_{f}$ and
$x_{g}=p_{2j}^{k+1}p_{2j+1}^{l+2}q_{g}$ for some $q_{f},q_{g}$ not divisible
by $p_{2j}$ or $p_{2j+1}$. But it is impossible that both $k+2=l+1$ and
$k+1=l+2$ hold, so $x_{f}$ and $x_{g}$ can not both belong to $A$.

(d) The set $2{\mathbb{N}}+1$ of odd numbers is A-piecewise syndetic, but is
clearly not a MAX-set.

(e) Let $A$ be a set not in NMAX; we may assume $A\in{\cal V}_{N}$. Suppose
that $A$ is A-pcws. Then there is finite
$F=\\{t_{1},t_{2},\dots,t_{f}\\}\subset{\mathbb{N}}$ such that, for
$B=\bigcup_{t\in F}(A-t)$, the set $\\{B-k:k\in{\mathbb{N}}\\}$ has the finite
intersection property. Since $A^{c}$ is ${\mathbb{N}}$-free, by Proposition
4.9 there are mutually prime $n_{1},n_{2},\dots,n_{f}\in A^{c}$. Let
$n=n_{1}n_{2}\dots n_{f}$; then there is $m\in\bigcap_{1\leq i\leq n}(B-i)$.
This means that $\\{m+1,m+2,\dots,m+n\\}\subseteq B$. By the Chinese remainder
theorem, the system

$x\equiv-t_{j}(\mbox{mod }n_{j}),\;\;j=1,2,\dots,f$

has a solution $x\in\\{m+1,m+2,\dots,m+n\\}$. Then $n_{j}\mid x+t_{j}$ and
$n_{j}\in A^{c}$ imply $x+t_{j}\in A^{c}$, since $A^{c}\in{\cal U}_{N}$.
However, since $x\in B$, $x+t_{j}\in A$ for some $1\leq j\leq f$; a
contradiction.

(f) Let $\langle a_{n}:n\in{\mathbb{N}}\rangle$ be a sequence in
${\mathbb{N}}$ such that $a_{n}>n+\sum_{i=1}^{n-1}a_{i}$ for $n>1$. Then
$A:=FS(\langle a_{n}\rangle)$ is an A-IP set. Assume that, for some
$n\in{\mathbb{N}}$ and $B=\bigcup_{t\leq n}(A-t)$, the family
$\\{B-k:k\in{\mathbb{N}}\\}$ has the finite intersection property. Then there
is $x\in\bigcap_{k\leq 2^{n-1}}(B-k)$, which means that $B$ contains $2^{n-1}$
successive numbers. But each of them is a nonempty sum of some of $a_{n}$
minus some $t$. Hence at least $n$ elements of $\\{a_{n}:n\in{\mathbb{N}}\\}$
must be ”used” to get these elements, say
$a_{i_{1}}<a_{i_{2}}<\dots<a_{i_{n}}<\dots$ If $y$ is the smallest among them
which contains $a_{i_{m}}$ for some $m\geq n$, say
$y=a_{i_{m}}+a_{j_{1}}+\dots+a_{j_{l}}-t$ for some $t\leq n$, then $y-1$ can
not belong to $B$, since any element of that set not using $a_{i_{k}}$ for
$k\geq m$ is no larger than $a_{1}+a_{2}+\dots+a_{i_{m}-1}<a_{i_{m}}-i_{m}\leq
y-1$. $\Box$

###### Theorem 5.2

For subsets of ${\mathbb{N}}$ the following implications hold:

(a) MAX*$\Rightarrow$M-central* but M-central*$\not\Rightarrow$MAX*.

(b) MAX*$\Rightarrow$A-IP* but A-IP*$\not\Rightarrow$MAX*.

(c) A-thick$\not\Rightarrow$MAX* and MAX*$\not\Rightarrow$A-thick.

(d) NMAX*$\Rightarrow$A-thick.

Proof. (a) We use Remark 1.2. M-central$\Rightarrow$MAX directly implies that
MAX*$\Rightarrow$M-central*. If M-central*$\Rightarrow$MAX* were true, by
duality we would have MAX$\Rightarrow$M-central, which is not true.

(b) By Remark 1.2, MAX*$\Rightarrow$A-IP* follows from Theorem 5.1(a). We can
use duality to prove A-IP*$\not\Rightarrow$MAX* as well, but here is a direct
proof. Since $\Gamma$ is not $+$-closed (Proposition 4.6), there are ${\cal
F},{\cal G}\in\Gamma$ such that ${\cal F}+{\cal G}\notin\Gamma$. By Lemmas 4.7
and 4.3(c), ${\cal F}+{\cal G}\in MAX$. $\Gamma$ is topologically closed, so
there is $A\in{\cal F}+{\cal G}$ such that $\overline{A}\cap\Gamma=\emptyset$.
This means that $A^{c}$ is an A-IP* set, but not a MAX*-set.

(c) We construct finite sets $F_{n}$ and elements $a_{n}\in{\mathbb{N}}$ by
recursion on $n$. Let $F_{0}=\emptyset$ and $a_{0}=1$. For $n>1$, let $F_{n}$
be any set of $n$ successive natural numbers larger than $a_{n-1}$, and let
$a_{n}$ be larger than all elements of $F_{n}$ and divisible by $n$. Then the
set $\bigcup_{n<\omega}F_{n}$ is A-thick but not in MAX* (it is disjoint from
$\\{a_{n}:n<\omega\\}$, so it does not contain a set in ${\cal U}$).

To show that MAX*$\not\Rightarrow$A-thick, note just that $2{\mathbb{N}}$ is a
MAX*-set but does not even contain an interval of length 2.

(d) If $X$ is a NMAX*-set, we can assume that $X\in{\cal U}_{N}$. By
Proposition 4.9 there is an infinite strong antichain
$A=\\{a_{m}:m\in{\mathbb{N}}\\}\subseteq X$. Let $n\in{\mathbb{N}}$ be given;
we are looking for an interval of length $n$ in $X$. Consider the system of
congruences

$x\equiv-m(\mbox{mod }a_{m}),\;\;m=1,2,\dots,n.$

Since $a_{m}$ are pairwise mutually prime, this system has a solution
$x\in{\mathbb{N}}$ by the Chinese remainder theorem. Then, for each
$m\in\\{1,2,\dots,n\\}$, $x+m$ is divisible by $a_{m}$, so $a_{m}\in X$
implies $x+m\in X$. $\Box$

There are several direct corollaries of the results above.

###### Corollary 5.3

(a) A-pcws$\not\Rightarrow$A-IP.

(b) MAX$\not\Rightarrow$A-pcws.

(c) MAX$\not\Rightarrow$M-pcws.

(d) M-central*$\not\Rightarrow$A-thick.

(e) MAX$\not\Rightarrow$M-IP.

(f) M-IP$\not\Rightarrow$M-J.

###### Corollary 5.4

(a) $\overline{K(\beta{\mathbb{N}},\cdot)}\subset MAX$.

(b) $K(\beta{\mathbb{N}},+)\not\subseteq MAX$.

Proof. (a) By Lemma 4.5 we only need to show that the inclusion is strict. By
Corollary 5.3(c) there is a set $A$ belonging to an ultrafilter ${\cal F}\in
MAX$ which is not M-piecewise syndetic. This means that $A$ is not an element
of any ultrafilter in $K(\beta{\mathbb{N}},\cdot)$. Since $\overline{A}$ is an
open set, it is disjoint from $\overline{K(\beta{\mathbb{N}},\cdot)}$, so
${\cal F}\notin\overline{K(\beta{\mathbb{N}},\cdot)}$.

(b) By Theorem 5.1(d) there is $A\subseteq{\mathbb{N}}$ which is A-piecewise
syndetic set but not a MAX-set. Then there is an ultrafilter ${\cal
F}\in\overline{A}\cap K(\beta{\mathbb{N}},+)$, but there is no ultrafilter
from $MAX$ in $\overline{A}$. $\Box$

##  6 Open problems and remarks

The $\leq_{fe}$ relation for sets has a nice nonstandard characterization: by
(an analogon of) [3], Proposition 4.3, $A\leq_{fe}B$ holds if and only if
$x\cdot A\subseteq{{}^{*}B}$ for some $x\in{{}^{*}{\mathbb{N}}}$. It would be
very useful if such a result could be found for the ultrafilter relation, but
we were not able to find it so far.

###### Problem 6.1

Find a nice nonstandard characterization of $\leq_{fe}$ for ultrafilters.

The only examples of ultrafilters such that ${\cal F}\leq_{fe}{\cal G}$ but
${\cal F}\mathchar 13613\relax_{M}{\cal G}$ we found so far are above $L$. So
it is natural to ask the following.

###### Problem 6.2

Does ${\cal F}\leq_{fe}{\cal G}$ imply ${\cal F}\mid_{M}{\cal G}$ for ${\cal
F},{\cal G}\in L$?

###### Problem 6.3

(a) Find a nice characterization of $\leq_{fe}$-minimal ultrafilters.

(b) Is every $\mid_{M}$-minimal ultrafilter also $\leq_{fe}$-minimal?

If the answer to (b) is yes, then a solution to (a) would also provide a nice
characterization of $\mid_{M}$-minimal ultrafilters.

In Section 2 we considered upper bounds of chains in
$(\beta{\mathbb{N}}/=_{fe},\leq_{fe})$, but did not resolve the analogous
questions for lower bounds.

###### Problem 6.4

Does every $\leq_{fe}$-decreasing chain in $\beta{\mathbb{N}}/=_{fe}$ has a
lower bound? If yes, does it always have a greatest lower bound?

Concerning Corollary 5.4 and Theorem 4.7 we have the following question.

###### Problem 6.5

(a) Is MAX the $\cdot$-ideal of $\beta{\mathbb{N}}$ generated by $\Gamma$?

(b) Is MAX the $+$-closure of $\Gamma$ or the $+$-closure of
$\overline{K(\beta{\mathbb{N}},\cdot)}$?

Finally, the last problem deals with a missing implication on the diagram from
Section 5.

###### Problem 6.6

Does A-J$\Rightarrow$NMAX?

The author gratefully acknowledges financial support of the Science Fund of
the Republic of Serbia (call PROMIS, project CLOUDS, grant no. 6062228) and
Ministry of Education, Science and Technological Development of the Republic
of Serbia (grant no. 451-03-9/2021-14/200125).

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