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arXiv:math/0701198v2 [math.LO] 12 May 2007
Generating self-map monoids of infinite sets∗
Zachary Mesyan
February 2, 2008
Abstract
Let Ω be a countably infinite set, S= Sym(Ω) the group of permutations of Ω, and
E= Self(Ω) the monoid of self-maps of Ω. Given two subgroups G1, G2⊆S, let us
write G1≈SG2if there exists a finite subset U⊆Ssuch that the groups generated
by G1∪Uand G2∪Uare equal. Bergman and Shelah showed that the subgroups
which are closed in the function topology on Sfall into exactly four equivalence classes
with respect to ≈S. Letting ≈denote the obvious analog of ≈Sfor submonoids of E,
we prove an analogous result for a certain class of submonoids of E, from which the
theorem for groups can be recovered. Along the way, we show that given two subgroups
G1, G2⊆Swhich are closed in the function topology on S, we have G1≈SG2if and
only if G1≈G2(as submonoids of E), and that clS(G)≈clE(G) for every subgroup
G⊆S(where clS(G) denotes the closure of Gin the function topology in Sand clE(G)
its closure in the function topology in E).
1 Introduction
Let Ω be a countably infinite set, let S= Sym(Ω) denote the group of all permutations
of Ω, and let E= Self(Ω) denote the monoid of self-maps of Ω. (Here “E” stands for
“endomap.”) Given two subgroups G1, G2⊆S, let us write G1≈SG2if there exists a finite
subset U⊆Ssuch that the group generated by G1∪Uis equal to the group generated by
G2∪U. In [4] Bergman and Shelah show that the subgroups of Sthat are closed in the
function topology on Sfall into exactly four equivalence classes with respect to the above
equivalence relation. (A subbasis of open sets in the function topology on Sis given by the
sets {f∈S: (α)f=β}(α, β ∈Ω).This topology is discussed in more detail in Sections 8
and 11.) In this note we investigate properties of an analogous equivalence relation ≈defined
for monoids. Two submonoids M1, M2⊆Ewill be considered equivalent if and only if there
exists a finite set U⊆Esuch that the monoid generated by M1∪Uis equal to the monoid
generated by M2∪U.
We will show that given two subgroups G1, G2⊆Sthat are closed in the function
topology on S, we have G1≈SG2if and only if G1≈G2(as submonoids of E). Writing
clS(G) for the closure of the subgroup G⊆Sin the function topology in Sand clE(G) for
its closure in the function topology in E, we will show that clS(G)≈clE(G).
∗2000 Mathematics Subject Classification numbers: 20M20 (primary), 20B07 (secondary).
1
Our main goal will be to classify into equivalence classes certain closed submonoids of E.
In general, unlike the case of groups, the closed submonoids of Efall into infinitely many
equivalence classes. To see this, let Ω = ω, the set of natural numbers, and for each positive
integer nlet Mnbe the submonoid of Egenerated by all maps whose images are contained
in {0,1,...,n}. It is easy to see that such submonoids are closed in the function topology.
Now, let nbe a positive integer and consider a monoid word g1g2. . . gkin elements of E,
where at least one of the gi∈Mn. Then the image of Ω under g1g2. . . gkhas cardinality
at most n. Hence if for some finite subset U⊆Ewe have that Mn+1 ⊆ hMn∪Ui(the
monoid generated by Mn∪U), then all the elements of Mn+1 whose images have cardinality
n+ 1 must be in hUi. This is impossible, since there are uncountably many such elements.
Therefore, if n6=mare two positive integers, then Mn6≈ Mm.
Classifying all closed submonoids of Einto equivalence classes appears to be a very
difficult task, so we will for the most part focus on submonoids that have large stabilizers
(i.e., submonoids Msuch that for any finite set Σ ⊆Ω the pointwise stabilizer of Σ in Mis
≈M) and the property that the subset of a.e. injective maps is dense (with respect to the
function topology). Examples of submonoids that have large stabilizers and dense sets of a.e.
injective maps include subgroups of S⊆E, as well as their closures in the function topology
in E. Another class of examples of monoids with these properties arises from preorders on
Ω. Given such a preorder ρ, let E(ρ) denote the submonoid of Econsisting of all maps f
such that for all α∈Ω, one has (α, (α)f)∈ρ. We will show that the submonoids E(ρ) have
large stabilizers and dense sets of a.e. injective maps. (They are also closed in the function
topology on E.)
The Bergman-Shelah theorem can be recovered from our classification of the submonoids
described above. A surprising aspect of this classification is that the submonoids in question
fall into five equivalence classes, rather than the four predicted by the Bergman-Shelah
theorem. Throughout, we will also give a number of examples demonstrating various unusual
features of our equivalence relation.
Conditions under which a submonoid M⊆Esatisfies M≈Ehave been considered
before. For instance, Howie, Ruˇskuc, and Higgins show in [8] that S≈E. (We give a shorter
proof of this fact below.) These authors and Mitchell also exhibit various submonoids that
are 6≈ Ein [7]. Related questions, but with other kinds of objects in place of E, are discussed
in [5], [7], and [11], as well as in papers referenced therein.
2 Chains
Let Ω be an arbitrary infinite set, and set E= Self (Ω), the monoid of self-maps of Ω.
Elements of Ewill be written to the right of their arguments. If U⊆Eis a subset, then we
will write hUito denote the submonoid generated by U. The cardinality of a set Γ will be
denoted by |Γ|. If Σ ⊆Ω and U⊆Eare subsets, let U{Σ}={f∈U: (Σ)f⊆Σ}.
Definition 1. Let Mbe a monoid that is not finitely generated. Then the cofinality c(M)of
Mis the least cardinal κsuch that Mcan be expressed as the union of an increasing chain
of κproper submonoids.
The main goal of this section is to show that c(E)>|Ω|, which will be needed later on.
2
This section is modeled on Sections 1 and 2 of [2], where analogous statements are proved
for the group of all permutations of Ω. Ring-theoretic analogs of these statements are proved
in [10].
Lemma 2. Let U⊆Eand Σ⊆Ωbe such that |Σ|=|Ω|and the set of self-maps of Σ
induced by U{Σ}is all of Self(Σ). Then E=gU h, for some g, h ∈E.
Proof. Let g∈Ebe a map that takes Ω bijectively to Σ, and let h∈Ebe a map whose
restriction to Σ is the right inverse of g. Then E=gU{Σ}h.
We will say that Σ ⊆Ω is a moiety if |Σ|=|Ω|=|Ω\Σ|. A moiety Σ ⊆Ω is called
full with respect to U⊆Eif the set of self-maps of Σ induced by members of U{Σ}is all of
Self(Σ). The following two results are modeled on group-theoretic results in [9].
Lemma 3 (cf. [2, Lemma 3]).Let (Ui)i∈Ibe any family of subsets of Esuch that Si∈IUi=E
and |I| ≤ |Ω|. Then Ωcontains a full moiety with respect to some Ui.
Proof. Since |Ω|is infinite and |I| ≤ |Ω|, we can write Ω as a union of disjoint moieties Σi,
i∈I. Suppose that there are no full moieties with respect to Uifor any i∈I. Then in
particular, Σiis not full with respect to Uifor any i∈I. Hence, for every i∈Ithere exists
a map fi∈Self(Σi) which is not the restriction to Σiof any member of (Ui){Σi}. Now, if
we take f∈Eto be the map whose restriction to each Σiis fi, then fis not in Uifor any
i∈I, contradicting Si∈IUi=E.
Proposition 4. c(E)>|Ω|.
Proof. Suppose that (Mi)i∈Iis a chain of submonoids of Esuch that Si∈IMi=Eand
|I| ≤ |Ω|. We will show that E=Mifor some i∈I.
By the preceding lemma, Ω contains a full moiety with respect to some Mi. Thus,
Lemma 2 implies that E=hMi∪ {g, h}i for some g, h ∈E. But, by the hypotheses on
(Mi)i∈I,Mi∪ {g, h} ⊆ Mjfor some j∈I, and hence E=hMi∪ {g, h}i ⊆ Mj, since Mjis a
submonoid.
This result is proved by a very different method in [3].
3 Equivalence classes
Throughout this note we will be primarily interested in submonoids of E= Self(Ω).
However, we begin this section with a definition applicable to submonoids of an arbitrary
monoid.
Definition 5. Let Mbe a monoid, κan infinite cardinal, and M1, M2submonoids of M.
We will write M14κ,M M2if there exists a subset U⊆Mof cardinality < κ such that
M1⊆ hM2∪Ui. If M14κ,M M2and M24κ,M M1,we will write M1≈κ,M M2, while if
M14κ,M M2and M264κ,M M1, we will write M1≺κ,M M2. The subscripts Mand κwill be
omitted when their values are clear from the context.
3
It is clear that 4κ,M is a preorder on submonoids of M, and hence ≈κ,M is an equivalence
relation. This equivalence relation and many of the results below are modeled on those in [4].
Ring-theoretic analogs of these results can be found in [11].
We record the following result for future use.
Theorem 6 (Sierpi´nski, cf. [12], [1], [7]).Every countable subset of Eis contained in a
subsemigroup generated by two elements of E.
Proposition 7. Let M1, M2⊆Ebe submonoids.
(i) M14ℵ0M2if and only if M14ℵ1M2(and hence M1≈ℵ0M2if and only if M1≈ℵ1M2).
(ii) M≈ℵ0Eif and only if M≈|Ω|+E(where |Ω|+is the successor cardinal of |Ω|).
Proof. (i) follows from Theorem 6. (ii) follows from Proposition 4. For, if M≈|Ω|+E, then
among subsets U⊆Eof cardinality ≤ |Ω|such that hM∪Ui=E, we can choose one of least
cardinality. Let us write U={fi:i∈ |U|}. Then the submonoids Mi=hM∪ {fj:j < i}i
(i∈ |U|) form a chain of ≤ |Ω|proper submonoids of E. If |U|were infinite, this chain would
have union E, contradicting Proposition 4. Hence Uis finite, and M≈ℵ0E.
Definition 8. Let E≤⊆Self(ω)denote the submonoid of all maps decreasing with respect
to the usual ordering of ω. Specifically, f∈E≤if and only if for all α∈ω,(α)f≤α
The following result will be our main tool for separating various equivalence classes of
submonoids of Self(Ω) throughout the paper.
Theorem 9. (i) Let κbe a regular infinite cardinal ≤ |Ω|and T⊆Self(Ω) a subset. If
|(α)T|< κ for all α∈Ω, then hTi ≺|Ω|+Self(Ω).
(ii) Let T⊆Self(ω)be a subset. If there exists a finite λsuch that |(α)T| ≤ λfor all
α∈ω, then hTi ≺ℵ0E≤.
Proof. (i) Let U⊆Self(Ω) be a subset of cardinality ≤ |Ω|. We will show that Self(Ω) 6⊆
hT∪Ui. Without loss of generality we may assume that 1 ∈T∩U. For all j∈ωand
u0,...,uj∈U, we define
(1) B(u0,...,uj) = {u0t0. . . ujtj:t0,...,tj∈T}.
Then the monoid hT∪Uican be written as the union of the sets B(u0,...,uj) (using the
assumption that 1 ∈T∩U).
Next, we show by induction on jthat for all α∈Ω, j∈ω, and u0,...,uj∈U, we have
|(α)B(u0,...,uj)|< κ. If j= 0, |(α)B(u0)|=|((α)u0)T|< κ, by our hypothesis on T. Now,
(α)B(u0,...,uj+1) = ((Σ)uj+1)T, where Σ = (α)B(u0,...,uj). Assuming inductively that
|Σ|< κ, and hence |(Σ)uj+1|< κ, the set (α)B(u0,...,uj+1) can be written as the union of
< κ sets of cardinality < κ. By the regularity of κ,|(α)B(u0,...,uj+1)|< κ.
Let us write Ω = Sj∈ωΩj, where the union is disjoint and each Ωjhas cardinality |Ω|.
Also, for each n∈ωlet hn: Ωn→Qn
j=0 Ube a surjection. By the previous paragraph,
there is a map f∈Self(Ω) such that for all α∈Ωj, (α)f∈Ω\(α)B(u0,...,uj), where
(u0,...,uj) = (α)hj. We conclude the proof by showing that f /∈ hT∪Ui.
4
Suppose that f∈ hT∪Ui. Then f∈B(u0,...,uj) for some j∈ωand u0,...,uj∈U.
Let α∈Ωjbe such that (u0,...,uj) = (α)hj. Then (α)f∈(α)B(u0,...,uj), contradicting
our definition of f. Hence f /∈ hT∪Ui.
(ii) Let T⊆Self(ω) and λ∈ωbe such that |(α)T| ≤ λfor all α∈ω, and let U⊆Self(ω)
be a finite subset. We note that for all α∈ω,|(α)(T∪U)| ≤ λ+|U|<ℵ0. Hence T∪U
satisfies our hypotheses on T. Therefore, to show that E≤6⊆ hT∪Uifor all Tand U, it
suffices to show that E≤6⊆ hTifor all Tas in the statement.
Let f∈E≤be any element such that
(2) (λj+ 1)f /∈(λj+ 1)Tjfor all j≥1,
where Tj={t0. . . tj−1:t0,...,tj−1∈T}. Such a map exists, since |(λj+ 1)Tj| ≤ λj, while
there are λj+ 1 possible values for (λj+ 1)f. Then f /∈Tjfor all j≥1, and hence f /∈ hTi.
It remains to be shown that hTi4ℵ0E≤.
We can find a set ω′⊆ωand a collection of disjoint sets ∆α⊆ω(α∈ω′) such that
|ω′|=|ω|,|∆α|=λ, and ∆α⊆(α)E≤. (Specifically, we can take ω′={λ, 2λ, 3λ, . . . }and
∆iλ ={(i−1)λ, (i−1)λ+ 1,(i−1)λ+ 2,...,iλ−1}for i≥1.) Now, let g∈Self(ω) be
an injective map from ωto ω′, and let h∈Self(ω) be a map that takes each ∆(α)g(α∈ω′)
onto (α)T. Then T⊆gE≤h, and hence hTi ⊆ hE≤∪ {g, h}i.
While we will not use the following two results in the future, they are of interest in their
own right.
Corollary 10. Let κbe a regular infinite cardinal ≤ |Ω|and {Ti}i∈κsubsets of E= Self(Ω)
such that |(α)Ti|< κ for all α∈Ωand i∈κ. Then hSi∈κTii ≺κE.
Proof. Suppose that hSi∈κTii ≈κE. Then there is a set U⊆Eof cardinality < κ such that
E=hSi∈κTi∪Ui. For each j∈κlet Nj=hSi≤jTi∪Ui. Then hSi∈κTi∪Ui=Si∈κNi,
and so E=Si∈κNi. Hence, by Proposition 4, E=Nnfor some n∈κ. For each α∈Ω, set
|(α)Si≤nTi|=λα. Then each λα< κ, and for all α∈Ω, we have that |(α)Si≤nTi∪U| ≤
λα+|U|< κ. Thus, E=Nncontradicts Theorem 9.
In view of Proposition 7(ii), ≺κcan be replaced with ≺|Ω|+in the above corollary. This
result can be viewed as a generalization to arbitrary Ω of [7, Corollary 2.2]. In [7] a subset
T⊆Self(ω) is said to be dominated (by U) if there exists a countable subset U⊆Self(ω)
having the property that for each f∈Tthere exists h∈Usuch that (α)f≤(α)hfor all
α∈ω. Rewritten using our notation, Corollary 2.2 states that if T⊆Self(ω) is a dominated
subset, then hTi ≺ℵ1Self(ω). This can be deduced from Corollary 10 as follows. Let
U⊆Self(ω) be a countable subset that dominates T⊆Self (ω), and write U={hi:i∈ω}.
Then T=Si∈ωTi, where Ti⊆Tis a subset dominated by {hi}. For all α∈ω, we have that
|(α)Ti| ≤ (α)hi+ 1 <ℵ0. Corollary 10 then implies that hTi=hSi∈ωTii ≺ℵ0Self(ω), which
is equivalent to hTi ≺ℵ1Self(ω).
The next result shows that if the κin the statement of Theorem 9 is assumed to be
uncountable, then a stronger conclusion can be obtained (with less work).
Proposition 11. Let κbe a regular uncountable cardinal ≤ |Ω|and T⊆Ea subset.
(i) If |(α)T|< κ for all α∈Ω, then |(α)hTi| < κ for all α∈Ω.
5
(ii) If there exists ℵ0≤λ < κ such that |(α)T| ≤ λfor all α∈Ω, then |(α)hTi| ≤ λfor
all α∈Ω.
Proof. For each β∈Ω, let λβ< κ be such that |(β)T| ≤ λβ. Also, let α∈Ω be any element.
Then, by definition,
(3) (α)hTi=
∞
[
j=1
(α)Tj.
We claim that (α)Tj< κ for all j≥1. This is true, by hypothesis, for j= 1. Assuming
inductively that Σ = (α)Tj−1has cardinality < κ, we have
(4) |(α)Tj|=|(Σ)T|=|[
σ∈Σ
{(σ)f:f∈T}| ≤ X
σ∈Σ
λσ.
This sum has < κ summands, each < κ; therefore |(α)Tj|< κ, by the regularity of κ.
Finally, (α)Tj< κ (j≥1) implies that |(α)hTi| < κ, since κis uncountable.
If there exists ℵ0≤λ < κ such that |(α)T| ≤ λfor all α∈Ω, then each λβcan be taken
to be λ. Let us assume inductively that for some j > 1, Σ = (α)Tj−1has cardinality ≤λj−1.
Then, the above argument shows that
(5) |(α)Tj| ≤ X
σ∈Σ
λ≤X
λj−1
λ=λj.
Since ℵ0≤λ, we have |λj|=λfor each j≥1. Therefore |(α)hTi| ≤ Pωλ=λ.
We conclude the section by noting that S= Sym(Ω), the group of all permutations of Ω,
is equivalent to E= Self(Ω), with respect to our equivalence relation. This result is known
(cf. [8, Theorem 3.3]), but we provide a quick proof, for the convenience of the reader.
Theorem 12 (Howie, Ruˇskuc, and Higgins).There exist g1, g2∈Esuch that E=g1Sg2.
In particular, E≈ℵ0S.
Proof. Since Ω is infinite, we can write Ω = Sα∈ΩΣα, where the union is disjoint, and for
each α∈Ω, |Σα|=|Ω|. Let g1∈Ebe an injective map such that |Ω\(Ω)g1|=|Ω|, and let
g2∈Ebe the map that takes each Σαto α.
Now, let f∈Ebe any element. For each α∈Ω, let ∆αdenote the preimage of αunder
f. Let h∈Ebe an injective self-map that embeds ∆αin Σα, for each α∈Ω, and such that
for some α∈Ω, |Σα\(∆α)h|=|Ω|. Then f=hg2. Also, since g1and hare both injective
and |Ω\(Ω)h|=|Ω|=|Ω\(Ω)g1|, there is a permutation ¯
h∈Ssuch that for all α∈Ω,
(α)g1¯
h= (α)h. Hence, we have f=g1¯
hg2∈g1Sg2.
4 The countable case
From now on we will restrict our attention to the case where Ω is countable. It will often
be convenient to assume that Ω = ω, the set of natural numbers. However, we will continue
using the symbol Ω; partly in order to distinguish the role of the set as the domain of our
6
maps from its role as an indexing set in some of the proofs, and partly because in Section 7
we will be interested in arbitrary orderings of the set. The symbols ≺ℵ0,E ,4ℵ0,E , and ≈ℵ0,E
will henceforth be written simply as ≺,4, and ≈, respectively.
We will say that a set Aof disjoint nonempty subsets of Ω is a partition of Ω if the
union of the members of Ais Ω. If U⊆E(= Self(Ω)) and Ais a partition of Ω, let
us define U(A)={f∈U: (Σ)f⊆Σ for all Σ ∈A}. Also, if U⊆Eand Σ ⊆Ω, let
U(Σ) ={f∈U: (α)f=αfor all α∈Σ}. (The notation U(A)can be considered an extension
of the notation U(Σ).)
The main aim of this section is to show that given two subgroups G1, G2⊆S(= Sym(Ω))
that are closed in the function topology, we have G1≈G2if and only if G1and G2are
equivalent as subgroups, in the sense of [4]. First, we need two preliminary results.
Proposition 13. Let Abe a partition of Ω. Then S(A)≈E(A).
Proof. If Ahas an infinite member, then E(A)≈E, by Lemma 2, and S(A)≈S, by a similar
argument. Hence, the result follows from Theorem 12. Let us, therefore, assume that all
members of Aare finite, and let us write A={Ai:i∈ω}. Further, let ni=|Ai|, for each
i∈ω, and write Ai={a(i, 0), a(i, 1),...,a(i, ni−1)}. Let B={Bi:i∈ω}be a partition
of Ω such that for each i∈ω,|Bi|=n2
i. By Theorems 13, 15, and 16 of [4], S(A)≈S(B).
We will show that E(A)4S(B).
For each i∈ω, write Bi={b(i, j, k) : j, k ∈ {0,1,...,ni−1}}. Let g1∈Ebe the
endomorphism that maps each Aiinto Bivia a(i, j)7→ b(i, j, 0), and let g2∈Ebe the
endomorphism that maps each Bionto Aivia b(i, j, k)7→ a(i, k).
Consider any element h∈E(A), and for each a(i, j)∈Ω write (a(i, j))h=a(i, cij ), for
some cij ∈ {0,1,...,ni−1}. Let ¯
h∈S(B)be any permutation such that for each i∈ωand
j∈ {0,1,...,ni−1}, (b(i, j, 0))¯
h=b(i, j, cij ) (e.g., we can define ¯
hby b(i, j, k)7→ b(i, j, k +
cij (mod ni−1))). Then, given any a(i, j)∈Ω, we have (a(i, j))g1¯
hg2= (b(i, j, 0))¯
hg2=
(b(i, j, cij ))g2=a(i, cij ) = (a(i, j))h, and hence E(A)⊆g1S(B)g2.
In the above proof, we called on results from [4] to deduce that S(A)≈S(B). This was
done primarily in the interests of space, since while it is not very difficult to show that
S(A)≈S(B)directly, several different cases would need to be considered.
Lemma 14. Let Abe a partition of Ωinto finite sets such that there is no common finite
upper bound on the cardinalities of the members of A. Then E(A)≈E≤.(See Definition 8
for the notation E≤.)
Proof. To prove that E≤4E(A), we will construct g, h ∈Esuch that E≤⊆gE(A)h. By our
hypotheses on A, we can find {Bi∈A:i∈ω}, consisting of disjoint sets, such that each
|Bi| ≥ i. Let us write each Bias {b(i, 0), b(i, 1),...,b(i, mi−1)}, where mi=|Bi|. Define
g∈Eby (i)g=b(i, 0) for all i∈Ω, and define h∈Eby (b(i, j))h=jfor all i∈ωand
j < mi(hcan be defined arbitrarily on elements not of the form b(i, j)). Then E≤⊆gE(A)h.
To prove that E(A)4E≤, write A={Ai:i∈ω}, and let g∈Ebe any injective map
such that for all i∈ωand a∈Ai, (a)gis greater than all the elements of Ai(⊆ω). Now,
let f∈E(A)be any element. Then for all a∈Ω, (a)f < (a)g, since a∈Aifor some i∈ω.
Thus, we can find a map h∈E≤such that for all a∈Ω, ((a)g)h= (a)f, since gis injective.
Therefore, E(A)⊆gE≤.
7
Theorem 15. Let G1and G2be subgroups of S= Sym(Ω) ⊆Ethat are closed in the
function topology on S. Let us write G1≈SG2if G1and G2are equivalent as groups (i.e.,
if the group generated by G1∪Uis equal to the group generated by G2∪U, for some finite
set U⊆S). Then G1≈SG2if and only if G1≈G2.
Proof. To show the forward implication, suppose that U⊆Sis a finite subset such that the
group generated by G1∪Uis equal to the group generated by G2∪U. Letting U−1be the set
consisting of the inverses of the elements of U, we see that hG1∪(U∪U−1)i=hG2∪(U∪U−1)i.
Hence G1≈SG2implies G1≈G2.
For the converse, let Abe a partition of Ω into finite sets such that there is no common
finite upper bound on the cardinalities of the members of A, and let Bbe a partition of Ω
into 2-element sets. By the main results of [4], every closed subgroup of Sis ≈Sto exactly
one of S,S(A),S(B), or {1}. We finish the proof by showing that these four groups are 6≈ to
each other.
By Proposition 13, S(A)≈E(A), and by the previous lemma, the latter is ≈E≤. By
Theorem 12, S≈E. Part (i) of Theorem 9 then implies that S(A)≺S, and part (ii) of that
theorem implies that S(B)≺S(A). Also, {1} ≺ S(B), since S(B)is uncountable. Thus, we
have {1} ≺ S(B)≺S(A)≺S.
5 An example
The goal of this section is to show that the partial ordering 4of submonoids of E=
Self(Ω) is not a total ordering, i.e., that there are submonoids M, M ′⊆Esuch that M64M′
and M′64M. In case the reader wishes to skip this section, we note that nothing in
subsequent sections will depend on the present discussion.
As in Section 1, upon identifying Ω with ω, let M2⊆Ebe the submonoid generated by
all maps whose images are contained in {0,1,2}. Let Σ1,Σ2⊆Ω be disjoint infinite subsets,
such that {0,1} ⊆ Σ1,{2,3} ⊆ Σ2, and Σ1∪Σ2= Ω. Let M′
3⊆Ebe the submonoid
generated by all maps that take Σ1to {0,1}and Σ2to {2,3}. We will show that M264M′
3
and M′
364M2.
Using the same argument as in Section 1, it is easy to see that M′
364M2. (If g1g2. . . gk
is any word in elements of E, where at least one of the gi∈M2, then the image of Ω
under g1g2. . . gkhas cardinality at most 3. Hence, if M′
3⊆ hM2∪Uifor some finite subset
U⊆E, then all the elements of M′
3whose images have cardinality 4 must be in hUi. This
is impossible, since there are uncountably many such elements.)
Next, let U⊆Ebe a finite set. We will show that M26⊆ hM′
3∪Ui. We begin by
characterizing the elements of hM′
3∪Ui. Let H⊆Ebe the (countable) set of all maps that
fix Ω\ {0,1,2,3}elementwise. Now, consider any word f=g1g2. . . gkin elements of E, such
that g1∈M′
3. Since the image of g1is contained in {0,1,2,3},fcan be written as g1h, where
his some element of H. Hence, any element f∈ hM′
3∪Uican be written as f=g1g2h,
where g1∈ hUi,g2∈M′
3, and h∈H. (Here we are using that fact that 1 ∈ hUi ∩ M′
3∩H.)
We note that each g∈ hUieither takes infinitely many elements of Ω to Σ1or takes
infinitely many elements to Σ2, since Σ1∪Σ2= Ω. For each such g, let Γg⊆Ω denote either
the set of those elements that gtakes to Σ1or the set of those elements that gtakes to Σ2-
8
whichever is infinite. Set F={Γg:g∈ hUi}. Since hUiis countable, so is F, and hence we
can write it as F={∆i:i∈ω}.
Next, let us construct for each i∈ωa triplet of distinct elements ai, bi, ci∈∆i, such
that the sets {ai, bi, ci}are disjoint. We take a0, b0, c0∈∆0to be any three distinct elements
(which must exist, since ∆0is infinite). Let 0 ≤jbe an integer, and assume that the elements
ai, bi, ci∈∆ihave been picked for all i≤j. Let aj+1, bj+1 , cj+1 ∈∆j+1 \Si≤j{ai, bi, ci}be
any three distinct elements. (Again, this is possible, by the fact that ∆j+1 is infinite.)
Now, let f∈M2be an element that takes each set {ai, bi, ci}bijectively to {0,1,2}, such
that f /∈ hUi. A self-map with these properties exists, since there are uncountably many
maps that take each {ai, bi, ci}bijectively to {0,1,2}, and hUiis countable. We finish the
proof by showing that f /∈ hM′
3∪Ui. Suppose, on the contrary, that f∈ hM′
3∪Ui. Then
f=g1g2h, where g1∈ hUi,g2∈M′
3, and h∈H, by the above characterization. Since
f /∈ hUi, we may assume that g26= 1. Let ∆i∈Fbe the set corresponding to g1(i.e., Γg1).
Then, by the above construction, we can find three distinct elements ai, bi, ci∈∆isuch that
ftakes {ai, bi, ci}bijectively to {0,1,2}. On the other hand, by choice of ∆i,g1either takes
{ai, bi, ci}to Σ1or takes {ai, bi, ci}to Σ2. In either case, |({ai, bi, ci})g1g2h| ≤ 2, since g2
takes each of Σ1and Σ2to a 2-element set. Hence f6=g1g2h; a contradiction. We therefore
conclude that f /∈ hM′
3∪Ui.
In summary, we have
Proposition 16. The partial ordering 4of submonoids of Eis not a total ordering.
6 Four lemmas
The results of this section (except for the first) are close analogs of results in [4]. We will
use them in later sections to classify various submonoids of E= Self(Ω) into equivalence
classes; Lemmas 18 -20 will be our main tools for showing that submonoids are ≈to each
other. The proofs of these three lemmas are, for the most part, simpler than those of
their group-theoretic analogs (namely [4, Lemma 10], [4, Lemma 12], and [4, Lemma 14],
respectively).
Lemma 17. Let M⊆Ebe a submonoid. Then M4E(A), where A={Aα:α∈Ω}is any
partition of Ωsuch that for each α∈Ω,|Aα|=|(α)M|.
Proof. Let g∈Ebe a map such that for all α∈Ω, (α)g∈Aα, and let h∈Ebe a map such
that for all α∈Ω, hmaps (Aα) onto (α)M. Then M⊆gE(A)h.
Lemma 18. Let Mbe a submonoid of E, and suppose there exist a sequence (αi)i∈ω∈Ωω
of distinct elements and a sequence of nonempty subsets Di⊆Ωi(i∈ω),such that
(i) For each i∈ωand (β0,...,βi−1)∈Di,there exist infinitely many elements β∈Ω
such that (β0,...,βi−1, β)∈Di+1;and
(ii) If (βi)i∈ω∈Ωωhas the property that (β0,...,βi−1)∈Difor each i∈ω, then there
exists g∈Msuch that for all i∈ω,βi= (αi)g, and the elements βiare all distinct.
Then there exist f, h ∈Esuch that E=f Mh. In particular, M≈E.
9
Proof. For each i∈ωand (β0,...,βi−1)∈Di,let
(6) Γ(β0,...,βi−1) = {β∈Ω : (β0,...,βi−1, β)∈Di+1}.
By (i), each Γ(β0,...,βi−1) is an infinite subset of Ω. Since for each i∈ω, Ωiis countable,
Si∈ωΩiis countable as well, and therefore, so is Si∈ωDi⊆Si∈ωΩi. Thus, there are only
countably many sets of the form Γ(β0,...,βi−1). By a standard inductive construction, we
can find a collection {Λ(β0,...,βi−1) : i∈ω, (β0,...,βi−1)∈Di}of disjoint infinite sets,
such that each Λ(β0,...,βi−1)⊆Γ(β0,...,βi−1).
Next, we define a map f∈Eby
(7) (i)f=αifor all i∈ω(= Ω).
Also, let h∈Ebe any map that takes each Λ(β0,...,βi−1) surjectively to Ω. We will show
that E=fM h.
Let g∈Ebe any element. We define recursively a sequence (βi)i∈ω∈Ωωas follows: for
each i∈ωlet βi∈Λ(β0, . . . , βi−1) be such that (βi)h= (i)g. Since each Λ(β0,...,βi−1)⊆
Γ(β0,...,βi−1), our sequence (βi)i∈ωhas the property that (β0,...,βi−1)∈Difor each i∈ω.
Thus, by (ii), there exists ¯g∈Msuch that for all i∈ω,βi= (αi)¯g. For each i∈ω(= Ω),
we then have that (i)f¯gh = (αi)¯gh = (βi)h= (i)g, and therefore g=f¯gh.
The next argument uses the same basic idea, but it is more complicated.
Lemma 19. Let Mbe a submonoid of E, and suppose there exist a sequence (αi)i∈ω∈Ωω
of distinct elements, an unbounded sequence of positive integers (Ni)i∈ω,and a sequence of
nonempty subsets Di⊆Ωi(i∈ω),such that
(i) For each i∈ωand (β0,...,βi−1)∈Di,there exist at least Nielements β∈Ωsuch
that (β0,...,βi−1, β)∈Di+1 ;and
(ii) If (βi)i∈ω∈Ωωhas the property that (β0,...,βi−1)∈Difor each i∈ω, then there
exists g∈Msuch that for each i∈ω,βi= (αi)g, and the elements βiare all distinct.
Then there exist f, h ∈Esuch that E≤⊆f Mh. In particular, E≤4M.
Proof. We will construct recursively integers i(−1) < i(0) < . . . < i(j)< . . . , and for each
j≥0 a subset Ci(j)⊆Di(j).
Set i(−1) = −1 and i(0) = 0, and let Ci(0) =D0be the singleton consisting of the empty
string. Now assume inductively for some j≥1 that i(0),...,i(j−1) and Ci(0),...,Ci(j−1)
have been constructed. Let i(j) be an integer such that Ni(j)> j ·|Ci(j−1)|+Pj−1
k=0 |Ci(k)|. Let
Ci(j)⊆Di(j)be a finite subset that for each (β0,...,βi(j−1)−1)∈Ci(j−1) contains jelements
of the form (β0,...,βi(j−1)−1,...,βi(j)−2, β), such that the elements βare distinct from each
other and from all elements that occur as last components of elements of Ci(0),...,Ci(j−1).
Our choice of i(j) and condition (i) makes this definition possible. (Actually, it would have
sufficed to pick i(j) so that Ni(j−1)+1Ni(j−1)+2 . . . Ni(j)> j · |Ci(j−1)|+Pj−1
k=0 |Ci(k)|.)
Once the above integers and subsets have been constructed, let us use the sets Ci(j)to
construct subsets Fi(j)⊆Ωj. Set Fi(0) =Ci(0). For each element (β0,...,βi(j)−1)∈Ci(j)with
j≥1, we define a sequence (γi(0),...,γi(j−1)) by setting γi(k)=βi(k+1)−1(0 ≤k≤j−1); i.e.,
10
we drop the βkthat do not occur as last components of elements of Ci(0),...,Ci(j−1). For
each j≥1, we then let Fi(j)consist of the tuples (γi(0),...,γi(j−1)). Also, for each element
(γi(0),...,γi(j−1) )∈Fi(j)let
(8) Γ(γi(0),...,γi(j−1) ) = {β∈Ω : (γi(0),...,γi(j−1), β )∈Fi(j+1)}.
By construction, each |Γ(γi(0),...,γi(j−1))| ≥ j+ 1; for simplicity, we will assume that this
is an equality, after discarding some elements if necessary. Let h∈Ebe a map such that
(9) htakes each Γ(γi(0),...,γi(j−1)) onto {0,...,j}.
Such a map exists, since the sets Γ(γi(0),...,γi(j−1)) are all disjoint. Also, let f∈Ebe
defined by
(10) (j)f=αi(j)for all j∈ω(= Ω).
We finish the proof by showing that E≤⊆fM h.
Let g∈E≤be any element. We first construct recursively a sequence (γi(j))j∈ωsuch
that for each j∈ω, (γi(0),...,γi(j−1))∈Fi(j). Let γi(0) be the unique element of Γ(). (We
note that (γi(0))h= 0 = (0)g, by definition of h.) Assuming that γi(0),...,γi(j−1) have
been defined, let γi(j)∈Γ(γi(0),...,γi(j−1)) be such that (γi(j))h= (j)g. (Such an element
exists, by our definition of hand the fact that for all k∈ω, (k)g≤k.) Since the sequence
(γi(j))j∈ω∈Ωωhas the property that (γi(0),...,γi(j−1))∈Fi(j)for each j∈ω, there exists
¯g∈Msuch that for all i∈ω,γi(j)= (αi(j))¯g. This follows from (ii), since (γi(j))j∈ω
is a subsequence of some (βi)i∈ωas in (ii). Hence, for each j∈ω(= Ω), we have that
(j)f¯gh = (αi(j))¯gh = (γi(j))h= (j)g, and therefore g=f¯gh.
Lemma 20. Let Mbe a submonoid of E. Suppose there exist three sequences (αi)i∈ω,(βi)i∈ω,
(γi)i∈ω∈Ωωof distinct elements, such that (βi)i∈ωand (γi)i∈ωare disjoint, and for every
element (δi)i∈ω∈Qi∈ω{βi, γi} ⊆ Ωω, there exists g∈Msuch that for all i∈ω,δi= (αi)g.
Then there exist f, h ∈Esuch that E(A)⊆f Mh, where Ais a partition of Ωinto 2-element
sets. In particular E(A)4M.
Proof. Write A={Ai:i∈ω}, where for each i∈ω, Ai={ai, bi}. Let f∈Ebe the map
defined by (ai)f=α2iand (bi)f=α2i+1, and let h∈Ebe a map that for each i∈ωtakes
{β2i, β2i+1}to aiand {γ2i, γ2i+1}to bi. Then E(A)⊆f M h.
7 Submonoids arising from preorders
Definition 21. Given a preorder ρon Ω, let E(ρ)⊆E(= Self(Ω)) denote the subset
consisting of all maps fsuch that for all α∈Ωone has (α, (α)f)∈ρ.
Clearly, subsets of the form E(ρ) are submonoids. The submonoids E(A)(where Ais
a partition of Ω) are of this form, as is E≤. The goal of this section is to classify such
submonoids into equivalence classes. To facilitate the discussion, let us divide them into five
types.
Definition 22. Let ρbe a preorder on Ω. For each α∈Ωset ∆ρ(α) = {β∈Ω : (α, β)∈ρ}
(the “principal up-set generated by α”). We will say that
11
The preorder ρis of type 1 if there is an infinite subset Γ⊆Ωsuch that for all α∈Γ,
∆ρ(α)is infinite.
The preorder ρis of type 2 if the cardinalities of the sets ∆ρ(α) (α∈Ω) have no
common finite upper bound, but ∆ρ(α)is infinite for only finitely many α.
The preorder ρis of type 3 if there is a number n∈ωsuch that |∆ρ(α)| ≤ nfor all
but finitely many α∈Ω, and there are infinitely many α∈Ωsuch that |∆ρ(α)|>1.
The preorder ρis of type 4 if |∆ρ(α)|= 1 for all but finitely many α∈Ω.
Let us further divide preorders of type 3into two sub-types. We will say that
The preorder ρis of type 3a if it is of type 3and, in addition, there are infinite families
{αi}i∈ω,{βi}i∈ω⊆Ωconsisting of distinct elements, such that for each i∈ω,αi6=βi
and (αi, βi)∈ρ.
The preorder ρis of type 3b if it is of type 3and, in addition, there is a finite set
Γ⊆Ωsuch that for all but finitely many α∈Ω,∆ρ(α)⊆Γ∪ {α}.
It is clear that every preorder on Ω falls into exactly one of the above five types. Further,
if ρis a preorder of type 2 or 3, and Σ = {α∈Ω : |∆ρ(α)|=ℵ0}, then E(ρ)≈E(ρ)(Σ).
(The notation E(ρ)(Σ) is defined at the beginning of Section 4.) For, if α∈Ω\Σ, then
∆ρ(α)∩Σ = ∅, since ∆ρ(α) is finite. This shows that (Ω\Σ)E(ρ)∩Σ = ∅. Hence, if f∈E(ρ)
is any element, then f∈E(ρ)(Σ) U, where U=E(Ω\Σ). Therefore, E(ρ)⊆ hE(ρ)(Σ) ∪Ui,
and E(ρ)≈E(ρ)(Σ). (Since Σ is finite, Uis countable. Theorem 6 then allows us to replace
Uby a finite set.) We will use this observation a number of times in this section.
Lemma 23. Let ρbe a preorder on Ω, and let Abe a partition of Ωinto 2-element sets.
(i) If ρis of type 1, then E(ρ)≈E.
(ii) If ρis of type 2, then E(ρ)≈E≤.
(iii) If ρis of type 3a, then E(ρ)≈E(A).
Proof. If ρis of type 1, respectively type 2, then E(ρ) clearly satisfies the hypotheses of
Lemma 18, respectively Lemma 19. Hence, E≈E(ρ), respectively E≤4E(ρ). Further,
if ρis of type 3a, then upon removing some elements if necessary, we may assume that the
{αi}i∈ωand {βi}i∈ωprovided by the definition are disjoint. E(ρ) then satisfies the hypotheses
of Lemma 20 (with αi=γi). Hence E(A)4E(ρ).
To finish the proof, we must show that E(ρ)4E≤if ρis of type 2, and that E(ρ)4E(A)
if ρis of type 3a. In either case, we may assume that for all α∈Ω, ∆ρ(α) is finite, by the
remarks following Definition 22. Now, E(ρ)4E(B), where Bis a partition of Ω into finite
sets, by Lemma 17, and in the case where ρis of type 3a, these finite sets can all be taken
to have cardinality n, for some n∈ω. If ρis of type 2, then we have E(ρ)4E(B)≈E≤,
by Lemma 14. If ρis of type 3a, then E(A)≈E(B), by Proposition 13 and [4, Theorem 15].
Hence E(ρ)4E(A).
12
Definition 24. Given γ∈Ω, let ργbe the preorder on Ωdefined by (α, β)∈ργ⇔β∈ {α, γ}.
It is clear that the preorders ργare of type 3b. Further, if α, β ∈Ω are any two elements,
then E(ρα)≈E(ρβ). More specifically, if g∈Eis any permutation of Ω that takes αto β,
then E(ρα) = gE(ρβ)g−1. (Given γ∈Ω, an element of gE(ρβ)g−1either fixes γor takes it
to α. Hence gE(ρβ)g−1⊆E(ρα), and similarly g−1E(ρα)g⊆E(ρβ). Conjugating the latter
expression by g, we obtain E(ρα)⊆gE(ρβ)g−1.)
Let E3bdenote the monoid generated by {E(ρα) : α∈Ω}, and let g∈Ebe a permutation
which is transitive on Ω. Then, by the previous paragraph, E3b⊆ hE(ργ)∪{g}i for any γ∈E.
Hence, for all γ∈E,E(ργ)≈E3b. As an aside, we note that E3bis closed under conjugation
by permutations of Ω. (Given any permutation gand any α∈Ω, gE(ρβ)g−1=E(ρα), where
β= (α)g. Thus gE3bg−1contains all the generators of E3b, and therefore E3b⊆gE3bg−1.
Conjugating by g−1, we obtain g−1E3bg⊆E3b.)
Lemma 25. Let ρand ρ′be preorders on Ωof type 3b. Then E(ρ)≈E(ρ′).
Proof. Let ρbe a preorder on Ω of type 3b. Then there is a β∈Ω and an infinite set Σ ⊆Ω
such that all α∈Σ, we have (α, β)∈ρ. Let f1∈Ebe a map that takes Ω bijectively to Σ,
while fixing β(which, we may assume, is an element of Σ), and let f2∈Ebe a right inverse
of f1. Then E(ρβ)⊆f1E(ρ)f2, and hence E3b4E(ρ). We conclude the proof by showing
that E(ρ)4E3b.
By the remarks following Definition 22, we may assume that for all α∈Ω, ∆ρ(α) is
finite. Let Γ ⊆Ω be a finite set such that for all α∈Ω, ∆ρ(α)⊆Γ∪ {α}, and write
Γ = {α0, α1,...,αn}. We will first show that E(ρ)(Γ) ⊆E3b.
Let h∈E(ρ)(Γ) be any element. Then we can write Ω as a disjoint union Λ0∪Λ1∪
· · · ∪ Λn∪Λ, where for all β∈Λi, (β)h=αi, and hacts as the identity on Λ. For each
i∈ {0,1,...,n}, let gi∈E(ραi) be the map that takes Λi\ {α0, α1,...,αn}to αiand fixes
all other elements. Then h=g0g1. . . gn∈E3b, and hence E(ρ)(Γ) ⊆E3b.
For each g∈Self(Γ) let fg∈Ebe such that fgacts as gon Γ and as the identity
elsewhere, and set V={fg:g∈Self(Γ)}. Now, let h∈E(ρ) be any element, and define
¯
h∈E(ρ)(Γ) by (α)¯
h= (α)hfor all α /∈Γ. Noting that, by definition of Γ, (Γ)h⊆Γ, there is
an element fg∈Vthat agrees with hon Γ. Then h=fg¯
h∈V E(ρ)(Γ) ⊆V E3b, and hence
E(ρ)4E3b.
We will say that a map f∈Eis finitely-many-to-one (or, more succinctly, fm-to-one) if
the preimage of each element of Ω under fis finite.
Lemma 26. Let ρand ρ′be preorders on Ωof types 3aand 3b, respectively. Then E(ρ′)≺
E(ρ).
Proof. Let Abe a partition of Ω into 2-element sets, and let us fix an element γ∈Ω. By
the previous two lemmas, it suffices to show that E(ργ)≺E(A). We begin by proving that
E(A)64E(ργ).
For each finite subset Σ ⊆Ω let fΣ∈Ebe the map defined by
(11) (α)fΣ=γif α∈Σ
αif α /∈Σ.
13
Now, let U⊆Ebe a finite subset, and consider a monoid word g=g0g1...gi−1gigi+1 . . . gn
in elements of E(ργ)∪U. Suppose that gis fm-to-one and that the element giis in E(ργ).
Let Γ ⊆Ω be the preimage of γunder gi. Then Σ := ((Ω)g0...gi−1)∩Γ must be finite, and
so we have g=g0. . . gi−1fΣgi+1 . . . gn. In a similar fashion, assuming that gis fm-to-one, we
can replace every element of E(ργ) occurring in the word gby an element of the form fΣ, for
some finite Σ ⊆Ω. Considering that all elements of E(A)are fm-to-one, we conclude that if
h∈E(A)∩ hE(ργ)∪Ui, then h∈ h{fΣ: Σ ⊆Ω finite} ∪ Ui. But, the latter set is countable
and hence cannot contain all of E(A). Therefore E(A)64E(ργ).
It remains to show that E(ργ)4E(A). Write A={Ai:i∈Ω}, and for each i∈Ω set
Ai={αi1, αi2}. Let g1∈Ebe the map defined by (i)g1=αi1(i∈Ω), and let g2∈Ebe
defined by (αi1)g2=αi1and (αi2)g2=γ. Then E(ργ)⊆g1E(A)g2.
Theorem 27. Let ρand ρ′be preorders on Ω. Then E(ρ)≈E(ρ′)if and only if ρand ρ′
are of the same type.
Proof. The above lemmas, in conjunction with Theorem 9, give the desired conclusion if ρ
and ρ′are each of type 1, 2, 3a, or 3b. The result then follows from the fact that ρis of type
4 if and only if E(ρ) is countable.
8 The function topology
From now on we will be concerned with submonoids that are closed in the function
topology on E, so let us recall some facts about this topology.
Regarding the infinite set Ω as a discrete topological space, the monoid E= Self(Ω),
viewed as the set of all functions from Ω to Ω, becomes a topological space under the function
topology. A subbasis of open sets in this topology is given by the sets {f∈E: (α)f=β}
(α, β ∈Ω).The closure of a set U⊆Econsists of all maps fsuch that, for every finite
subset Γ ⊆Ω,there exists an element of Uagreeing with fat all members of Γ.It is easy
to see that composition of maps is continuous in this topology. Given a subset U⊆E, we
will write clE(U) for the closure of Uin E.
The following lemma will be useful later on. It is an analog of [4, Lemma 8], with monoids
in place of groups and forward orbits in place of orbits. (Given an element α∈Ω and a
subset U⊆E, we will refer to the set (α)Uas the forward orbit of αunder U.) The proof
is carried over from [4] almost verbatim.
Lemma 28. Let M⊆Ebe a submonoid. Then
(i) clE(M)is also a submonoid of E.
(ii) Mand clE(M)have the same forward orbits in Ω.
(iii) If Γis a finite subset of Ω,then clE(M)(Γ) = clE(M(Γ)).
Proof. Statement (i) follows from the fact that composition of maps is continuous.
From the characterization of the closure of a set in our topology, we see that for α, β ∈Ω,
the set clE(M) will contain elements taking αto βif and only if Mdoes, establishing (ii).
14
Given any subset Γ ⊆Ω, the elements of clE(M(Γ)) fix Γ elementwise, by (ii). Hence,
clE(M)(Γ) ⊇clE(M(Γ)). To show clE(M)(Γ) ⊆clE(M(Γ)), assume that Γ ⊆Ω is finite, and let
f∈clE(M)(Γ). Then every neighborhood of fcontains elements of M, since f∈clE(M).
But, since ffixes all points of the finite set Γ, every sufficiently small neighborhood of f
consists of elements which do the same. Hence, every such neighborhood contains elements
of M(Γ), and so f∈clE(M(Γ)).
9 Large stabilizers
Let us say that a submonoid Mof E= Self(Ω) has large stabilizers if for each finite
subset Σ ⊆Ω, M(Σ) ≈M. For example, all subgroups of Shave large stabilizers, by [4,
Lemma 2], as do submonoids of the form E(A), where Ais a partition of Ω. More generally,
we have
Proposition 29. Let ρbe a preorder on Ω. Then E(ρ)has large stabilizers.
Proof. Let Σ ⊆Ω be finite. Then E(ρ)(Σ) is still a submonoid of the form E(ρ′), where
ρ′is a preorder on Ω of the same type as ρ. The desired conclusion then follows from
Theorem 27.
The following lemma gives another class of submonoids that have large stabilizers. The
proof is similar to the one for [4, Lemma 2]
Lemma 30. Let G⊆Sym(Ω) ⊆Ebe a subgroup. Then clE(G)has large stabilizers.
Proof. Let Σ ⊆Ω be finite, and take f∈clE(G). Then there is a sequence of elements of
Gthat has limit f. Eventually, the elements of this sequence must agree on the members of
Σ, and hence, must lie in some right coset of G(Σ), say the right coset represented by g∈G.
Then f∈clE(G(Σ))g= clE(G)(Σ)g, by Lemma 28. Letting R⊆Gbe a set of representatives
of the right cosets of G(Σ), we have clE(G)⊆ hclE(G)(Σ) ∪Ri. Now, Ris countable, since Ω
is countable and Σ is finite, and hence clE(G)≈clE(G)(Σ), by Theorem 6.
Not all submonoids of E, however, have large stabilizers. Given a natural number n, the
monoid Mn(generated by all maps whose images are contained in {0,1,...,n}) mentioned in
Section 1 is an easy example of this. (For any finite Σ ⊆Ω (= ω) such that Σ∩{0,1,...,n}=
∅, (Mn)(Σ) ={1} 6≈ Mn.)
Definition 31. Let us say that a map f∈Eis a.e. injective if there exists some finite set
Γ⊆Ωsuch that fis injective on Ω\Γ.
In much of the sequel we will be concerned with submonoids where the a.e. injective
maps form dense subsets (viewing Eas topological space under the function topology). A
submonoid M⊆Ehas this property if and only if for every finite Σ ⊆Ω and every f∈M,
there is an a.e. injective map g∈Mthat agrees with fon Σ. Clearly, a.e. injective maps
form dense subsets in submonoids of Ethat consist of injective maps, such as subgroups of
Sym(Ω) and their closures in the function topology. It is easy to see that this is also the
case for the submonoids E(ρ), since given any g∈E(ρ) and any finite Σ ⊆Ω, we can find
15
an f∈E(ρ) that agrees with gon Σ and acts as the identity elsewhere. From now on we
will abuse language by referring to submonoids in which the a.e. injective maps form dense
subsets as submonoids having dense a.e. injective maps.
Our goal will be to classify into equivalence classes submonoids of Ethat are closed
in the function topology, and have large stabilizers and dense a.e. injective maps. By the
remarks above, examples of such submonoids include closures in the function topology in E
of subgroups of Sym(Ω) and submonoids of the form E(ρ). These equivalence classes will be
shown to be represented by E,E≤,E(A),E(ργ), and {1}, respectively (where Ais a partition
of Ω into 2-element sets, and γ∈Ω). Most of the work will go into showing that a given
monoid from the above list is 4M, for a monoid Msatisfying an appropriate condition. In
the first three cases (and to some extent in the fourth) our arguments will follow a similar
pattern. We will construct sequences of the form g0, g0g1, g0g1g2,..., and use closure in the
function topology to conclude that such sequences have limits in the appropriate monoid
M. The subsets consisting of these limits will then satisfy the hypotheses of one of the
Lemmas 18 -20, giving us the desired conclusion.
Proposition 32. Let M⊆Ebe a submonoid that is closed in the function topology and has
dense fm-to-one (and hence a.e. injective)maps. If M(Σ) has an infinite forward orbit for
every finite Σ⊆Ω, then M≈E.
Proof. This proof closely follows that of [4, Theorem 11]. We begin by recursively construct-
ing for each j≥0 an element αj∈Ω and a finite subset Kj⊆M, consisting of fm-to-one
maps (“fm-to-one” is defined directly preceding Lemma 26), indexed
(12) Kj={g(k0, k1, . . . , kr−1) : k0, k1,...,kr−1, r ∈ω, r +k0+...+kr−1=j}.
Let α0be any element that has an infinite forward orbit under M. Assume inductively that
α0,...,αj−1have been defined, and let Γj={α0,...,αj−1} ∪ {0,...,j −1}. We then take
αj∈Ω to be any element that has an infinite forward orbit under M(Γj).
Now we construct the sets Kj. If j= 0, we have only one element to choose, g(), and
we take this to be the identity element 1 ∈M. Assume inductively that the sets Kihave
been defined for all nonnegative i < j. Let us fix arbitrarily an order in which the elements
of Kjare to be constructed. When it is time to define g(k0, k1,...,kr−1), let us write
g′=g(k0, k1,...,kr−2), noting that this is a member of Kj−kr−1−1and hence already defined
(and fm-to-one). We set g(k0, k1,...,kr−1) = hg′, where h∈M(Γr−1)is chosen so that the
image of αr−1under hg′is distinct from the images of α0,...,αr−2under the finitely many
elements of K0∪...∪Kj−1, and also under the elements of Kjthat have been constructed so
far. This is possible, since (αr−1)M(Γr−1)is infinite, by the choice of αr−1, and (αr−1)M(Γr−1)g′
is infinite, since g′is fm-to-one. Further, hcan be chosen to be fm-to-one, by our hypothesis
on M, making g(k0, k1, . . . , kr−1) fm-to-one as well. We note that the images of α0,...,αr−2
and 0,...,r−2 under hg′will be the same as their images under g′, since elements of M(Γr−1)
fix {α0,...,αr−2} ∪ {0, . . . , r −2}= Γr−1.
Once the elements of each set Kjare constructed, we have monoid elements g(k0,...,ki−1)
for all i, k0,...,ki−1∈ω. We can thus define, for each i∈ω,
(13) Di={((α0)g, . . . , (αi−1)g) : g=g(k0,...,ki−1) for some k0,...,ki−1∈ω}.
16
Any two elements of the form g(k0,...,ki) with indices k0,...,ki−1the same, but different
last indices ki, act differently on αi, so the sets Disatisfy condition (i) of Lemma 18. Suppose
that (βi)∈Ωωhas the property that for every ithe sequence (β0,...,βi−1) is in Di. By
construction, the elements βiare all distinct. Also, we see inductively that successive strings
(),(β0),..., (β0,...,βi−1),... must arise from unique elements of the forms g(), g(k0), ...,
g(k0,...,ki−1), .... By including {0,...,r −1}in Γr−1above, we have ensured that the
elements of this sequence agree on larger and larger subsets of ω= Ω, which have union Ω.
Thus, this sequence converges to a map g∈E, which necessarily sends (αi) to (βi). Since
Mis closed, g∈M; which establishes condition (ii) of Lemma 18. Hence, that lemma tells
us that M≈E.
In the above proposition, the hypothesis that M(Σ) has an infinite forward orbit for every
finite Σ ⊆Ω is not necessary for M≈E. For example, let E>⊆Edenote the submonoid
generated by maps that are strictly increasing with respect to the usual ordering of ω= Ω
(i.e., maps f∈Esuch that for all α∈Ω, (α)f > α). This submonoid is closed in the
function topology, and its a.e. injective maps form a dense subset. It is also easy to see that
the submonoid E>satisfies the hypotheses of Lemma 18, and hence is ≈E, but given any
finite Σ ⊆Ω, (E>)(Σ) ={1}. However, for submonoids M⊆Ethat have large stabilizers,
the condition that M(Σ) has an infinite forward orbit for every finite Σ ⊆Ω is necessary for
M≈E.
Corollary 33. Let M⊆Ebe a submonoid that is closed in the function topology, and has
large stabilizers and dense fm-to-one maps. Then M≈Eif and only if M(Σ) has an infinite
forward orbit for every finite Σ⊆Ω.
Proof. If there is a finite set Σ ⊆Ω such that M(Σ) has no infinite forward orbits, then
M(Σ) ≺E, by Theorem 9(i). Since Mhas large stabilizers, this implies that M≺E. The
converse follows from the previous proposition.
Let us now turn to submonoids whose stabilizers have finite forward orbits. We will say
that a fm-to-one map f∈Eis boundedly finitely-many-to-one (abbreviated bfm-to-one) if
there is a common finite upper bound on the cardinalities of the preimages of elements of Ω
under f.
Proposition 34. Let M⊆Ebe a submonoid that is closed in the function topology and has
dense bfm-to-one (and hence a.e. injective)maps. If for every finite Σ⊆Ωthe cardinalities
of the forward orbits of M(Σ) have no common finite bound, then E≤4M.
Proof. Again, this proof closely follows that of [4, Theorem 13]. Let us fix an unbounded
sequence of positive integers (Ni)i∈ω. We begin by recursively constructing for each j≥0
an element αj∈Ω and a finite subset Kj⊆M, consisting of bfm-to-one maps, indexed
(14) Kj={g(k0, k1, . . . , kj−1) : 0 ≤ki< Ni(0 ≤i < j)}.
Let the 1-element set K0={g()}consist of the identity map 1 ∈M. Assume inductively
that for some j≥0 the elements α0,...,αj−1and the sets K0,...,Kjhave been defined,
and let Γj={α0,...,αj−1} ∪ {0,...,j −1}. Now, let us choose αj∈Ω so that for each
g=g(k0, k1,...,kj−1)∈Kj, the forward orbit of αjunder M(Γj)ghas cardinality at least
17
(15) (j+ 1) ·(
j+1
X
i=0
|Ki|) = (j+ 1) ·(
j
X
i=0
|Ki|+N0N1...Nj).
This is possible, since each such gis bfm-to-one, and since Kjis finite.
Next, let us fix arbitrarily an order in which the elements of Kj+1 are to be constructed.
When it is time to construct g(k0, k1,...,kj), let us write g′=g(k0, k1,...,kj−1)∈Kj. We
define g(k0, k1,...,kj) = hg′, where h∈M(Γj)is chosen so that the image of αjunder hg′is
distinct from the images of α0,...,αj−1under the elements of K0∪...∪Kjand also under
the elements of Kj+1 that have been constructed so far. Our choice of αjmakes this possible.
Further, hcan be chosen to be bfm-to-one, by our hypothesis on M, making g(k0, k1,...,kj)
bfm-to-one as well. We note that the images of α0,...,αj−1and 0,...,j−1 under hg′will be
the same as their images under g′, since elements of M(Γj)fix {α0,...,αj−1}∪{0,...,j−1}=
Γj.
Once the sets Kjare constructed, for each i∈ω, let
(16) Di={((α0)g, (α1)g, . . . , (αi−1)g) : g∈Ki}.
Any two elements of the form g(k0,...,ki) with indices k0,...,ki−1the same but differ-
ent last indices kiact differently on αi, so the sets Disatisfy condition (i) of Lemma 19.
Suppose that (βi)∈Ωωhas the property that for every ithe sequence (β0,...,βi−1) is in
Di. By construction, the elements βiare all distinct. Also, we see inductively that succes-
sive strings (),(β0),..., (β0,...,βi−1),... must arise from unique elements of the forms g(),
g(k0), ...,g(k0,...,ki−1), .... The elements of the above sequence agree on the successive
sets {0,...,j−1}(having union ω= Ω), and hence the sequence converges to a map g∈E,
which must send (αi) to (βi). Since Mis closed, we have that g∈M, establishing condition
(ii) of Lemma 19. Hence, that lemma tells us that E≤4M.
As with Proposition 32, the hypothesis that for every finite Σ ⊆Ω the cardinalities of
the forward orbits of M(Σ) have no common finite bound is not necessary for E≤4M. For
instance, the submonoid E<⊆Egenerated by maps that are strictly decreasing with respect
to the usual ordering of ω= Ω (i.e., maps f∈Esuch that for all α∈Ω\ {0}, (α)f < α, and
(0)f= 0) is closed in the function topology, has dense a.e. injective maps, and satisfies the
hypotheses of Lemma 19. Therefore E≤4E<, but given any finite Σ ⊆Ω, (E<)(Σ) ={1}.
Corollary 35. Let M⊆Ebe a submonoid that is closed in the function topology, and has
large stabilizers and dense bfm-to-one maps. Then M≈E≤if and only if for every finite
Σ⊆Ωthe cardinalities of the forward orbits of M(Σ) have no common finite bound, and
there exists a finite set Σ⊆Ωsuch that all forward orbits of M(Σ) are finite.
Proof. If M≈E≤, then for every finite Σ ⊆Ω, the cardinalities of the forward orbits of
M(Σ) have no common finite bound. For, otherwise, there would be some finite Σ ⊆Ω for
which M(Σ) would satisfy the hypotheses of Theorem 9(ii), implying that M≈M(Σ) ≺E≤.
Also, there must be a finite set Σ ⊆Ω such that all forward orbits of M(Σ) are finite, since
otherwise we would have M≈E, by Proposition 32, contradicting Theorem 9(i).
Conversely, if for every finite Σ ⊆Ω, the cardinalities of the forward orbits of M(Σ)
have no common finite bound, then E≤4M, by the previous proposition. Suppose that,
in addition, there exists a finite set Σ ⊆Ω such that all forward orbits of M(Σ) are finite.
18
Then M(Σ) 4E(A), where Ais a partition of Ω into finite sets, by Lemma 17. By the large
stabilizer hypothesis and Lemma 14, we have M≈M(Σ) 4E(A)≈E≤. Hence M≈E≤.
The rest of the section is devoted to the more intricate case of submonoids with stabilizers
whose forward orbits have a common finite bound.
Lemma 36. Let M⊆Ebe a submonoid that has large stabilizers, and let Abe a partition
of Ωinto 2-element sets. If there exists a finite set Σ⊆Ωand a positive integer nsuch that
the cardinalities of all forward orbits of M(Σ) are bounded by n, then M4E(A).
Proof. Given a finite set Σ ⊆Ω and a positive integer nas above, M(Σ) 4E(B), where Bis a
partition of Ω into sets of cardinality n, by Lemma 17. But, E(A)≈E(B), by Proposition 13
and [4, Theorem 15]. Hence M≈M(Σ) 4E(A).
The following proof is based on that of [4, Theorem 15], but it is more complicated.
Lemma 37. Let M⊆Ebe a submonoid that is closed in the function topology and has dense
a.e. injective maps. If for all finite Γ,∆⊆Ωthere exists α∈Ωsuch that (α)M(∆) 6⊆ Γ∪{α},
then E(A)4M(where Ais a partition of Ωinto 2-element sets).
Proof. We may assume that there exist finite sets Γ,∆⊆Ω and a positive integer nsuch
that for all α∈Ω, |(α)M(∆) \Γ| ≤ n. For, otherwise E≤4M, by Proposition 34, and
E(A)≺E≤, by Theorem 9(ii).
Let m > 1 be the largest integer such that for all finite Γ,∆⊆Ω, there exists α∈Ω
such that |(α)M(∆) \Γ| ≥ m. Then there exist finite sets Γ′,∆′⊆Ω such that for all α∈Ω,
|(α)M(∆′)\Γ′| ≤ m. Now, M′=M(∆′)has the property that for all finite Γ,∆⊆Ω there
exists α∈Ω such that (α)M′
(∆) 6⊆ Γ∪ {α}, and for all finite Γ,∆⊆Ω, with Γ′⊆Γ, the
maximum of the cardinalities of the sets (α)M′
(∆) \Γ (α∈Ω) is m. From now on we will be
working with M′in place of M.
A consequence of the above considerations is that if for some α∈Ω and finite ∆,Γ⊆Ω,
with Γ′⊆Γ, we have |(α)M′
(∆) \Γ|=m, then (α)M′
(∆) \Γ = (α)M′\Γ′, since (α)M′\Γ′
cannot have cardinality larger than m. Thus
(17) If Γ,∆⊆Ω are finite subsets, with Γ′⊆Γ, and α∈Ω has the property that
|(α)M′
(∆) \Γ|=m, then for every g∈M′that embeds (α)M′
(∆) \Γ in Ω \Γ′, we
have ((α)M′
(∆) \Γ)g= (α)M′
(∆) \Γ.
(This is because for such an element g, ((α)M′
(∆) \Γ)g⊆(α)M′\Γ′, and the latter set is
equal to (α)M′
(∆) \Γ.)
We now construct recursively, for each j≥0, elements αj, βj, γj∈Ω and a finite subset
Kj⊆M′, consisting of a.e. injective maps, indexed
(18) Kj={g(k0, k1, . . . , kj−1) : (k0, k1,...,kj−1)∈ {0,1}j}.
Let the 1-element set K0={g()}consist of the identity map 1 ∈M′. Assume induc-
tively that for some j≥0 the elements α0,...,αj−1, β0,...,βj−1, γ0,...,γj−1and the sets
K0,...,Kjhave been defined. Let ∆j={α0,...,αj−1} ∪ {1,...,j−1}, and let Γj⊆Ω be a
finite set, containing Γ′∪ {β0,...,βj−1, γ0,...,γj−1}, such that all elements of K0∪···∪Kj
19
embed Ω \Γjin Ω \Γ′. (Since K0∪ · · · ∪ Kjis finite and consists of elements that are a.e.
injective, we can find a finite set Γj, containing Γ′∪ {β0,...,βj−1, γ0,...,γj−1}, such that all
elements of K0∪ · · · ∪ Kjare injective on Ω \Γj. We can then enlarge this Γjto include the
finitely many elements that are mapped to Γ′by K0∪ · · · ∪ Kj.) Now, let us choose αj∈Ω
so that |(αj)M′
(∆j)\Γj|=m, and let βj, γj∈(αj)M′
(∆j)\Γjbe two distinct elements. We
note that βjand γjare distinct from β0,...,βj−1, γ0,...,γj−1, since the latter are elements
of Γj.
Next, let us construct the elements g(k0,...,kj−1,0), g(k0,...,kj−1,1) ∈Kj+1. Writing
g′=g(k0, k1,...,kj−1)∈Kj, we define g(k0,...,kj−1,0) = hg′and g(k0,...,kj−1,1) = f g′,
where h, f ∈M′
(∆j)are chosen so that (αj)hg′=βjand (αj)fg′=γj. It is possible to find
such hand f, by (17), using the fact that g′embeds Ω \Γjin Ω \Γ′. By the hypothesis that
a.e. injective maps form a dense subset in M(and hence in M′), we may further assume that
fand gare a.e. injective. As before, the images of α0,...,αj−1and 1,...,j −1 under hg′
and fg′will be the same as their images under g′, and these two maps will be a.e. injective
(as composites of a.e. injective maps).
Given an infinite string (ki)i∈ω∈ {0,1}ω, the elements of the sequence g(), g(k0), ...,
g(k0,...,ki−1), ... agree on successive sets {1,...,j−1}, and hence the sequence converges
in E. Since Mis closed, the limit belongs to M(in fact, M′). Considering our definition of
the elements αi,βi,γi, the submonoid Msatisfies the hypotheses of Lemma 20, and hence
E(A)4M.
In the previous lemma we were concerned with monoids under which elements of Ω, for
the most part, had disjoint forward orbits. Next, we give a similar argument, but adjusted
for monoids under which forward orbits can fall together.
Lemma 38. Let M⊆Ebe a submonoid that is closed in the function topology and has
dense a.e. injective maps. If for every finite Σ⊆Ω,M(Σ) 6={1}, then E(ργ)4Mfor some
γ∈Ω.
Proof. We may assume that there exist finite sets Γ,∆⊆Ω such that for all α∈Ω,
(α)M(∆) ⊆Γ∪ {α}. For, otherwise, the previous lemma implies that E(A)4M, where
Ais a partition of Ω into 2-element sets (and E(ργ)4E(A)for all γ∈Ω, by Lemmas 23,
25, and 26). Now, for every finite Σ ⊆Ω, we have that (M(∆))(Σ) =M(∆∪Σ) 6={1}. Also,
M(∆) is closed in the function topology, and a.e. injective maps form a dense subset in M(∆) .
Hence, we may replace Mwith M(∆) and thus assume that
(19) For every α∈Ω, (α)M⊆Γ∪ {α}.
We now construct recursively for each j≥0 an element αj∈Ω and a finite subset Kj⊆M,
consisting of a.e. injective maps, indexed
(20) Kj={g(k0, k1, . . . , kj−1) : (k0, k1,...,kj−1)∈ {0,1}j}.
Let the 1-element set K0={g()}consist of the identity map 1 ∈M. Assume inductively that
for some j≥0 the elements α0,...,αj−1and the sets K0,...,Kjhave been defined. Since
these sets are finite and consist of a.e. injective maps, there is a finite set ∆j⊆Ω, such that
all members of K0∪· · ·∪Kjare injective on Ω\∆j. We note that every h∈K0∪· · ·∪Kj, since
20
it is injective on Ω \∆j, must act as the identity on all but finitely many elements of Ω \∆j,
by (19). Let Γj⊆Ω be the finite set consisting of ∆j∪Γ∪ {α0,...,αj−1} ∪ {1,...,j −1}
and those elements of Ω \∆jthat are taken to Γ by members of K0∪ · · · ∪ Kj. We then
take αj∈Ω\Γjto be any element such that |(αj)M(Γj)|>1.
Next, let us construct the elements g(k0,...,kj−1,0), g(k0,...,kj−1,1) ∈Kj+1. Writing
g′=g(k0, k1,...,kj−1)∈Kj, we define g(k0,...,kj−1,0) = g′, noting that (αj)g′=αj, by
definition of Γj. Also, let us define g(k0,...,kj−1,1) = f g′, where f∈M(Γj)is chosen so that
(αj)fg′∈Γ and fis a.e. injective. (It is possible to find an a.e. injective f∈M(Γj)such that
(αj)f∈Γ, by our hypotheses that |(αj)M(Γj)|>1, that for every α∈Ω, (α)M⊆Γ∪ {α},
and that a.e. maps form a dense subset in M. Then (αj)f g′∈Γ, by (19).) We note that the
images of α0,...,αj−1and 1,...,j−1 under f g′will be the same as their images under g′,
and that fg′will be a.e. injective.
Given an infinite string (ki)i∈ω∈ {0,1}ω, the elements of the sequence g(), g(k0), ...,
g(k0,...,ki−1), ... agree on successive sets {1,...,j−1}, and hence the sequence converges
in E. Since Mis closed, the limit belongs to M. Now, let us fix some γ∈Γ. Also, let
h1∈Ebe a map that takes Ω bijectively to {αi}i∈ω, let h2∈Ebe the map that takes Γ
to γand fixes all other elements of Ω, and let h3∈Ebe a right inverse of h1that fixes γ.
Then E(ργ)⊆h1Mh2h3.
Lemma 39. Let M⊆Ebe a submonoid that has large stabilizers. Assume that there exist
finite sets Γ,∆⊆Ωsuch that for all α∈Ω,(α)M(∆) ⊆Γ∪ {α}. Then M4E(ργ) (where
γ∈Ωis any fixed element).
Proof. Since Mhas large stabilizers, we may replace Mwith M(∆) and assume that for all
α∈Ω, (α)M⊆Γ∪ {α}. Let us define a preorder ρon Ω by (α, β)∈ρif and only if
β∈(α)M. Then M⊆E(ρ), and ρis of type 3b. Hence, by Lemma 25, M4E(ργ).
The following proposition summarizes the previous four lemmas.
Proposition 40. Let M⊆Ebe a submonoid that is closed in the function topology, and has
large stabilizers and dense a.e. injective maps. Assume that there exists a finite set Σ⊆Ω
and a positive integer nsuch that the cardinalities of all forward orbits of M(Σ) are bounded
by n, but for every finite Σ⊆Ω,M(Σ) 6={1}. Let Abe a partition of Ωinto 2-element sets
and γ∈Ω. Then
(i) M≈E(A)if and only if for all finite Γ,∆⊆Ωthere exists α∈Ωsuch that (α)M(∆) 6⊆
Γ∪ {α}, and
(ii) M≈E(ργ)if and only if there exist finite sets Γ,∆⊆Ωsuch that for all α∈Ω,
(α)M(∆) ⊆Γ∪ {α}.
10 Main results
Combining the results of the previous section, we obtain our desired classification.
21
Theorem 41. Let M⊆Self(Ω) be a submonoid that is closed in the function topology, and
has large stabilizers and dense a.e. injective maps. Then Mfalls into exactly one of five
possible equivalence classes with respect to ≈, depending on which of the following conditions
Msatisfies:
(i) M(Σ) has an infinite forward orbit for every finite Σ⊆Ω.
(ii) There exists a finite set Σ⊆Ωsuch that all forward orbits of M(Σ) are finite, but for
every finite Σ⊆Ωthe cardinalities of the forward orbits of M(Σ) have no common
finite bound.
(iii) There exists a finite set Σ⊆Ωand a positive integer nsuch that the cardinalities of all
forward orbits of M(Σ) are bounded by n, but for all finite Γ,∆⊆Ωthere exists α∈Ω
such that (α)M(∆) 6⊆ Γ∪ {α}.
(iv) There exist finite sets Γ,∆⊆Ωsuch that for all α∈Ω,(α)M(∆) ⊆Γ∪ {α}, but for
every finite Σ⊆Ω,M(Σ) 6={1}.
(v) There exists a finite set Σ⊆Ωsuch that M(Σ) ={1}.
Proof. Every submonoid of E= Self(Ω) clearly satisfies exactly one of the above five con-
ditions. By Corollary 33, Corollary 35, Proposition 40, and Theorem 9, the first four of
these conditions describe disjoint ≈-equivalence classes, when considering submonoids that
are closed in the function topology, and have large stabilizers and dense a.e. injective maps.
Now, let M, M ′⊆Ebe submonoids that satisfy (v). Then we can find finite sets Σ,Γ⊆Ω
such that M(Σ) ={1}and M′
(Γ) ={1}. If Mand M′have large stabilizers, this implies that
M≈ {1} ≈ M′. Finally, for submonoids of Ethat satisfy the hypotheses of the theorem,
those that are described by condition (v) are countable (since they are ≈ {1}) and hence are
≺those submonoids that satisfy one of the other four conditions.
One can recover Theorem 27 from this by noting that when restricting to submonoids of
the form E(ρ), the above five cases exactly correspond to the five types of preorders on Ω
identified in Definition 22.
If we instead restrict our attention to submonoids M⊆Econsisting of injective maps,
then case (iv) of the above theorem cannot occur, and case (iii) can be stated more simply.
Corollary 42. Let M⊆Self(Ω) be a submonoid that has large stabilizers, is closed in the
function topology, and consists of injective maps. Then Mfalls into exactly one of four
possible equivalence classes with respect to ≈, depending on which of the following conditions
Msatisfies:
(i) M(Σ) has an infinite forward orbit for every finite Σ⊆Ω.
(ii) There exists a finite set Σ⊆Ωsuch that all forward orbits of M(Σ) are finite, but for
every finite Σ⊆Ωthe cardinalities of the forward orbits of M(Σ) have no common
finite bound.
(iii) There exists a finite set Σ⊆Ωand a positive integer nsuch that the cardinalities of
all forward orbits of M(Σ) are bounded by n, but for every finite Σ⊆Ω,M(Σ) 6={1}.
22
(iv) There exists a finite set Σ⊆Ωsuch that M(Σ) ={1}.
Let Einj ⊆Ebe the submonoid consisting of injective maps, and write ≈inj to denote
≈ℵ0,E inj . One might wonder whether the above four ≈-classes are also ≈inj-classes. They
are not. For example, let E≥⊆Edenote the submonoid of maps that are increasing with
respect to the usual ordering of ω= Ω (i.e., maps f∈Esuch that for all α∈Ω, (α)f≥α),
and set Einj
≥=Einj ∩E≥. Then Einj and Einj
≥are both closed in the function topology
(in E), since limits of sequences of injective (respectively, increasing) maps are injective
(respectively, increasing). Also, these two submonoids have large stabilizers. For, let Σ ⊆Ω
be any finite set. Upon enlarging Σ, if necessary, we may assume that Σ = {0,1,...,n−1}
for some n∈ω. Let f∈Ebe defined by (i)f=i+nfor all i∈ω, and let g∈Ebe defined
by (i)g=i−nfor i≥n(and arbitrarily on Σ). Then Einj ⊆fE inj
(Σ)gand Einj
≥⊆f(Einj
≥)(Σ)g.
(For, given h∈Einj, respectively Einj
≥, we can find ¯
h∈Einj
(Σ), respectively (Einj
≥)(Σ), such
that (i+n)¯
h= (i)h+nfor all i∈ω. Then h=f¯
hg.) This shows that Einj and Einj
≥
satisfy the hypotheses of the above corollary. They both clearly satisfy condition (i), and
hence Einj ≈Einj
≥. On the other hand, Einj 6≈inj Einj
≥. For, suppose that Einj =hEinj
≥∪Ui
for some finite U⊆Einj. Then every element of Einj can be written as a word f0f1. . . fm
in elements of Einj
≥∪U. Let f=f0f1...fi−1fifi+1 . . . fmbe such a word, and suppose that
fi∈Einj
≥\{1}. As an increasing injective non-identity element, fiis necessarily not surjective
(e.g., if j∈ωis the least element that is not fixed by fi, then j /∈(Ω)fi, since only elements
that are ≤jcan be mapped to jby fi). Since fi+1 . . . fmis injective, this implies that f
cannot be surjective either. Thus, if f∈ hEinj
≥∪Uiis surjective, then f∈ hUi, which is
absurd, since Uis finite.
We can further show that Einj and Einj
≥satisfy the analog of the large stabilizer con-
dition in Einj (i.e., for each finite subset Σ ⊆Ω, Einj
(Σ) ≈inj Einj and (Einj
≥)(Σ) ≈inj Einj
≥),
demonstrating that the obvious analog of the previous corollary for Einj in place of Eis
not true. Let Σ ⊆Ω be a finite subset. For each injective map f: Σ →Ω, let us pick
a permutation gf∈Einj that agrees with fon Σ, and let Ube the (countable) set con-
sisting of these gf. Now, let f∈Einj be any map. Then f g−1
f∈Einj
(Σ), which implies
that f∈ hEinj
(Σ) ∪Ui. Now, as a countable set of permutations, Ucan be embedded in a
subgroup of Sym(Ω) generated by two permutations, by [6, Theorem 5.7], and hence in a
submonoid of Sym(Ω) ⊆Einj generated by four elements. Therefore Einj ≈inj Einj
(Σ). To show
that Einj
≥≈inj (Einj
≥)(Σ), we will employ a similar method, though now we will assume, for
simplicity, that Σ = {0,1, . . . , n −1}for some n∈Ω, and require a more specific definition
of the elements gf. For each injective f: Σ →Ω, let us define a map gf∈Einj to agree
with fon Σ, fix all elements not in Σ ∪(Σ)f, and act on (Σ)fin any way that turns gfinto
a permutation of Ω. As before, let Ube the set consisting of these elements gf. Now, let
f∈Einj
≥be any map, and let h∈(Einj
≥)(Σ) be such that hagrees with fon Ω \Σ. We note
that for any α∈Ω\Σ, (α)h /∈Σ∪(Σ)f, since fis injective and increasing. Hence f=hgf,
and therefore f∈ h(Einj
≥)(Σ) ∪Ui. As before, this implies that Einj
≥≈inj (Einj
≥)(Σ).
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11 Groups
We recall that the group S= Sym(Ω) inherits from E= Self(Ω) the function topology
but is not closed in Ein this topology. For instance, using cycle notation for permutations
of Ω = ω, we see that the sequence (0,1),(0,1,2),...,(0,...,n),... converges to the map
i7→ i+ 1, which is not surjective. Thus we need a different notation for the closure of a set
of permutations of Ω in S; given a subset U⊆S, let clS(U) = clE(U)∩S. It is easy to see
that given a subset U⊆S, clE(clS(U)) = clE(U).
Let G⊆Sbe a subgroup, and let Σ ⊆Ω be finite. Then the forward orbits of clE(G)(Σ) =
clE(G(Σ)) coincide, by Lemma 28, with the forward orbits of G(Σ) , which are simply the
(group-theoretic) orbits of G(Σ) in Ω. In particular, by the above remark, the forward orbits
of clE(G)(Σ) = clE(clS(G))(Σ) coincide with the orbits of clS(G)(Σ).
We are now ready to prove that while clS(G) and clE(G) may be different for a given
group G, they are ≈-equivalent.
Proposition 43. Let Gbe a subgroup of S⊆E. Then clS(G)≈clE(G).
Proof. Let A, B, C, and Dbe partitions of Ω such that Aconsists of only one set, Bconsists
of finite sets such that there is no common finite upper bound on their cardinalities, C
consists of 2-element sets, and Dconsists of 1-element sets. By the main results of [4],
clS(G) is ≈Sto exactly one of S=S(A),S(B),S(C), or {1}=S(D)(see Theorem 15 for the
notation ≈S), and by Corollary 42, clE(G) is ≈to exactly one of E=E(A),E(B),E(C), or
{1}=E(D). Moreover, by the above remarks about (forward) orbits, clS(G)≈SS(X)for
some X∈ {A, B, C, D}if and only if clE(G)≈E(X). But, S(X)≈E(X), by Proposition 13,
and clS(G)≈SS(X)if and only if clS(G)≈S(X), by Theorem 15. Hence clS(G)≈clE(G).
We have referred to the main results of Bergman and Shelah in [4] while proving our
results. However, it is interesting to note that the Bergman-Shelah theorems can be recovered
from them.
Theorem 44 (Bergman and Shelah).Let G⊆Sym(Ω) be a subgroup that is closed in the
function topology on S. Then Gfalls into exactly one of four possible equivalence classes with
respect to ≈S(as defined in Theorem 15), depending on which of the following conditions G
satisfies:
(i) G(Σ) has an infinite orbit for every finite Σ⊆Ω.
(ii) There exists a finite set Σ⊆Ωsuch that all orbits of G(Σ) are finite, but for every
finite Σ⊆Ωthe cardinalities of the orbits of G(Σ) have no common finite bound.
(iii) There exists a finite set Σ⊆Ωand a positive integer nsuch that the cardinalities of
all orbits of G(Σ) are bounded by n, but for every finite Σ⊆Ω,G(Σ) 6={1}.
(iv) There exists a finite set Σ⊆Ωsuch that G(Σ) ={1}.
Proof. Let G1, G2⊆Sym(Ω) be two subgroups that are closed in the function topology.
By the remarks at the beginning of the section, G1and G2fall into the same one of the
four classes above if and only if clE(G1) and clE(G2) fall into the same one of the four
24
classes in Corollary 42 if and only if clE(G1)≈clE(G2). By the previous proposition,
clE(G1)≈clE(G2) if and only if G1≈G2, which, by Theorem 15, occurs if and only if
G1≈SG2.
Acknowledgements
The author is grateful to George Bergman, whose numerous comments and suggestions
have led to vast improvements in this note, and also to the referee for mentioning related
literature and suggesting ways to simplify some of the arguments.
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Department of Mathematics
University of Southern California
Los Angeles, CA 90089
USA
Email: mesyan@usc.edu
26
