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Topological transformation monoids
Z. Mesyan, J. D. Mitchell, and Y. Pe´resse
September 27, 2018
Abstract
We investigate semigroup topologies on the full transformation monoid ΩΩ of an
infinite set Ω. We show that the standard pointwise topology is the weakest Hausdorff
semigroup topology on ΩΩ, show that this topology is the unique Hausdorff semigroup
topology on ΩΩ that induces the pointwise topology on the group Sym(Ω) of all per-
mutations of Ω, and construct |Ω| distinct Hausdorff semigroup topologies on ΩΩ. In
the case where Ω is countable, we prove that the pointwise topology is the only Polish
semigroup topology on ΩΩ. We also show that every separable semigroup topology on
ΩΩ is perfect, describe the compact sets in an arbitrary Hausdorff semigroup topol-
ogy on ΩΩ, and show that there are no locally compact perfect Hausdorff semigroup
topologies on ΩΩ when |Ω| has uncountable cofinality.
Keywords: transformation monoid, topological semigroup, semitopological semigroup,
pointwise topology, Polish space
2010 MSC numbers: 08A35, 20M20, 54H15
1 Introduction
Recall that a topological group is a group G together with a topology on G that makes
the multiplication and inversion operations continuous. We shall refer to topologies of this
sort as group topologies. The discrete and trivial topologies are group topologies on every
group, but the question of finding interesting group topologies has received a great deal of
attention in the literature. We begin with a brief overview of this literature, to motivate our
work in this paper.
Markov [20] sparked significant interest in the subject when he asked whether there exist
infinite groups with no nondiscrete Hausdorff group topologies. This question was answered
in the affirmative by Shelah [27], assuming the continuum hypothesis, and, in ZFC, by
Olshanskii [25] in the same year. There are even infinite groups where every quotient of
every subgroup has no nondiscrete Hausdorff group topologies [17].
A substantial portion of the literature on topological groups has focused on Polish groups,
which arise in many areas of mathematics, particularly in descriptive set theory. These are
topological groups where the topology is Polish, that is, completely metrizable and separable.
Compared to the situation explored by Markov, it is easy to find examples both of Polish
groups and of topological groups that are not Polish. Specifically, since by the Cantor-
Bendixson Theorem [15, Theorem 6.4], every Polish space has cardinality either at most
1
ℵ0 or equal to 2
ℵ0 , any topological group with cardinality greater than 2ℵ0 is not Polish.
Moreover, there is no nondiscrete Polish group topology on any free group. (This follows
from the result of Dudley [8] that every homomorphism from a complete metric group to
a free group, with the discrete topology, is continuous.) On the other end of the spectrum
are Polish groups with infinitely many non-homeomorphic Polish group topologies. Perhaps
the simplest such example is that of the additive group R of real numbers, which is a
Polish group with respect to the standard topology. As vector spaces over the field Q of
the rational numbers, R and Rn are isomorphic for every positive integer n, and so R and
Rn are isomorphic as additive groups also. But Rn is homeomorphic to Rm if and only if
m = n, and so there are infinitely many non-homeomorphic Polish group topologies on R.
In contrast, Shelah [28] showed that it is consistent with the Zermelo-Fraenkel axioms of
set theory, without the axiom of choice, that every Polish group has a unique Polish group
topology. Of course, in the above example, the axiom of choice was needed to show that R
and Rn are isomorphic.
One of the most extensively studied topological groups is the group Sym(Ω) of all per-
mutations of a set Ω, which has a natural group topology, known as the pointwise topology.
This topology is Polish in the case where Ω is countable. Gaughan [10] showed that ev-
ery Hausdorff group topology on Sym(Ω) contains the pointwise topology, and that there is
no nondiscrete locally compact Hausdorff group topology on Sym(Ω). The latter answered
problem 96 in the Scottish Book [21], posed by Ulam. Kechris and Rosendal [16] showed that
any homomorphism from Sym(Ω) into a separable group is continuous, which together with
Gaughan’s result about the pointwise topology implies that there is a unique Polish group
topology on Sym(Ω), when Ω is countable. Moreover, Rosendal [26] extended Gaughan’s
local compactness result to show that there are not even any non-trivial homomorphisms
from Sym(Ω) into a locally compact Polish group. Gaughan’s result regarding the pointwise
topology was also recently extended, by Banakh, Guran, and Protasov [1], to any subgroup
of Sym(Ω) containing the permutations with finite support.
There are many further examples of groups with a unique Polish group topology, such
as the group of isometries of Minkowski spacetime (the usual framework for special relativ-
ity) [14]; see also [9, 13]. Furthermore, there is a wealth of examples of infinite groups with
no Polish group topologies [6, 19]. Additional references include [5, 7, 11].
Topological semigroups, that is, semigroups with topologies that make multiplication
continuous, have received somewhat less attention in the literature than topological groups.
However, some notable recent papers on the topic include [2, 3, 4].
In this paper we focus on the natural semigroup analogue of Sym(Ω), namely the semi-
group ΩΩ of all functions from Ω to Ω, which also has a natural pointwise topology. Our
goal is to explore the pointwise topology on ΩΩ in detail, along with semigroup topologies
on ΩΩ in general. In Section 2 we give a brief review of the basics of topological semigroups,
along with some methods for constructing topologies on semigroups. We then show in Theo-
rem 3.1 that the pointwise topology is the weakest T1 (and hence also Hausdorff) semigroup
topology on ΩΩ. More generally, the same holds for the topology induced by the pointwise
topology on any subsemigroup of ΩΩ that contains all the elements of ranks 1 and 2. This is
analogous to the result of Gaughan regarding the pointwise topology on Sym(Ω) mentioned
above.
Using Theorem 3.1 we show in Proposition 3.3 that the pointwise topology is the unique
2
T1 semigroup topology on Ω
Ω that induces the pointwise topology on Sym(Ω). From this
we conclude in Theorem 3.4 that, analogously to a result about Sym(Ω) stated above, the
pointwise topology is the only Polish semigroup topology on ΩΩ, when Ω is countably infinite.
We also show in Theorem 4.4 that if Ω is infinite, then there are either no isolated points or
2cf(|Ω|) isolated points in any semigroup topology on ΩΩ, where cf(|Ω|) denotes the cofinality
of |Ω|. In particular, every separable semigroup topology on ΩΩ is perfect (i.e., has no
isolated points), when Ω is infinite (Corollary 4.5).
In Section 6 we describe the compact sets in an arbitrary T1 semigroup topology on Ω
Ω
(Proposition 6.2). We then show that there are no locally compact perfect T1 semigroup
topologies on ΩΩ when |Ω| has uncountable cofinality (Theorem 6.4). This is a partial
analogue of Gaughan’s local compactness result mentioned above.
Along the way we give various examples to illustrate our results. For instance, we con-
struct |Ω| distinct Hausdorff semigroup topologies on ΩΩ, for Ω infinite (Proposition 4.1).
We also construct a natural perfect Hausdorff semigroup topology on ΩΩ for Ω of regular
cardinality, which is separable when Ω is countable, and has a very different flavour from
the pointwise topology (Proposition 5.1). The paper concludes with an open question.
2 Topological semigroups
A topological semigroup is a semigroup S together with a topology on S, such that the
semigroup multiplication, viewed as a function m : S × S → S, is continuous; where the
topology on S × S is the corresponding product topology. A semigroup is semitopological
if for every s ∈ S the maps rs : S → S and ls : S → S induced by right, respectively left,
multiplication by s are continuous with respect to the relevant topology. It is a standard
and easily verified fact that every topological semigroup is semitopological.
Next we give several simple but useful methods for constructing topologies on semigroups.
Lemma 2.1. Let f : S1 → S2 be a homomorphism of semigroups. Suppose that S2 is a
topological semigroup with respect to a topology T2, and let T1 be the least topology on S1
such that f is continuous. Then T1 = {(A)f
−1 : A ∈ T2}, and S1 is a topological semigroup
with respect to T1.
Proof. Noting that (
⋃
i∈I Ai)f
−1 =
⋃
i∈I(Ai)f
−1 and (
⋂
i∈I Ai)f
−1 =
⋂
i∈I(Ai)f
−1 for any
collection {Ai : i ∈ I} of subsets of S2, it follows that {(A)f
−1 : A ∈ T2} is a topology on S1.
Clearly, this topology is contained in T1, and f is continuous with respect to it. Therefore
T1 = {(A)f
−1 : A ∈ T2}.
Next, let mi : Si×Si → Si denote the multiplication map on Si, for i ∈ {1, 2}, and define
fc : S1×S1 → S2×S2 by (x, y)fc = ((x)f, (y)f) for all x, y ∈ S1. Then, viewing S1×S1 and
S2 × S2 as topological spaces in the product topologies induced by T1 and T2, respectively,
fc is continuous, since both of its coordinate functions, namely f : S1 → S2, are continuous.
(See, e.g., [24, Exercise 18.10].)
To show that m1 is continuous, let A ∈ T1, and let A
′ ∈ T2 be such that A = (A
′)f−1.
Then
(x, y) ∈ (A)m−11 ⇔ xy ∈ (A
′)f−1 ⇔ (x)f(y)f ∈ A′
⇔ ((x)f, (y)f) ∈ (A′)m−12 ⇔ (x, y) ∈ (A
′)m−12 f
−1
c
3
for all x, y ∈ S1, and hence (A)m
−1
1 = (A
′)m−12 f
−1
c . Since the composite fcm2 : S1×S1 → S2
of continuous functions is continuous, and A′ ∈ T2, it follows that (A)m
−1
1 is open in S1×S1.
Hence m1 is continuous, and therefore S1 is a topological semigroup with respect to T1.
Lemma 2.2. Let S be a semigroup.
(1) Given an ideal I of S, let T1 = {A : A ⊆ S \ I}∪ {S} and T2 = T1 ∪ {A∪ I : A ∈ T1}.
Then S is a topological semigroup with respect to T1 and T2.
(2) Let T1 and T2 be topologies on S, and let T3 be the topology generated by T1 ∪ T2. If
S is a topological semigroup with respect to T1 and T2, then the same holds for T3.
Proof. Let m : S × S → S denote the multiplication map on S.
(1) It is easy to see that T1 is closed under unions and (finite) intersections, and is hence
a topology on S. Since I is an ideal, (A)m−1 ⊆ (S \ I)× (S \ I), and hence (A)m−1 is open
in the product topology on S × S induced by T1, for any A ∈ T1 \ {S}. It follows that S is
a topological semigroup with respect to T1.
Next, let f : S → S/I be the natural homomorphism. Then the elements of T2 are
precisely the preimages under f of the subsets open in the discrete topology on S/I. Thus
S is a topological semigroup with respect to T2, by Lemma 2.1.
(2) Let A ∈ T3. Then A =
⋃
i(Bi ∩ Ci) for some Bi ∈ T1 and Ci ∈ T2. Thus
(A)m−1 =
⋃
i
((Bi ∩ Ci)m
−1) =
⋃
i
((Bi)m
−1 ∩ (Ci)m
−1).
Since S is a topological semigroup with respect to T1 and T2, for each i we can write
(Bi)m
−1 =
⋃
j(Bij×B
′
ij) and (Ci)m
−1 =
⋃
l(Cil×C
′
il), for some Bij , B
′
ij ∈ T1 and Cil, C
′
il ∈ T2.
Thus
(A)m−1 =
⋃
i
⋃
j
⋃
l
((Bij ×B
′
ij) ∩ (Cil × C
′
il)) =
⋃
i
⋃
j
⋃
l
((Bij ∩ Cil)× (B
′
ij ∩ C
′
il)),
since it is easy to see that
(X1 × Y1) ∩ (X2 × Y2) = (X1 ∩X2)× (Y1 ∩ Y2)
for any sets X1, X2, Y1, Y2. Hence (A)m
−1 is open in the product topology on S×S induced
by T3. It follows that S is a topological semigroup with respect to T3.
The two obvious topologies on a semigroup S can be viewed as extremal cases of the
constructions in Lemma 2.2(1). Specifically, if I = S, then T1 = T2 is the trivial topology,
while if I = ∅, then T1 = T2 is the discrete topology. As we shall see in Proposition 4.1, the
two topologies T1 and T2 can also be distinct from each other.
Recall that a topological space X is T1 if for any two distinct points x, y ∈ X there is an
open neighbourhood of x that does not contain y, and X is T2, or Hausdorff, if for any two
distinct points x, y ∈ X there are open neighbourhoods U and V of x and y, respectively,
such that U ∩ V = ∅.
We conclude this section with an observation that, while interesting, will not be used in
the rest of the paper.
4
Proposition 2.3. Let S be a topological semigroup with respect to a topology T , define
Ax =
⋂
{U : U ∈ T and x ∈ U}
for every x ∈ S, and let
ρS = {(x, y) : Ax = Ay} ⊆ S × S.
Then the following hold.
(1) The relation ρS is a congruence on S.
(2) If T is T1, then ρS = {(x, x) : x ∈ S}.
(3) The topology T is contained in the least topology on S with respect to which the natural
homomorphism S → S/ρS is continuous, where S/ρS is endowed with the discrete
topology.
Proof. (1) It is clear that ρS is an equivalence relation on S, and so it suffices to show that
(xs, ys), (sx, sy) ∈ ρS for all s ∈ S and (x, y) ∈ ρS. Thus suppose that (x, y) ∈ ρS, and
let U is an open neighbourhood of xs, for some s ∈ S. By the continuity of multiplication,
there exist open neighbourhoods V and W of x and s, respectively, such that VW ⊆ U . But
Ax = Ay, and so, in particular, y ∈ V . Thus ys ∈ VW ⊆ U , which shows that ys ∈ Axs. By
symmetry, also xs ∈ Ays, giving (xs, ys) ∈ ρS. The proof that (sx, sy) ∈ ρS is dual.
(2) If T is T1, then clearly Ax = {x} for all x ∈ S, which implies that ρS = {(x, x) : x ∈
S}.
(3) Let f : S → S/ρS be the natural homomorphism defined by (s)f = s/ρS, and let
P = {(V )f−1 : V ⊆ S/ρS}
be the least topology such that f is continuous (by Lemma 2.1). It is straightforward to
verify that U ∈ P if and only if U is a union of ρS-classes. Hence to prove that T ⊆ P, it
suffices to show that if x ∈ V ∈ T , then y ∈ V for all y ∈ S such that (x, y) ∈ ρS. But if
(x, y) ∈ ρS, then every open neighbourhood of x is also an open neighbourhood of y, and
hence y ∈ V , as required.
3 The pointwise topology
Given a set Ω, we denote by ΩΩ the full transformation monoid of Ω, consisting of all
functions from Ω to Ω, under composition. The pointwise (or function, or finite) topology
on ΩΩ has a base of open sets of the following form:
[σ : τ ] = {f ∈ ΩΩ : (σi)f = τi for all 0 ≤ i ≤ n},
where σ = (σ0, σ1, . . . , σn) and τ = (τ0, τ1, . . . , τn) are sequences of elements of Ω, and n ∈ N
(the set of the natural numbers). It is straightforward to see that this coincides with the
product topology on ΩΩ =
∏
ΩΩ, where each component set Ω is endowed with the discrete
topology. As a product of discrete spaces, this space is Hausdorff. It is well-known and easy
5
to see that ΩΩ is a topological semigroup with respect to the pointwise topology. If Ω is
finite, then ΩΩ is discrete in this topology. Finally, it is a standard fact that if Ω is countable,
then the pointwise topology on ΩΩ is Polish, i.e., separable and completely metrisable (see,
e.g., [15, Section 3.A, Example 3]).
We are now ready to prove an analogue of a result of Gaughan [10, Theorem 1] for
infinite symmetric groups, which shows that the pointwise topology is the weakest Hausdorff
semigroup topology on ΩΩ.
Recall that the rank of a function f ∈ ΩΩ is the cardinality of the image (Ω)f of f .
Theorem 3.1. Let Ω be a set, let S be a subsemigroup of ΩΩ that contains all the trans-
formations of rank at most 2, and let T be a topology on S with respect to which S is a
semitopological semigroup. Then the following are equivalent.
(1) T is Hausdorff.
(2) T is T1.
(3) Every set of the form [σ : τ ] ∩ S is open in T . (I.e., T contains the topology induced
on S by the pointwise topology on ΩΩ.)
(4) Every set of the form [σ : τ ] ∩ S is closed in T .
Proof. We may assume that Ω is infinite, since each of the conditions (1)–(4) is equivalent to
T being discrete in the case where Ω is finite. We note, however, that our arguments below
only require that Ω contains at least 2 elements. We shall prove that (1)⇒ (2)⇒ (3)⇒ (1)
and (3)⇒ (4)⇒ (2).
(1)⇒ (2) This is a tautology.
(2)⇒ (3) Assuming that T is T1, we start by showing that for any α, β ∈ Ω, the set
Fα,β = {f ∈ S : (α)f 6= β}
is closed in T . Let γ ∈ Ω \ {α}, let f ∈ S be the constant function with image α, and define
g ∈ S by
(δ)g =
{
γ if δ = β
α if δ 6= β
.
Since T is T1, the singleton {f} is closed, and since S is semitopological with respect to T ,
the composite lf ◦ rg : S → S of multiplication maps is continuous. Hence
(f)(lf ◦ rg)
−1 = {h ∈ S : fhg = f} = Fα,β
must be closed in T .
Let σ = (σ0, σ1, . . . , σn) and τ = (τ0, τ1, . . . , τn) be arbitrary sequences of elements of Ω,
of the same finite length. Then
S \ ([σ : τ ] ∩ S) =
n⋃
i=0
Fσi,τi ,
is closed, being a finite union of closed sets. Thus [σ : τ ] ∩ S is open in T .
6
(3)⇒ (1) Since the pointwise topology on ΩΩ is Hausdorff, so is the topology it induces
on S, and hence so is any topology on S that contains this induced topology.
(3)⇒ (4) Let σ = (σ0, σ1, . . . , σn) and τ = (τ0, τ1, . . . , τn) be arbitrary finite sequences of
elements of Ω, of length n + 1. Then
S \ ([σ : τ ] ∩ S) =
⋃
φ 6=τ
([σ : φ] ∩ S),
where φ = (φ0, φ1, . . . , φn) runs over all sequences of elements of Ω of length n + 1, distinct
from τ . Thus, if each [σ : φ] ∩ S is open in T , then so is S \ ([σ : τ ] ∩ S). In this case
[σ : τ ] ∩ S must be closed in T .
(4)⇒ (2) Let f ∈ S be any function. Then
{f} =
⋂
α∈Ω
{g ∈ S : (α)g = (α)f} =
⋂
α∈Ω
([(α) : ((α)f)] ∩ S).
Thus, if (4) holds, then {f} is closed in T . Since f ∈ S was arbitrary, it follows that T is
T1.
We shall show in Proposition 4.1 that ΩΩ has many Hausdorff semigroup topologies
different from the pointwise one.
Next we give an example of a semitopological subsemigroup S of ΩΩ with a topology
that is T1 but not Hausdorff, to show the necessity of the hypothesis on S in Theorem 3.1.
Given a set X we denote by |X| the cardinality of X .
Example 3.2. Let Ω be an infinite set, let S be an infinite subsemigroup of ΩΩ consisting of
bijections (so S is a subsemigroup of Sym(Ω)), and let T be the cofinite topology on S. That
is, T consists of precisely ∅ and the cofinite subsets of S (i.e., X ⊆ S such that |S \X| < ℵ0).
Clearly, T is T1. (In fact, the cofinite topology is the weakest T1 topology on any set.) On
the other hand, since S is infinite, the intersection of any two nonempty elements of T is
infinite, and hence T is not Hausdorff. We shall show that S is semitopological with respect
to T .
Let U ∈ T \ {S} be a nonempty open set, write S \ U = {g0, . . . , gn} for some n ∈ N,
and let f ∈ S. Since f is a bijection, and hence invertible both on the left and the right (in
Sym(Ω)), for each i ∈ {0, . . . , n} there can be at most one h ∈ S such that fh = gi, and at
most one h ∈ S such that hf = gi. Hence
(U)l−1f = {h ∈ S : fh ∈ U} = {h ∈ S : fh /∈ {g0, . . . , gn}},
and therefore |S \ (U)l−1f | ≤ n + 1. Similarly, |S \ (U)r
−1
f | ≤ n + 1. It follows that
(U)l−1f , (U)r
−1
f ∈ T for all U ∈ T , and hence S is semitopological with respect to T .
Given a set Ω, the pointwise topology on the group Sym(Ω) of all permutations of Ω is the
subspace topology induced on Sym(Ω) by the pointwise topology on ΩΩ. It is well-known
and easy to see that Sym(Ω) is a topological group with respect to the pointwise topology.
(That is, Sym(Ω) is a topological semigroup with respect to this topology, and the inversion
map Sym(Ω) → Sym(Ω) is continuous.) Moreover, as with ΩΩ, if Ω is countable, then the
pointwise topology on Sym(Ω) is Polish (see, e.g., [15, Section 9.A, Example 7]).
It turns out that the pointwise topology on ΩΩ is the only T1 semigroup topology that
induces the pointwise topology on Sym(Ω).
7
Proposition 3.3. Let Ω be a set, and let P denote the pointwise topology on ΩΩ. Suppose
that ΩΩ is a semitopological semigroup with respect to a T1 topology T , and that T induces
a topology on Sym(Ω) that is contained in the pointwise topology. Then T = P.
Proof. We may assume that Ω is infinite, since otherwise the only T1 topology on Ω
Ω is the
discrete topology, which coincides with P in this case.
Write Ω =
⋃
α∈ΩΣα, where the union is disjoint, and |Σα| = |Ω| for each α ∈ Ω. Let
g1 ∈ Ω
Ω be an injective function such that |Ω \ (Ω)g1| = |Ω|, and let g2 ∈ Ω
Ω be the function
that takes each Σα to α. Then it is easy to show that g1Sym(Ω)g2 = Ω
Ω (see the proof of [22,
Theorem 12] for the details).
Let U ∈ T be a nonempty open set, and let f ∈ U . Also let h ∈ Sym(Ω) be such
that g1hg2 = f . Then h ∈ (U)(lg1 ◦ rg2)
−1 ∈ T , and hence there is an open neighbourhood
V ⊆ Sym(Ω) ∩ (U)(lg1 ◦ rg2)
−1 of h in the topology induced on Sym(Ω) such that
V ⊇ {g ∈ Sym(Ω) : (β0)g = (β0)h, . . . , (βn)g = (βn)h},
for some distinct β0, . . . , βn ∈ Ω. Let Ξ = ({β0, . . . , βn})g
−1
1 , and let
W = {g ∈ ΩΩ : (α)g = (α)g1hg2 for all α ∈ Ξ}.
Then for each g ∈ W , there exists p ∈ V such that (α)g1p ∈ Σ(α)g for all α ∈ Ω, and hence
g = g1pg2 ∈ g1V g2. Thus f ∈ W ⊆ g1V g2 ⊆ U . Since f ∈ U was arbitrary and W ∈ P, this
implies that U ∈ P, and hence T ⊆ P. Finally, since T was assumed to be T1, we conclude
that T = P, by Theorem 3.1.
Recall that a topological space is separable if it contains a countable dense subset. Kechris
and Rosendal showed in [16, Theorem 6.26] that the pointwise topology is the only nontriv-
ial separable group topology on Sym(Ω), when Ω is countably infinite. In particular, the
pointwise topology is the only Polish group topology on Ω in this situation, since any Polish
topology must be Hausdorff, and hence nontrivial. Using Proposition 3.3, we can prove the
analogous statement for ΩΩ.
Theorem 3.4. Let Ω be a countably infinite set. Then the pointwise topology is the only
Polish topology on ΩΩ with respect to which it is a semitopological semigroup.
Proof. Let P denote the pointwise topology on ΩΩ, and let T be any Polish topology on ΩΩ
with respect to which it is a semitopological semigroup. Then, in particular, T is Hausdorff
and so T ⊇ P, by Theorem 3.1.
It is a standard fact (see, e.g., [15, Theorem 3.11]) that a subspace of a Polish space is
Polish if and only if it is Gδ (i.e., a countable intersection of open sets). Since Sym(Ω) is
Polish in the topology induced by P, it follows that Sym(Ω) is a Gδ subspace of Ω
Ω with
respect to P. From the fact that T ⊇ P, we conclude that Sym(Ω) is also a Gδ subspace of
ΩΩ with respect to T , and hence Sym(Ω) is a Polish space in the topology induced by T .
Since ΩΩ is a semitopological semigroup with respect to T , it is easy to see that the same
holds for Sym(Ω) with respect to the topology induced by T , and therefore it is a Polish
semigroup in this topology.
According to a result of Montgomery [23, Theorem 2], if a group is a semitopological
semigroup with respect to a Polish topology, then it is a topological group with respect to
8
that topology. Hence Sym(Ω) is a Polish group in the topology induced by T , and therefore
this topology must be the pointwise topology on Sym(Ω), by [16, Theorem 6.26]. Finally,
Proposition 3.3 implies that T = P.
We note that [16, Theorem 6.26] and Proposition 3.3 imply a stronger statement than
what is used in the final paragraph of the above proof. Specifically, the pointwise topology
is the only T1 semitopological semigroup topology on Ω
Ω that induces a separable group
topology on Sym(Ω), when Ω is countably infinite. To complement this observation and
Theorem 3.4, we record the following fact.
Corollary 3.5. Let Ω be an uncountable set. Then ΩΩ does not admit a separable T1
topology with respect to which it is a semitopological semigroup.
Proof. Suppose that ΩΩ is a semitopological semigroup with respect to a T1 topology T ,
and that X ⊆ ΩΩ is countable and dense in T . By Theorem 3.1, T contains the pointwise
topology, and hence each subset of ΩΩ of the form [σ : τ ] contains an element of X . Letting
α ∈ Ω be any element, we can find some β ∈ Ω \ (α)X , since |(α)X| ≤ ℵ0. Hence [(α) :
(β)] ∩X = ∅, producing a contradiction.
4 Topologies with isolated points
It is well-known (see [18, Lemma 1]) that the proper ideals of ΩΩ are precisely the
subsemigroups of the form
Iλ = {f ∈ Ω
Ω : |(Ω)f | < λ},
where λ is a cardinal satisfying 1 ≤ λ ≤ |Ω|. This fact allows us to construct |Ω| distinct
Hausdorff semigroup topologies on ΩΩ (containing the pointwise topology). These topologies,
along with all the others considered in this section, have isolated points, i.e., points x such
that {x} is open.
Proposition 4.1. Let Ω be an infinite set, let TP denote the pointwise topology on Ω
Ω, and
for each nonempty proper ideal Iλ of Ω
Ω, let
TI1
λ
= {A : A ⊆ ΩΩ \ Iλ} ∪ {Ω
Ω}
and
TI2
λ
= TI1
λ
∪ {A ∪ Iλ : A ∈ TI1
λ
}.
Also let TP∪Ii
λ
be the topology on ΩΩ generated by TP and TIi
λ
(i ∈ {1, 2}, 1 < λ ≤ |Ω|).
Then the following diagram consists of Hausdorff topologies with respect to which ΩΩ is a
topological semigroup:
TP ⊂ · · · ⊂ TP∪I2
λ
⊂ · · · ⊂ TP∪I2
3
⊂ TP∪I2
2
= ∪ ∪ ∪
TP ⊂ · · · ⊂ TP∪I1
λ
⊂ · · · ⊂ TP∪I1
3
⊂ TP∪I1
2
(1 < λ ≤ |Ω|). Moreover, TP∪I2
2
is the discrete topology.
9
Proof. By Lemma 2.2(1), ΩΩ is a topological semigroup with respect to each TIi
λ
, and hence,
by Lemma 2.2(2), the same holds for each TP∪Ii
λ
. Moreover, the topologies TP∪Ii
λ
are Haus-
dorff, since they contain the pointwise topology TP .
It is easy to see that for any 1 < λ ≤ |Ω| the topology TP∪I1
λ
consists of all sets of the
form A ∪ B, where A ∈ TP and B ⊆ Ω
Ω \ Iλ, while the topology TP∪I2
λ
consists of all sets
of the form (A ∩ Iλ) ∪ B, where A ∈ TP and B ⊆ Ω
Ω \ Iλ. Thus, TP∪I1
λ
⊂ TP∪I2
λ
for each
1 < λ ≤ |Ω|. Moreover, given two cardinals 1 < λ < κ ≤ |Ω| and i ∈ {1, 2}, Iλ ⊆ Iκ and
(A ∩ Iκ) ∪ B = (A ∩ Iλ) ∪ (B ∪ (A ∩ (Iκ \ Iλ)))
imply that TP∪Iiκ ⊆ TP∪Iiλ . Noting that any f ∈ Iκ \ Iλ is isolated in TP∪Iiλ, to conclude that
TP∪Iiκ ⊂ TP∪Iiλ it suffices to show that f ∈ Iκ \ Iλ is not isolated in TP∪Iiκ. But this follows
immediately from the fact that A∩ Iκ is either empty or infinite, for any A ∈ TP and κ ≥ 3,
and hence any open set in TP∪Iiκ containing f is infinite.
Finally, TP∪I2
2
is the discrete topology, since I2 ∩ [(α) : (α)] ∈ TP∪I2
2
is the singleton
set consisting of the constant function with value α, for any α ∈ Ω, and all non-constant
elements of ΩΩ are isolated in TP∪I2
2
, by definition.
In addition to describing the ideals of ΩΩ, Mal’cev [18] classified all the congruences ρ
on this semigroup. Thus, one can obtain additional Hausdorff semigroup topologies on ΩΩ
by putting the discrete topology on ΩΩ/ρ, and considering the topology generated by the
pointwise topology together with the one induced on ΩΩ via Lemma 2.1 (with f : ΩΩ → ΩΩ/ρ
taken to be the natural projection). However, the topologies obtained this way typically,
though not always, coincide with ones described in the previous proposition. Since non-Rees
congruences on ΩΩ tend to be complicated to describe, we shall not discuss the resulting
topologies in further detail here.
All the topologies on ΩΩ constructed in Proposition 4.1 have 2|Ω| isolated points, since
for example, all the surjective elements of ΩΩ are isolated in these topologies. Our next goal
is to show that any semigroup topology on ΩΩ with isolated points also must have “many”
of them. We begin with a couple of lemmas.
Lemma 4.2. Let Ω be a set, and suppose that ΩΩ is a semitopological semigroup with
respect to some topology. Also suppose that f, g ∈ ΩΩ have the property that there exist
b0, . . . , bn ∈ Ω
Ω and injective a0, . . . , an ∈ Ω
Ω, such that
(Ω)b0 ∪ · · · ∪ (Ω)bn = Ω
and
b0ga0 = · · · = bngan = f.
If f is isolated, then so is g.
Proof. Suppose that i ∈ {0, . . . , n} and that h ∈ ΩΩ is such that bihai = f . Then bihai =
bigai, and since ai is injective, bih = big. In other words, h and g agree on (Ω)bi. Hence if
bihai = f for all i ∈ {0, . . . , n}, then h = g, since (Ω)b0 ∪ · · · ∪ (Ω)bn = Ω. Thus
n⋂
i=0
(f)(lbi ◦ rai)
−1 = {g},
and therefore if f is isolated, then so is g.
10
A kernel class of an element f ∈ ΩΩ is a nonempty set of the form
(β)f−1 = {α ∈ Ω : (α)f = β}.
The kernel ker(f) of f ∈ ΩΩ is the collection of the kernel classes of f . Finally, the kernel
class type of f is the collection {fκ : 1 ≤ κ ≤ |Ω|}, where fκ is the cardinality of the set of
kernel classes of f of size κ.
Lemma 4.3. Let Ω be a nonempty set, and suppose that ΩΩ is a semitopological semigroup
with respect to some topology. Also let f, g ∈ ΩΩ be such that there is a bijection p from the
set of kernel classes of g to the set of kernel classes of f , with the property that |(Σ)p| ≥ |Σ|
for each kernel class Σ of g. If f is isolated, then so is g.
In particular, if an element of ΩΩ is isolated, then so is every other element of ΩΩ with
the same kernel class type.
Proof. We may assume that |Ω| ≥ 2, since otherwise f = g.
Since the kernel classes of f and g are in one-to-one correspondence, |(Ω)f | = |(Ω)g|,
and hence we can write (Ω)f = {βι : ι ∈ Γ} and (Ω)g = {γι : ι ∈ Γ}, for some index set Γ.
Moreover, by the hypothesis on p, we may choose the βι and γι such that |(βι)f
−1| ≥ |(γι)g
−1|
for all ι ∈ Γ. Let b ∈ ΩΩ be any surjection such that ((βι)f
−1) b = (γι)g
−1 for all ι ∈ Γ.
Also, let a0, a1 ∈ Ω
Ω be any elements that take γι to βι for all ι ∈ Γ, and that have constant
value δ0, δ1 ∈ Ω, respectively, on Ω \ (Ω)g, where δ0 6= δ1. Then bga0 = bga1 = f . We shall
show that g is the only element of ΩΩ with this property.
Suppose that bha0 = f for some h ∈ Ω
Ω. Then (α)bh = (α)bg for any α ∈ Ω satisfying
(α)f 6= δ0. Similarly, if bha1 = f and (α)f 6= δ1, then (α)bh = (α)bg. Thus if both
bha0 = f and bha1 = f , then bh = bg, and hence h = g, since b is surjective and therefore
left-invertible. Thus, we have shown that
(f)(lb ◦ ra0)
−1 ∩ (f)(lb ◦ ra1)
−1 = {g},
and therefore if f is isolated, then so is g.
The final claim is immediate.
The cofinality cf(κ) of a cardinal κ is the least cardinal λ such that κ is the union of
λ cardinals, each smaller than κ. A cardinal κ is regular if cf(κ) = κ. It is a standard
and easily verified fact that cf(κ) ≤ κ for every cardinal κ. Let us also recall that for
an infinite cardinal κ, the cofinality cf(κ) is necessarily infinite. (Otherwise it would be
the case that κ = λ0 + · · · + λn for some n ∈ N and infinite cardinals λi < κ. But∑n
i=0 λi = max{λ0, . . . , λn}, since the λi are infinite, contradicting λi < κ.)
We are now ready for the main result of this section.
Theorem 4.4. Let Ω be an infinite set, let κ = cf(|Ω|), and suppose that ΩΩ is a semitopo-
logical semigroup with respect to some topology. Then there are either no isolated points or
at least 2κ isolated points in ΩΩ.
In particular, if |Ω| is regular, then there are either no isolated points or 2|Ω| isolated
points in ΩΩ.
11
Proof. Suppose that f ∈ ΩΩ is isolated. If f has at least κ distinct kernel classes, then the
set of those g ∈ ΩΩ with the same kernel class type as f has cardinality at least 2κ. (Since
κ is infinite, one can obtain 2κ such g by permuting κ of the elements in the image of f .)
Hence, there are at least 2κ isolated points in ΩΩ, by Lemma 4.3. Let us therefore assume
that f has strictly fewer than κ distinct kernel classes, and so, in particular, Ω \ (Ω)f 6= ∅,
since κ ≤ |Ω|.
It cannot be the case that every kernel class of f has cardinality strictly less than |Ω|,
since then the union of these (fewer than κ) kernel classes would have cardinality strictly
less than |Ω|, by the definition of “cofinality”. This would contradict the fact that the union
of the kernel classes of any element of ΩΩ is Ω. Therefore, there must be a kernel class Σ of
f such that |Σ| = |Ω|. Let (Σ)f = β, and partition Σ as Σ = Λ0 ∪ Λ1, such that Λ0 and Λ1
are nonempty. Let g ∈ ΩΩ be defined by
(α)g =


(α)f if α ∈ Ω \ Σ
β if α ∈ Λ0
γ if α ∈ Λ1
,
where γ is any value in Ω \ (Ω)f (which exists by assumption). Let b0, b1 ∈ Ω
Ω be any
functions that act as the identity on Ω \ Σ, and where (Σ)b0 = Λ0 and (Σ)b1 = Λ1. Also let
a0 ∈ Ω
Ω be the identity function, and let a1 ∈ Ω
Ω be the transposition interchanging β and
γ. Then
(Ω)b0 ∪ (Ω)b1 = Λ0 ∪ Λ1 ∪ (Ω \ Σ) = Ω,
and f = b0ga0 = b1ga1. Hence g is isolated, by Lemma 4.2. Since there are 2
|Σ| = 2|Ω|
ways to partition Σ into Λ0 and Λ1, resulting in 2
|Ω| ≥ 2κ functions g, the desired conclusion
follows.
For the final claim, we recall the fact that 2λ = µλ for any infinite cardinal λ and any
cardinal µ satisfying 2 ≤ µ ≤ λ (see, e.g., [12, Lemma 5.6]). Thus 2|Ω| = |Ω||Ω| = |ΩΩ|, since
Ω is assumed to be infinite. Therefore, if |Ω| is regular and there are isolated points in ΩΩ,
then there are precisely 2cf(|Ω|) = 2|Ω| of them.
Recall that a topology is perfect if it has no isolated points.
Corollary 4.5. Let Ω be an infinite set. Then every separable topology on ΩΩ, with respect
to which ΩΩ is a semitopological semigroup, is perfect.
Proof. Suppose that ΩΩ is a semitopological semigroup with respect to a separable topology
T . By Theorem 4.4, if T has isolated points, then it must have at least 2cf(|Ω|) ≥ 2ℵ0 of
them, since as noted above, cf(|Ω|) ≥ |Ω| ≥ ℵ0. But since T is separable, it can have at
most countably many isolated points, and therefore T must be perfect.
We conclude this section with another result about isolated points. It sheds additional
light on the topologies constructed in Proposition 4.1 and strengthens Lemma 4.3.
Proposition 4.6. Let Ω be an infinite set, and suppose that ΩΩ is a semitopological semi-
group with respect to some topology. Let f, g ∈ ΩΩ be such that f is isolated. If either of the
following conditions holds, then g is also isolated.
12
(1) |(Ω)f | ≤ |(Ω)g| < ℵ0.
(2) |(Ω)f | = |(Ω)g| ≥ ℵ0, and there is an injection p from the set Kg of kernel classes
of g to the set Kf of kernel classes of f , with the property that |(Σ)p| ≥ |Σ| for each
Σ ∈ Kg.
Proof. Suppose that (1) holds. Let Ξ0, . . . ,Ξn be the kernel classes of f , where we may
assume that |Ξ0| = |Ω|, and let Υ0, . . . ,Υm be the kernel classes of g. Also, write (Ξi)f = αi
and (Υj)g = βj for all i ∈ {0, . . . , n} and j ∈ {0, . . . , m}. For each j ∈ {0, . . . , m} let
aj ∈ Ω
Ω be an injection such that (βj)aj = α0 and
{α0, . . . , αn} ⊆ ({β0, . . . , βm})aj.
Such functions exist, since n ≤ m. Also for each j ∈ {0, . . . , m} let bj ∈ Ω
Ω be such that
(Ξ0)bj = Υj, and (Ξi)bj ⊆ Υl, where βl = (αi)a
−1
j , for each i ∈ {1, . . . , n}. Such functions
exist, since |Ξ0| = |Ω|. Then bjgaj = f for each j ∈ {0, . . . , m}, and
Ω ⊇
m⋃
i=0
(Ω)bj ⊇
m⋃
j=0
Υj = Ω,
which implies that
(Ω)b0 ∪ · · · ∪ (Ω)bm = Ω.
Therefore g is isolated, by Lemma 4.2.
Now suppose that (2) holds. Upon replacing f with another function having the same
kernel, we may assume, by Lemma 4.3, that |Ω\ (Ω)f | = |Ω|. Since the image of g is infinite,
we can write
Kg = {Ξι : ι ∈ Γ} ∪ {Υζ : ζ ∈ ∆},
where the union is disjoint, and |Γ| = |∆| = |(Ω)g|. Let b0 ∈ Ω
Ω be such that ((Ξι)p)b0 = Ξι
for each ι ∈ Γ, and b0 takes Kf \ ({Ξι : ι ∈ Γ})p injectively into {Υζ : ζ ∈ ∆}. (That is, b0
maps all the points in an element of Kf \ ({Ξι : ι ∈ Γ})p to some Υζ .) Such a transformation
exists, since |(Ξι)p| ≥ |Ξι| for each Ξι, and |∆| = |(Ω)f |. Similarly, let b1 ∈ Ω
Ω be such
that ((Υζ)p)b1 = Υζ for each ζ ∈ ∆, and b1 takes Kf \ ({Υζ : ζ ∈ ∆})p injectively into
{Ξι : ι ∈ Γ}. Then
Ω ⊇ (Ω)b0 ∪ (Ω)b1 ⊇
⋃
ι∈Γ
Ξι ∪
⋃
j∈∆
Υζ = Ω,
giving (Ω)b0 ∪ (Ω)b1 = Ω. Moreover,
ker(b0g) = ker(b1g) = ker(f),
and since |Ω \ (Ω)f | = |Ω|, there exist injective a0, a1 ∈ Ω
Ω such that b0ga0 = b1ga1 = f .
Hence, by Lemma 4.2, g is isolated.
13
5 Topologies obtained by restricting images
Given a set Ω and collection {Σα : α ∈ Ω} of subsets of Ω, we identify
∏
α∈ΩΣα with
{f ∈ ΩΩ : (α)f ∈ Σα for all α ∈ Ω}.
Next we construct a perfect Hausdorff topology on ΩΩ, different from the pointwise topology,
by declaring certain sets of the form
∏
α∈ΩΣα open.
Proposition 5.1. Let Ω be an infinite set, let
B =
{∏
α∈Ω
Σα : |{α ∈ Ω : β /∈ Σα}| < |Ω| for all β ∈ Ω
}
,
and let T = 〈B〉 denote the topology on ΩΩ generated by B. Then the following hold.
(1) B is closed under finite intersections, and in particular is a base for T .
(2) Any base for T must have strictly more than |Ω| elements.
(3) T strictly contains the pointwise topology. In particular, T is Hausdorff.
(4) T is perfect.
(5) There is a subset of ΩΩ of cardinality |
∑
κ<|Ω| |Ω|
κ| that is dense in T . In particular,
if Ω is countable, then T is separable.
(6) No nonempty element of B is contained in a compact subset of ΩΩ.
(7) If |Ω| is regular, then ΩΩ is a topological semigroup with respect to T .
Proof. (1) Let n ∈ N, and let
∏
α∈ΩΣi,α ∈ B for i ∈ {0, . . . , n}. Then
n⋂
i=0
(∏
α∈Ω
Σi,α
)
=
∏
α∈Ω
( n⋂
i=0
Σi,α
)
.
Now ∣∣∣∣
{
α ∈ Ω : β /∈
n⋂
i=0
Σi,α
}∣∣∣∣ ≤
n∑
i=0
|{α ∈ Ω : β /∈ Σi,α}| <
n∑
i=0
|Ω| = |Ω|
for any β ∈ Ω. Hence
⋂n
i=0(
∏
α∈ΩΣi,α) ∈ B, showing that B is closed under finite intersec-
tions. Since the elements of B cover ΩΩ (e.g., because ΩΩ ∈ B), it follows that B is a base
for T = 〈B〉.
(2) Let {Xα : α ∈ Ω} ⊆ T be a collection of nonempty open subsets of Ω
Ω. We shall
construct sets Σα ⊆ Ω such that
∏
α∈ΩΣα ∈ B and Xβ 6⊆
∏
α∈ΩΣα for all β ∈ Ω, which
implies that {Xα : α ∈ Ω} cannot be a base for T , and hence that any base for T must have
strictly more than |Ω| elements.
Without loss of generality, we may assume that Ω is a cardinal. Since, by (1), B is a base
for T , for each β ∈ Ω we can find a nonempty
∏
α∈ΩΣα,β ∈ B such that
∏
α∈ΩΣα,β ⊆ Xβ.
14
Next we shall construct recursively a function f : Ω→ Ω. Assuming that (α)f is defined for
all α < β ∈ Ω, let (β)f be such that (β)f 6= (α)f for all α < β, and such that β ∈ Σ(β)f,β .
Such (β)f ∈ Ω exists, since
|{(α)f : α < β} ∪ {α ∈ Ω : β /∈ Σα,β}| < Ω.
Now for each α ∈ Ω let
Σα =
{
Ω \ {β} if α = (β)f
Ω if α /∈ (Ω)f
.
This is well-defined, since f is injective, and clearly
∏
α∈ΩΣα ∈ B. Moreover,
∏
α∈ΩΣα,β 6⊆∏
α∈ΩΣα for all β ∈ Ω, since β ∈ Σ(β)f,β \ Σ(β)f . Hence Xβ 6⊆
∏
α∈ΩΣα for all β ∈ Ω, as
desired.
(3) Let σ = (σ0, σ1, . . . , σn) and τ = (τ0, τ1, . . . , τn) be two finite sequences of elements of
Ω of the same length, and let
Σα =
{
{τi} if α = σi for some i ∈ {0, . . . , n}
Ω if α ∈ Ω \ {σ0, . . . , σn}
.
Then [σ : τ ] =
∏
α∈ΩΣα ∈ T , showing that T contains the pointwise topology, and is hence
Hausdorff. Since the sets of the form [σ : τ ] constitute a base for the pointwise topology of
cardinality |Ω|, we conclude from (2) that T strictly contains the pointwise topology.
(4) If
∏
α∈ΩΣα ∈ B, then for any β, γ ∈ Ω there must be infinitely many α ∈ Ω such that
β, γ ∈ Σα. Thus each nonempty element of B is (uncountably) infinite, and hence by (1), T
has a base consisting of infinite sets. In particular, no point can be isolated in T .
(5) Fix α ∈ Ω. Given Γ ⊆ Ω, where 0 < |Γ| < |Ω|, let XΓ ⊆ Ω
Ω consist of all f ∈ ΩΩ
such (Ω \ Γ)f = α. Then letting X be the union of the XΓ, we see that X is dense in T .
Moreover
|X| =
∑
Γ
|XΓ| =
∑
Γ
|ΩΓ| =
∑
κ<|Ω|
|Ω|κ.
In the case where Ω is countable, |X| = ℵ0, which implies that T is separable.
(6) Let
∏
α∈ΩΣα ∈ B be nonempty, and suppose that
∏
α∈ΩΣα ⊆ X for some compact
X ⊆ ΩΩ. Then, by the definition of B, we can find an infinite Γ ⊆ Ω and distinct γβ ∈ Ω,
such that γβ ∈ Σβ and |Σβ| ≥ 2 for all β ∈ Γ. For each β ∈ Γ let
Ξα,β =
{
{γβ} if α = β
Ω if α 6= β
,
and let
∆β =
{
Ω \ {γβ} if β ∈ Γ
Ω if β /∈ Γ
.
Also let
Y =
{∏
α∈Ω
∆α
}
∪
{∏
α∈Ω
Ξα,β : β ∈ Γ
}
.
Then
∏
α∈Ω Ξα,β ∈ B for each β ∈ Γ, and
∏
α∈Ω∆α ∈ B. Moreover, Y is an open cover
of ΩΩ, and hence also of X , since for each f ∈ ΩΩ \
∏
α∈Ω∆α there is some β ∈ Γ such
15
that (β)f = γβ, and hence f ∈
∏
α∈Ω Ξα,β. However, it is easy to see that
∏
α∈ΩΣα is not
contained in the union of any finite collection of elements of Y , and hence neither is X ,
contradicting X being compact. Therefore
∏
α∈Ω Σα is not contained in any compact subset
of ΩΩ.
(7) Without loss of generality we may assume that Ω is a (regular) cardinal, and so in
particular, it is well-ordered. Let
∏
α∈ΩΣα ∈ B and f, g ∈ Ω
Ω be such that fg ∈
∏
α∈ΩΣα.
We shall construct
∏
α∈Ω Γα,
∏
α∈Ω∆α ∈ B such that f ∈
∏
α∈Ω Γα, g ∈
∏
α∈Ω∆α, and(∏
α∈Ω
Γα
)(∏
α∈Ω
∆α
)
⊆
∏
α∈Ω
Σα. (†)
The existence of such sets implies that the multiplication map on ΩΩ is continuous with
respect to T , and hence that ΩΩ is a topological semigroup.
For each α ∈ Ω let
Γα = {(α)f} ∪ {β ∈ Ω : (β)g ∈ Σα and β < α}
and
∆α =
⋂
{β∈Ω:α∈Γβ}
Σβ .
Then clearly f ∈
∏
α∈Ω Γα. If β /∈ Γα for some α, β ∈ Ω, then either α ≤ β or (β)g /∈ Σα.
Since
|{α ∈ Ω : α ≤ β} ∪ {α ∈ Ω : (β)g /∈ Σα}| < Ω
for all β ∈ Ω, we see that |{α ∈ Ω : β /∈ Γα}| < Ω for all β ∈ Ω, and hence
∏
α∈Ω Γα ∈ B.
To show that g ∈
∏
α∈Ω∆α, let α ∈ Ω, and let β ∈ Ω be such that α ∈ Γβ. Then either
α = (β)f , or (α)g ∈ Σβ. In both cases, (α)g ∈ Σβ , and hence (α)g ∈ ∆α, as desired.
To prove that
∏
α∈Ω∆α ∈ B we note that for any β ∈ Ω,
|{α ∈ Ω : β /∈ ∆α}| = |{α ∈ Ω : ∃γ ∈ Ω (α ∈ Γγ ∧ β /∈ Σγ)}| =
∣∣∣ ⋃
{γ∈Ω:β /∈Σγ}
Γγ
∣∣∣ < Ω,
since |{γ ∈ Ω : β /∈ Σγ}| < Ω, |Γγ| < Ω for each γ ∈ Ω, and Ω is regular.
Finally, let γ ∈ Ω, f ′ ∈
∏
α∈Ω Γα, and g
′ ∈
∏
α∈Ω∆α. Then (γ)f
′ ∈ Γγ, and so
(γ)f ′g′ ∈ ∆(γ)f ′ =
⋂
{β∈Ω:(γ)f ′∈Γβ}
Σβ ⊆ Σγ .
Hence f ′g′ ∈
∏
α∈ΩΣα, which shows that (†) holds.
The next result suggests that making sets of the form
∏
α∈Ω Σα open, other than those in
the above construction, tends to result in semigroup topologies on ΩΩ that are not perfect.
More specifically, only non-perfect semigroup topologies on ΩΩ have open sets
∏
α∈ΩΣα
not of the above form that contain constant functions. In particular, no perfect semigroup
topology on ΩΩ can have an open set of the form
∏
α∈ΩΣ, where Σ is a proper subset of Ω.
Proposition 5.2. Let Ω be an infinite set, and suppose that ΩΩ is a semitopological semi-
group with respect to some topology.
16
(1) Suppose that there exist β ∈ Ω, a constant function f ∈ ΩΩ, and an open neighbourhood
U of f , such that U ⊆
∏
α∈ΩΣα and |{α ∈ Ω : β /∈ Σα}| = |Ω|. Then f is isolated.
(2) Let {Σα : α ∈ Ω} be a collection of disjoint nonempty subsets of Ω, and suppose that
U ⊆
∏
α∈Ω Σα is open. Then every element of U is isolated.
Proof. (1) Write (Ω)f = γ, set Γ = {α ∈ Ω : β ∈ Σα}, and let g ∈ Ω
Ω be defined by
(α)g =
{
β if α ∈ Ω \ {γ}
γ if α = γ
.
Then
(U)r−1g ⊆
(∏
α∈Ω
Σα
)
r
−1
g ⊆ {y ∈ Ω
Ω : (α)y = γ for all α ∈ Ω \ Γ},
and f ∈ (U)r−1g . If Γ = ∅, then this implies that (U)r
−1
g = {f}, and this set is open, since U
is. Hence we may assume that Γ 6= ∅. Since |Ω \ Γ| = |Ω|, there exists an h ∈ ΩΩ such that
(Ω \ Γ)h = Γ. Then
(U)r−1g ◦ l
−1
h ⊆
(∏
α∈Ω
Σα
)
r
−1
g ◦ l
−1
h ⊆ {y ∈ Ω
Ω : (α)y = γ for all α ∈ Γ},
and f ∈ (U)r−1g ◦ l
−1
h . Since U is open, it follows that (U)r
−1
g ∩ ((U)r
−1
g ◦ l
−1
h ) = {f} is open.
(2) We may assume that U 6= ∅, since otherwise there is nothing to prove. Let f ∈ U be
arbitrary. For each α ∈ Ω we wish to construct Ξα ⊆ Ω, such that the following properties
are satisfied:
(a) |Σα| ≤ |Ξα| for all α ∈ Ω,
(b) Σα ∩ Ξα = {(α)f} for all α ∈ Ω,
(c) Ξα ∩ Ξβ = ∅ for all distinct α, β ∈ Ω,
(d)
⋃
α∈Ω Ξα = Ω.
First, partition Ω as Ω =
⋃
α∈Ω Λα, where |Λα| = |Ω| for each α ∈ Ω. Then |(
⋃
β∈Λα
Σβ) \
Σα| = |Ω| for each α ∈ Ω, since the Σα are disjoint and |Λα| = |Ω|. Hence for each α ∈ Ω we
can construct Ξα that satisfies (a) and (b) by choosing |Σα| elements from (
⋃
β∈Λα
Σβ) \Σα,
along with (α)f . Since the Λα are disjoint, the sets so constructed also satisfy (c). Now add
any remaining elements of Ω to the Ξα in any way that preserves Σα ∩ Ξα = {(α)f} for all
α ∈ Ω. The resulting sets Ξα will then satisfy (a), (b), (c), and (d).
Since the Ξα satisfy (a), (b), and (c), we can find a g ∈ Ω
Ω such that (Ξα)g = Σα and
(α)fg = (α)f for all α ∈ Ω. If h ∈ ΩΩ is such that hg ∈ U , then (α)hg ∈ Σα, and hence
(α)h ∈ (Σα)g
−1 = Ξα, for all α ∈ Ω (since the Σα are disjoint and
⋃
α∈Ω Ξα = Ω). Thus
(U)r−1g ⊆
∏
α∈Ω Ξα, and therefore U ∩ (U)r
−1
g = {f}, in light of condition (b). Since U is
open, it follows that f is isolated.
17
6 Compactness
Our next goal is to describe the compact sets in an arbitrary T1 semigroup topology on
ΩΩ. We begin with a characterisation of the compact sets in the pointwise topology on ΩΩ.
If Ω is countable, in which case the pointwise topology on ΩΩ is Polish, and hence metrisable,
this characterisation can be obtained from the standard fact that a subset of a metric space
is compact if and only if it is complete and totally bounded. Proving the fact in question
for arbitrary Ω is also easy, but we provide the details for the convenience of the reader.
Given a topological space X , a subset U of X is nowhere dense if X \ U is dense in X ,
where U denotes the closure of U .
Lemma 6.1. Let Ω be a set, let X ⊆ ΩΩ, and let T denote the pointwise topology on ΩΩ.
Then the following hold.
(1) X is compact in T if and only if X is closed in T and |(α)X| < ℵ0 for all α ∈ Ω.
(2) If Ω is infinite and X is compact in T , then X is nowhere dense.
Proof. (1) We may assume that Ω is nonempty, since otherwise every subset of ΩΩ is both
compact and closed in T .
Suppose that X is compact. It is a standard fact that in a Hausdorff space every compact
subset is closed (see, e.g., [24, Theorem 26.3]), and hence X is closed in T . Now, let α ∈ Ω.
Then
X =
⋃
β∈Ω
([(α) : (β)] ∩X).
Since X compact in T ,
X =
n⋃
i=0
([(α) : (βi)] ∩X)
for some β0, . . . , βn ∈ Ω and n ∈ N. Hence (α)X ⊆ {β0, . . . , βn}, giving |(α)X| < ℵ0.
Conversely, suppose that X is closed in T and |(α)X| < ℵ0 for all α ∈ Ω. For each
α ∈ Ω let Σα = (α)X , and let Y =
∏
α∈ΩΣα ⊆ Ω
Ω consist of all functions f ∈ ΩΩ such
that (α)f ∈ Σα. Then X ⊆ Y . Since every closed subset of a compact set is compact
(see, e.g., [24, Theorem 26.2]), it suffices to show that Y is compact. Now the subspace
topology on Y induced by T is precisely the product topology on
∏
α∈ΩΣα resulting from
endowing each Σα with the discrete topology. Since each Σα is finite, and hence compact, Y
is a product of compact sets. Therefore Y compact, by Tychonoff’s theorem (see, e.g., [24,
Theorem 37.3]), as desired.
(2) If X is compact, then by (1), for each α ∈ Ω the set (α)X = Σα is finite. As before
we have X ⊆ Y =
∏
α∈ΩΣα. Now let σ and τ be any two sequences of elements of Ω of the
same finite length. Since Ω is infinite, we can find some α ∈ Ω \σ, and some f ∈ [σ : τ ] such
that (α)f /∈ Σα. Then f ∈ Ω
Ω \ Y ⊆ ΩΩ \X . Since sets of the form [σ : τ ] constitute a base
for T , this implies that ΩΩ \X = ΩΩ \X is dense in ΩΩ, and hence X is nowhere dense.
Proposition 6.2. Let Ω be a set, and suppose that ΩΩ is a semitopological semigroup with
respect to some T1 topology. If X ⊆ Ω
Ω is compact, then X is closed and |(α)X| < ℵ0 for
all α ∈ Ω.
18
Proof. By Theorem 3.1, if T is a T1 topology with respect to which Ω
Ω is a semitopological
semigroup, then T contains the pointwise topology. Hence a set compact in such a topology
T is also compact in the pointwise topology. The desired conclusion now follows from
Lemma 6.1(1).
In general, it is not the case that all subsets of ΩΩ (for Ω infinite) that are closed and
satisfy |(α)X| < ℵ0 for all α ∈ Ω are compact in a T1 semigroup topology. For example,
consider the discrete topology, where only finite sets are compact.
Recall that a topological space is locally compact if every element has an open neigh-
bourhood that is contained in some compact set. The remainder of this section is devoted
to showing that locally compact semigroup topologies on ΩΩ are generally not well-behaved.
Corollary 6.3. Let Ω be an infinite set, and suppose that ΩΩ is a semitopological semigroup
with respect to some locally compact T1 topology T . Then any base for T must have strictly
more than |Ω| elements.
Proof. Suppose that there is a base B for T such that |B| ≤ |Ω|. Since T is locally compact,
for every f ∈ ΩΩ there exists some Uf ∈ B and a compact set Xf ⊆ Ω
Ω such that f ∈ Uf ⊆
Xf . Hence
⋃
f∈ΩΩ Xf = Ω
Ω. Since |B| ≤ |Ω|, we can find some subset {Xα : α ∈ Ω} of
{Xf : f ∈ Ω
Ω}, such that
⋃
α∈ΩXα = Ω
Ω. In view of each Xα being compact, it follows from
Proposition 6.2 that (α)Xα is finite for each α ∈ Ω. But since Ω is infinite, we can find some
f ∈ ΩΩ such that (α)f /∈ (α)Xα for all α ∈ Ω, which contradicts
⋃
α∈ΩXα = Ω
Ω. Thus any
base for T must have > |Ω| elements.
We observe that while the topology on ΩΩ constructed in Proposition 5.1 satisfies the
conclusion of the previous result, it is not locally compact, since Proposition 5.1(6) implies
that no element of ΩΩ has an open neighbourhood that is contained in a compact set.
We are now ready to give a partial analogue of the result [10, Section 4] of Gaughan that
there is no nondiscrete locally compact Hausdorff group topology on Sym(Ω).
Theorem 6.4. Let Ω be an infinite set such that |Ω| has uncountable cofinality, and suppose
that ΩΩ is a semitopological semigroup with respect to a locally compact topology T . Then
T is either not perfect or not T1.
Proof. Seeking a contradiction, suppose that T is perfect and T1, and let f ∈ Ω
Ω be a
constant function. Since T is locally compact, there must be a compact set X and an
open neighbourhood U of f such that U ⊆ X . Letting Σα = (α)X for each α ∈ Ω, we
have U ⊆
∏
α∈ΩΣα. Since T is perfect, by Proposition 5.2(1), it must be the case that
|{α ∈ Ω : β /∈ Σα}| < |Ω| for each β ∈ Ω. Then letting Γ ⊆ Ω be a countably infinite set,
|{α ∈ Ω : Γ 6⊆ Σα}| ≤
∣∣∣ ⋃
β∈Γ
{α ∈ Ω : β /∈ Σα}
∣∣∣ < |Ω|,
since |Ω| is assumed to have uncountable cofinality. Hence Γ ⊆ Σα for some α ∈ Ω, making
Σα infinite. But since T is T1, by Proposition 6.2, |Σα| < ℵ0 for all α ∈ Ω, producing the
desired contradiction.
In light of Theorem 6.4 we ask the following.
Question 6.5. Does there exist an infinite set Ω and a perfect locally compact T1 (or Haus-
dorff) topology on ΩΩ with respect to which it is a topological semigroup?
19
Acknowledgement
We are grateful to Slawomir Solecki for bringing to our attention a result of Mont-
gomery [23, Theorem 2], that plays a crucial role in the proof of Theorem 3.4.
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Z. Mesyan, Department of Mathematics, University of Colorado, Colorado Springs, CO,
80918, USA
Email: zmesyan@uccs.edu
J. D. Mitchell, Mathematical Institute, North Haugh, St Andrews, Fife, KY16 9SS, Scotland
Email: jdm3@st-and.ac.uk
Y. Pe´resse, University of Hertfordshire, Hatfield, Hertfordshire, AL10 9AB, UK
Email: y.peresse@herts.ac.uk
21