@
Appl. Gen. Topol. 20, no. 2 (2019), 419-430

doi:10.4995/agt.2019.11605

c© AGT, UPV, 2019

On proximal fineness of topological groups in

their right uniformity

Ahmed Bouziad

Département de Mathématiques, Université de Rouen, UMR CNRS 6085, Avenue de l’Université,

BP.12, F76801 Saint-Étienne-du-Rouvray, France. (ahmed.bouziad@univ-rouen.fr)

Communicated by J. Galindo

Abstract

A uniform space X is said to be proximally fine if every proximally
continuous function defined on X into an arbitrary uniform pace Y is
uniformly continuous. We supply a proof that every topological group
which is functionally generated by its precompact subsets is proximally
fine with respect to its right uniformity. On the other hand, we show
that there are various permutation groups G on the integers N that are
not proximally fine with respect to the topology generated by the sets
{g ∈ G : g(A) ⊂ B}, A, B ⊂ N.

2010 MSC: 22A05; 54E15.

Keywords: uniform space; topological group; proximal continuity; proxi-
mally fine group; symmetric group; o-radial space.

1. Introduction

A function f : X → Y between two uniform spaces is said to be proximally
continuous if for every bounded uniformly continuous function g : Y → R,
the composition function g ◦ f : X → R is uniformly continuous; the reals
R being equipped with the usual metric. A uniform space X is said to be
proximally fine if every proximally continuous function defined on X into an
arbitrary uniform space Y is uniformly continuous. It is well-known that metric
spaces and uniform products of metric spaces are proximally fine; however,
there are many uniform spaces that are not proximally fine although they are

Received 02 April 2019 – Accepted 30 July 2019

http://dx.doi.org/10.4995/agt.2019.11605


A. Bouziad

topologically well-behaved (some may be locally compact and even discrete).
We refer the reader to Hušek’s recent paper [12] for more information about
proximally fineness of general uniform spaces. We are interested here in the fine
proximal condition of the topological groups when these spaces are endowed
with their right uniformity. This subject seems to have no specific literature,
although there are some questions which could have been naturally addressed
in this setting, such as the Itzkowitz problem for metric and/or locally compact
groups [13] (the link between Itzkowitz problem and proximity theory was made
later in [6]).

In view of the close relationship between the right uniformity of topological
groups and the system of neighborhoods of their identity, it is reasonable to
expect that the proximal fineness of a given topological group G is satisfied
provided that a not too much restrictive condition is imposed on the topology
of G. This feeling is heightened by Corollary 2.6 in this note asserting that
it is sufficient to assume that G is functionally generated (in Arkhangel’skǐı’s
sense) by its precompact subsets. In fact, this is still true under much less
restrictive conditions (Theorem 2.4 below). Let us mention that one part of
Theorem 2.4 was asserted without proof in [5]1. The main subject of this note
can be examined in the context of G-sets (or G-spaces) without significantly
altering its essence, so the main result is stated and established in a somewhat
more general form (Theorem 2.5).

Some examples of non-proximally fine groups are given in Section 3. To
do that, we consider the topology τ on NN of uniform convergence when (the
target set) N is endowed with the Samuel uniformity. We show in Corollary 3.4
that for any permutation group H on N for which all but finitely many orbits
are finite and uniformly bounded, the only group topology on H which is finer
than τ and proximally fine is the discrete topology. The question of whether
there is an abelian (or at least SIN) group which is non-proximally fine group
is left open.

2. Main result

As usual, if (X,U) is a uniform space, where U is the uniform neighborhoods
of the diagonal of X, then for A ⊂ X and U ∈ U, U[A] stands for the set of
y ∈ X such that (x,y) ∈ U for some x ∈ A. For undefined terms we refer to
the books [8] and [16]. One of the tools used here is Katetov’s extension theo-
rem of uniformly continuous bounded real-valued functions [14]. The following
statement is also well-known (see [8]); its proof is outlined here for the sake of
completeness and because it can be adapted to show a result in the same spirit
that will be used in the proof Theorem 2.5.

Proposition 2.1. Let f : X → Y , where (X,U) and (Y,V) are two uniform
spaces. Then the following are equivalent:

(1) f : X → Y is proximally continuous,

1The author is indebted to Professor Michael D. Rice for his interest in this result and its

proof, which motivated the present work.

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On proximal fineness of topological groups

(2) for every A ⊂ X and V ∈ V, there exists U ∈ U such that f(U[A]) ⊂
V [f(A)].

Proof. To show that 1) implies 2), let d be a bounded uniformly continuous
pseudometric on Y such that (x,y) ∈ V whenever d(x,y) < 1 ([8]) and consider
the uniformly continuous function φ on Y defined by φ(y) = d(y,f(A)). Since
φ ◦ f is uniformly continuous, there is U ∈ U such that (x,y) ∈ U implies
|φ(f(x)) −φ(f(y))| < 1. Then f(U[A]) ⊂ V [f(A)].

For the converse, let φ : Y → R be a bounded uniformly continuous function
and let ε > 0. Define V = {(x,y) ∈ Y ×Y : |φ(x) −φ(y)| < ε/3}. Then V ∈V
and since φ is bounded, there is a finite set F ⊂ Y so that Y = V [F]. Let
Az = f

−1(V [z]), z ∈ F, and choose U ∈ U such that f(U[Az]) ⊂ V [f(Az)]
for each z ∈ F . Then |φ ◦ f(x) − φ ◦ f(y)| ≤ ε for each (x,y) ∈ U. Indeed,
there exists z ∈ F such that x ∈ Az so (f(x),z) ∈ V . Since y ∈ U[x] we have
f(y) ∈ f(U[Az]) ⊂ V [f(Az)] thus there exists t ∈ Az such that (f(y),f(t)) ∈ V .
As t ∈ Az we also have (f(t),z) ∈ V . Therefore, (f(x),f(y)) ∈ V ◦V ◦V . �

Let us say that a topological group G is proximally fine if the uniform space
(G,Ur) is proximally fine, where Ur is the right uniformity of G. It is equivalent
to say that (G,Ul) is proximally fine, where Ul is the left uniformity of G.
Recall that a basis of Ur (respectively, Ul) is given by the sets of the form
{(g,h) ∈ G×G : gh−1 ∈ V} (respectively, {(g,h) ∈ G×G : g−1h ∈ V}) as V
runs over the set V(e) of neighborhoods of the unit e of G.

Proposition 2.2. Let G be a topological group, (Y,U) a uniform space and let
f : G → Y be a function. For g ∈ G, let ψg : G → Y be the function defined by
ψg(h) = f(gh). Then, the following are equivalent:

(1) f is uniformly continuous, G being equipped with the right uniformity,
(2) the function ψ : g ∈ G → ψg ∈ Y G is continuous when Y G is endowed

with the uniformity of uniform convergence on G.

Proof. To show that (1) implies (2), suppose that f is right uniformly contin-
uous and let U ∈ U. There is a neighborhood V of the unit e in G such that
xy−1 ∈ V implies (f(x),f(y)) ∈ U. Let g ∈ G. Then V g is a neighborhood of
g in G and for each h ∈ V g and x ∈ G we have hx(gx)−1 ∈ V . It follows that
(f(hx),f(gx)) ∈ U, so (ψh(x),ψg(x)) ∈ U for every x ∈ G.

To show that (2) implies (1), let U ∈ U and choose V ∈ V(e) such that
(ψg(x),ψe(x)) ∈ U for every g ∈ V and x ∈ G. Then, for every g,h ∈ G
such that h ∈ V g we have (ψhg−1 (x),ψe(x)) ∈ U for each x ∈ G, equivalently,
(f(hx),f(gx)) ∈ U, for every x ∈ G. In particular, (f(h),f(g)) ∈ U. �

It is possible to expand substantially the framework of the starting topic
of this note without major changes as follows: Let G be a topological group
and let X be a G-set, that is, X is a nonempty set for which there is a map
∗ : G×X → X satisfying (gh)∗x = g∗(h∗x) for every g,h ∈ G and x ∈ X. The
function ∗ is called a left action of G on X. To simplify, write gx in place of g∗x
and UA in place of U ∗A if U ⊂ G and A ⊂ X. No topology will be required

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A. Bouziad

on X. Let (Y,V) be a uniform space and f : X → Y a function. It is consistent
with the definitions given above to say that a f is right uniformly continuous if
for each V ∈V, there is U ∈V(e) such that (f(gx),f(hx)) ∈ V for each x ∈ X,
whenever gh−1 ∈ U. Similarly, the function f is said to be right proximally
continuous if for each bounded uniformly continuous function φ : Y → R, the
function φ◦f is right uniformly continuous. It is easy to check (see the proof of
Proposition 2.2) that f is right uniformly continuous if and only if the function
ψ : g ∈ G → ψ(g) ∈ Y X (where ψ(g)(x) = f(gx)) is continuous when Y X is
endowed with the uniform convergence. Similarly, a simple adaptation of the
proof of Proposition 2.1 shows that f is right proximally continuous if and only
if for each A ⊂ GX and V ∈V, there is U ∈V(e) such that f(UA) ⊂ V [f(A)].

The following properties are required for Theorem 2.5; we are formulating
them separately to reduce the proof to its essential components. Let X be a
G-set and f : X → Y a right proximally continuous, as defined above. Then,
for each A ⊂ G:

(c1) for every x ∈ X, the function g ∈ G → f(gx) ∈ Y is continuous,
(c2) if ψ|A is right uniformly continuous, then ψ|A is right uniformly con-

tinuous,
(c3) if a ∈ A is a point of continuity of ψ|A, then a is a point of continuity

of ψ|A.

We check the validity of these properties for the benefit of the reader. Let
V ∈ V. For every x ∈ X and g ∈ G, there is U ∈ V(e) such that f(Ugx) ⊂
V [f(gx)]. Since Ug is a neighborhood of g in G, (c1) holds. Property (c2) and
(c3) follows from (c1). For, if x ∈ X, U ∈V(e) are such that (f(ax),f(bx)) ∈ V
for every a,b ∈ A with a ∈ UUb, then (c1) implies that f(gx),f(hx)) ∈ V 2 for
each g,h ∈ A such that g ∈ Uh. Taking x arbitrary in X gives (c2). For (c3),
let U be an open neighborhood of the unit in G such that (f(gx),f(ax)) ∈ V
for ever g ∈ Ua∩A and x ∈ X. Since Ua∩A ⊂ Ua∩A, it follows from (c1)
that for every g ∈ Ua∩A and x ∈ X, (f(gx),f(ax)) ∈ V 2.

To establish Theorem 2.5 we also need the next key lemma; this is a well-
known tool in the theory of proximity spaces (most often with W 4 instead of
W 3).

Lemma 2.3. Let X be a set and W be a symmetric binary relation on X.
Then, for every infinite cardinal η and for every sequence (xn,yn)n<η ⊂ X×X
such that (xn,yn) 6∈ W 3 for each n < η, there is a cofinal set A ⊂ η such that
(xn,ym) 6∈ W for every n,m ∈ A.

Proof. Replacing η by its cofinality cf(η), we may suppose that η is regular.
Let M ⊂ η be a maximal set satisfying (xn,ym) 6∈ W for every n,m ∈ M. If M
is cofinal in η, the proof is finished, so suppose that M is not cofinal in η. For
each j ∈ M, let Aj = {n < η : (xn,yj) ∈ W}, Bj = {n < η : (xj,yn) ∈ W},
Cj = Aj∪Bj and C = ∪j∈MCj. The maximality of M implies that η ⊂ M∪C.
Since η is regular, there is j ∈ M such that Cj is cofinal in η, therefore Aj or
Bj is cofinal in η. We suppose that it is Aj, the other case is similar. Let

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On proximal fineness of topological groups

n,m ∈ Aj. Then (xn,yj) ∈ W and (xm,yj) ∈ W , hence (xn,ym) 6∈ W since
(xm,ym) 6∈ W 3 (recall that W is symmetric). Similarly, (xm,yn) 6∈ W . �

We will now specify a few topological concepts that will be used in what
follows. The first is a variant of Herrlich’s notion of radial spaces [11]. Radial
spaces were characterized by A.V. Arhangel’skǐı [3] as follows: A space X is
radial if and only if for each x ∈ X and A ⊂ X such that x ∈ A, there is B ⊂ A
of regular cardinality |B| such that x ∈ C for every C ⊂ B having the same
cardinality as B. Let us say that a subset A of X is relatively o-radial in X if
for every collection (Oi)i∈I of open sets in X and x ∈ A such that

x ∈∪i∈IOi ∩A\∪i∈IOj,

there is a set J ⊂ I of regular cardinality such that x ∈ ∪j∈LOj whenever
L ⊂ J and |L| = |J|. If the set J can always be chosen countable, then A is
said to be relatively o-Malykhin in X. All closures are taken in X.

Every almost metrizable (in particular, Čech-Complete) group is o-Malykhin
(in itself). More generally, every inframetrizable group [16] is o-Malykhin, see
[6].

Let X be a topological space and M a collection of subspaces of X. Fo-
llowing Arhangel’skǐı [2], the space X is said to be functionally generated by
the collection M if for every discontinuous function f : X → R, there exists
A ∈M such that the restriction f|A : A → R of f to the subspace A of X has
no continuous extension to X. The space X is said to be strongly functionally
generated by M if for every discontinuous function f : X → R, there exists
A ∈ M such that the restriction f|A : A → R of f to the subspace A of X is
discontinuous.

The following is the main result of this note. The statement corresponding
to the case (2) was asserted (without proof) in [5].

Theorem 2.4. Let G be a topological group satisfying at least one of the fo-
llowing:

(1) G is functionally generated by the sets A ⊂ G such that AA−1 is rela-
tively o-radial in G,

(2) G is strongly functionally generated by the sets A ⊂ G such that A is
relatively o-radial in G.

Then G is proximally fine.

According to Proposition 2.2 and keeping the above notations, Theorem 2.4
is obtained from the following general result by considering the left action of
G on itself.

Theorem 2.5. Let G be a topological group and suppose that for each discon-
tinuous bounded function α : G → R, there is a set A ⊂ G having at least one
of the following conditions:

(1) α|A has no continuous extension to G and AA
−1 is relatively o-radial

in G,

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A. Bouziad

(2) α|A is discontinuous at some point of A and A is relatively o-radial in

G.

Let X be a G-set, (Y,V) a uniform space and let f : X → Y be a right proxi-
mally continuous. Then f : X → Y is right uniformly continuous.

Proof. We have to show that the function ψ : g ∈ G → ψ(g) ∈ Y X (where
ψ(g)(x) = f(gx)) is continuous. We proceed by contradiction by supposing that
ψ is not continuous. Then, there is a bounded uniformly continuous function
θ : Y X → R such that θ◦ψ is not continuous (see [8]). Let A ⊂ G satisfying at
least one of the conditions (1) and (2) with respect to the function θ◦ψ. In case
(1), there is no compatible uniformity on G making uniformly continuous the
function θ◦ψ|A; for, otherwise, Katetov’s theorem would give us a continuous
extension of θ ◦ψ|A. In particular, ψ|A is not right uniformly continuous. As
remarked above, it follows from (c2) that ψ|A is not right uniformly continuous.
There is then an open and symmetric W ∈ U such that for every V ∈ V(e),
there exist aV ∈ V , gV ∈ A and xV ∈ X satisfying aV gV ∈ A and

(f(aV gV xV ),f(gV xV )) 6∈ W 6.(2.1)

For each V ∈V(e), let hV = gV xV and define

OV = {g ∈ G : (f(ghV ),f(aV hV )) ∈ W}.

Since the functions g ∈ G → f(ghV ) ∈ Y , V ∈ V(e), are continuous by the
property (c1), each OV is open in G. Since aV ∈ OV ∩ Ag−1V ⊂ AA

−1 and
aV ∈ V for each V ∈V(e), it follows that

e ∈
⋃

V∈V(e)

OV ∩ (AA−1).(2.2)

We also have e 6∈ OV , for each V ∈ V(e). Indeed, otherwise, there exist V ∈
V(e) and g ∈ OV such that (f(ghV ),f(hV )) ∈ W . Therefore (f(aV hV ),f(hV )) ∈
W 2 which contradicts (2.1).

Since AA−1 is relatively o-radial in G, in view of (2.2), there is a set Γ ⊂V(e)
of regular cardinal such that for each set I ⊂ Γ of the same cardinal as Γ, we
have

e ∈
⋃
V∈I
{g ∈ G : (f(ghV ),f(aV hV )) ∈ W}.(2.3)

By Lemma 2.3 and (2.1), there is I ⊂ Γ such that |I| = |Γ| (since |Γ| is regular)
and (f(aUhU ),f(hV )) 6∈ W 2 for every U,V ∈ I. Since f is right proximally
continuous, there exists V ∈V(e) such that

f(V{hU : U ∈ I}) ⊂ W [f({hU : U ∈ I})].(2.4)

By (2.3) applied to I, there is U1 ∈ I such that V ∩ OU1 6= ∅. Let g ∈ V
be such that (f(ghU1 ),f(aU1hU1 )) ∈ W . By (2.4), there is U2 ∈ I so that
(f(ghU1 ),f(hU2 )) ∈ W . It follows that (f(aU1hU1 ),f(hU2 )) ∈ W 2, which is a
contradiction. Therefore, ψ is continuous in case (1).

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On proximal fineness of topological groups

In case (2), A is o-radial in G and θ ◦ψ|A is discontinuous at some point
a ∈ A. Since θ is continuous, ψ|A is necessarily discontinuous at a. It follows
from the property (c3) that ψ|A is discontinuous at a. Let W ∈U be symmetric
and open such that for every V ∈ V(e), there exist aV ∈ V and xV ∈ G, such
that aV a ∈ A and (f(aV axV ),f(axV )) 6∈ W 6. Taking hV = axV for each
V ∈V(e), we have

e ∈
⋃

V∈V(e)

{g ∈ G : (f(ghV ),f(hV )) ∈ W}∩ (Aa−1).

It is easy to see that Aa−1 is relatively o-radial in G, therefore the proof can
be continued and concluded in the same way as in the first case. It should be
noted that Katetov’s theorem was not used in this case. �

Let A ⊂ G, where G is a topological group. It is proved in [6] that AA−1 is
relatively o-Malykhin in G, provided that A is left and right precompact. Thus
Theorem 2.4 yields:

Corollary 2.6. Every topological group G which is functionally generated by
the collection of its precompact subsets is proximally fine.

In view of the role played by locally compact groups in many areas of math-
ematics, it is worth mentioning the following particular case of Corollary 2.6.

Corollary 2.7. Every locally compact topological group is proximally fine.

Recall that the topological group G is said to SIN (or with small invariant
neighborhoods of the identity) if the left uniformity Ul and the right uniformity
Ul of G are equal. It is well-known and easy to see that the property of being
SIN for G is equivalent to the inclusion Ur ⊂Ul or, equivalently, to the uniform
continuity of the identity map from (G,Ul) to (G,Ur). Following Protasov [15],
the group G is said to be FSIN (or functionally balanced) if every bounded right
uniformly continuous function f : G → R is left uniformly continuous. Clearly,
the property of being FSIN for G is equivalent to the proximal continuity of the
identity map from (G,Ul) to (G,Ur). Consequently, every proximally fine FSIN
group is SIN. The question whether every FSIN group is SIN is called Itzkowitz
problem and is still open. We refer the reader to [5] for more information; see
also [17] for a very recent contribution to this topic.

Since every proximally fine FSIN group is SIN, it follows from Theorem 2.4
that every FSIN group is SIN provided that it is strongly functionally generated
by its relatively o-radial subsets. This result has already been formulated (im-
plicitly and without proof) in [5]. This is supplemented by the next corollary
of Theorem 2.4 :

Corollary 2.8. Every FSIN group G which is functionally generated by the
sets A ⊂ G such that AA−1 is relatively o-radial in G is a SIN group.

To conclude this section, we would like to take this opportunity to comment
on the parenthesized question of [5, Question 6] whether every bounded topo-
logical group is FSIN. The answer is of course no, since FSIN is a hereditary

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A. Bouziad

property (by Katetov’s theorem) and every topological group is isomorphic
both algebraically and topologically to a subgroup of a bounded group [10]
(see also [9]).

3. Examples

In this section, we give some examples of non-proximally fine Hausdorff
topological groups and examine their behavior towards the FSIN property.
The set of positive integers is denoted by N and U is the Samuel uniformity
of the uniform discrete space N. The uniformity U is sometimes called the
precompact reflection of the uniform discrete space N. A basis of U is given
by the sets ∪i≤nAi × Ai, where A1, . . .An is a partition of the integers. Let
NN be endowed with the uniformity V of uniform convergence when N (the
target space) is equipped with the uniformity U. Let G denote the permutation
group of the set N of positive integers and let S be the normal subgroup of G
given by finitary permutations g ∈ G; that is, g ∈ S if and only if the set
supp(g) = {x ∈ N : g(x) 6= x} is finite. It is easy to see that G (hence
S) is a topological group when equipped with the topology τ induced by the
uniformity V. More precisely, G is a non-Archimedean group, since a basis of
neighborhoods of its unit is given by the subgroups of G of the form

Hπ = {g ∈ G : g(Ai) = Ai, i = 1, . . . ,n},

where π = {A1, . . . ,An} is a finite partition of N. Another approach (which we
will not adopt here) is to consider G as the group of all auto-homeomorphisms
of βN equipped with the compact-open topology, where βN is the Stone-Čech-
compactification of the integers.

The so-called natural Polish topology τ0 on G, given by pointwise conver-
gence, is coarser than τ and for every partition π of N the set Hπ is τ0-closed.
This is to say that τ0 is a cotopology for τ in the sense of [1]; in particular, (G,τ)
is submetrizable and a Baire space. In what follows, unless otherwise stated,
the groups G and S will be systematically considered under the topology τ.

For later use, we check that the quotient group G/S (with the quotient
topology) is Hausdorff, that is, S is closed in G. Let g ∈ G \ S. A simple
induction allows to construct an infinite set A ⊂ N such that g(A) ⊂ N \ A
(alternatively, A is obtained from Lemma 2.3 applied to the set {(x,g(x)) : x ∈
supp(g)}). Let π = {A,N \ A}. Then gHπ is a τ-neighborhood of g and for
each h ∈ Hπ, gh(A) = g(A) ⊂ N\A. Hence gHπ ∩S = ∅.

Recall that for a topological group H, the lower uniformity Ul ∧Ur on H
is called the Roelcke uniformity and has base consisting of the sets {(x,y) ∈
H ×H : x ∈ V yV}, V ∈V(e) (see [16]). Every (right) proximally fine group is
proximally fine with respect to the Roelcke uniformity; therefore, the following
shows in a strong way that the groups G and S are not proximally fine:

Proposition 3.1. Let k ∈ N and φ : G → N be the evaluation function
φ(g) = g(k), N being equipped with the discrete uniformity.

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On proximal fineness of topological groups

(1) The function φ : G → N is left uniformly continuous and right proxi-
mally continuous. In particular, φ is Roelcke-proximally continuous.

(2) If N is endowed with the uniformity U, then φ is right uniformly con-
tinuous. Conversely, if V is a uniformity on N such that the restriction
of φ|S : S → (N,V) is right uniformly continuous, then V ⊂U.

In particular, G and S are not Roelcke proximally fine.

Proof. 1) Clearly, φ is left uniformly continuous with respect to the natural
topology τ0; since τ0 ⊂ τ, φ is left uniformly continuous. To show that φ
is right proximally continuous, let L ⊂ G and put π0 = {A,N \ A}, where
A = {g(k) : g ∈ L}, and let us verify that φ(Hπ0L) ⊂ φ(L). Proposition 2.1
will then conclude the proof. Let h0 ∈ Hπ0 and g0 ∈ L. Then h0(g0(k)) ∈
{g(k) : g ∈ L}, hence we can write h0(g0(k)) = g(k) for some g ∈ L, thus
φ(h0g0) ∈ φ(L).
2) φ : G → (N,U) is right uniformly continuous, since for every partition
π = {A1, . . . ,An} of N, we have (g(k),h(k)) ∈ ∪i≤nAi × Ai provided that
hg−1 ∈ Hπ. For the converse, suppose that V is a uniformity on N satisfying
V 6⊂U and let V ∈V\U. We will check that for any partition π = {A1, . . . ,An}
of N, there are g,h ∈ S having g ∈ Hπh and (φ(g),φ(h)) 6∈ V . We may suppose
that A1 = {k} and that A2 contains two elements a and b such that (a,b) 6∈ V .
Define g ∈ S by g(k) = a, g(a) = k and g(x) = x for x /∈ {k,a}. Define
also h ∈ S by h(k) = b, h(a) = k, h(b) = a and h(x) = x otherwise. Then
gh−1 ∈ Hπ, but (φ(g),φ(h)) 6∈ V . �

The next statements 3.2 and 3.4 give some extremal properties showing that
the above examples provided by Proposition 3.1 are somehow optimal. For a
subgroup H of G and L ⊂ N, let H(L) stand for the pointwise stabilizers of L
in H (that is, the set of h ∈ H such that h(x) = x for all x ∈ L).

Let us recall that a topological group H is said to be strongly FSIN if every
real-valued right uniformly continuous function on H is left uniformly contin-
uous.

Proposition 3.2. Let H be subgroup of G and F ⊂ N a finite set. Let τ1 be a
group topology on H and for each k ∈ F , let φk : g ∈ H → g(k) ∈ N, where N
is endowed with the discrete uniformity.

(1) If for each k ∈ F , φk : (H,τ1) → N is right proximally continuous,
then τ is coarser than τ1 on H(N\HF).

(2) If for each k ∈ F , φk : (H,τ1) → N is right uniformly continuous, then
τ1 is discrete on H(N\HF).

(3) If τ0|H ⊂ τ1 and (H,τ1) is strongly FSIN, then τ1 is discrete on
H(N\HF).

Proof. 1) We first show that for a given A ⊂ N, there is a τ1-neighborhood
VA of the unit (in H) such that f(A ∩ (HF)) ⊂ A for every f ∈ VA. To
do that, for each k ∈ F, define Lk = {g ∈ H : g(k) ∈ A}. According to
Proposition 2.1, there is a τ1-neighborhood VA of the unit such that for every
k ∈ F, φk(VALk) ⊂ φk(Lk). Let n ∈ A ∩ (HF) and f ∈ VA. Choose g ∈ H

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A. Bouziad

and k ∈ F such that g(k) = n. Then g ∈ Lk, hence φk(fg) ∈ φk(Lk) and thus
f(n) ∈ A. This shows that f(A∩ (HF)) ⊂ A for every f ∈ VA. It follows that
for any partition π = {A1, . . . ,A1} of N, there is a τ1-neighborhood of the unit
in H(N\HF), namely V = H(N\HF) ∩VA1 ∩ . . .∩VAn , such that V ⊂ Hπ. Since
τ1 is a group topology, it follows that τ is coarser than τ1 on H(N\HF).

2) Suppose that τ1 is not discrete on H(N\HF) and let V be τ1-neighborhood
of the unit in H. We will show that there are g,h ∈ H and k ∈ F such that
gh−1 ∈ V and g(k) 6= h(k), contradicting the right uniform continuity of φl for
at least one l in the finite set F . Since for each k ∈ F , φk is τ1-continuous,
there is f ∈ V ∩ H(N\HF) such that f(k) = k for all k ∈ F and f(a) 6= a for
some a ∈ N. Then a ∈ HF , hence there are h ∈ H and k ∈ F such that
h(k) = a. Taking g = fh, we get gh−1 ∈ V and φk(g) 6= φk(h).

3) If (H,τ1) is strongly FSIN, then the functions φk, k ∈ F, are τ1-right uni-
formly continuous on H because they are left uniformly continuous on (G,τ0)
and τ0|H ⊂ τ1. It follows form (2) that τ1 is discrete on H(N\HF). �

Lemma 3.3. Let H be a subgroup of G, m ∈ N and L ⊂ N such that |L∩K| ≤
m for each orbit K of the action of H on N. Then H(L) ∈ τ.

Proof. There is a finite partition A0, . . . ,Am of N such that A0 = N \ L and
|Ai ∩K| ≤ 1 for each 1 ≤ i ≤ m and every orbit K. Then Hπ ⊂ H(L). Indeed,
if f ∈ Hπ and x ∈ K ∩ Ai where 1 ≤ i ≤ m, then f(x) ∈ K and f(x) ∈ Ai,
thus f(x) = x since |Ai ∩K| ≤ 1. �

Corollary 3.4. Let H be a subgroup of G for which all but finitely many orbits
are finite and uniformly bounded. Then the discrete topology is the only group
topology on H that is both proximally fine and finer than τ|H.

Proof. If τ1 is a proximally fine group topology on H finer than τ, then the
evualuation mappings φk : H → N, k ∈ N, are right uniformly continuous (with
respect to τ1). It follows from Proposition 3.2(2) that H(N\HF) is τ1-discrete,
where F ⊂ N is a finite set such that the cardinals of all orbits Hn, n 6∈ F,
are finite and uniformly bounded. By Lemma 3.3, H(N\HF) is τ-open hence
τ1-open, consequently, τ1 is discrete. �

Similarly, the next result follows from Proposition 3.2(3) and Lemma 3.3.

Corollary 3.5. Let H be subgroup of G for which all but finitely many orbits
are finite and uniformly bounded. If (H,τ) is strongly FSIN, then (H,τ) is
discrete. Moreover, if H has finitely many orbits and (H,τ0) is strongly FSIN,
then H is a (closed) discrete subgroup of (G,τ0).

It is plain that strongly FSIN groups are FSIN, but it still isn’t known
whether the converse is true or not (see Question 3 in [5]). It follows from
Proposition 3.1 that the groups G and S are not strongly FSIN, but this does
not allow us to conclude that they are not FSIN, because none of the functions
φk, k ∈ N, is bounded. The next result shows that the groups G and S as well
as their quotient G/S are not FSIN.

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On proximal fineness of topological groups

For a topological group H, let R(H), respectively U(H), stand for the real
Banach spaces of bounded right uniformly continuous and of bounded right
and left uniformly continuous functions on H. As usual, c denotes the cardinal
of R.

Proposition 3.6. The groups G, S and G/S are not FSIN. Moreover, the
density character of the quotient Banach space R(G/S)/U(G/S) is at least 2c.

Proof. We shall exhibit, in (1) below, a real-valued bounded function which
is left uniformly continuous on G but not right uniformly continuous when
restricted to S. It will follow that G and S are not FSIN. As for G/S, our
strategy is as follows: For each nonprincipal ultrafilter p on N, we shall give
in (2) a bounded right uniformly continuous φp defined on G which is not left
uniformly continuous. This function is in addition constant on every coset of
S, so it factorizes to G/S. Then, we show that for each bounded left uniformly
continuous function ψ on G, we have ||ψp+ψq+ψ|| ≥ 1 for any distinct nonprin-
cipal ultrafilters p, q. Since the quotient map G → G/S is both left and right
uniformly continuous, this will imply that the Banach space R(G/S)/U(G/S)
contains a uniformly discrete set of cardinal 2c (as βN\N, see [8]).
(1) Let χ : G →{0, 1} be the function defined by χ(f) = 1 if f(1) ≤ f(2) and
χ(f) = 0 otherwise. Then, clearly, χ is left uniformly continuous. Let us show
that it is not right uniformly continuous on S. Let A1, . . . ,An be a partition
of N. We may suppose that A1 = {a,b} with a 6= b. Define f and g in S
by f(1) = g(2) = a, f(2) = g(1) = b, and f(x) = g(x) for x 6∈ {1, 2}. Then
f−1(Ai) = g

−1(Ai) for each i = 1, . . . ,n, but χ(f) 6= χ(g).

(2) Let p be a nontrivial ultrafilter on N and fix an infinite A ⊂ N such that N\A
is infinite. Let ψp : G →{0, 2} be the function given by ψp(f) = 2 if f−1(A) ∈
p. Clearly, the function ψp is bounded and right uniformly continuous. To show
that ψp is constant on every coset of S, let g ∈ G and f ∈ S. Then, for every
B ⊂ N, g−1(B) \ supp(f) ⊂ (gf)−1(B). Thus, taking B = A if g−1(A) ∈ p or
B = N\A if not, we get that ψp(gf) = ψp(g).

Let p and q be two distinct nonprincipal ultrafilters on N and let us verify
that ||ψp +ψq +ψ|| ≥ 1 for each bounded left uniformly continuous ψ : G → R.
It will follow that the quotient R(G/S)/U(G/S) contains a norm 1 discrete set
of cardinal |βN\N|. Let ε > 0 and π = {B1, . . . ,Bn} be a partition of N such
that |ψ(f) −ψ(g)| < ε for every f,g ∈ G such that g ∈ fHπ. We may suppose
that B1 = C∪D with C ∈ p, D ∈ q and C∩D = ∅. Write again D = D1∪D2,
where D1 and D2 are infinite, disjoint and D1 ∈ q. Finally, let {E,F,K} be
a partition of of N \ A, with E and F infinite and |K| = |N \ B1|. There are
certainly f,g ∈ G such that f(C) = A, f(D) = E ∪F , g(C) = F, g(D1) = E,
g(D2) = A and f = g on N \B1. We have f−1g ∈ Hπ (i.e, f(Bi) = g(Bi) for
each i ≤ n), ψp(f) +ψq(f) = 2 and ψp(g) +ψq(g) = 0. Since |ψ(f)−ψ(g)| < ε,
it follows that |ψp(f) + ψq(f) + ψ(f)| ≥ 1−ε or |ψp(g) + ψq(g) + ψ(g)| ≥ 1−ε.
Since ε is arbitrary, we have ||ψp + ψq + ψ|| ≥ 1. �

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A. Bouziad

Let us mention that the corollary of 3.6 that the Hausdorff group G/S is
not discrete (being not FSIN) was established and used by T. Banakh et al. in
[4] to answer a question by D. Dikranjan and A. Giordano Bruno in [7].

Knowing that the symmetric group G and its finitary subgroup S are highly
nonabelian (their centers are trivial) and taking Corollary 3.5 into account, we
are naturally led to conclude by asking the following:

Question 3.7. Is there a Hausdorff topological group that is abelian (or at
least SIN) and non-proximally fine?

Acknowledgements. The author would like to thank the reviewer for the
thoughtful comments and efforts towards improving the paper.

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