Balanced and functionally balanced $P$-groups

In relation to Itzkowitz's problem, we show that a $\mathfrak c$-bounded $P$-group is balanced if and only if it is functionally balanced. We prove that for an arbitrary $P$-group, being functionally balanced is equivalent to being strongly functionally balanced. A special focus is given to the uniform free topological group defined over a uniform $P$-space. In particular, we show that this group is (functionally) balanced precisely when its subsets $B_n,$ consisting of words of length at most $n,$ are all (resp., functionally) balanced.


Introduction and preliminaries
A topological group G is balanced if its left and right uniformities coincide. Recall that the left uniformity L G of a topological group G is formed by the sets U L := {(x, y) ∈ G 2 : x −1 y ∈ U }, where U is a neighborhood of the identity element of G. The right uniformity R G is defined analogously. A topological group G is called functionally balanced [14] in case every bounded left-uniformly continuous real-valued function on G is also right-uniformly continuous. Omitting the term "bounded" we obtain the definition of a strongly functionally balanced group. In the sequel we extend these definitions, in a natural way, to include also the symmetric subsets of a topological group (see Definition 2.9). The question of whether every strongly functionally balanced group is balanced was raised by Itzkowitz [5]. This longstanding problem is still open.
Nevertheless, it is known that a functionally balanced group is balanced whenever G is either locally compact [5,6,14], metrizable [14] or locally connected [7]. Recall that a topological group is non-archimedean if it has a local base at the identity consisting of open subgroups. A strongly functionally balanced non-archimedean group is balanced in case it is ℵ 0 -bounded [4] or strongly functionally generated by the set of all its subspaces of countable o-tightness [16]. For more known results concerning Itzkowitz's problem we refer the reader to the survey paper [2].
The class of all non-archimedean groups contains the class of all P -groups (see Definition 2.1). We prove that a P -group is functionally balanced if and only if it is strongly functionally balanced (Corollary 2.11). This gives a positive answer to [2,Question 3] for P -groups. One of the main result we obtain is that a c-bounded P -group is balanced if and only if it is functionally balanced (Theorem 2.13). So, a negative solution to Itzkowitz's problem cannot be found in the class of c-bounded P -groups.
A uniform space that is closed under countable intersection is called a uniform P -space (see also Definition 2.1). Such a space is necessarily non-archimedean, which means that it possesses a base of equivalence relations (Lemma 2.3). In Section 3 we discuss the coincidence of some universal free objects over the same uniform P -space.
For a free group F (X), over a nonempty set X, we denote by B n its subset containing all words of length not greater than n. In Section 4 we show that the uniform free topological group F (X, U ), over a uniform P -space, is (functionally) balanced if and only if B n is (resp., functionally) balanced for every n ∈ N. Hopefully, this theorem can be useful in providing a negative solution to Itzkowitz's problem.
Given a symmetric subset B of a topological group G we denote by L B G the trace of the left uniformity L G on B. That is, ε ∈ L B G if and only if there exists δ ∈ L G such that δ ∩ (B × B) = ε. The uniformity R B G is the trace of R G . In case {A n } n∈N is a countable collection of subsets of G, we write L n G (R n G ) instead of L An G (resp., R An G ). The character of G is the minimum cardinal of a local base at the identity. For a uniform space (X, U ), the weight w(X, U ) denotes the minimal cardinality of a base of (X, U ). For ε ∈ U and a ∈ X we let ε[a] := {x ∈ X : (a, x) ∈ ε}. All topological groups and uniform spaces in this paper are assumed to be Hausdorff. Unless otherwise is stated the uniformity of a topological group G is the two-sided uniformity, that is, the supremum L G ∨ R G . Finally, TGr, NA and NA b denote, respectively, the classes of all topological groups, non-archimedean groups and non-archimedean balanced groups.
2. P -groups and uniform P -spaces Definition 2.1. (see [1], for example) A P -space is a topological space in which the intersection of countably many open sets is still open. A topological group which is a P -space is called a P -group. A uniform P -space (X, U ) is a uniform space in which the intersection of countably many elements of U is again in U .
Proof. Let U be a neighborhood of the identity element e. We have to show that U contains an open subgroup H. For every n ∈ N there exists a symmetric neighborhood W n such that W n n ⊆ U. Since G is a P -group the set W = ∩ n∈N W n is a neighborhood of e. Let H be the subgroup generated by W. Clearly, H is open and H ⊆ U.
Proof. Let ε ∈ U . We will find an equivalence relation δ ∈ U such that δ ⊆ ε. For every n ∈ N there exists a symmetric entourage δ n ∈ U such that δ n n ⊆ ε. Since (X, U ) is a uniform P -space, the equivalence relation δ = m∈N (∩ n∈N δ n ) m ⊆ ε is an element of U .
(1) A topological group G is called τ -bounded if for every neighborhood U of the identity, there exists a set F of cardinality not greater than τ such that F U = G.

Lemma 2.5. Let τ be an infinite cardinal. Let G be a topological group in which the intersection of any family of cardinality at most τ of open sets is open. If
It is easy to see that N = ∩ x∈F xHx −1 .
Since |F | ≤ τ, this intersection must be open and applying Lemma 2.2 completes the proof.
Proof. The "if" part is trivial for every uniform space (even if it is not a uniform Pspace). We prove the "only if" part. Since f : (X, U ) → R is uniformly continuous, for every n ∈ N there exists ε n ∈ U such that (

uniformly continuous if and only if there exists an open subgroup
) is a uniform P -space. Now use Lemma 2.7 and the definition of the left (resp., right) uniformity to conclude the proof. Definition 2.9. We say that a symmetric subset B of a topological group G is: (1) balanced if the left and right uniformities of G coincide on B. ( G and clearly ε has at most c equivalence classes. By (3) In case the non-archimedean group is a P -group it suffices to require c-boundedness, as it follows from the following theorem: Theorem 2.13. Let G be a c-bounded P -group. Then G is balanced if and only if it is functionally balanced.
Proof. If H is a subgroup of index at most c, then ε := {(t, s)| t −1 s ∈ H} has at most c equivalence classes. So, in case G is a c-bounded P -group, condition (3) of Theorem 2.10 is equivalent to the coincidence of the left and right uniformities. This completes the proof.
For Ω = TGr the universal object F Ω (X, U ) is the uniform free topological group of (X, U ). This group was invented by Nakayama [9] and studied, among others, by Numella [11] and Pestov [12,13]. In particular, Pestov described its topology. Let (X, U ) be a non-archimedean uniform space.
(1) For Ω = NA we obtain the free non-archimedean group F N A .
(2) In case Ω = NA b , the universal object is the free non-archimedean balanced group F b N A . These groups were defined and studied by Megrelishvili and the author in [8].
is simply the subgroup generated by V.
(2) [8, Definition 4.9.2] As a particular case in which every ψ is a constant function we obtain the setε : ε ∈ B} is a local base at the identity element of F b N A (X, U ), the uniform free non-archimedean balanced group. Remark 3.6. Let (X, U ) be a non-archimedean uniform space. By the universal properties of the universal objects it is clear that: (1) the topology of F N A(X, U ) is the maximal non-archimedean group topology on F (X) that is coarser than the topology of F (X, U ). (2) the topology of F b N A (X, U ) is the maximal non-archimedean balanced group topology on F (X) that is coarser than the topology of F (X, U ). In particular, if F (X, U ) is non-archimedean, then F (X, U ) coincides with F N A(X, U ). If F (X, U ) is also balanced, then these groups coincide also with F b N A (X, U ). Theorem 3.7. Let (X, U ) be a uniform space. Suppose that there exists an infinite cardinal τ such that i∈I ε i ∈ U for any family of entourages (1) : Let I be an arbitrary set with |I| ≤ τ. For every i ∈ I let {ψ i n } be a sequence of elements from U F (X) (see Theorem 3.5.1). We define a function ϕ as follows. For every w ∈ F (X) let ϕ(w) = i∈I,n∈N ψ i n (w). By our assumption on the cardinal τ, we have ϕ ∈ U F (X) . (2): It is known that the universal morphism i : (X, U ) → F (X, U ) is a uniform embedding and that i(X) algebraically generates F (X, U ). Since (X, U ) is τ -narrow, we obtain by [1, Theorem 5.1.19] (see also [1, Exercise 5.1.a]) that F (X, U ) is τ -bounded.
By item (1) and Lemma 2.5, we conclude that the non-archimedean group F (X, U ) is also balanced and so we have F (X, Omitting the τ -narrowness assumption from Theorem 3.7.2, we obtain the following counterexample. Example 3.8. By [10, Example 3.14], for every cardinal τ > ℵ 1 , there exists a Hausdorff uniform P -space such that w(X, U ) = ℵ 1 < τ < χ(F (X, U )). In view of Theorem 3.7.1 and [8, Theorem 4.16.1], we have F (X, U ) = F N A (X, U ) = F b N A (X, U ). As corollaries we obtain the following two results of Nickolas and Tkachenko. Proof. If (X, U ) is an ℵ 0 -narrow uniform P -space, then by Theorem 3.7.2 we have On the other hand, by [8,Theorem 4.16.1], we have χ(F b N A (X, U )) = w(X, U ). We conclude that χ(F (X, U )) = w(X, U ).

The subsets B n
Theorem 4.1. Suppose that τ is a topological P -group topology on a free group F (X) and µ is either the right, left or two-sided uniformity of (F (X), τ ). Then, ε ∈ µ if (and only if ) ε ∩ (B n × B n ) ∈ µ ∩ (B n × B n ) ∀n ∈ N.
Corollary 4.2. Let (X, U ) be a uniform P -space and µ be either the right, left or twosided uniformity of F (X, U ). Then, ε ∈ µ if (and only if ) ε ∩ (B n × B n ) ∈ µ ∩ (B n × B n ) ∀n ∈ N.
Proof. By Theorem 3.7, F (X, U ) is a P -group. Now the proof follows from Theorem 4.1.  (1) B n is strongly functionally balanced for every n ∈ N.
(2) B n is functionally balanced for every n ∈ N.