CC BY
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Vladikavkaz Mathematical Journal 2021, Volume 23, Issue 1, P. 20-31
УДК 517.98
DOI 10.46698/a8913-4331-4311-d
TOPOLOGICAL LATTICE RINGS WITH THE AM-PROPERTY
O. Zabeti1
1University of Sistan and Baluchestan P.O. Box 98135-674, Zahedan, Iran E-mail: o.zabeti@gmail.com
Abstract. Motivated by the recent definition of the AM-property in locally solid vector lattices [O. Zabeti, doi: 10.1007/s41980-020-00458-7], in this note, we try to investigate some counterparts of those results in the category of all locally solid lattice rings. In fact, we characterize locally solid lattice rings in which order bounded sets and bounded sets agree. Furthermore, with the aid of the AM-property, we find conditions under which order bounded group homomorphisms and different types of bounded group homomorphisms coincide. Moreover, we show that each class of bounded order bounded group homomorphisms on a locally solid lattice ring X has the Lebesgue or the Levi property if and only if so is X.
Key words: locally solid lattice ring, bounded group homomorphism, AM-property, Levi property, Lebesgue property.
Mathematical Subject Classification (2010): 13J25, 06F25.
For citation: Zabeti, O. Topological Lattice Rings with the AM-Property, Vladikavkaz Math. J., 2021, vol. 23, no. 1, pp. 20-31. DOI: 10.46698/a8913-4331-4311-d.
1. Motivation and Preliminaries
Let us start with some motivation. In general, combination between different aspects of mathematics usually arise more efficient results and applications. For example, a topological vector space is a combination between linear algebra and topology. Moreover, a locally solid vector lattice is a powerful connection between ordered sets, linear algebra, and topology. These notions have been studied sufficiently because many classical examples in functional analysis fit in this category. Nevertheless, there are other topological algebraic structures that possess ordered structure, too. This leads us to the theory of ordered groups and ordered rings. When we add appropriate topological connections to them, we obtain more fruitful structures, for example locally solid lattice groups and locally solid lattice rings. These concepts are almost unexplored with respect to the locally solid vector lattices although there are many applicable examples of them that fail to have either a vector space or a topological vector space structure for example the discrete topology, the box topology on product spaces, the multiplicative group S1, the integers, and so on.
So, it is of independent interest to discover these phenomena. Recently, a suitable reference regarding lattice ordered groups has been announced in [1]. Furthermore, lattice ordered rings is partially considered in [2].
© 2021 Zabeti, O.
On the other hand, it is shown in [3] that there are several types of bounded group homomorphisms between topological rings; with respect to the suitable topologies, each class of them forms a topological ring, too. Moreover, when X is a locally solid lattice ring, each class of bounded order bounded group homomorphisms, also, forms a locally solid lattice ring. This is done in [4], recently. Before, we proceed with some preliminaries, let us again present another detailed motivation. It is worthwhile to mention that although it might seem at the first glance that there is no advantage in topological groups and topological rings with respect to the topological vector spaces, but there are some less considered facts about them. For example, we know that the discrete topology is the most powerful topology but the only topological vector space with the discrete topology is the zero one. On the other hand, any group with the discrete topology forms a topological group. Furthermore, the box topology is important in the product spaces because of finer neighborhoods with respect to the product topology and also to construct counterexamples; nevertheless, product of topological vector spaces with the box topology is not a topological vector space but this happens for the product of topological groups.
The known Hahn-Banach theorem that relies on the scalar multiplication, appears in many situations when we are dealing with locally convex spaces. The bad news is that we lack it in the category of all topological groups and there is no fruitful tool we can use it instead. Furthermore, many results regarding the AM-property and applications utilize this theorem in their nature. So, we can not expect those results in the setting of topological groups, directly. The good news is that when we are working with topological rings, the multiplication is another handy tool in this direction which turns out to be the right object for our purpose. In fact, the main aim of this note, is to characterize rings and also group homomorphisms in which bounded and order bounded notions agree. This is done by using the concept " AM-property" that is defined at first in [5] in the category of all locally solid vector lattices. Moreover, as an application, we show that each class of bounded order bounded group homomorphisms defined on a locally solid lattice ring X, has the Lebesgue or the Levi property if and only if so is X. The lattice structures for these classes of homomorphisms have been obtained recently in [4].
Suppose G is a topological group. A set B C G is said to be bounded if for each neighborhood U at zero, there is a positive integer n with B C nU in which nU = {xi + ... + Xn : Xi € U}.
A lattice group (l-group) G is called order complete if every non-empty bounded above subset of G has a supremum. G is Archimedean if nx ^ y for each n € N implies that x ^ 0. It is easy to see that every order complete l-group is Archimedean. A set S C G is called solid if x € G, y € S and |x| ^ |y| imply that x € S. Also, recall that a group topology t on an l-group G is referred to as locally solid if it has a local basis at zero consisting of solid sets.
Suppose G is a locally solid l-group. A net (xa) C G is said to be order convergent to x € G if there exists a net (zp) (possibly over a different index set) such that zp l 0 and for every there is an a0 with |xa — x| ^ zp for each a ^ a0. A set A C G is called order closed if it contains limits of all order convergent nets which lie in A. Keep in mind that topology t on a locally solid l-group (G, t) is referred to as Fatou if it has a local basis at zero consists of solid order closed neighborhoods. Observe that a locally solid l-group (G, t) is said to have the Levi property if every t-bounded upward directed set in G+ has a supremum. Finally, recall that a locally solid l-group (G, t) possesses the Lebesgue property if for every net (ua) in G, ua l 0 implies that ua A 0. For undefined expressions and related topics, see [1, 6].
Now, suppose X is a topological ring. A set B C X is called bounded if for each zero neighborhood W, there is a zero neighborhood V with VB C W and BV C W. A lattice ring
(l-ring) is a ring that is also a lattice where the ring multiplication and the lattice structure are compatible via the inequality |xy| ^ |x||y|. By a topological l-ring, we mean a topological ring which is an l-ring, simultaneously. Moreover, observe that a locally solid l-ring is a topological l-ring that possesses a local basis consisting of solid sets. Also, note that since in this case, the underlying topological group is also locally solid, all of the properties regarding locally solid l-groups, mentioned above, can be transformed directly to the category of all locally solid l-rings; because in this case, order structure in a ring and the underlying group is the same, just, we need to replace boundedness in some statements with the one related to topological rings. Moreover, note that by an ideal I of an l-ring X, we mean a solid subring of X.
Suppose X is a locally solid l-ring. Then, it is called a Birkhoff and Pierce ring (/-ring) if it satisfies in this property: a A b = 0 and c > 0 imply that ca A b = ac A b = 0. For ample facts regarding this subject, see [2].
For a brief but informed context related to topological lattice rings, we refer the reader to [4].
2. Main Results
Observation. Suppose G is an Archimedean l-group. For every subset A, by Av, we mean the set of all finite suprema of elements of A; more precisely, Av = (ai V ... V an : n € N, ai € A}. It is obvious that A is order bounded above in G if and only if so is Av and in this case, when the supremum exists, sup A = sup Av. Moreover, put AA = (a1 A ... A an : n € N,ai € A}. It is easy to see that A is order bounded below if and only if so is AA and inf A = inf AA (when the infimum exists). Observe that Av can be viewed as an upward directed set in G and AA can be considered as a downward directed set.
Suppose G is a locally solid l-group. We say that G has the AM -property provided that for every bounded set B C G, Bv is also bounded. It is worthwhile to mention that when B is bounded and solid, Bv is bounded if and only if BA is bounded; this follows from the fact that G is locally solid and x1 A ... A xn = — ((-x1) V ... V (—xn)) for any n € N and for any xi € B. One can consider this definition exactly for Archimedean l-rings. Note that when the ring multiplication is zero, every locally solid l-ring possesses the AM-property. This definition was originally defined in [5] for locally solid vector lattices.
Let us first prove a version of [7, Theorem 3.1] for topological rings.
Theorem 1. Let (Xa)aeA be a family of topological rings and X = na€A Xa with the product topology and pointwise addition and multiplication. Then B C X is bounded if and only if there exists a family of subsets (Ba)aeA such that each Ba C Xa is bounded and B C naeA Ba.
< Suppose B C X is bounded. Put
Ba = {x € Xa : 3 y = (yg) € B and x is a-th coordinate of y}.
Each Ba is bounded. For, if Ua is a zero neighborhood in Xa, put U = Ua x g=a Xg. Indeed, U is a zero neighborhood in X. Therefore, there is a zero neighborhood V with VB C U. Suppose Va is the a-th component of V; it is clear that VaBa C Ua.
For the converse, assume that there is a net (Ba)aeA of bounded sets with Ba C Xa such that B C a&A Ba. It is enough to show that Ha&A Ba is bounded. Assume that U is an arbitrary zero neighborhood in X. So, U = Ha^A Ua in which Ua = Xa for all but finitely many a; namely, Uai = Xai for i € (1,2,...,n}. Find zero neighborhoods Vai with VaiBai C Uai. Put V = nn=1 Vai x[]g={ai)...;Q,n} Xg. It is now easy to see that V(IIaeA Ba) C U, as claimed. >
Proposition 1. Suppose (Xa)aeA is a family of locally solid l-rings. Put X = Xa
with the product topology, pointwise ordering and pointwise addition and multiplication. Then X has the AM -property if and only if so is Xa for each a € A.
< First, assume that each Xa has the AM-property. Suppose B C X is bounded. By Theorem 1, there exists a net (Ba)aeA such that for each a, Ba C Xa is bounded and B C J}Ba. We show that Bv is also bounded. Let W be an arbitrary zero neighborhood in X. So, there are zero neighborhoods (Uai)ie{1;...)n} such that W = nn=1 Uai x
ripeA-{ai,...,an} Xp.
Observe that each x € B is a net x = (xp)peA with xp € Bp. Now, consider the set {x1,..., xm} C B in which m € N is fixed but arbitrary. It is enough to show that x1 V... Vxm is also bounded. Note that
x1 V ... V xm = (xp) V ... V (xm) = (xp V ... V xm)pSA,
where xp € Bp for each j € {1,...m}. For each i € {1,...n}, Bai has the AM-property
so that choose zero neighborhoods (Va.)n=1 such that V«.(x«. V ... V xm) C Ua.. Put
V = in= 1 V«i X npeA-{a1,...,a„} Xp. Then, it can be easily seen that V(x1 V ... V xm) C W, as claimed.
For the converse, fix an a € A and suppose B C Xa is bounded. Put D = B x npeA-{a}{0p}, in which 0p denotes the zero element in Xp. Since singletons are bounded in a topological ring, we conclude that by Theorem 1, D is bounded in X. By the assumption, Dv is also bounded in X. It is easy to see that Dv = Bv x ripeA-{a}{0p} so that Bv is bounded in Xa, as desired. >
Proposition 2. Suppose (Xa)aeA is a family of locally solid l-rings. Put X = «€a Xa with the product topology, pointwise ordering and pointwise addition and multiplication. If each Xa has the Levi property, then so is X.
< Suppose (xp)peB is a bounded increasing net in X. We need to show that its supremum exists. Observe that for each xp = (x«) €a. Since X has the product topology, we conclude that the net is pointwise bounded; more precisely, for each fixed a, the net (x«)peB is bounded and also increasing in Xa so that it has a supremum by the assumption, namely, y« = sup{(xa)peB}. Now, it can be easily seen that y = (y«)agA = sup{(xa)aeA,peB}. >
Observe that Proposition 2, can be restated exactly for locally solid l-groups, too. Moreover, when we consider the box topology, we have the following observations. Just, recall that the product of any family of topological groups with respect to the box topology is again a topological group (see [8, Chapter 3, Exercise 9]).
Proposition 3. Suppose (Ga)aeA is a family of locally solid l-groups whose singletons are bounded. Put G = H«€a Ga with the box topology, pointwise ordering and poinwise addition. If each Ga has the AM property, then so is G.
< Suppose B C X is bounded. By [7, Theorem 3.4], there exists a family (ai)i=1)...)n of indices such that B C (Hn=1 Bai) x[[peA-{«1 «n}{0p}. Consider a set {x1,... ,xm} in B. For each j = 1,..., m, we can write xj = (xpj)peA, where for ^ € A — {a1,..., an}, xpj = 0p and xjai € Bai for i = 1,..., n. Therefore, x1 V ... V xm = (xp1 V ... V xpm)peA. Thus, this supremum is the net consisting of (xai j V ... V xan j) in the j — th-place for j = 1,..., m and for other terms, zero. By the assumption, we conclude that Bv is also bounded. >
Furthermore, by considering this point that when a set in the product space is bounded in the box topology, it is bounded in the product topology and compatible with Proposition 2, we have the following.
Corollary 1. Suppose (Ga)a€A is a family of locally solid l-groups. Put G = Ha€A Ga with the box topology, pointwise ordering and pointwise addition. If each Ga has the Levi property, then so is G.
Now, we recall some notes about bounded group homomorphisms between topological rings; for a detailed exposition on this concept, see [3, 4].
Definition 1. Let X and Y be topological rings. A group homomorphism T : X — Y is said to be
(1) nr-bounded if there exists a zero neighborhood U C X such that T(U) is bounded in Y.
(2) br-bounded if for every bounded set B C X, T(B) is bounded in Y.
The set of all nr-bounded (br-bounded) homomorphisms from a topological ring X to a topological ring Y is denoted by Homnr(X, Y) (Hombr(X, Y)). The set of all continuous homomorphisms from X into Y will be denoted by Homcr (X, Y).
Homnr(X, Y) is equipped with the topology of uniform convergence on some zero neighborhood; note that a net (Sa) of nr-bounded homomorphisms converges uniformly on a neighborhood U C X to a homomorphism S if for each zero neighborhood V C Y there exists an ao such that for each a ^ a0, (Sa — S)(U) C V. Hombr(X, Y) is allocated to the topology of uniform convergence on bounded sets; observe that a net (Sa) of br-bounded homomorphisms uniformly converges to a homomorphism S on a bounded set B C X if for each zero neighborhood V C Y there is an a0 with (Sa — S)(B) C V for each a ^ a0. Homcr (X, Y) is assigned with the topology of cr-convergence; a net (Sa) of continuous homomorphisms cr-converges to a homomorphism S if for each zero neighborhood W C Y, there is a neighborhood U C X such that for every zero neighborhood V C Y there exists an a0 with (Sa — S)(U) C VW for each a ^ a0.
Each class of bounded homomorphisms as well as continuous homomorphisms between topological rings can possess a topological ring structure (see [3] for more information). Moreover, bounded order bounded homomorphisms between topological lattice rings can have lattice structures, using a kind of the Riesz-Kantorovich formulae, this is investigated in [4].
Remark 1. It is known that every zero neighborhood in a topological vector space is absorbing so that singletons are bounded. This useful fact relies on the scalar multiplication that we lack in topological groups, certainly. Therefore, we can not expect in a topological group that singletons are bounded, in general. For example, consider the additive group R with the usual topology and the additive group Z with the discrete topology. Put G = R x Z. It is easy to see that (0,1) is not bounded in G. But in many classical groups, singletons are bounded; for example when G is a connected topological group ( see [8, Chapter 3, Theorem 6]. Moreover, suppose G is a locally convex topological vector space. So, we have two notions for boundedness in G; when G is considered as a topological group and when it is considered as a topological vector space. It is easy to see that these notions agree. Now, suppose a locally solid l-group G has this mild property. So, we prove that in this case, order bounded sets are bounded. But in general, this is not true, consider [1, Example 4.2].
Lemma 1. Suppose G is a locally solid l-group whose singletons are bounded. Then, every order bounded set in G is bounded.
< Suppose [u, v] is an order interval in G and U is an arbitrary neighborhood at zero in G. There is a positive integer n with (|u| + |v|) € nU. So, for each u ^ w ^ v, since |w| ^ |u| + |v| and U is solid, we conclude that w € nU, as claimed. >
It is known that every singleton in a topological ring is bounded. So, we have the following observation, too.
Lemma 2. Suppose X is a locally solid l-ring. Then, every order bounded set in X is bounded.
< Suppose [u, v] is an order interval in X and W is an arbitrary zero neighborhood. There is a zero neighborhood V C X with V(|u| + |v|) C W. So, for each u ^ x ^ v, since |x| ^ |u| + |v| and W is solid, we conclude that Vx C W. >
Now, we improve [6, Proposition 2]; in fact, the underlying topological group need not be connected, just, it suffices to have boundedness condition for singletons. The proof is essentially the same. We provide it for the convenience of the reader.
Proposition 4. Suppose X is a topological ring that singletons in the underlying topological group are bounded. Then, we have the following.
(i) If B C X is bounded in the sense of the underlying topological group, then B is bounded.
(ii) If, in addition, X possesses a unity and B C X is bounded, then, it is bounded in the sense of the topological group.
< (i). Consider X as a topological group and assume that B C X is bounded. Furthermore, suppose W is an arbitrary zero neighborhood. There is a zero neighborhood V with VV C W. Find positive integer n such that B C nV. Choose zero neighborhood Vo with nVo C V. Therefore, VoB C nVoV C VV C W. Similarly, BVo C W.
(ii). Now, consider X as a topological ring and suppose B C X is bounded. For an arbitrary zero neighborhood W , there is a neighborhood V with VV C W, BV C W and VB C W. We claim there exists n € N such that B C nW. Suppose on a contrary, for any n € N, B ^ nW. So, there exists a sequence (xn) C B such that xn € nW. Moreover, since singletons are bounded in X, one can find m € N with 1 € mV. Thus, xm € mVB C mW a contradiction. >
Remark 2. Note that being unital is a sufficient condition in Proposition 4; in many classical spaces such as lp for 1 ^ p ^ oo, c0 and c00, it can be verified that notions of boundedness in the sense of topological vector space, underlying topological group, and topological ring (while they are considered with the pointwise multiplication) agree.
Let us first consider, as an application of the AM-property, a useful fact about locally solid l-groups.
Proposition 5. Suppose G is an order complete locally solid l-group whose singletons are bounded. Then, the following are equivalent.
(i) G possesses the AM and Levi properties.
(ii) Every order bounded set in G is bounded and vice versa.
< (i) A (ii). The direct implication is trivial by Lemma 1. For the converse, assume that B C G is bounded; W.L.O.G, we may assume that B is solid, otherwise, consider the solid hull of B which is again bounded. So, B+ = {x € B,x ^ 0} is also bounded. Assume that (B+)v is the set of all finite suprema of elements of B+. By the AM-property, (B+)v is also bounded. In addition, (B+)v can be considered as an increasing net in G+. So, by the Levi property, sup(B+ )v exists. But in this case, supB+ also exists and sup(B+)v = supB+. Put y = sup B+. Therefore, for each x € B+, x ^ y; now, it is clear from the relation B C B+ — B+ that B is also order bounded.
(ii) A (i). Suppose B C G is bounded so that order bounded. Now, it is clear that Bv is also order bounded and therefore bounded by Lemma 1, again.
Suppose D is an upward directed bounded set in G+. So, it is order bounded. Now, D has a supremum since G is order complete. >
Assume that H = Z with the discrete topology. It is a locally solid l-group. The only bounded set is the singleton zero and other singletons are never bounded. So, H possesses the Levi and AM properties. Nevertheless, note that every non-zero singleton is order bounded but not bounded. This justifies importance of the above observation (boundedness of singletons in a topological group). Moreover, using Proposition 4, we obtain the following result for locally solid l-rings.
Corollary 2. Suppose X is an order complete locally solid l-ring with unity such that singletons in the underlying topological group are bounded. Then, the following are equivalent.
(i) X possesses the AM and Levi properties.
(ii) Every order bounded set in X is bounded and vice versa.
But the surprising point here is that it is not necessary for locally solid l-ring X to be unital; more precisely, we improve Corollary 2. The main idea of the proof is essentially as the same as the proof of Proposition 5.
Theorem 2. Suppose X is an order complete locally solid /-ring. Then, the following are equivalent.
(i) X possesses the AM and Levi properties.
(ii) Every order bounded set in X is bounded and vice versa.
< (i) — (ii). The direct implication is trivial by Lemma 2. For the converse, assume that B C X is bounded; W.L.O.G, we may assume that B is solid, otherwise, consider the solid hull of B which is again bounded by [4, Lemma 5]. So, B+ = (x € B,x ^ 0} is also bounded. Assume that (B+)v is the set of all finite suprema of elements of B+. By the AM -property, (B+)v is also bounded. In addition, (B+)v can be considered as an increasing net in X+. So, by the Levi property, sup(B+)v exists. But in this case, supB+ also exists and sup(B+)v = supB+. Put y = supB+. Therefore, for each x € B+, x ^ y; now, it is clear from the relation B C B+ — B+ that B is also order bounded.
(ii) - (i). Suppose B C X is bounded so that order bounded. Now, it is clear that Bv is also order bounded and therefore bounded by Lemma 2, again.
Suppose D is an upward directed bounded set in X+. So, it is order bounded. Now, D has a supremum since X is order complete. >
Observe that order completeness is essential in the assumptions of Theorem 2 and can not be removed. Consider the ring X = C[0,1] with the pointwise multiplication; it possesses the AM-property. Also, boundedness and order boundedness notions agree in X by [3, Proposition 2.1], Proposition 4 and also using this fact that in C(K)-spaces, boundedness and order boundedness coincide. But it does not have the Levi property.
Corollary 3. Suppose X is a locally solid l-ring whose singletons in the underlying topological group are bounded and Y is a locally solid /-ring that possesses the AM and Levi properties. Then, for a group homomorphism T : X — Y, we have the following observations.
(i) If T is nr-bounded, then T is order bounded.
(ii) If T is br-bounded, then T is order bounded.
(iii) If T is continuous, then T is order bounded.
< (i). Suppose T is nr-bounded. So, there is a zero neighborhood U C X such that T(U) is bounded. Assume that B C X is bounded in the sense of the underlying topological group. Thus, there exists a positive integer n with B C nU so that T(B) C nT(U). This implies that T(B) is bounded. Now, suppose A C X is order bounded so that bounded in the sense of the topological group. Using previous argument, we conclude that T(A) is bounded in Y. Thus, Theorem 2 yields that T(A) is order bounded, as claimed.
(ii). Suppose A C X is order bounded. Therefore, it is bounded by Lemma 2. By the assumption, T(A) is also bounded in Y. Therefore, Theorem 2 results in order boundedness of T(A).
(iii). Now, suppose T is continuous. By [9, Remark 2.4], T is bb-bounded in the sense that it maps bounded sets to bounded sets while we consider boundedness in the topological group setting. Now, suppose A C X is order bounded so that bounded in the sense of the underlying topological group by Lemma 1. This results in boundedness of T(A) in Y (again in the topological group sense). By Proposition 4 (i), we conclude that T(A) is bounded and by Theorem 2, order bounded, as we wanted. >
By considering Corollary 3 and [4, Lemma 4, Lemma 5, Lemma 6], we have the following observations.
Corollary 4. Suppose X is a locally solid f -ring that possesses the AM, Fatou, and Levi properties and singletons in the underlying topological group are bounded. Then Homnr(X) is a lattice ring.
Corollary 5. Suppose X is a locally solid f -ring that possesses the AM, Fatou, and Levi properties and singletons in the underlying topological group are bounded. Then Hombr(X) is a lattice ring.
Corollary 6. Suppose X is a locally solid f -ring that possesses the AM, Fatou, and Levi properties and singletons in the underlying topological group are bounded. Then Homcr(X) is a lattice ring.
Proposition 6. Suppose X is a locally solid f-ring that possesses the AM and Levi properties and Y is any locally solid l-ring. Then, every order bounded group homomorphism T : X A Y is br-bounded.
< Suppose B C X is bounded. By Theorem 2, B is also order bounded. By the assumption, T(B) is order bounded so that bounded by Lemma 2. >
Remark 3. We can not expect Proposition 6 for either nr-bounded group homomorphisms or continuous group homomorphisms. Consider the identity group homomorphism on RN. It is order bounded but not an nr-bounded group homomorphism by [3, Example 2.1]; observe that Rn has the AM and Levi properties by Proposition 1 and Proposition 2.
Furthermore, suppose X is the additive group l^ with the absolute weak topology, pointwise ordering and pointwise addition and multiplication and Y is l^ with the uniform norm topology, pointwise ordering and pointwise addition and multiplication. Then, the identity group homomorphism I from X to Y is order bounded but not continuous. Observe that X possesses the Levi and AM properties.
Before, we proceed with another application of the AM-property, we have the following useful observation. Recall that Homb(X, Y) is the ring of all order bounded group homomorphisms from an l-ring X into an l-ring Y.
Lemma 3. Suppose X is a locally solid f -ring and Y is a locally solid f -ring that possesses the Fatou property and is order complete. Then we have the following.
(i) Homnr(X, Y) is an ideal of Homb(X, Y).
(ii) Hombr(X, Y) is an ideal of Homb(X, Y).
(iii) Hombr(X, Y) is an ideal of Homb(X, Y).
< (i). Assume |T| ^ |S| where T is order bounded and S € Hom[jr(X, Y). There exists a zero neighborhood U C X such that S(U) is bounded. So, for each zero neighborhood W C Y, there is a zero neighborhood V C Y with VS(U) C W. Since U is solid, for any y € U, y+,y-, |y| € U. Fix any x € U+. Then |T|(x) ^ |S|(x). In addition, by [4, Theorem 1],
|S|(x) = sup(|S(u)| : |u| ^ x}. Since U is solid and V is order closed, we conclude that V|S|(x) C W so that V|T|(x) C W. Since |T(x)| < |T|(x), we see that V|T(x)| C W. So, VT(x) C W. Therefore, VT(U+) C W. Since U C U+ — U+, we conclude that T(U) is also bounded.
(ii). It is similar to the proof of (i). Just, observe that for a bounded set B C X, W.L.O.G, we may assume that B is solid; otherwise, consider the solid hull of B which is also bounded by [4, Lemma 5].
(iii). Assume |T| ^ |S| where T is order bounded and S € Homj?r(X, Y). Choose arbitrary zero neighborhood W C Y. There is a zero neighborhood V with V — V C W. Find any neighborhood U such that S(U) C V. Fix any x € U+. Then, |T|(x) ^ |S|(x). In addition, by [4, Theorem 1], |S|(x) = sup(|S(u)| : |u| ^ x}. Since U is solid and also V and W are order closed, we conclude that |S|(x) € V so that |T|(x) € V. Since |T(x)| ^ |T|(x), we see that |T(x)| € V. So, T(x) € V. Therefore, T(U+) C V. Since U C U+ — U+, we conclude that T(U) C T(u+) — T(U+) C V — V C W, as desired. >
As a consequence, we state a domination property for each class of bounded order bounded group homomorphisms.
Corollary 7. Suppose X is a locally solid /-ring and Y is a locally solid /-ring that possesses the Fatou property and is order complete. Moreover, assume that T, S : X — Y are group homomorphisms such that 0 ^ T ^ S. Then we have the following.
(i) If S € Hombr(X, Y) then T € Hom^r(X, Y).
(ii) If S € Hombr(X, Y) then T € Hom^r(X, Y).
(iii) If S € Hombr(X, Y) then T € HomJ?r(X, Y).
Theorem 3. Suppose X is an order complete locally solid /-ring with unity and the Fatou property. Then Hom^(X) has the Levi property if and only if so is X.
< Suppose (Ta) is a bounded increasing net in Hom^.(X)+. Therefore, for every bounded set B C X, it follows that (Ta(B)) is uniformly bounded for each a. Thus, for each x € X+, the net (Ta(x)) is bounded and increasing in X so that it has a supremum, namely, ax. Define T : X+ — X+ via T(x) = ax. It is an additive map; it is easy to see that ax+y ^ ax + ay. For the converse, fix any a0. For each a ^ a0, we have Ta(x) ^ ax+y — Ta(y) ^ ax+y — Tao(y) so that ax ^ ax+y — Tao (y). Since a0 was arbitrary, we conclude that ax + ay ^ ax+y. By [4, Lemma 1], it extends to a positive group homomorphism T : X — X. We need to show that T € Hom ¡br(X). It is clear that T is order bounded. Suppose W is an arbitrary zero neighborhood in X. There is a zero neighborhood V with VTa(B) C W. This means that VT(B) C W since W has the Fatou property and also using [2, Theorem 3.15].
For the converse, assume that (xa) is a bounded increasing net in X+. Define Ta : X — X with Ta(x) = xxa. It is easy to see that each Ta is br-bounded as well as order bounded. Fix a bounded set B C X. Suppose W C X is an arbitrary zero neighborhood. Since the net (xa) is bounded, there exists a zero neighborhood V C X such that V(Bxa) C W for each a. It follows that (Ta) is bounded and increasing in Hom ¡br(X). Thus, by the assumption, Ta t T for some T € Hombr(X). Therefore, Ta(1) t T(1); that is xa t T(1), as claimed. >
Observe that by a locally bounded topological ring, we mean a topological ring X with a bounded zero neighborhood.
Lemma 4. Suppose X is a locally bounded order complete locally solid /-ring with unity and the Fatou property whose singletons are bounded in the sense of the underlying topological group. Then Homnr(X) = Hom br(X).
< Assume that X is locally bounded and a group homomorphisms T on X is nr-bounded. So, there exists a zero neighborhood U C X such that T(U) is bounded in X. Suppose B C X
is bounded. By Proposition 4 it is bounded also in the sense of the underlying topological group. Find positive integer n with B C nU so that T(B) C nT(U). This means that T is br-bounded. Furthermore, by the assumption, there exists a bounded zero neighborhood V C X. Now, suppose a group homomorphism T on X is br-bounded so that T(V) is also bounded in X. This shows that T is also nr-bounded, as claimed. >
Compatible with Lemma 4 and Theorem 3, we have the following.
Corollary 8. Suppose X is a locally bounded order complete locally solid f-ring with unity and the Fatou property whose singletons are bounded in the sense of the underlying topological group. Then Homnr(X) has the Levi property if and only if so is X.
Theorem 4. Suppose X is an order complete locally solid f -ring with unity and the Fatou property. Then Hombr (X) has the Levi property if and only if so is X.
< Suppose (Ta) is a bounded increasing net in Hombr(X)+. This implies that the set (Ta) is equicontinuous in the sense that for each zero neighborhood W C X, there is a zero neighborhood U such that Ta(U) C W for each a. So, for each x € X+, the net (Ta(x)) is bounded and increasing in X so that has a supremum, namely, ax. Define T : X+ a X+ via T(x) = ax. It is an additive map. By [4, Lemma 1], it extends to a positive group homomorphism T : X A X. We need to show that T € Hombr(X). It is clear that T is order bounded. Moreover, it can be easily seen that T(U) C W since W has the Fatou property.
For the converse, assume that (xa) is a bounded increasing net in X+. Define Ta : X a X via Ta(x) = xxa. It is easy to see that each Ta is continuous as well as order bounded. For an arbitrary zero neighborhood W C X, there is a zero neighborhood U such that U(xa) C W. It follows that (Ta) is bounded and increasing. Thus, by the assumption, Ta t T for some T € Hombr(X). Therefore, T«(1) t T(1); that is xa t T(1), as claimed. >
In this step, we recall a ring version of [10, Theorem 1.35]. The proof is essentially the same.
Lemma 5. Suppose X is an l-ring and I is an ideal of X. Then for a set D C I+, D ^ 0 in X if and only if D ^ 0 in I.
Proposition 7. Suppose X is an order complete locally solid f -ring with unity and the Fatou property. If Homj^(X) has the Lebesgue property then so is X.
< Suppose (xa) is a net in X such that xa ^ 0. Define Ta : X a X with Ta(x) = xxa. It is easy to see that each Ta is br-bounded as well as order bounded. First, note that by using [4, Theorem 1], we conclude that Ta ^ 0 in Homb(X) if and only if Ta(x) ^ 0 for each x € X+. Furthermore, observe that by Lemma 3 and Lemma 5, we conclude that Ta ^ 0 in Hombr(X). So, by the assumption, Ta a 0 uniformly on bounded sets. Therefore, Ta(1) a 0 in X; this means (xa) is a null net in X, as claimed. >
By using Lemma 4 and Proposition 7, one may consider the following.
Corollary 9. Suppose X is a locally bounded order complete locally solid f-ring with unity and the Fatou property whose singletons are bounded in the sense of the underlying topological group. If Homnr(X) has the Lebesgue property then so is X.
For the converse of Proposition 7, we have the following.
Theorem 5. Suppose X is a locally solid f -ring that possesses AM and Levi properties and Y is an order complete locally solid f -ring. If Y has the Lebesgue property, then so is Homb(X, Y).
< First, observe that by Proposition 6, Homb(X, Y) = Hombr(X, Y). Suppose (Ta)ae/ is a net in Hombr(X, Y) such that Ta I 0. Choose a bounded set B C X; W.L.O.G, we may
assume that B is solid, otherwise, consider the solid hull of B which is certainly bounded by [4, Lemma 5]. By Corollary 3, B is order bounded. The remaining part of the proof has the same line as in [5, Theorem 5]. Put A = (Ta(x),a € I, x € B+}. Again, W.L.O.G, assume that B+ = [0, u], in which u € X+. Define A = I x [0, u]. Certainly, A is a directed set while we consider it with the lexicographic order, namely, (a, x) ^ (5, y) if a < 5 or a = 5 and x ^ y. In notation, A = (yA)AeA ^ 0. So, by considering AA, one can assume A as a decreasing net in Y+. Therefore, it has an infimum. We claim that A ^ 0; otherwise, there is a 0 = y € Y+ such that yA ^ y for each A € A. Therefore, for each a and each x € B+, Ta(x) ^ y which is in contradiction with Ta ^ 0. By the assumption, yA — 0 in Y. Therefore, for an arbitrary zero neighborhood V C Y, there exists a A0 = (a0,x0) such that yA € V for each A ^ A0. Suppose A = (a,x). So, for each a > a0 and for each x € B+, Ta(x) € V. Since B C B+ — B+, we conclude that Ta — 0 in Hombr(X, Y). >
Remark 4. Observe that hypotheses in Theorem 5 are essential and can not be removed. Consider locally solid l-ring X = c0 with the norm topology, pointwise ordering and pointwise addition and multiplication. It possesses the AM-property and its topology is Lebesgue but it fails to have the Levi property. Suppose (Pn) is the sequence of the pointwise group homomorphisms on X, namely Pn((xm)) = (x^...,xn, 0,...). Each Pn is br-bounded and Pn t I, where I is the identity group homomorphism on X. But Pn ^ I uniformly on the unit ball of X.
Moreover, consider Y = li with the norm topology, pointwise ordering and pointwise addition and multiplication; it has the Lebesgue and the Levi properties but it fails to have the AM-property. Again, if (Pn) is the sequence of the pointwise group homomorphisms on Y, Pn t I but certainly not in the topology of uniform convergence on bounded sets.
Just observe that by Remark 2, the notions of boundedness in topological vector space and topological ring setting coincide.
Proposition 8. Suppose X is an order complete locally solid /-ring with unity and the Fatou property. If HomJ?r (X) has the Lebesgue property then so is X.
< Suppose (xa) is a net in X such that xa ^ 0. Define Ta : X — X with Ta(x) = xxa. It is easy to see that each Ta is continuous as well as order bounded. Observe that by Lemma 3 and Lemma 5, we conclude that Ta ^ 0 in HomJ?r(X). So, by the assumption, Ta — 0 in the cr-convergence topology. Therefore, Ta(1) — 0 in X; this means (xa) is a null net in X, as claimed. >
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Received May 17, 2019 Omid Zabeti
University of Sistan and Baluchestan, P.O. Box 98135-674, Zahedan, Iran, Assistant Professor E-mail: o.zabeti@gmail.com https://orcid.org/0000-0001-6875-2418
Владикавказский математический журнал 2021, Том 23, Выпуск 1, С. 20-31
ТОПОЛОГИЧЕСКИЕ РЕШЕТОЧНО УПОРЯДОЧЕННЫЕ КОЛЬЦА С АМ-СВОЙСТВОМ
Забети О.1
1 Университет Систана и Белуджистана, Иран, Захедан, P.O. Box 98155-987 E-mail: o.zabeti@gmail.com
Аннотация. В этой заметке предпринята попытка исследования AM-свойства, введенного недавно автором для локально солидных векторных решеток, в категории локально солидных решеточно упорядоченных колец. Фактически, получена характеризация локально солидных решеточно упорядоченных колец, в которых совпадают классы ограниченных и порядково ограниченных множеств. Кроме того, с помощью AM-свойства найдены условия, при которых совпадают порядково ограниченные гомоморфизмы групп и разные типы ограниченных групповых гомоморфизмов. Показано также, что каждый класс ограниченных групп порядково ограниченных гомоморфизмов на локально солидном решеточно упорядоченном кольце X обладает свойством Лебега или Леви тогда и только тогда, когда таковым является X.
Ключевые слова: локально солидное решеточно упорядоченное кольцо, ограниченный групповой гомоморфизм, AM-свойство, свойство Леви, свойство Лебега.
Mathematical Subject Classification (2010): 45G10, 65R20.
Образец цитирования: Zabeti, O. Topological lattice rings with the AM-property // Владикавк. мат. журн.—2021.—Т. 23, № 1.—C. 20-31 (in English). DOI: 10.46698/a8913-4331-4311-d.
