A family of totally ordered groups with some special properties

Let K be a ﬁeld with a Krull valuation | | and value group G 6 = { 1 } , and let B K be the valuation ring. Theories about spaces of countable type and Hilbert-like spaces in [1] and spaces of continuous linear operators in [2] require that all absolutely convex subsets of the base ﬁeld K should be countably generated as B K -modules. By [1] Prop. 1.4.1, the ﬁeld K is metrizable if and only if the value group G has a coﬁnal sequence. We prove that for any ﬁxed cardinality ℵ κ , there exists a metrizable ﬁeld K whose value group has cardinality ℵ κ . The existence of a coﬁnal sequence only depends on the choice of some appropriate ordinal α which has cardinality ℵ κ and which has coﬁnality ω . By [2] Prop. 1.4.4, the condition that any absolutely convex subset of K be countably generated as a B K -module is equivalent to the fact that the value group has a coﬁnal sequence and each element in the completion G # is obtained as the supremum of a sequence of elements of G . We prove that for any ﬁxed uncountable cardinal ℵ κ there exists a metrizable ﬁeld K of cardinality ℵ κ which has an absolutely convex subset that is not countably generated as a B K -module. We prove also that for any cardinality ℵ κ > ℵ 0 for the value group the two conditions (the whole group has a coﬁnal sequence and every subset of the group which is bounded above has a coﬁnal sequence) are logically independent.

recall the principal features concerning to ordinals and cardinals.Readers may refer to [3] for additional information.
A linear ordering ≤ of a set A is a well-ordering if every nonempty subset of A has a smallest element.A set T is an ordinal number if every element of T is a subset of T and T is well-ordered with respect to the membership-relation (∈).
We use small Greeks letters to denote ordinal numbers.The class of all ordinals is denoted by Ord; it is easy to see that Ord is not a set.The relation on Ord defined by α < β if and only if α ∈ β is a well-ordering of the class Ord.Thus, 0 = φ is the first ordinal, and for each ordinal number α, α = {β : β < α}, we have that α+1 = α∪{α} = inf{β : β > α} is an ordinal called the successor of α.If α is not a successor, then α = sup {β : β < α} = β<α β and it is called a limit ordinal.We also consider 0 as the first limit ordinal.Every well-ordered set is isomorphic to a unique ordinal number.The finite ordinals are denoted by 0, 1, 2, . . ., n, . . .and they correspond to the ordertype of the natural numbers.The order-type of the set of natural numbers is denoted by ω and it is the first limit ordinal different from 0, and the first infinite ordinal.It is important to note that ω, ω + 1 and ω + ω represent different order-types, although all of them are isomorphic as sets with the set of natural numbers.For example, ω + 1 represents the order-type of the set N ∪ {∞} such that n < ∞ for all n ∈ N ; ω + ω represents the order-type of the set { n n + 1 : n ∈ N } ∪ {1 + n n + 1 : n ∈ N } with the usual ordering on the rational numbers.That is to say "two copies of ω".The lexicographical order on N × N is represented by ω copies of ω.The first uncountable ordinal is denoted by ω 1 .Also, ω 1 , ω 1 + n, ω 1 + ω, ω 1 + ω 1 are different as ordered sets but all of them are isomorphic as sets.With this fact in mind, cardinals numbers are defined.Two sets have the same cardinality if there exists a bĳection between them.The cardinality of a set A is denoted by |A|.Thus, for example, |ω| = |ω + ω|.
The class of all cardinals is denoted by Card.Every infinite cardinal is a limit ordinal.Converse is not true.For example ω + ω is a limit ordinal but it is not a cardinal number.We use ℵ α to denote the cardinal number and ω α to denote its order-type.Thus, ℵ 0 = ω 0 = ω; ℵ α+1 = ω α+1 = ℵ + α ; and ℵ α = ω α = sup {ω β : β < α}, if α is a limit.In this case we say that ℵ α is a limit cardinal.The rules for addition and multiplication of infinite cardinals are quite simple: Definition 1.2: Let α > 0 a limit ordinal.The cofinality of α, cf (α) is defined as the smallest limit ordinal λ such that there exists an increasing family of ordinals indexed by λ, {α ε : ε < λ} with sup For example, for any ordinal α, we have cf (ω α+ω ) = ω because sup n<ω {ω α+n } = ω α+ω .On the other hand, cf (ω 1 ) = ω 1 since there does not exist a countable family of ordinals {α n } such that sup There are arbitrarily large singular cardinals.Using the axiom of choice, it can be proven that every ℵ κ+1 is a regular cardinal.

The construction of the group Γ α
Let I be a totally ordered set.For each index i ∈ I, let G i be a totally ordered multiplicative group with unit element 1 G i .The direct product i∈I G i of the family {G i } i∈I consists of all functions f : In [4], the Hahn product is defined as the subgroup of the direct product consisting of all functions f such that supp(f ) is a well-ordered set and it is denoted by where k is the first element of I such that f (k) = g(k).We will define groups Γ α as special Hahn products.
Let α be an ordinal, {G β } β<α a family of abelian multiplicatively written totally ordered groups of rank 1. (This means that each G β is a subgroup of (0, ∞), •, ≤ ).The group Γ α is a subset of the direct product of the family {G β } β<α defined by: with componentwise multiplication.We define the degree of f as deg(f ) = max{supp(f )}.The ordering on Γ α is defined by f > 1 if and only if f (degf ) > 1.We say that Γ α is antilexicographically ordered.Furthermore, for every element b ∈ G β , we define the element For example, Γ ω = n<ω G n is used in [1] with a countable family {G n } of cyclic groups.Note that Γ ω has not a "last copy" of G n .On the other hand, the group Γ ω+1 is also a subgroup of the direct product of a countable family of groups, but in this case we do have a "last copy", the group G ω .Consider the set I, ≺ where I = {β ∈ Ord : β < α} with the inverse ordering on Ord.That is to say β ≺ γ if and only if γ < β for all β, γ ∈ I.It is clear that I, ≺ is not well-ordered, but the well-ordered subsets of I are precisely the finite ones.Therefore the groups Γ α are Hahn products over I, ≺ .
A convex subgroup H of a totally ordered group G is called principal if there is an element g ∈ G such that H is the smallest convex subgroup of G containing g.By [4], every convex subgroup H which is not principal is equal to the union of all principal convex subgroups of G contained in H.In this case H is called a limit convex subgroup.For every β < α we define the sets: For convenience, we put H * 0 = {1}.The next proposition justifies this notation.

Proposition 2.1: H β is a principal convex subgroup generated by any element
Therefore, H β is a principal convex subgroup because it is the smallest convex subgroup that contains f .Now let H be a principal convex subgroup generated by some element g ∈ H. Then g ∈ H deg(g) which means that H = H deg(g) .

Corollary 2.2:
It is immediate because if H is not a principal convex subgroup, from the above proposition H = H β for all β < α, and by [4], H is the union of principal convex subgroups, that is to say H = γ<β H γ , for some β < α.This Hence, the order-type of the set of all principal convex subgroups of Γ α , ordered by inclusion, is α.Proposition 2.4: Let α be an infinite limit ordinal.Then the following are equivalent: i) There exists an increasing sequence of principal convex subgroups {H βn : If there exists such a sequence then there exists an increasing sequence {β n : n < ω} such that sup{β n : n < ω} = α.Because each group G β is a multiplicative subgroup of R + , each of them has a cofinal sequence.Let a βn be the n-th element in a cofinal sequence of G βn , for all n ∈ N. Then the sequence {χ (a βn ,βn) : In infinite rank it is possible to have a quasidiscrete group which has quasidense subgroups.(See [2], example at the end of 1.2).

Completions of linearly ordered sets
Completions of totally ordered groups are important in order to obtain supremum and infimum of subsets of them.In this section we define the Dedekind completion of arbitrary linearly ordered sets.We study the extension of mappings between two such sets (or groups) to their completions.A linearly ordered set X is called Dedekind complete if each nonempty subset of X that is bounded above has a supremum.A subset A of a linearly ordered set X is called dense (in X) if for each s ∈ X: where X # is a complete linearly ordered set and i : X → X # is a strictly increasing and dense mapping.This completion satisfies the following universal property: Proposition 3.1: Let (X # , i) be a Dedekind completion of a linearly ordered set X. Then for every linearly ordered Dedekind complete set Y and every strictly increasing and dense mapping ϕ : X → Y , there is only one strictly increasing function ψ such that ϕ = ψ • i.
Proof: Let Y be a linearly ordered complete set and ϕ a strictly increasing and dense mapping from X to Y .We define ψ(s) = sup Y {ϕ(x) : x ∈ X ∧ i(x) ≤ s} for each s ∈ X # .Then for all x ∈ X we have ϕ(x) = (ψ•i)(x).Furthermore, if s < t in X # , by density there exist a, b ∈ X such that s ≤ i(a) < i(b) ≤ t and then ψ(s) ≤ ϕ(a) < ϕ(b) ≤ ψ(t), hence ψ is strictly increasing.
For uniqueness, let δ : X # → Y another strictly increasing mapping such that δ • i = ϕ.Then, there is an s ∈ X # with δ(s) = ψ(s).By the definition of ψ we have ψ(s) < δ(s) and by density of ϕ there is an a ∈ X such that ψ(s) ≤ ϕ(a) < δ(s) or ψ(s) < ϕ(a) ≤ δ(s).In the first case if s < i(a) then δ(s) ≤ δ(i(a)) = ϕ(a) a contradiction, therefore i(a) < s from which it follows that ϕ(a) = ψ(s).But s / ∈ X, then there is a b ∈ X with i(a) < i(b) < s and ϕ(a) = ϕ(b), but that is not possible by injectivity of ϕ.A similar argument, taking into consideration that s = inf X # {x ∈ X : i(x) ≥ s} works in the case ψ(s) < ϕ(a) ≤ δ(s).
Therefore, ψ is unique.In particular, for each s ∈ X # we have: The above proposition proves that all completion are order isomorphic and therefore the canonical completion by Dedekind cuts with the natural embedding i is a Dedekind completion.As usual, we shall identify i with the inclusion and we shall write x instead of i(x) for all x ∈ X.From now on we say complete instead Dedekind complete.Proposition 3.2: Let Y ⊆ X and Y # , X # their completions.Then Y # can be embedded in X # through a strictly increasing mapping τ : Y # → X # such that τ (y) = y for all y ∈ Y .
Proof: Let τ (s) = sup X # {y ∈ Y : y ≤ s} for all s ∈ Y # .Then τ (y) = y for all y ∈ Y and s < t in Y # implies there are y 1 , y 2 ∈ Y such that s ≤ y 1 < y 2 ≤ t.Therefore τ (s) ≤ y 1 and τ (t) ≥ y 2 , hence τ (s) < τ (t) that is to say τ is strictly increasing.Remark 3.3: Actually, the mapping τ of the above proposition is not necessarily unique.For instance, let (2,3] where a ∈ [1, 2] is arbitrary.Hence, there are infinite mappings which extend IdQ.Uniqueness is obtained if Y is dense in the convex hull of Y in X. Definition 3.4: [1] Let X be a totally ordered set and let Y ⊆ X.The X-convex hull of Y is defined by: then the mapping τ of the above proposition is unique. This means that there exist y 1 , y 2 ∈ Y such that y 1 < s < y 2 .Hence, y 1 < τ (s) < δ(s) < y 2 .Therefore there is an element y ∈ Y such that τ (s) ≤ y < δ(s).But s < y implies δ(s) ≤ y and y < s implies y = τ (s).Hence τ is unique.

Corollary 3.6: Let H be a convex subgroup of a totally ordered group G.
Then there is only one strictly increasing mapping τ : totally ordered groups and let τ : G 1 → G 2 a surjective homomorphism of totally ordered groups.Then there exists an increasing mapping τ Remark 3.8: Since Ker τ is a convex subgroup of G 1 we can restate the proposition 3.7 in the following way: Let H be a convex subgroup of a totally ordered group G and π : G → G/H the canonical projection.Then there exists an increasing mapping π # : G # → (G/H) # which extends the projection π.
It is well known that for every totally ordered group Γ there exists a valued field K such that Γ is its value group.For our purposes we shall use the construction indicated in [5] of such a K.
Let F be an arbitrary field and let R = {f : Γ → F : supp(f ) is finite}.With the operations + and • defined by: R becomes a domain.Let v : R → Γ be the mapping defined by v(f ) = max{γ ∈ Γ : f (γ) = 0} and let K be the field of fractions of R. We extend v to K by letting v(f Proof: It is enough to consider the group Γ α of the above proposition which has a cofinal sequence (hence it has a coinitial sequence).Then we use the construction of [5] in order to obtain a field K with value group Γ α and, by [1] Theorem 1.4.1, the metrizability of this field is guaranteed.

Theorem 4.3:
For any fixed uncountable cardinal ℵ κ , there exists a group Γ α with |α| = ℵ κ and an element s ∈ Γ # α which is not the supremum of some countable subset of Γ α .

Corollary 4.4:
For any fixed uncountable cardinal ℵ κ there exists a field K with a Krull valuation which is metrizable and contains absolutely convex subsets which are not countably generated as B K -modules.
We claim that the set B(0, s) − , where s = sup H * ω1 is not countably generated as a B K -module.In fact, let B(0, s) − be generated as a B Kmodule by k 1 , k 2 , . . .∈ K.By [1] Let α be an ordinal and consider the group Γ α .The condition 'Γ α has a cofinal sequence' will be denoted M and the condition 'every element of Γ # α is the supremum of a sequence of elements of Γ α ' will be called S. The above results can be restated as follows: i) If α < ω 1 then Γ α satisfies M and S.
ii) If α = ω 1 then Γ α satisfies S but does not satisfy M. iii) If α > ω 1 and α is a succesor or cf (α) = ω then Γ α satisfies M but does not satisfy S. iv) If α > ω 1 and cf (α) ≥ ω 1 then Γ α does not satisfy neither M nor S. Now, let I be an arbitrary linearly ordered set and let {G i } i∈I be a family of totally ordered multiplicative groups of rank 1.Let Γ I = {f ∈ i∈I G i : supp(f ) is finite} with componentwise multiplication and antilexicographically ordered.We prove that Γ I has a cofinal sequence if and only if I has a cofinal sequence or I has a last element.
With respect to the second one -every s ∈ Γ # I is the supremum of a sequence of elements of Γ I -we show that it is equivalent to a condition on the index set I, that sup I # {deg(f ) : f ≤ s} should be equal to the supremum of a sequence in I.A straightforward proof shows: Proof: (→) We only need to prove that if k ∈ I # \ I then there exists a sequence {i n } n<ω such that k = sup I # {i n } n<ω .For each i ∈ I, i < k, let g i ∈ G i , g i > 1 arbitrary.Let s = sup Γ I # {χ (g i ,i) : i < k}.By hypothesis, there exists a sequence {f n } ⊆ Γ I such that s = sup Γ I # {f n : n < ω}.

Corollary 4 . 5 :Theorem 4 . 6 :
The group Γ I has a cofinal sequence if and only if I has a last element or has a cofinal sequence.For every s ∈ Γ # I there exists a sequence {g n } n<ω ⊆ Γ I such that s = sup Γ # I {g n } n<ω if and only if for all k ∈ I # there exists a sequence {i n } n<ω such that k = sup I # {i n } n<ω .
It is easy to see that v is a Krull valuation of K with v(K) = Γ.For any fixed cardinal ℵ κ , there exists a group Γ α , of cardinality ℵ κ which has a cofinal sequence.Proof: Consider the ordinal α = ω κ + ω.Then |α| = ℵ κ and cf (α) = ω.Therefore, the group Γ α has cardinality ℵ κ and, by Proposition 2.4, it has a cofinal sequence.For any fixed infinite cardinal ℵ κ there exists a Krull valued field of cardinality ℵ κ which is metrizable.