ULTRAMETRIC FIXED POINT THEOREMS KATARZYNA AND FRANZ-VIKTOR KUHLMANN
Abstract. We present a general fixed point theorem which can be seen as the quintessence of the principles of proof for Banach’s Fixed Point Theorem and ultrametric fixed point theorems. It works in a minimal setting, not involving any metrics, only based on the notion of “ball” and the property of “spherical completeness”. We demonstrate its applications to the metric and the ultrametric cases, and (in two possible ways) to ordered abelian groups and fields.
1. Introduction
What is the common denominator of Banach’s Fixed Point Theorem and its ultrametric analogues as developed in [5, 6, 7, 4]? Is there a general principle of proof that works for both worlds, the (ordinary) metric and the ultrametric, and beyond? In this paper, we give an answer to these questions, based on the notions of “ball” and “spherical completeness” that we borrow from the ultrametric world.
While investigating spaces of real places, we found that in certain topo-logical spaces, it may be much easier and more natural to define “balls” than to define the “distance” between two elements. For example, if we are dealing with quotient topologies, like in the case of spaces of real places where the topology is induced by the Harrison topology of spaces of or-derings, balls may come up naturally as images of certain open or closed sets in the inducing topology. Therefore, we will work with ball spaces (X,B), that is, sets X with a set B of distinguished subsets, which we call
balls. We require no further structure on these spaces. We do not even
need a topology generated in some way by the balls. But let us mention that the way we formulate our theorems, the balls should be considered closed, rather than open, in such a topology, because singletons appear and are important. One can reformulate everything in an “open ball” approach, but this makes the exposition less elegant.
Date: December 29, 2011.
2000 Mathematics Subject Classification. Primary 54A05; Secondary 12J15, 12J25, 13A18, 54C10.
The research of the first author was partially supported by a sabbatical grant from the Silesian University at Katowice, Poland. The research of the second author was partially supported by a Canadian NSERC grant.
We found the idea of centering the attention on balls, rather than metrics, in the paper [8]. But there, ball spaces carry much more structure and the conditions for the fixed point theorem are unnecessarily restrictive.
Here are the notions that we take over from the ultrametric world. A nest
of balls in (X,B) is any nonempty collection of balls in B that is totally
ordered by inclusion. The ball space (X,B) will be called spherically
complete if every nest of balls has a nonempty intersection.
Consider a mapping f : X → X. We will write fx for f(x) and fix for
the image of x under the i-th iteration of f , that is, f0x = x, f1x = f (x)
and fi+1x = f (fix). The mapping f will be called strictly contracting
on orbits if there is a mapping
X ∋ x 7→ Bx ∈ B
such that for all x∈ X, the following conditions hold:
(SC1) x∈ Bx,
(SC2) Bf x⊆ Bx, and if x̸= fx, then Bfix ⊄= Bx for some i≥ 1,
and it will be called self-contractive if in addition, it satisfies:
(SC3) ifN ⊆ B is a nest of balls of the form Bx such that Bx ∈ N implies
Bf x ∈ N , and if z ∈ ∩
N , then Bz ⊆ ∩
N .
Note that (SC1) and (SC2) imply that fix∈ Bx for all i≥ 0. Here is our Fixed Point Theorem:
Theorem 1. Every self-contractive mapping on a spherically complete ball space has a fixed point.
This theorem is a corollary to the following more general result which only needs spherical completeness restricted to nests of the form as described in (SC3). Given a self-contractive mapping f on (X,B), we will say that a nest of balls is an f -nest if it consists of balls of the form Bx and satisfies the condition Bx ∈ N ⇒ Bf x∈ N .
Theorem 2. Take a mapping f on a ball space which is strictly contracting on orbits. If for every f -nest N in this ball space there is some z ∈ ∩N such that Bz ⊆
∩
N , then f has a fixed point.
For the proof, see the next section. In later sections, we discuss how to derive Banach’s Fixed Point Theorem, the ultrametric fixed point theo-rems and fixed point theotheo-rems for contractive mappings on nonarchimedean ordered abelian groups and fields.
Let us derive from Theorem 1 an attractor theorem which is modeled after the ultrametric attractor theorem in [3]. We consider two ball spaces (X,B) and (X′,B′) and a mapping φ : X → X′. Take an element z′ ∈ X′. If there is a mapping f : X → X which is strictly contracting on orbits, and a mapping
such that for all x∈ X, the following conditions hold:
(AT1) z′ ∈ Bx′ and φ(Bx)⊆ Bx′,
(AT2) if φ(x)̸= z′, then Bf x′ ⊄= B′x,
then z′ will be called a weak f -attractor for φ. If in addition f is self-contractive, then z′ will be called an attractor for φ.
Theorem 3. (Attractor Theorem 1)
Take a mapping φ : X → X′ and an attractor z′ ∈ X′ for φ. If (X,B) is spherically complete, then z′ ∈ φ(X).
Indeed, by Theorem 1, f has a fixed point z. But by condition (AT2),
f z = z implies that φ(z) = z′. The following version of the Attractor Theorem follows in a similar way from Theorem 2:
Theorem 4. (Attractor Theorem 2)
Take a mapping φ : X → X′ and a weak f -attractor z′ ∈ X′ for φ. If for every f -nest N in this ball space there is some z ∈ ∩N such that Bz ⊆
∩
N , then z′ ∈ φ(X).
2. Proof of Theorem 2
Take a self-contractive mapping f on the ball space (X,B). For every
x∈ X, the set {Bfix | i ≥ 0} is an f-nest. The set of all f-nests is partially
ordered by inclusion. Clearly, the union over an ascending chain of f -nests is again an f -nest (observe that the cardinality of this union is bounded by the cardinality of the power set of X, so the union is a set). Hence by Zorn’s Lemma, there is a maximal f -nest N . By the assumption of Theorem 2, there is some z ∈∩N such that Bz ⊆
∩
N . We wish to show
that z is a fixed point of f . If we would have that z ̸= fz, then by (SC2),
Bfiz ⊄= Bz ⊆
∩
N for some i ≥ 1, and the f-nest N ∪ {Bfkz | k ∈ N}
would properly containN . But this would contradict the maximality of N . Hence, z is a fixed point of f , so f has at least one fixed point.
3. Banach’s Fixed Point Theorem
Banach’s Fixed Point Theorem states that every strictly contracting ping on a complete metric space (X, d) has a unique fixed point. A map-ping f : X → X is called strictly contracting if there is a positive real number C < 1 such that d(f x, f y) ≤ Cd(x, y) for all x, y ∈ X. We will show now how Banach’s Fixed Point Theorem can be derived from The-orem 2. We work in the ball space (X,B) where B consists of all balls
{y ∈ X | d(x, y) ≤ r} for x ∈ X and r ∈ R≥0. Our aim is not to provide a new proof of Banach’s Fixed Point Theorem; the existing ones are much shorter. Our aim here is to show how to convert the problem from metric to ball space and to pave the way for the case of nonarchimedean ordered groups and fields that we will treat in Section 5.
We will prove the existence of a fixed points under the slightly more general assumption that f is
1) contracting, that is, d(f x, f y)≤ d(x, y) for all x, y ∈ X, and
2) strictly contracting on orbits, that is, there is a positive real number
C < 1 such that d(f x, f2x)≤ Cd(x, fx) for all x ∈ X.
Take any x∈ X. Then
d(x, fix) ≤ d(x, fx) + d(fx, f2x) + . . . + d(fi−1x, fix) ≤ d(x, fx)(1 + C + C2+ . . . + Ci−1) ≤ d(x, fx) ∞ ∑ i=0 Ci = d(x, f x) 1− C . Hence if we set Bx := { y∈ X | d(x, y) ≤ d(x, f x) 1− C } , then fix ∈ B
x for i ≥ 0. In particular, x ∈ Bx, hence (SC1) holds. We wish to show that Bf x ⊆ Bx. Take any y∈ Bf x. Then
d(x, y) ≤ d(x, fx) + d(fx, y) ≤ d(x, fx) + d(f x, f 2x) 1− C ≤ d(x, fx) + C 1− C d(x, f x) = d(x, f x) 1− C . Thus, y ∈ Bx, which proves our assertion.
Since C < 1, there is some i≥ 1 such that
Ci 1− C < 1 2 . Then d(fix, fi+1x) 1− C ≤ Ci 1− C d(x, f x) < 1 2d(x, f x) ,
which implies that x and f x cannot both lie in Bfix. Therefore, Bfix ⊄= Bx
and we have now proved that (SC2) holds.
Next, we show that also (SC3) holds. Take an f -nest N and assume that z ∈ ∩N . Pick any Bx ∈ N . Since f is contracting, d(fi+1x, f z) ≤
d(fix, z) = d(z, fix). Using that z ∈ N ⊆ Bfix for all i, and we compute
for i > 0: d(z, f z) ≤ d(z, fix) + d(fix, fi+1x) + d(fi+1x, f z) ≤ d(fix, fi+1x) 1− C + d(f i x, fi+1x) + d(f ix, fi+1x) 1− C = 3− C 1− Cd(f i x, fi+1x) ≤ 3− C 1− C C i d(x, f x) .
Since limi→∞Ci = 0, we find that f z = z and hence Bz ⊆ ∩
N . (In fact,
∩
It remains to show that if (X, d) is complete, then ∩N ̸= ∅. We have seen above that the radii of the balls Bfix converge to 0. Thus the balls Bfix
form a coinitial chain inN , and completeness implies that∩N =∩{Bfix |
i≥ 0} ̸= ∅.
Now the existence of a fixed point follows from Theorem 2.
Note that if f is strictly contracting, then the fixed point is unique. In-deed, if there were distinct fixed points x, y, then d(x, y) = d(f x, f y) <
d(x, y), a contradiction.
4. Ultrametric fixed point theorems
Let (X, d) be an ultrametric space. That is, d is a map from X× X to a partially ordered set Γ with first element 0, satisfying that for all x, y, z ∈ X and all γ ∈ Γ,
(U1) d(x, y) = 0 if and only if x = y,
(U2) if d(x, y)≤ γ and d(y, z) ≤ γ, then d(x, z) ≤ γ, (U3) d(x, y) = d(y, x) (symmetry).
(U2) is the ultrametric triangle law; if Γ is totally ordered, it can be replaced by
(UT) d(x, z)≤ max{d(x, y), d(y, z)}.
We obtain a ball space (X,Bd) from (X, d) by taking B to be the set of all
B(x, y) := {z ∈ X | d(x, z) ≤ d(x, y)} .
It follows from the ultrametric triangle law that B(x, y) = B(y, x) and that
(1) B(x, y)⊆ B(z, t) if and only if y ∈ B(z, t) and d(x, y) ≤ d(z, t) .
Two elements γ and δ of Γ are comparable if γ ≤ δ or γ ≥ δ. Hence if
d(x, y) and d(z, y) are comparable, then B(x, y) ⊆ B(z, y) or B(z, y) ⊆ B(x, y). If d(x, y) > d(z, y), then in addition, x /∈ B(z, y) and thus,
B(z, y)⊄= B(x, y). We note:
(2) d(x, y) > d(z, y) =⇒ B(z, y) ⊄= B(x, y) .
If Γ is totally ordered and B1 and B2 are any two balls with nonempty
intersection, then B1 ⊆ B2 or B2 ⊆ B1.
The ultrametric space (X, d) is called spherically complete if the cor-responding ball space is spherically complete.
A mapping f : X → X is called strictly contracting on orbits if for all x ∈ X, x ̸= fx implies d(fx, f2x) < d(x, f x). The following theorem
appeared in [7]:
Theorem 5. (Strong Ultrametric Fixed Point Theorem)
Take a spherically complete ultrametric space (X, d) and a mapping f : X → X. If f is strictly contracting on orbits and has the property that for all
x, z ∈ X, d(z, fx) ≤ d(fx, f2x) implies d(z, f z) ≤ d(x, fx), then f has a
fixed point.
Proof: Our theorem follows from Theorem 1 once we have shown that
f is self-contractive on the ball space (X,Bd). We define
Bx := B(x, f x)
and observe that x∈ Bx and Bf x ⊄= Bx because f x∈ Bx and d(f x, f2x) <
d(x, f x). To show that (SC3) holds, take an f -nest N and any z ∈ ∩N .
We have to show that Bz ⊆ ∩
N , that is, Bz ⊆ Bx for all Bx ∈ N . For this, it suffices that f z ∈ Bx since we know that z ∈ Bx and d is ultrametric. Since z ∈ ∩N ⊆ Bf x = B(f x, f2x), we have that d(z, f x) ≤ d(fx, f2x). By assumption, this implies d(z, f z)≤ d(x, fx), which shows that fz ∈ Bx.
2
A mapping f : X → X is called contracting if d(fx, fy) ≤ d(x, y) for all x, y ∈ X. It is shown in [7] that the following theorem follows from Theorem 5. We leave it as an exercise to the reader to deduce it directly from Theorem 1.
Theorem 6. (Ultrametric Fixed Point Theorem)
Every contracting mapping on a spherically complete ultrametric space which is strictly contracting on orbits has a fixed point.
Take nonempty ultrametric spaces (X, d) and (X′, d′) and a mapping
φ : X → X′. An element z′ ∈ Y′ is called attractor for φ if for every
y∈ X such that z′ ̸= φy, there is an element t ∈ X which satisfies: (UAT1) d′(φt, z′) < d′(φy, z′),
(UAT2) φ(B(y, t))⊆ B(φy, z′),
(UAT3) if x ∈ X such that d′(φy, z′) < d′(φx, z′) and φ(B(x, y)) ⊆
B(φx, z′), then d(x, y) and d(y, t) are comparable.
Condition (UAT1) says that the approximation φy of z′ from within the image of φ can be improved, and condition (UAT2) says that this can be done in a somewhat continuous way. Condition (UAT3) is always satisfied when the value set of (X, d) is totally ordered; for this reason, it does not appear as a condition in the attractor theorem of [3].
Theorem 7. Assume that z′ ∈ X′ is an attractor for φ : X → X′ and that
(X, d) is spherically complete. Then z′ ∈ φ(X).
Proof: For x ∈ X, we define Bx′ := B′(φx, z′). Then we define a mapping f : X → X as follows. If φx = z′, we set f x = x. Otherwise, we choose some t∈ X that satisfies (UAT1), (UAT2) and (UAT3); then we set f x := t and Bx := B(x, f x). Then z′ ∈ Bx′ by definition, and φ(Bx) =
Now assume that φx ̸= z′; then f x ̸= x. By (UAT1), we have that
d′(φf x, z′) < d′(φx, z′), which by (2) implies that Bf x′ = B(φf x, z′) ⊄=
B(φx, z′) = Bx′ . Thus, (AT2) holds.
We also see that φx /∈ B(φfx, z′). But by what we have already shown,
φ(Bf x)⊆ Bf x′ , so x /∈ Bf x. By (UAT3), d(x, f x) and d(f x, f2x) are compa-rable, so Bx ⊆ Bf x or Bf x⊆ Bx. Since x /∈ Bf x, it follows that Bf x⊄= Bx, which proves that f is strictly contracting on orbits.
We have shown that z′ is a weak f -attractor for φ. Our theorem will thus follow from Theorem 4 once we have proved that the second condition of that theorem is also satisfied. Take an f -nest N ; by the spherical completeness of (X, d), there is some z ∈∩N . We have to show that Bz ⊆
∩
N , that is, Bz ⊆ Bx for all Bx ∈ N . Since z ∈
∩ N ⊆ Bf x, we have that φz ∈ φ(Bf x) ⊆ Bf x′ = B(φf x, z′) . It follows that (3) d′(φz, z′) ≤ d′(φf x, z′) < d′(φx, z′) and (4) φ(B(x, z)) ⊆ φ(Bx) ⊆ Bx′ = B(φx, z′) .
Hence by (UAT3) with z in place of y, d(x, z) and d(z, f z) are comparable, and we find that B(x, z) ⊆ Bz or Bz ⊆ B(x, z) . On the other hand, we have already shown that φx /∈ Bf x′ = B(φf x, z′), hence φx /∈ φ(Bz) by (4). Therefore, x /∈ Bz, which yields that Bz ⊆ B(x, z) ⊆ Bx. 2
Note that our condition (UAT3) is somewhat stronger than condition (8) of [PR] because we do not start with a given mapping f (which is called g in [PR]), but construct it in our proof. Rewritten in our present notation, condition (8) of [PR] states:
(UAT3’) if d′(φz, z′) < d′(φx, z′) and B(z, f z) ∩ B(x, fx) ̸= ∅, then
d(z, f z) < d(x, f x).
For the convenience of the reader, we show how (UAT3’) implies the second condition of Theorem 4:
For z ∈ ∩N , we have to show that Bz ⊆ Bx for all Bx ∈ N . As in the above proof, one shows that (3) holds. Since z ∈∩N ⊆ Bx = B(x, f x), we have that B(z, f z)∩ B(x, fx) ̸= ∅. Hence by (UAT3’), d(z, fz) < d(x, fx), which implies that Bz ⊆ Bx.
Observe that the condition “d(z, f z) < d(x, f x)” in (UAT3’) can be replaced by the weaker condition that d(z, f z) and d(x, f x) are comparable.
5. Application to ordered abelian groups and fields Take a (totally) ordered abelian group (G, <). We take the balls in G to be the sets of the form
B(g, r) := {z ∈ G | |g − z| ≤ r}
for arbitrary g ∈ G and nonnegative r ∈ G, which we will call order balls. We set B := {B(g, r) | g ∈ G, 0 ≤ r ∈ G}.
Two elements a, b ∈ G are called archimedean equivalent if there is some n∈ N such that n|a| ≥ |b| and n|b| ≥ |a|. The ordered group (G, <) is archimedean ordered if all nonzero elements are archimedean equivalent. If 0 ≤ a < b and na < b for all n ∈ N, then “a is infinitesimally smaller than b” and we will write a≪ b.
We call a mapping f : G → G o-contracting if |fx − fy| ≤ |x − y| for all x, y ∈ X, and we call it strictly o-contracting on orbits if there is a positive rational number mn < 1 with m, n ∈ N such that n|fx − f2x| ≤
m|x − fx| for all x ∈ X. Note that n|fx − f2x| ≤ m|x − fx| holds if and
only if |fx − f2x| ≤ m
n|x − fx| in the divisible hull of G with the unique extension of the ordering.
The following theorem is a consequence of Theorem 1:
Theorem 8. Suppose that the ball space (G,B) associated with the ordered abelian group (G, <) is spherically complete. Then every o-contracting map-ping on G which is strictly o-contracting on orbits has a fixed point.
Proof: We compute in the divisible hull of G. We choose C ∈ Q such that mn < C < 1. With d(x, y) := |x − y|, the computations of Section 3
remain valid, and we may define the balls Bx in exactly the same way. They will then again satisfy (SC1) and (SC2). However, we have to work a little bit harder to show that (SC3) holds. In the present case, we cannot conclude that d(z, f z) = 0. But for given Bx ∈ N we can conclude that
d(z, f z) ≤ 3− C
1− C C
id(x, f x)
for all i, which means that d(z, f z)≪ d(x, fx). By choosing i so large that 3− C
(1− C)2 C
i ≤ C − m
n ,
we see that for every y ∈ Bz,
d(z, y) ≤ d(z, f z) 1− C ≤ 3− C (1− C)2 C id(x, f x) ≤ (C− m n ) d(x, f x) .
Using that z∈∩N ⊆ Bf2x, we compute: d(x, y) ≤ d(x, fx) + d(fx, f2x) + d(f2x, z) + d(z, y) ≤ (1 + m n ) d(x, f x) + C 2 1− C d(x, f x) + ( C−m n ) d(x, f x) = (1 + C)d(x, f x) + C 2 1− C d(x, f x) = d(x, f x) 1− C .
Hence y ∈ Bx. We have proved that Bz ⊆ Bx, as required. Now the
assertion of our theorem follows from Theorem 1. 2
We leave it to the reader to formulate a corresponding version of Theo-rem 2.
In the case of an ordered field (K, <), we may actually give a slightly more general definition of “strictly contracting on orbits” by requiring that there is an element C ∈ K such that 0 < C < 1 and |fx − f2x| ≤ C|x − fx|
for all x ∈ X, as this condition in fact implies the condition of the original definition. Working in the ordered additive group of the field, one then immediately obtains from Theorem 8 the corresponding version for ordered fields.
For the conclusion of this section, we wish to prove a characterization of those ordered abelian groups and ordered fields which are spherically complete with respect to the order balls. For more information on the notions we will work with, we refer the reader to chapter 2 of [4]. But note that there, valuations are written in the Krull style, which inverts the ordering on the (additively written) value groups, which also affects the definition of the Hahn product given below.
We define the natural valuation of (G, <) as follows. We denote by va the archimedean equivalence class of a. The set of archimedean equivalence classes is ordered as follows: va < vb if and only if |a| < |b| and a and
b are not archimedean equivalent, that is, n|a| < |b| for all n ∈ N. We
write 0 := v0 ; this is the minimal element in the totally ordered set of equivalence classes. The map a 7→ va is a group valuation on G, i.e., it satisfies vx = 0⇔ x = 0 and the ultrametric triangle law
(UT) v(x− y) ≤ max{vx, vy}.
The natural valuation induces an ultrametric defined by d(x, y) := v(x− y) and hence an ultrametric ball space, with the balls defined as in Section 4. In order to distinguish them from the order balls, we will write them as
Bu(x, y).
If (K, <) is a (totally) ordered field, then we consider the natural valuation on its ordered additive group and define va· vb := v(ab). This turns the set of archimedean classes into a multiplicatively written ordered abelian group, with neutral element v1 and inverses (va)−1 = v(a−1) . In this way,
v becomes a field valuation (with multiplicatively written value group). It
is the finest valuation on K which is compatible with the ordering.
Lemma 9. Take an ordered abelian group G. If G is spherically complete w.r.t. the order balls, then it is spherically complete w.r.t. the ultrametric balls.
Proof: Take a nest N of ultrametric balls Bu(xi, yi), i ∈ I. If this nest contains a smallest ball, then its intersection is nonempty and there is nothing to show. So we assume that it does not contain a smallest ball. That is, for every i ∈ I there is j ∈ I such that Bu(xj, yj) ⊄= Bu(xi, yi), which implies that v(xj − yj) < v(xi− yi) since these balls are ultrametric. For each i∈ I, we pick such an index j and define an order ball
Bi := B(xj,|xi− yi|) ⊆ Bu(xi, yi) .
Since v(xj − yj) < v(xi − yi) implies that |xj − yj| ≪ |xi − yi|, we find that Bu(xj, yj) ⊆ Bi. It follows that {Bi | i ∈ I} is a nest of balls and its intersection is equal to the intersection ofN . Since G is spherically complete w.r.t. the order balls by assumption, this intersection is nonempty. 2
Lemma 10. An archimedean ordered abelian group is spherically complete w.r.t. the order balls if and only if it is isomorphic to the additive group of the integers or of the reals with the canonical ordering.
Proof: First, observe that every nest N of order balls in the integers contains a smallest ball, so the integers are spherically complete w.r.t. the order balls.
Next, take a nest N of order balls in the reals. Pick a ball B in this nest and consider the nestN0 of all balls inN that are contained in B; this nest
has the same intersection as N . The order ball B is compact in the order topology of R. All balls in N0 being closed in the order topology, N0 is a
centered system of closed subsets of B, in the sense of [1, Proposition 2, p. 57]. By this proposition, the intersection of N0 is nonempty. We have
proved that every ordered abelian group which is isomorphic to the additive ordered group of the integers or the reals is spherically complete w.r.t. the order balls.
For the converse, we use that by the Theorem of H¨older, every archimedean ordered abelian group G can be embedded in the additive ordered group of the reals. We identify G with its image under this embedding and show that if G is spherically complete w.r.t. the order balls, but not isomorphic to Z, then G = R. Since G is not isomorphic to Z, it is dense in R. Hence for every a∈ R there is a sequence (gi)i∈N in G converging to a. By passing to a subsequence if necessary, we may assume that for each i∈ N,
So Bi, i ∈ N, is a nest of order balls in G. Since G is spherically complete w.r.t. the order balls, this nest has a nonempty intersection. As this inter-section can only consist of the element a, we find that a ∈ G. 2 Take any ordered abelian group G and an element a ∈ G, a ̸= 0. Then
Ba := {g ∈ G | vg ≤ va} and Ba◦ := {g ∈ G | vg < va} are convex subgroups of G. It follows that Ca := Ba/Ba◦ is an archimedean ordered group, called the archimedean component of G at a. If va = γ, we will also write Cγ for Ca; note that Cγ does not depend on the choice of the elment a for which va = γ.
Proposition 11. An ordered abelian group is spherically complete w.r.t. the order balls if and only if it is spherically complete w.r.t. the ultrametric balls and all of its archimedean components are isomorphic to the ordered additive groups of Z or R.
Proof: Assume that the ordered abelian group G is spherically complete w.r.t. the order balls. Then by Lemma 9 it is spherically complete w.r.t. the ultrametric balls. Take any nonzero a ∈ G and a nest of order balls
B(hi, si), i ∈ I, in Ca. For each i ∈ I, choose elements gi, ri ∈ G such that hi = gi+ Ba◦ and si = ri+ B◦a. Then all ri are positive and B(gi, ri),
i∈ I, is a nest of order balls in G. By our assumption on G, its intersection
is nonempty. Hence also the intersection of the balls B(hi, si), i ∈ I, is nonempty. We have shown that Ca is spherically complete w.r.t. the order balls, and from Lemma 10 it now follows that Cais isomorphic to the ordered additive groups of Z or R.
For the converse, assume that G is spherically complete w.r.t. the ultra-metric balls and all of its archimedean components are isomorphic to the ordered additive groups of Z or R. Take a nest B(gi, ri), i ∈ I, of order balls in G. We distinguish two cases.
First, assume that the set vri, i ∈ I, has no smallest element. Then for every i ∈ I there is j ∈ I such that B(gj, rj)⊄= B(gi, ri) and vrj < vri. For each i ∈ I, we pick such an index j and define an ultrametric ball
Biu := Bu(gj, gj+ rj) ⊇ B(gj, rj) .
Since v(gj + rj − gj) = vrj < vri implies that |gj + rj − gj| ≪ ri, we find that Bui ⊆ B(gi, ri). It follows that {Bui | i ∈ I} is a nest of balls and its intersection is equal to the intersection of the balls B(gi, ri). Since
G is spherically complete w.r.t. the ultrametric balls by assumption, this
intersection is nonempty.
Next, assume that the set vri, i ∈ I, has a smallest element vri0. Set
r = ri0 and I0 :={i ∈ I | vri = vr}. If i ∈ I0 and j ∈ I \ I0, then B(gi, ri)⊄=
B(gj, rj). Therefore, the intersection of the nest B(gi, ri), i∈ I, is equal to that of the nest B(gi, ri), i∈ I0. Subtracting gi0 from all elements, we shift
v-value ≤ vri0, so we can project the balls to the archimedean component
Cr, where the images of the balls will form a nest B(gi− gi0+ Br◦, ri+ Br◦),
i ∈ I0, of order balls. By assumption, Cr is isomorphic to the ordered additive groups of Z or R. Hence by Lemma 10, the induced nest has a nonempty intersection. Hence the same holds for the nest B(gi, ri), i∈ I.
2
Given a totally ordered index set Γ and for every γ ∈ Γ an arbitrary abelian group Cγ, we define a group called the Hahn product, denoted by
H
γ∈ΓCγ. Consider the product∏
γ∈ΓCγ and an element c = (cγ)γ∈Γ of this group. Then the support of c is the set of all γ ∈ Γ for which
cγ ̸= 0 . As a set, the Hahn product is the subset of ∏
γ∈ΓCγ containing all elements whose support is a anti-wellordered subset of Γ, that is, every nonempty subset of the support has a maximal element). In particular, the support of every element c in the Hahn product has a maximal element γ0,
which enables us to define a group valuation by setting vc = γ0. The Hahn
product is a subgroup of the product group. Indeed, the support of the sum of two elements is contained in the union of their supports, and the union of two anti-wellordered sets is again anti-wellordered.
If the components Cγ are (not necessarily archimedean) ordered abelian groups, we obtain the ordered Hahn product, also called
lexicograph-ical product, where the ordering is defined as follows. Given an element c = (cγ)γ∈Γ, let γ0 be the maximal element of its support. Then we take
c > 0 if and only if cγ0 > 0. If all Cγ are archimedean ordered, then the
valuation v of the Hahn product coincides with the natural valuation of the ordered Hahn product. Every ordered abelian group G can be embedded in the Hahn product with its set of archimedean classes as index sets and its archimedean components as components. Then G is spherically complete w.r.t. the ultrametric balls if and only if the embedding is onto.
In the case of an ordered field K, we know from [2, Theorem 6] that K can be embedded in the power series field with exponents in the value group and coefficients in the residue field of its natural valuation. (A nontrivial factor set may be needed, unless the residue field is closed under radicals, which for instance is the case if K is real closed.) Again, K is spherically complete w.r.t. the ultrametric balls if and only if the embedding is onto. If we forget about the multiplication on the power series field, we obtain the underlying Hahn product that corresponds to the ordered additive group of
K. The multiplication of K induces an isomorhism of all its components
to the additive ordered group of the residue field of K. Also, the latter is divisible and cannot be isomorphic toZ. So we obtain the following theorem from Proposition 11:
Theorem 12. An ordered abelian group is spherically complete w.r.t. the order balls if and only if it is isomorphic to a Hahn product with all its components being Z or R.
An ordered field is spherically complete w.r.t. the order balls if and only if it is isomorphic to a power series field (possibly with nontrivial factor set) with residue field R.
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Department of Mathematics & Statistics, University of Saskatchewan, 106 Wiggins Road, Saskatoon, SK, S7N 5E6, Canada
E-mail address: [email protected]
Institute of Mathematics, Silesian University, Bankowa 14, 40-007 Ka-towice, Poland
