An embedding, an extension, and an interpolation of ultrametrics
aa r X i v : . [ m a t h . M G ] A ug AN EMBEDDING, AN EXTENSION, AND ANINTERPOLATION OF ULTRAMETRICS
YOSHITO ISHIKI
Abstract.
The notion of the ultrametrics can be considered as azero-dimensional analogue of ordinary metrics, and it is expectedto prove ultrametric versions of theorems on metric spaces. In thispaper, we provide ultrametric versions of the Arens–Eells isomet-ric embedding theorem of metric spaces, the Hausdorff extensiontheorem of metrics, the Niemytzki–Tychonoff characterization the-orem of the compactness, and the author’s interpolation theoremof metrics and theorems on dense subsets of spaces of metrics. Introduction
Let X be a set. A metric d on X is said to be an ultrametric or a non-Archimedean metric if for all x, y, z ∈ X we have d ( x, y ) ≤ d ( x, z ) ∨ d ( z, y ) , (1.1)where the symbol ∨ stands for the maximum operator on R . Theinequality (1.1) is called the strong triangle inequality . We say that aset S is a range set if S ⊂ [0 , ∞ ) and 0 ∈ S . For a range set S , we saythat a metric d : X → [0 , ∞ ) on X is S -valued if d ( X ) ⊂ S . Note that[0 , ∞ )-valued ultrametrics are nothing but ultrametrics. The S -valuedultrametrics are studied as a special case and a reasonable restrictionof ultrametrics. Gao and Shao [13] studied R -valued Urysohn universalultrametric spaces for a countable range set R . Brodskiy, Dydak, Higesand Mitra [3] utilized ( { }∪{ n | n ∈ Z } )-valued ultrametrics for theirstudy on the 0-dimensionality in categories of metric spaces.The notion of the ultrametrics can be considered as a 0-dimensionalanalogue of ordinary metrics, and it is often expected to prove ultra-metric versions of theorems on metric spaces. In this paper, for everyrange set S , we provide S -valued ultrametric versions of the Arens–Eells isometric embedding theorem [1] of metric spaces, the Hausdorffextension theorem [17] of metrics, the Niemytzki–Tychonoff character-ization [27] of the compactness, and the author’s interpolation theoremof metrics and theorems on dense G δ subsets of spaces of metrics [22]. Date : August 25, 2020.2010
Mathematics Subject Classification.
Primary 54E35; Secondary 30L05,54E52.
Key words and phrases.
Ultrametric space, Baire space, Isometric embedding.The author was supported by JSPS KAKENHI Grant Number 18J21300.
In 1956, Arens and Eells [1] established the result which today wecall the Arens–Eells embedding theorem, stating that for every metricspace (
X, d ), there exist a real normed linear space V and an isometricembedding I : X → V such that(1) I ( X ) is closed in V ;(2) I ( X ) is linearly independent in V .Before stating our first main result, we introduce some basic notions.Let R be a commutative ring, and let V be an R -module. A subset S of V is said to be R -independent if for every finite subset { f , . . . , f n } of S ,and for all N , . . . , N n ∈ R , the identity P ni =1 N i f i = 0 implies N i = 0for all i . A function k ∗ k : V → [0 , ∞ ) is said to be an ultra-norm on V if the following are satisfied:(1) k x k = 0 if and only if x = 0;(2) for every x ∈ V , we have k − x k = k x k ;(3) for all x, y ∈ V , we have k x + y k ≤ k x k ∨ k y k .The pair ( V, k ∗ k ) is called an ultra-normed R -module (see also [49]).Megrelishvili and Shlossberg [28] have already proven an ultrametricversion of the Arens–Eells embedding theorem, stating that every ultra-metric space is isometrically embeddable into an ultra-normed Booleangroup (a Z / Z -module) as a closed set of it, which is a consequence oftheir study on free non-Archimedean topological groups and Booleangroups with actions from topological groups. By introducing modulestructures into universal ultrametric spaces of Lemin–Lemin type [24],we obtain a more general S -valued ultrametric version of the Arens–Eells embedding theorem. Theorem 1.1.
Let S be a range set. Let R be an integral domain,and let ( X, d ) be an S -valued ultrametric space. Then there exist an S -valued ultra-normed R -module ( V, k ∗ k ) , and an isometric embedding I : X → V such that (1) I ( X ) is closed in V ; (2) I ( X ) is R -independent in V .Moreover, if ( X, d ) is complete, then we can choose ( V, k ∗ k ) as acomplete metric space.Remark . Let R be an integral domain. Let t R be the trivial val-uation on R defined by t R ( x ) = 1 if x = 0; otherwise t R ( x ) = 0. Let V be a torsion-free R -module. Then every ultra-norm k ∗ k on V iscompatible with t R , i.e., for every r ∈ R and for every x ∈ V , we have k r · x k = t R ( r ) k x k . For every finite field, there exist no valuations onit except the trivial valuation. Thus we can consider that Theorem1.1 includes the Arens–Eells embedding theorem into normed spacesover all finite fields. The author does not know whether such an em-bedding theorem into normed spaces over all non-Archimedean valuedfields holds true or not. N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 3
Remark . There are various isometric embedding theorems from anultrametric space into a metric space with algebraic structures. Forinstance, Schikhof [37] constructed an isometric embedding from anultrametric space into a non-Archimedean valued field. The existenceof an isometric embedding from an ultrametric space into a Hilbertspace was first proven by Timan and Vestfrid [42] in a separable case,and was proven by A. J. Lemin [25] in a general setting. The papers[43], [47] and [12] also contain related results.For a range set S , and for a topological space X , we denote byUM( X, S ) (resp. M( X )) the set of all S -valued ultrametrics (resp. met-rics) on X that generate the same topology as the original one of X .We denote by UM( X ) the set UM( X, [0 , ∞ )). We say that a topo-logical space X is S -valued ultrametrizable (resp. ultrametrizable ) ifUM( X, S ) = ∅ (resp. UM( X ) = ∅ ). We say that X is completely S -valued ultrametrizable (resp. completely ultrametrizable ) if there existsa complete metric d ∈ UM(
X, S ) (resp. d ∈ UM( X )).We say that a range set S has countable coinitiality if there exists anon-zero strictly decreasing sequence { r i } i ∈ N in S with lim i →∞ r i = 0. Remark . For a range set S with the countable coinitiality, and fora topological space X , it is worth clarifying a relation between theultrametrizability and the S -valued ultrametrizability. In Proposition2.12, we show that these two properties are equivalent to each other.In 1930, Hausdorff [17] proved the extension theorem stating that forevery metrizable space X , for every closed subset A of X , and for everymetric e ∈ M( A ), there exists a metric D ∈ M( X ) with D | A = e .By using the Arens–Eells embedding theorem, Toru´nczyk [44] pro-vided a simple proof of the Hausdorff extension theorem. Due toToru´nczyk’s method and Theorem 1.1, we can prove an S -valued ul-trametric version of the Hausdorff extension theorem. Theorem 1.2.
Let S be a range set. Let X be an S -valued ultra-metrizable space, and let A be a closed subset of X . Then For every e ∈ UM(
A, S ) , there exists D ∈ UM(
X, S ) with D | A = e . Moreover,if X is completely metrizable and e ∈ UM(
A, S ) is complete, then wecan choose D as a complete S -valued ultrametric.Remark . There are several studies on extending a partial or con-tinuous ultrametrics (see [10], [37], [38], or [45]).In 1928, Niemytzki and Tychonoff [27] proved that a metrizable space X is compact if and only if all metrics in M( X ) are complete. Hausdorff[17] gave a simple proof of their characterization theorem by applyinghis extension theorem of metrics mentioned above. By using Haus-dorff’s argument and Theorem 1.2, we obtain an ultrametric version ofthe Niemytzki–Tychonoff theorem. YOSHITO ISHIKI
Corollary 1.3.
Let S be a range set with the countable coinitiality. Let X be an S -valued ultrametrizable space. Then the space X is compactif and only if for every ultrametric d ∈ UM(
X, S ) is complete. To state our next results, for a topological space X , and for a rangeset S , we define a function U D SX : UM( X, S ) → [0 , ∞ ] by assigning U D SX ( d, e ) to the infimum of ǫ ∈ S ⊔ {∞} such that for all x, y ∈ X wehave d ( x, y ) ≤ e ( x, y ) ∨ ǫ, and e ( x, y ) ≤ d ( x, y ) ∨ ǫ. The function
U D SX is an ultrametric on UM( X, S ) valued in CL( S ) ⊔{∞} , where CL( S ) stands for the closure of S in [0 , ∞ ). We also definea function D X : M( X ) × M( X ) → [0 , ∞ ] by D X ( d, e ) = sup ( x,y ) ∈ X | d ( x, y ) − e ( x, y ) | . The function D X is a metric on M( X ) valued in [0 , ∞ ]. Remark . Qiu [40] introduced the strong ǫ -isometry in the study onthe non-Archimedean Gromov–Hausdorff distance (see [51]). This con-cept is an analogue for ultrametric spaces of the ǫ -isometry in the studyon the ordinary Gromov–Hausdorff distance (see [4]). Roughly speak-ing, for a range set S , for an S -valued ultrametrizable space X , and for S -valued ultrametrics d, e ∈ UM(
X, S ), the inequality
U D SX ( d, e ) ≤ ǫ isequivalent to the statement that the identity maps 1 X : ( X, d ) → ( X, e )and 1 X : ( X, e ) → ( X, d ) are strong ǫ -isometries.The author [22] proved an interpolation theorem of metrics withthe information of D X (see [22, Theorem 1.1]). As its application,the author proved that for every non-discrete metrizable space X theset of all metrics in M( X ) with a transmissible property, which is ageometric property determined by finite subsets (see Definition 6.2), isdense G δ in the metric space (M( X ) , D X ) (see [22, Theorem 1.2]), andalso proved a local version of it (see [22, Theorem 1.3]).By using Theorems 1.1 and 1.2, we can prove an ultrametric versionof the author’s interpolation theorem.A family { H i } i ∈ I of subsets of a topological space X is said to be discrete if for every x ∈ X there exists a neighborhood of x intersectingat most single member of { H i } i ∈ I . For a range set S , and for a subset E of S , we denote by sup E the supremum of E taken in [0 , ∞ ], not in S .For C ∈ [1 , ∞ ), we say that S is C -quasi-complete if for every boundedsubset E of S , there exists s ∈ S with sup E ≤ s ≤ C · sup E . We saythat S is quasi-complete if S is C -quasi-complete for some C ∈ [1 , ∞ ).Note that a range set is 1-quasi-complete if and only if it is closed underthe supremum operator. N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 5
Theorem 1.4.
Let C ∈ [1 , ∞ ) , and let S be a C -quasi-complete rangeset. Let X be an ultrametrizable space, and let { A i } i ∈ I be a discretefamily of closed subsets of X . Then for every S -valued ultrametric d ∈ UM(
X, S ) , and for every family { e i } i ∈ I of ultrametrics with e i ∈ UM( A i , S ) for all i ∈ I , there exists an S -valued ultrametric m ∈ UM(
X, S ) satisfying the following: (1) for every i ∈ I we have m | A i = e i ; (2) sup i ∈ I U D SA i ( e i , d | A i ) ≤ U D SX ( m, d ) ≤ C · sup i ∈ I U D SA i ( e i , d | A i ) .Moreover, if X is completely metrizable, and if each e i ∈ UM( A i , S ) iscomplete, then we can choose m ∈ UM(
X, S ) as a complete metric. Similarly to [22], Theorem 1.4 enables us to prove theorems on dense G δ subsets of UM( X, S ). The definitions of the transmissible parame-ter, the transmissible property, and the S -ultra-singularity for a rangeset can be seen in Definitions 6.1, 6.2 and 6.3. Theorem 1.5.
Let S be a quasi-complete range set with the count-able coinitiality. Let G be an S -ultra-singular transmissible parame-ter. Then for every non-discrete ultrametrizable space X , the set ofall d ∈ UM(
X, S ) for which ( X, d ) satisfies the anti- G -transmissibleproperty is dense G δ in the space (UM( X, S ) , U D SX ) . For a property P on metric spaces, we say that a metric space ( X, d )satisfies the local P if every non-empty open metric subspace of X satisfies the property P . Theorem 1.6.
Let S be a quasi-complete range set with the countablecoinitiality. Let X be a second countable, locally compact locally non-discrete ultrametrizable space. Then for every S -ultra-singular trans-missible parameter G , the set of all d ∈ UM(
X, S ) for which ( X, d ) satisfies the local anti- G -transmissible property is a dense G δ set in thespace (UM( X, S ) , U D SX ) . For example, the doubling property is equivalent to a G -transmissibleproperty for some ultra-singular transmissible parameter G .The organization of this paper is as follows: In Section 2, we reviewthe basic or classical statements on S -valued ultrametric spaces and 0-dimensional spaces. We give proofs of some of them. In Section 3, weobserve that a construction of universal ultrametric spaces and isomet-ric embeddings of Lemin–Lemin-type [24] can be applied to S -valuedultrametric spaces for all range set S . We also discuss algebraic struc-tures on universal ultrametric spaces of Lemin–Lemin-type. After that,we prove Theorem 1.1. In Section 4, we prove Theorem 1.2 and Corol-lary 1.3 by following Toru´nczyk and Hausdorff’s methods, and by usingTheorem 1.1. In Section 5, we prove Theorem 1.4 by converting theauthor’s proof of [22, Theorem 1.1] into an S -valued ultrametric proofwith Theorems 1.1 and 1.2. In Section 6, we introduce transmissible YOSHITO ISHIKI property, originally defined in [22], and we prove Theorem 1.5 as anapplication of Theorem 1.4. In Section 7, we first show that UM(
X, S )is a Baire space for a range set S , and for a second countable locallycompact X (see Lemma 7.6). We next prove Theorem 1.6, which is alocal version of Theorem 1.5. Acknowledgements.
The author would like to thank Professor KoichiNagano for his advice and constant encouragement.2.
Preliminaries
In this paper, we denote by N the set of all positive integers. For a set E , we denote by card( E ) the cardinality of E . For a metric space ( X, d ),and for a subset A of X , we denote by δ d ( A ) the diameter of A . Wedenote by B ( x, r ) (resp. U ( x, r )) the closed (resp. open) ball centeredat x with radius r . We also denote by B ( A, r ) the set S a ∈ A B ( a, r ).To emphasize metrics under consideration, we sometimes denote by B ( x, r ; d ) (resp. U ( x, r ; d )) the closed (resp. open) ball in ( X, d ). For arange set S , we define S + = S \ { } .2.1. Modification of ultrametrics.Lemma 2.1.
Let S be a range set. Let ( X, d ) be an S -valued ultramet-ric space. Let ǫ ∈ S + . Then the function e : X → [0 , ∞ ) defined by e = min { d, ǫ } belongs to UM(
X, S ) .Proof. For all a, b, c ∈ R , we have ( a ∨ b ) ∧ c = ( a ∧ c ) ∨ ( b ∧ c ), where ∧ stands for the minimum operator on R . This leads to the lemma. (cid:3) Let (
X, d ) and (
Y, e ) be metric spaces. Define a function d × ∞ e on( X × Y ) by ( d × ∞ e )(( x, y ) , ( z, w )) = d ( x, z ) ∨ e ( y, w ) . It is well-known that d × ∞ e is a metric on X × Y , and it generates theproduct topology of X × Y . In the case of ultrametrics, we have: Lemma 2.2.
Let S be a range set. Let ( X, d ) and ( Y, e ) be S -valuedultrametric spaces. Then the metric d × ∞ e belongs to UM( X × Y, S ) . The following proposition is known as an amalgamation of ultramet-rics. The proof can be seen in, for example, [2, Theorem 2.2], or, byreplacing the symbol “+” with the symbol “ ∨ ” in the proof of [22,Propositions 3.3], we can prove the following proposition.For a mutually disjoint family { T i } i ∈ I of topological spaces, we con-sider that the set ` i ∈ I T i is equipped with the direct sum topology. Proposition 2.3.
Let S be a range set. Let ( X, d X ) and ( Y, d Y ) be S -valued ultrametric spaces. If X ∩ Y = ∅ , then for every r ∈ S + thereexists an S -valued ultrametric h ∈ UM( X ⊔ Y, S ) such that (1) h | X = d X ; N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 7 (2) h | Y = d Y ; (3) for all x ∈ X and y ∈ Y we have r ≤ h ( x, y ) . As a consequence of Proposition 2.3, we can construct a one-pointextension of an ultrametric space.
Corollary 2.4.
Let S be a range set. Let ( X, d ) be an S -valued ultra-metric space, and let o X . Then there exists an S -valued ultrametric D ∈ UM( X ⊔ { o } , S ) with D | X = d . Invariant metrics on modules.
Let R be a commutative ring,and let V be an R -module. We say that a metric on d on V is invariant ,or invariant under the addition if for all a, x, y ∈ V we have d ( x + a, y + a ) = d ( x, y ) . By the definitions of the ultra-norms and invariant metrics, we obtain:
Lemma 2.5.
Let R be a commutative ring, and let ( V, k ∗ k ) be anultra-normed R -module. Then the metric d on V defined by d ( x, y ) = k x − y k is an invariant ultrametric on V . Conversely, if ( V, d ) is a pairof an R -module V and an invariant ultrametric d , then the function k ∗ k : V → [0 , ∞ ) defined by k x k = d ( x, is an ultra-norm on V . Based on Lemma 2.5, in what follows, we will use a pair (
V, d ) ofan R -module and an invariant ultrametric d on V , rather than a pair( V, k ∗ k ) of V and an ultra-norm k ∗ k on V .By the definition of the ultra-norms, we obtain: Lemma 2.6.
Let R be a commutative ring, and let ( V, d ) be an ultra-normed R -module. Then the addition + : V × V → V and the inversion m : V → V defined by m ( x ) = − x are continuous with respect to thetopology induced from d . The next lemma is utilized in the proof of Theorem 1.1.
Lemma 2.7.
Let R be a commutative ring, and let ( V, d ) be an ultra-normed R -module. If for every non-zero r ∈ R and for every v ∈ V we have d ( r · v,
0) = d ( v, , then the completion of ( V, d ) becomes anultra-normed R -module which contains V as an R -submodule.Proof. We introduce an R -module structure into the completion ( W, D )of (
V, d ). For all x, y ∈ W , take sequences { x n } n ∈ N and { y n } n ∈ N in V such that x n → x and y n → y as n → ∞ . Then we define an additionon W by x + y = lim n →∞ ( x n + y n ) . Since d is an ultra-norm, this addition is well-defined.For every r ∈ R , we define a scalar multiplication on W by r · x = lim n →∞ r · x n . YOSHITO ISHIKI
By the assumption on the scalar multiplication on V and the ultra-norm, the scalar multiplication on W is well-defined. By these defini-tions, ( W, D ) becomes an ultra-normed R -module which contains V asan R -submodule. This finishes the proof. (cid:3) Basic properties of ultrametric spaces.
The next lemma fol-lows from the strong triangle inequality.
Lemma 2.8.
Let X be a set, and let w : X → R be a symmetric map.Then w satisfies the strong triangle inequality if and only if for all x, y, z ∈ X the inequality w ( x, z ) < w ( y, z ) implies w ( y, z ) = w ( x, y ) . By this lemma, we see that in an ultrametric space, every triangleis isosceles, and the side-length of the legs of the isosceles triangle isequal to or greater than the side-length of base.The property in the next proposition follows from the strong triangleinequality (see (12) in [7, Theorem 1.6]).
Proposition 2.9.
Let S be a range set, and let ( X, d ) be an S -valuedultrametric space. Then the completion of ( X, d ) is an S -valued ultra-metric space.Remark . By Proposition 2.9, for every separable ultrametric space(
X, d ), the set { d ( x, y ) | x, y ∈ X } is countable. This phenomenon isa reason why we consider S -valued ultrametrics for a range set S .We now prove that for every range set S with the countable coini-tiality, the ultrametrizability and the S -valued ultrametrizability areequivalent to each other. Lemma 2.10.
Let S be a range set with the countable coinitiality.Let { r ( i ) } i ∈ N be a non-zero strictly decreasing sequence in S such that r ( i ) → as i → ∞ . Put T = { } ∪ { r ( i ) | i ∈ N } . If UM(
X, S ) = ∅ ,then UM(
X, T ) = ∅ .Proof. Take h ∈ UM(
X, S ). Put d = min { h, r (1) } . Then by Lemma2.1, we have d ∈ UM(
X, S ). Define a symmetric function e : X → T by e ( x, y ) = ( r ( n ) if r ( n + 1) < d ( x, y ) ≤ r ( n );0 if d ( x, y ) = 0.Then e is a T -valued ultrametric. Take x ∈ X . For each s ∈ S , take n ∈ N with r ( n + 1) < s ≤ r ( n ). By the definition of e , we have U ( x, r ( n + 1); e ) ⊂ U ( x, s ; d ). On the other hand, for all n ∈ N , we alsohave U ( x, r ( n ); d ) ⊂ U ( x, r ( n ); e ). Thus the ultrametric e generatesthe same topology as X . (cid:3) Lemma 2.11.
Let X be a topological space. Let S and T be two rangesets, and let ψ : S → T be a bijective monotone map. Then for every d ∈ UM(
X, S ) the function ψ ◦ d belongs to UM(
X, T ) . N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 9
Proof.
Since ψ is monotone, the function ψ ◦ d satisfies the strong trian-gle inequality, and hence it is a T -valued ultrametric. For every x ∈ X and for every s ∈ S , we have U ( x, s ; d ) = U ( x, ψ ( s ); ψ ◦ d ). This impliesthat ψ ◦ d ∈ UM(
X, T ). (cid:3) Proposition 2.12.
Let S be a range set with the countable coinitiality,and let X be a topological space. Then X is ultrametrizable if and onlyif X is S -valued ultrametrizable.Proof. It suffices to show that if X is ultrametrizable, then UM( X, S ) = ∅ . Let { r ( i ) } i ∈ N be a non-zero strictly decreasing sequence in S suchthat r ( i ) → i → ∞ . Put T = { } ∪ { r ( i ) | i ∈ N } , and put A = { } ∪ { − i | i ∈ N } . Then there exists a bijective monotonemap ψ : A → T . Since UM( X ) = ∅ , Lemmas 2.10 and 2.11 implythat UM( X, T ) = ∅ . From UM( X, T ) ⊂ UM(
X, S ), the propositionfollows. (cid:3)
Remark . If S does not have countable coinitiality, Proposition 2.12does not hold true. In this case, all S -valued ultrametrizable spacesare discrete (see Lemma 2.16).A topological space X is said to be 0 -dimensional if for every pair ofdisjoint two closed subsets A and B of X , there exists a clopen subset Q of X with A ⊂ Q and Q ∩ B = ∅ . Such a space is sometimes also said tobe ultranormal . Note that a metric space ( X, d ) is 0-dimensional if andonly if every finite open covering of X has a refinement covering of X consisting of mutually disjoint finite open sets. This equivalence followsfrom the fact that the large inductive dimensions and the coveringdimensions coincide on metric spaces (see e.g., [33, Theorem 5.4]).The following was proven by de Groot [14] (see also [7]). Proposition 2.13.
All ultrametrizable spaces are -dimensional. We now clarify a relation between the complete S -valued ultrametriz-ability for a range set S and the complete metrizability. The proofsof the following lemma and proposition are S -valued ultrametric ana-logues of [50, Theorem 24.12]. Lemma 2.14.
Let S be a range set with the countable coinitiality. Let X be a completely S -valued ultrametrizable, and let G be an open subsetof X . Then G is completely S -valued ultrametrizable.Proof. Sine X is 0-dimensional, and since all open sets of metric spacesare F σ , there exists a sequence { O n } n ∈ N of clopen sets of X such that(1) for each n ∈ N , we have O n ⊂ O n +1 ;(2) G = S n ∈ N O n .Take a sequence of { a n } n ∈ N in the field Q of all 2-adic numbers suchthat for each m ∈ N the sum P ∞ i = m a i is convergent to a non-zero 2-adicnumber (for example, we can take a n = 2 n − n +1 ). Define a function F : X → Q by F ( x ) = P ∞ i =1 a i · χ O i ( x ), where χ O i is the characteristicfunction of O i . Then F is continuous. By the assumption on { a n } n ∈ N ,for every x ∈ G , we have F ( x ) = 0 and F | X \ G = 0. Take a complete S -valued ultrametric d ∈ UM(
X, S ). We denote by v : Q → Z ⊔{∞} the2-adic valuation on Q . Take a non-zero strictly decreasing sequence { r ( i ) } i ∈ N in S with lim i →∞ r ( i ) = 0. We put r ( ∞ ) = 0. Then a metric W : ( Q ) → S defined by W ( x, y ) = r ( v ( x − y ) ∨ Q , S ), and it is complete. Define a metric D on G by D ( x, y ) = W (cid:18) F ( x ) , F ( y ) (cid:19) ∨ d ( x, y ) . Since the function 1 /F is continuous on G , we have D ∈ UM(
G, S ).We next show that D is complete. Assume that { x n } n ∈ N is Cauchy in( G, D ). Then { /F ( x n ) } n ∈ N and { x n } n ∈ N are Cauchy in ( Q , W ) and( X, d ), respectively. Thus, there exist A ∈ Q and B ∈ X such that1 /F ( x n ) → A in ( Q , W ), and x n → B in ( X, d ). If B G , then wehave F ( x n ) →
0. This contradicts to 1 /F ( x n ) → A . Thus B ∈ G ,and hence D is complete. Therefore we conclude that G is completely S -valued ultrametrizable. (cid:3) Proposition 2.15.
Let S be a range set with the countable coinitiality.A topological space X is completely S -valued ultrametrizable if and onlyif X is completely metrizable and S -valued ultrametrizable.Proof. It suffices to show that if X is completely metrizable and S -valued ultrametrizable, then X is completely S -valued ultrametriz-able. Take a non-zero strictly decreasing sequence { r ( i ) } i ∈ N in S withlim i →∞ r ( i ) = 0. We put T = { } ∪ { r ( i ) | i ∈ N } . Then T is a rangeset with T ⊂ S . Note that T has countable coinitiality and it is aclosed set of [0 , ∞ ). By Proposition 2.12, we can take an ultrametric d ∈ UM(
X, T ). Let (
Y, D ) be a completion of (
X, d ). By Proposition2.9, the space (
Y, D ) is a T -valued ultrametric space. Since X is com-pletely metrizable, X is G δ in Y (see [11, Theorem 4.3.24]). Thus thereexists a sequence { G n } n ∈ N of open sets in Y such that X = T n ∈ N G n .By Lemmas 2.1 and 2.14, we can take a sequence { e n } n ∈ N of complete T -valued ultrametrics such that e n ∈ UM( G n , T ) and e n ( x, y ) ≤ r ( n )for all x, y ∈ G n and for all n ∈ N . Define an S -valued ultrametric D ∈ UM(
X, S ) by D ( x, y ) = sup n ∈ N e n ( x, y ) . Then D is complete. Since T is a closed set of [0 , ∞ ), we have D ∈ UM(
X, T ). By UM(
X, T ) ⊂ UM(
X, S ), we obtain a complete S -valuedultrametric D ∈ UM(
X, S ). (cid:3) If a range set does not have countable coinitiality, we obtain:
N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 11
Lemma 2.16.
Let S be a range set which does not have the countablecoinitiality. Then every S -valued ultrametric space ( X, d ) is discreteand complete.Proof. Since S does not have countable coinitiality, there exists r ∈ [0 , ∞ ) such that [0 , r ) ∩ S = { } . Thus for every x ∈ X we have U ( x, r ) = { x } . This implies the lemma. (cid:3) Combining Lemma 2.16 and Proposition 2.15, we obtain:
Proposition 2.17.
Let S be a range set. A topological space X is com-pletely S -valued ultrametrizable if and only if X is completely metrizableand S -valued ultrametrizable. Continuous functions on -dimensional spaces. The follow-ing theorem was stated in [9, Theorem 1.1], and a Lipschitz version ofit was proven in [3, Theorem 2.9].
Proposition 2.18.
Let X be an ultrametrizable space, and let A bea closed subset of X . Then there exists a retraction from X into A ;namely, there exists a continuous map r : X → A with r | A = 1 A . Corollary 2.19.
Let X be an ultrametrizable space, and let A be aclosed subset of X . Let Y be a topological space. Then every continuousmap f : A → Y can be extended to a continuous map from X to Y .Proof. By Proposition 2.18, there exists a retraction r : X → A . Put F = f ◦ r , then F is a continuous extension of f . (cid:3) Let Z be a metrizable space. We denote by C ( Z ) the set of all non-empty closed subsets of Z . For a topological space X we say that amap φ : X → C ( Z ) is lower semi-continuous if for every open subset O of Z the set { x ∈ X | φ ( x ) ∩ O = ∅ } is open in X .The following theorem is known as the 0-dimensional Michael con-tinuous selection theorem. This was stated in [31], essentially in [30](see also [29, Proposition 1.4]). Theorem 2.20.
Let X be a -dimensional paracompact space, and A a closed subsets of X . Let Z be a completely metrizable space. Let φ : X → C ( Z ) be a lower semi-continuous map. If a continuous map f : A → Z satisfies f ( x ) ∈ φ ( x ) for all x ∈ A , then there exists acontinuous map F : X → Z with F | A = f such that for every x ∈ X we have F ( x ) ∈ φ ( x ) . By the invariance of the ultra-norms under the addition, we have:
Proposition 2.21.
Let R be a commutative ring, and let ( V, d ) be anultra-normed R -module. Let x, y ∈ V . Then for every r ∈ (0 , ∞ ) wehave H ( B ( x, r ) , B ( y, r )) ≤ d ( x, y ) , where H is the Hausdorff distance induced from d . Proof.
For every w ∈ B ( y, r ), the invariance of d under the additionimplies that x + w − y ∈ B ( x, r ) and d ( w, x + w − y ) = d ( x, y ).Thus, B ( y, r ) ⊂ B ( B ( x, r ) , d ( x, y )). Similarly, we obtain B ( x, r ) ⊂ B ( B ( y, r ) , d ( x, y )). Therefore, H ( B ( x, r ) , B ( y, r )) ≤ d ( x, y ). (cid:3) Corollary 2.22.
Let X be a topological space, Let R be a commutativering, and let ( V, d ) be an ultra-normed R -module. Let H : X → V bea continuous map and r ∈ (0 , ∞ ) . Then a map φ : X → C ( V ) definedby φ ( x ) = B ( H ( x ) , r ) is lower semi-continuous.Proof. For every open subset O of V , and for every point a ∈ X with φ ( a ) ∩ O = ∅ , choose u ∈ φ ( a ) ∩ O and l ∈ (0 , ∞ ) with U ( u, l ) ⊂ O . ByProposition 2.21 and the continuity of H , we can take a neighborhood N of the point a in X such that for every x ∈ N we have H ( φ ( x ) , φ ( a )) ≤ d ( H ( x ) , H ( a )) < l. Then we have φ ( x ) ∩ U ( u, l ) = ∅ , and hence φ ( x ) ∩ O = ∅ . Thereforethe set { x ∈ X | φ ( x ) ∩ O = ∅ } is open in X . (cid:3) The following theorem is known as the Stone theorem on the para-compactness, which was proven in [39].
Theorem 2.23.
All metrizable spaces are paracompact.
By Theorem 2.23 and Proposition 2.13, we can apply Theorem 2.20to all ultrametrizable space.2.5.
Baire spaces.
A topological space X is said to be Baire if anintersection of every countable family of dense open subsets of X isdense in X .The following is known as the Baire category theorem. Theorem 2.24.
Every completely metrizable space is a Baire space.
Since G δ subset of completely metrizable space is completely metriz-able (see, e.g. [50, Theorem 24.12]), we obtain the following: Lemma 2.25.
Every G δ subset of a completely metrizable space is aBaire space. An embedding theorem of ultrametric spaces
In this section, we prove Theorem 1.1.3.1.
Proof of Theorem 1.1.
We first discuss general algebraic facts.
Lemma 3.1.
Let R be a commutative ring, and let V be an R -module.Let P be an R -independent set of V , and let Q be a subset of P . Let H be an R -submodule of V generated by Q . Then H ∩ P = Q .Proof. By the definition of H , first we have Q ⊂ H ∩ P . Since P is R -independent, we have ( P \ Q ) ∩ H = ∅ . Thus every x ∈ H ∩ P mustbelong to Q , and hence we conclude that H ∩ P ⊂ Q . (cid:3) N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 13
Let R be a commutative ring. Let X be a set, and let o X . Wedenote by F( R, X, o ) the free R -module M satisfying that(1) X ⊔ { o } ⊂ M ;(2) o is the zero element of M ;(3) X is an R -independent generator of M .Note that by the construction of free modules, F( R, X, o ) uniquelyexists up to isomorphism.For two sets A , B , we denote by Map( A, B ) the set of all maps from A into B . Let R be a commutative ring, and let V be an R -module.Let E be a set. Then the set Map( E, V ) becomes an R -module withthe coordinate-wise addition and scalar multiplication. Note that thezero element of Map( E, V ) is the zero function of Map(
E, V ); namelythe constant function valued at the zero element of V . In what follows,the set Map( E, V ) will be always equipped with this module structure.We next discuss a construction of universal ultrametric spaces ofLemin–Lemin type [24]. Let S be a range set. Let M be a set, andlet o ∈ M be a fixed base point. A map f : S + → M is said to be eventually o -valued if there exists C ∈ S + such that for every q > C we have f ( q ) = o . We denote by L( S, M, o ) the set of all eventually o -valued maps from S + to M . Define a metric ∆ on L( S, M, o ) by∆( f, g ) = sup { q ∈ S + | f ( q ) = g ( q ) } . Note that ∆ takes values in the closure CL( S ) of S in [0 , ∞ ).The next lemma follows from the definitions of L( S, M, o ) and ∆.
Lemma 3.2.
For every range set S , for every set M and for everypoint o ∈ M , the space (L( S, M, o ) , ∆) is a complete CL( S ) -valuedultrametric space. In the next theorem, we review the Lemin–Lemin construction [24] ofembeddings into their universal spaces in order to obtain more detailedinformation of their construction.
Theorem 3.3.
Let S be a range set. Let ( X ⊔ { o } , d ) be an S -valuedultrametric space with o X . Let K be a set with X ⊔ { o } ⊂ K . Thenthere exists an isometric embedding L : X → L( S, K, o ) such that (1) for every q ∈ S + we have L ( o )( q ) = o ; (2) for every x ∈ X the function L ( x ) is valued in X ⊔ { o } ; (3) for all x, y ∈ X we have (0 , d ( x, y )] ∩ S + = { q ∈ S + | L ( x )( q ) = L ( y )( q ) } . Proof.
Let X ⊔ { o } = { x ( α ) } α<κ be an injective index with x (0) = o ,where κ is a cardinal. By following the Lemin–Lemin’s way [24], weconstruct an isometric embedding L : X → L( S, K, o ) by transfiniteinduction. Put L ( x (0)) = o . Let γ < κ . Assume that an isometricembedding L : { x ( α ) | α < γ } → L( S, K, o ) is already defined. Set D γ = inf { d ( x ( α ) , x ( γ )) | α < γ } . Case 1. (There exists an ordinal β < γ with D γ = d ( x ( β ) , x ( γ )). )We define an eventually o -valued map L ( x ( γ )) : S + → K by L ( x ( γ ))( q ) = ( x ( γ ) if q ∈ (0 , D γ ]; L ( x ( β ))( q ) if q ∈ ( D γ , ∞ ).Case 2. (No ordinal β < γ satisfies D γ = d ( x ( β ) , x ( γ )). ) Take asequence { α n } n ∈ N with α n < γ and d ( x ( α n ) , x ( γ )) < D γ + 1 /n for all n ∈ N . We define an eventually o -valued map L ( x ( γ )) : S + → K by L ( x ( γ ))( q ) = ( x ( γ ) if q ∈ (0 , D γ ]; L ( x ( α n ))( q ) if D γ + 1 /n < q .Similarly to [24], the map L : X ⊔ { o } → L( S, K, o ) is well-defined andisometric, and we see that the conditions (1) and (2) are satisfied.We now prove the condition (3). Note that for each α < κ , thefunction L ( x ( α )) is valued in { x ( β ) | β ≤ α } . Let γ < κ . Assumethat for all α, β < γ , the condition (3) is satisfied for x = x ( α ) and y = x ( β ). We prove that for every α < γ , the condition (3) is satisfiedfor x = x ( α ) and y = x ( γ ).In Case 1, by the definition of D γ and β < γ , we have d ( x ( β ) , x ( γ )) ≤ d ( x ( α ) , x ( γ )). From this inequality and the strong triangle inequality(or Lemma 2.8), it follows that d ( x ( α ) , x ( β )) ≤ d ( x ( α ) , x ( γ )). Thus,by the hypothesis of transfinite induction and the definition of L ( x ( γ )),we conclude that the condition (3) is satisfied.In Case 2, by the definition of D γ , we have D γ < d ( x ( α ) , x ( γ )), andfor all sufficiently large n ∈ N , we obtain d ( x ( α n ) , x ( γ )) < d ( x ( α ) , x ( γ )).Lemma 2.8 implies that d ( x ( α n ) , x ( α )) = d ( x ( α ) , x ( γ )). Since onthe set ( D γ + 1 /n, d ( x ( α ) , x ( α n ))] the function L ( x ( γ )) coincides with L ( x ( α n )), by the hypothesis of transfinite induction we have( D γ + 1 /n, d ( x ( α ) , x ( γ ))] ∩ S + ⊂ { q ∈ S + | L ( x ( α ))( q ) = L ( x ( γ ))( q ) } . By L ( x ( α ))( S + ) ⊂ { x ( β ) | β ≤ α } and L ( x ( γ )) | (0 ,D γ ] = x ( γ ), we alsohave (0 , D γ ] ∩ S + ⊂ { q ∈ S + | L ( x ( α ))( q ) = L ( x ( γ ))( q ) } . These imply the condition (3) for x = x ( α ) and y = x ( γ ). (cid:3) Remark . An ultrametric space X is said to be universal for a class ofultrametric space if every ultrametric space in the class is isometricallyembeddable into X . In [24], for each cardinal τ , Lemin and Leminconstructed a universal ultrametric space for the class of all ultrametricspaces of topological weight τ . There are several studies on universalultrametric spaces. Vaughan [46] studied universal ultrametric spacesof Lemin–Lemin-type. Vestfrid [47], Gao and Shao [13], and Wan [48]studied universal ultrametric spaces of Urysohn-type. N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 15
The following lemma plays a central role in the proof of our embed-ding theorem from ultrametric spaces into ultra-normed modules.
Lemma 3.4.
Let S be a range set. Let R be a commutative ring. If M is an R -module, then the following are satisfied: (1) the space L( S, M, becomes an R -submodule of Map( S + , M ) ; (2) the ultrametric ∆ on L( S, M, is invariant under the addition;namely, (L( S, M, , ∆) is ultra-normed; (3) if M is torsion-free, then for every r ∈ R \ { } , and for every x ∈ L( S, M, o ) , we have ∆( r · x,
0) = ∆( x, .Proof. The condition (1) follows from L(
S, M, ⊂ Map( S + , M ) andthe definition of the eventually 0-valued maps. We prove the condition(2). For all f, g, h ∈ L( S, M, q ∈ S + , we have f ( q ) = g ( q ) if and only if f ( q ) + h ( q ) = g ( q ) + h ( q ). Thus, by the definition,∆ is invariant under the addition. By the similar way, since R is anintegral domain, we see that the condition (3) is satisfied. (cid:3) Lemma 3.5.
Let R be a commutative ring. Let S be a range set.Let ( X ⊔ { o } , D ) be an S -valued ultrametric space with o X . Put M = F( R, X, o ) . Let L : ( X ⊔ { o } ) → L( S, M, o ) be an isometricembedding constructed in Theorem 3.3. Then L ( X ) is R -independentin the R -module L( S, M, o ) .Proof. In this proof, we denote by 0 M the zero element o of M . Let C = { x , . . . , x n } be an arbitrary finite subset of X . Assume that n X i =1 N i · L ( x i ) = 0 L , where N i ∈ R for all i and 0 L stands for the zero function of L( S, M.o ).Put b = min { ∆( L ( x ) , L ( y )) | x, y ∈ C ⊔ { o } and x = y } . Then b >
0. Take c ∈ S + with c < b . By the definition of ∆ andthe conditions (1) and (3) stated in Theorem 3.3, we see that for all i, j = 1 , . . . , n we have L ( x i )( c ) = L ( x j )( c ), and for each i we have L ( x i )( c ) = 0 M . By the definition of F( R, X, o ), we see that the set { L ( x )( c ) , . . . , L ( x n )( c ) } is R -independent in M . Since n +1 X i =1 N i · L ( x i )( c ) = 0 M , we have N i = 0 for all i . Thus { L ( x ) , . . . , L ( x n ) } is R -independentin L( S, M, o ). Since S = { x , . . . , x n } is arbitrary, we see that L ( X ) is R -independent in L( S, M, o ). (cid:3) Lemma 3.6.
Let R be a commutative ring. Let S be a range set.Let ( X ⊔ { o } , D ) be an S -valued ultrametric space with o X . Put M = F( R, X, o ) . Let L : ( X ⊔ { o } ) → L( S, M, o ) be an isometricembedding constructed in Theorem 3.3. Let Q be an R -submodule of L( S, M, o ) generated by L ( X ) . Then the metric ∆ | Q takes values inthe range set S .Proof. In this proof, we denote by 0 M the zero element o of M .By the invariance of ∆ under the addition, it suffices to show thatfor every x ∈ Q we have ∆( x, L ) ∈ S , where 0 L is the zero func-tion of L( S, N, o ). Take x ∈ Q . Then there exist a finite subset { x , . . . , x n } of X and a finite subset { N , . . . , N n } of R \ { } suchthat x = P ni =1 N i L ( x i ). Let p , p , . . . , p k be a sequence in S such that(1) p = 0;(2) p j < p j +1 for all j ;(3) { d ( x i , | i = 1 , . . . , n } ∪ { d ( x i , x j ) | i = j } = { p , . . . , p k } .For l ∈ { , . . . , k − } , we put I ( j ) = ( p j , p j +1 ] ∩ S , and we put I ( k ) =( p k , ∞ ) ∩ S . By the definition of { p j } kj =0 , and by the properties (2) and(3) of the map L stated in Theorem 3.3, we obtain:(A) for all a ∈ { , . . . , n } we have L ( x a ) = 0 M on I ( k );(B) for every a ∈ { , . . . , n } , and for every j ∈ { , . . . , k } , if thereexists c ∈ I ( j ) with L ( x a )( c ) = 0 M , then we have L ( x a ) = 0 L on I ( j );(C) for all a, b ∈ { , . . . , n } , and for every j ∈ { , . . . , k } , if thereexists c ∈ I ( j ) with L ( x a )( c ) = L ( x b )( c ), then we have L ( x a ) = L ( x b ) on I ( j ).Suppose that ∆( x, S . By the property (A), we can take j ∈{ , . . . , k − } such that ∆( x, ∈ I ( j ). By the definition of ∆, thereexists p ∈ I ( j ) with x ( p ) = 0 M , and we see that x ( p j +1 ) = 0 M . Put q = p j +1 . Take a subset { y , . . . , y m } of { x , . . . , x n } such that(a) L ( y )( q ) , . . . , L ( y m )( q ) are not equal to the zero element 0 M of M , and they are different to each other;(b) m is maximal in cardinals of all subsets of the set { x , . . . , x n } satisfying the property (a).By the properties (B) and (C), the set { L ( y ) | I ( j ) , . . . , L ( y m ) | I ( j ) } isa maximal R -independent subset of { L ( x ) | I ( j ) , . . . , L ( x n ) | I ( j ) } in the R -module Map( I ( j ) , M ). Then there exists a subset { C , . . . , C m } of R such that x | I ( j ) = m X l =1 C l L ( y l ) | I ( j ) . Since x ( q ) = 0 M , we have m X l =1 C l L ( y l )( q ) = 0 M . Since { L ( y )( q ) , . . . , L ( y m )( q ) } is a subset of X , it is R -independentin M . Thus we have C l = 0 for all l ∈ { , . . . , m } . This implies N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 17 that x = 0 M on I ( j ). This contradicts the existence of p ∈ I ( j ) with x ( p ) = 0 M . Therefore, ∆( x, L ) ∈ S . This completes the proof. (cid:3) Before proving Theorem 1.1, we recall that every free module onevery integral domain is torsion-free.
Proof of Theorem 1.1.
Let S be a range set. Let R be a commutativering, and let ( X, d ) be an ultrametric space.We first deal with the case where (
X, d ) is complete. Take o X .Put M = F( R, X, o ). Let ( X ⊔ { o } , D ) be a one-point extension of( X, d ) (see Corollary 2.4). Let L : ( X ⊔{ o } , D ) → (L( S, M, o ) , ∆) be anisometric embedding stated in Theorem 3.3. Let Q be an R -submoduleof L( S, M, o ) generated by L ( X ), and let ( V, Ξ) be the completionof ( Q, ∆ | Q ). By Lemmas 2.7, 3.6, and Proposition 2.9, the space( V, Ξ) is an S -valued ultra-normed R -module. Since complete metricsubspaces are closed in metric spaces, Lemma 3.5 implies that ( V, Ξ)and L : ( X, d ) → ( V, Ξ) satisfy the conditions (1) and (2) stated inTheorem 1.1. Moreover, the latter part of the theorem is also proven.In the case where (
X, d ) is not complete, let (
Y, e ) be the completionof (
X, d ). As in the above, we can take an ultra-normed R -module( W, D ) and an isometric embedding I : Y → W satisfying the con-ditions (1) and (2) in Theorem 1.1. Let H be an R -submodule of W generated by I ( X ). Since I ( Y ) is R -independent, Lemma 3.1 yields H ∩ I ( Y ) = I ( X ). Thus I ( X ) is closed in H , and hence ( H, D | H ) and I are desired ones. This completes the proof of Theorem 1.1. (cid:3) Remark . If a range set S is closed under the supremum operator,then we can replace the assumption that R is an integral domain inthe statement of Theorem 1.1 with the condition that R is a com-mutative ring. In this case, the space (L( S, M, o ) , ∆) is an S -valuedultrametric space, and in the proof of Theorem 1.1, we can use thespace (L( S, M, o ) , ∆) instead of the space ( V, Ξ).3.2.
Ultrametrics taking values in general totally ordered sets.
We say that an ordered set is bottomed if it has a minimal element. Let( T, ≤ T ) be a bottomed totally ordered set. Let X be a set. A function d : X × X → T is said to be a ( T, ≤ T ) -valued ultrametric on X if thefollowing are satisfied:(1) for all x, y ∈ X we have d ( x, y ) = 0 T if and only if x = y , where0 T stands for the minimal element of ( T, ≤ T );(2) for all x, y ∈ X we have d ( x, y ) = d ( y, x );(3) for all x, y, z ∈ X we have d ( x, y ) ≤ T d ( x, z ) ∨ T d ( z, y ), where ∨ T is the maximal operator of ( T, ≤ T ).Such general ultrametric spaces, or general metric spaces on whichdistances are valued in a totally ordered Abelian group are studied fora long time (see e.g., [35], [5], [36], [32] and [8]). The construction of universal ultrametric space of Lemin–Lemin-type mentioned above and the proof of Theorem 1.1 are still valid for( T, ≤ T )-valued ultrametric spaces for all bottomed totally ordered set( T, ≤ T ). For simplicity, and for necessity of our study, we omit thedetails of the proof of the following: Theorem 3.7.
Let ( T, ≤ T ) be a bottomed totally ordered set. Let R bean integral domain, and let ( X, d ) be a ( T, ≤ T ) -valued ultrametric space.Then there exist a ( T, ≤ T ) -valued ultra-normed R -module ( V, k∗k ) , andan isometric embedding I : X → V such that (1) I ( X ) is closed in V ; (2) I ( X ) is R -independent in V .Moreover, if ( X, d ) is complete, then we can choose ( V, k ∗ k ) as acomplete ( T, ≤ T ) -valued ultrametric space. For a bottomed totally ordered set ( T, ≤ T ), we define the coinitiality coi( T, ≤ T ) of T as the minimal cardinal κ > f : κ + 1 → T with f ( κ ) = 0 T such that forevery t ∈ T , there exists α < κ with f ( α ) ≤ t . Note that a range set S has countable coinitiality if and only if coi(CL( S ) , ≤ ) = ω . Somereaders may think our results such as Corollary 1.3 and Theorems 1.2–1.6 in this paper can be generalized for ( T, ≤ T )-valued ultrametricsfor a bottomed totally ordered set ( T, ≤ T ) satisfying coi( T, ≤ T ) > ω .Unfortunately, it seems to be quite difficult. Our proofs of Theorems1.2–1.6 require the extension theorem (Corollary 2.19) of continuousfunctions on ultrametric spaces. An analogue for ( T, ≤ T )-valued ultra-metric spaces of Corollary 2.19 seems not to hold true.4. An extension theorem of ultrametrics
In this section, by following the methods of Toru´nczyk [44] and Haus-dorff [17], we prove Theorem 1.2 and Corollary 1.3. Since Toru´nczyk’sproof of Lemma in [44] on real linear spaces does not depend on the co-efficient ring R , we can apply that method to all ultra-normed modulesover all commutative rings. Toru´nczyk used the Dugundji extensiontheorem in the proof of Lemma in [44]. Instead of the Dugundji exten-sion theorem, we use Corollary 2.19, which is an extension theorem forcontinuous functions on ultrametrizable spaces. Lemma 4.1.
Let R be a commutative ring. Let ( E, D E ) and ( F, D F ) be two ultra-normed R -modules. Let K and L be closed subsets of E and F , respectively. Let f : K → L be a homeomorphism. Let g : K × { } → { } × L be a homeomorphism defined by g ( x,
0) = (0 , f ( x )) ,where we consider K × { } ⊂ E × F and { } × L ⊂ E × F . Then thereexists a homeomorphism h : E × F → E × F with h | K ×{ } = g .Proof. By Corollary 2.19, we obtain a continuous map β : F → E whichis an extension of f − : L → K . Define a map J : E × F → E × F N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 19 by J ( x, y ) = ( x + β ( y ) , y ). Lemma 2.6 implies that the addition andthe inversion on E is continuous, and hence J is continuous. The map Q : E × F → E × F defined by Q ( x, y ) = ( x − β ( y ) , y ) is also continuous,and it is the inverse map of J , and hence J is a homeomorphism.Similarly, by Corollary 2.19, we obtain a continuous map α : E → F which is an extension of f : K → L . Define a map I : E × F → E × F by I ( x, y ) = ( x, y + α ( x )). Then I is a homeomorphism. Define ahomeomorphism h : E × F → E × F by h = J − ◦ I . Since for every x ∈ K we have I ( x,
0) = ( x, α ( x )) = ( x, f ( x )), we obtain h ( x,
0) = J − ( x, f ( x )) = Q ( x, f ( x )) = ( x − β ( f ( x )) , f ( x ))= ( x − f − ( f ( x )) , f ( x )) = (0 , f ( x )) = g ( x, , and hence h is an extension of g . (cid:3) Proof of Theorem 1.2.
Let S be a range set. Let X be an S -valuedultrametrizable space, and let A be a closed subset of X . Let e ∈ UM(
A, S ). Take d ∈ UM(
X, S ). Theorem 1.1 implies that there existan S -valued ultra-normed Z -module ( E, D E ) and a closed isometricembedding i : ( X, d ) → ( E, D E ). Similarly, there exist an S -valuedultra-normed Z -module ( F, D F ) and a closed isometric embedding j :( A, e ) → ( F, D F ).Since A is closed in X , the set i ( A ) is closed in E . Since i and j aretopological embeddings, i ( A ) and j ( A ) are homeomorphisms. Definea map f : i ( A ) → j ( A ) by f = j ◦ ( i | A ) − , and by applying Lemma4.1 to f , we obtain a homeomorphism h : E × F → E × F whichis an extension of the map g : i ( A ) × { } → { } × j ( A ) defined by g ( i ( a ) ,
0) = (0 , j ( a )).Let k : E → E ×{ } be a natural embedding defined by k ( x ) = ( x, H : X → E × F defined by H = h ◦ k ◦ i is a topologicalembedding. Define a metric D on X by D ( x, y ) = ( D E × ∞ D F )( H ( x ) , H ( y )) . Then D ∈ UM(
X, S ). Since for every a ∈ A we have H ( a ) = (0 , j ( a )),and since j : ( A, ρ ) → ( F, D F ) is an isometric embedding, we have D | A = e . This completes the proof of the former part.We next show the latter part. Assume that X is completely metriz-able, and e ∈ UM(
A, S ) is complete. Then by Proposition 2.17, wecan choose d ∈ UM(
X, S ) as a complete S -valued ultrametric. Thus,we can choose ( E, D E ) and ( F, D F ) as complete ultrametric spaces,and hence the metric space ( X, D ) can be regarded as a closed metricsubspace of the complete metric space ( E × F, D F × ∞ D E ). Therefore D is complete. This finishes the proof. (cid:3) Remark . In the proof of Theorem 1.2, for simplicity, we use Z -modules. The proof described above is still valid even if we use anyintegral domain as a coefficient ring. We next prove Corollary 1.3, which characterizes the compactness interms of the completeness of ultrametrics.
Lemma 4.2.
Let S be a range set with the countable coinitiality. Let M be a countable discrete space. Then there exists a non-complete S -valued ultrametric d ∈ UM(
M, S ) .Proof. Take a non-zero strictly decreasing sequence { a ( i ) } i ∈ N in S withlim i →∞ a ( i ) = 0. We may assume that M = N . Define a metric d on M by d ( n, m ) = ( a ( n ) ∨ a ( m ) if n = m ;0 if n = m. Then d ∈ UM(
M, S ) and d is non-complete. In particular, { n } n ∈ N isCauchy, and it does not have any limit point in ( M, d ). (cid:3) Proof of Corollary 1.3.
Assume that X is not compact. Then thereexists a closed countable discrete subset M of X . By Theorem 1.2 andLemma 4.2, we obtain a non-complete S -valued ultrametric D on X with D ∈ UM(
X, S ). This implies Corollary 1.3. (cid:3) An interpolation theorem of ultrametrics
In this section, we prove Theorem 1.4.5.1.
Amalgamations.
The following lemma is a specialized versionof [22, Proposition 3.2] for our study on S -valued ultrametrics. Lemma 5.1.
Let S be a range set. Let ( X, d X ) and ( Y, d Y ) be S -valuedultrametric spaces, and let Z = X ∩ Y . Assume that (A) Z = ∅ ; (B) d X | Z = d Y | Z ; (C) there exists s ∈ S + such that for every x ∈ X \ Z we have inf z ∈ Z d X ( x, z ) = s .Then there exists an S -valued ultrametric h on X ∪ Y such that (1) h | X = d X ; (2) h | Y = d Y .Proof. We define a symmetric function h : ( X ∪ Y ) → [0 , ∞ ) by h ( x, y ) = d X ( x, y ) if x, y ∈ X ; d Y ( x, y ) if x, y ∈ Y ;inf z ∈ Z ( d X ( x, z ) ∨ d Y ( z, y )) if ( x, y ) ∈ X × Y .Since d X | Z = d Y | Z , the function h is well-defined. By the definition, h satisfies the conditions (1) and (2). N EMBEDDING, AN EXTENSION, AND AN INTERPOLATION 21
We next prove that h satisfies the strong triangle inequality. In thecase where x, y ∈ X and z ∈ Y , for all a, b ∈ Z we have h ( x, y ) = d X ( x, y ) ≤ d X ( x, a ) ∨ d X ( a, b ) ∨ d X ( b, y )= d X ( x, a ) ∨ d Y ( a, b ) ∨ d X ( b, y ) ≤ ( d X ( x, a ) ∨ d Y ( a, z )) ∨ ( d Y ( z, b ) ∨ d X ( b, y )) , and hence we obtain h ( x, y ) ≤ h ( x, z ) ∨ h ( z, y ). In the case where x, z ∈ X and y ∈ Y , for all a ∈ Z we have h ( x, y ) ≤ d X ( x, a ) ∨ d Y ( a, y ) ≤ d X ( x, z ) ∨ ( d X ( z, a ) ∨ d Y ( a, y )) , and hence we have h ( x, y ) ≤ h ( x, z ) ∨ h ( z, y ). By replacing the role of X with that of Y , we see that h satisfies the strong triangle inequality.We now prove that h takes values in S . It suffices to show that forall x ∈ X \ Z and y ∈ Y \ Z , we have h ( x, y ) ∈ S . By the assumption(C) and the definition of h , we obtain s ≤ h ( x, y ). If s = h ( x, y ), then h ( x, y ) is in S . If s < h ( x, y ), by the assumption (C), there exists z ∈ Z with h ( x, z ) < h ( x, y ). Lemma 2.8 implies that h ( x, y ) = h ( z, y ). Since h ( z, y ) = d Y ( z, y ), we have h ( x, y ) ∈ S . This completes the proof. (cid:3) Let X and Y be two sets, and let τ : X → Y be a bijective map.For a metric d on Y , we denote by τ ∗ d the metric on X defined by( τ ∗ d )( x, y ) = d ( τ ( x ) , τ ( y )). Remark that the map τ is an isometryfrom ( X, τ ∗ d ) into ( Y, d ).The following Proposition 5.2 and Lemmas 5.3 and 5.4 are ultramet-ric versions of [22, Proposition 3.1, Lemma 3.4, Lemma 3.5].
Proposition 5.2.
Let S be a range set. Let X be an ultrametrizablespace. Let r ∈ S + and d, e ∈ UM(
X, S ) satisfy U D SX ( d, e ) ≤ r . Put X = X , and let X be a set with card( X ) = card( X ) and X ∩ X = ∅ . Let τ : X → X be a bijection. Then there exists an ultrametric h ∈ UM( X ⊔ X , S ) such that (1) h | X = d ; (2) h | X = ( τ − ) ∗ e ; (3) for every x ∈ X we have h ( x, τ ( x )) = r .Proof. We define a symmetric function h : ( X ⊔ X ) → [0 , ∞ ) by h ( x, y ) = d ( x, y ) if x, y ∈ X ; e ( x, y ) if x, y ∈ X ;inf a ∈ X ( d ( x, a ) ∨ r ∨ e ( τ ( a ) , y )) if ( x, y ) ∈ X × X .By the definition, for every x ∈ X , we have h ( x, τ ( x )) ≥ r , and h ( x, τ ( x )) ≤ d ( x, x ) ∨ r ∨ e ( τ ( x ) , τ ( x )) = r. Therefore for every x ∈ X we have h ( x, τ ( x )) = r . We now prove that h satisfies the strong triangle inequality. In thecase where x, y ∈ X and z ∈ X , for all a, b ∈ X , by U D SX ( d, e ) ≤ r we have h ( x, y ) = d ( x, y ) ≤ d ( x, a ) ∨ d ( a, b ) ∨ d ( b, y ) ≤ d ( x, a ) ∨ r ∨ e ( τ ( a ) , τ ( b )) ∨ d ( b, y ) ≤ d ( x, a ) ∨ r ∨ e ( τ ( a ) , z ) ∨ e ( τ ( b ) , z ) ∨ d ( b, y ) ≤ ( d ( x, a ) ∨ r ∨ e ( τ ( a ) , z )) ∨ ( d ( y, b ) ∨ r ∨ e ( τ ( b ) , z )) , and hence we obtain h ( x, y ) ≤ h ( x, z ) ∨ h ( z, y ). In the case where x, z ∈ X and y ∈ X , for all a ∈ X we have h ( x, y ) ≤ d ( x, a ) ∨ r ∨ e ( τ ( a ) , y ) ≤ d ( x, z ) ∨ ( d ( z, a ) ∨ r ∨ e ( τ ( a ) , y )) , and hence h ( x, y ) ≤ h ( x, z ) ∨ h ( z, y ). By replacing the role of X withthat of X , we see that h satisfies the strong triangle inequality. Bythe property (3), we also see that h ∈ UM( X ⊔ X ).We next prove that h takes values in S . It suffices to show that forall ( x, y ) ∈ X × X , we have h ( x, y ) ∈ S . By the definition of h , wehave r ≤ h ( x, y ). If r = h ( x, y ), then h ( x, y ) is in S . If r < h ( x, y ),by h ( x, τ ( x )) = r , we have h ( x, τ ( x )) < h ( x, y ). From Lemma 2.8, itfollows that h ( x, y ) = h ( τ ( x ) , y ). Since h ( τ ( x ) , y ) = e ( τ ( x ) , y ) ∈ S , weconclude that h takes values in S . (cid:3) Lemma 5.3.