The Range of Ultrametrics, Compactness, and Separability

Home , Finite topological space, Totally bounded space, Ultrametric space

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Definition 1.2 means that τ and d are compatible if and only if every B ∈ BX,d belongs to τ and every nonempty A ∈ τ can be written as the union of some family of elements of BX,d. If (X, τ) admits a compatible with τ ultrametric on X, then we say that the topological space (X, τ) (the topology τ) is ultrametrizable. Necessary and sufficient conditions under which topological spaces are ultrametrizable were found by De Groot [30, 31]. See also [4, 5, 7, 8, 43] for some future results connected with ultrametrizable topologies. We define the distance set D(X) of an ultrametric space (X, d) as the range of the ultrametric d: X×X → R+, D(X)= D(X, d)= {d(x, y): x, y ∈ X}. In the present paper we focus on interconnections between ultrametrizable topologies and related properties of distance sets of ultrametrics which are compatible with these topologies. The paper is organized as follows. Section 2 contains all necessary basic facts about metric spaces and some concepts related to totally ordered sets. Theorem 3.6 of Section 3 describes the distance sets of totally bounded ultrametric spaces. In particular, it is shown that the distance sets are order isomorphic for all infinite totally bounded ultrametric spaces. The main results of the paper are Theorems 4.7 and 4.12 from Section 4. Theorem 4.7 claims that an ultrametrizable infinite topological space (X, τ) is compact iff any two compatible with τ ultrametrics have the order isomorphic distance sets. In Theorem 4.12 we prove that an ultrametrizable topological space (X, τ) is separable iff every compatible with τ ultrametric has at most countable distance set.

2. Basic deﬁnitions and facts

2.1. Metrics and ultrametrics Let (X, d) be a metric space. A sequence

(xn)n∈N ⊆ X is a Cauchy sequence in (X, d) if, for every r > 0, there are x ∈ X and an integer n0 ∈ N such that xn ∈ Br(x) ∈ BX for every n > n0, i.e., (xn)n∈N eventually belongs to Br(x).

Remark 2.1. Here and later the symbol (xn)n∈N ⊆ X means that xn ∈ X holds for every n ∈ N.

Deﬁnition 2.2. A sequence (xn)n∈N of points in a metric space (X, d) is said to converge to a point a ∈ X,

lim xn = a, n→∞ if (xn)n∈N eventually belongs to every open ball B containing a. It is clear that every convergent sequence is a Cauchy sequence. Now we recall a deﬁnition of total boundedness.

Deﬁnition 2.3. A subset A of a metric space (X, d) is totally bounded if for every r > 0 there is a ﬁnite set {Br(x1),...,Br(xn)}⊆ BX such that

n A ⊆ Br(xi). i=1 [ The next basic concept is the concept of completeness.

2 Deﬁnition 2.4. A subset A of metric space (X, d) is complete if, for every Cauchy sequence (xn)n∈N ⊆ A, there is a point a ∈ A such that lim xn = a. n→∞ An important subclass of complete metric spaces is the class of compact metric spaces. Deﬁnition 2.5 (Borel—Lebesgue property). Let (X, d) be a metric space. A subset A of X is compact if every family F ⊆ BX satisfying the inclusion

A ⊆ B B[∈F contains a ﬁnite subfamily F0 ⊆F such that

A ⊆ B. B[∈F0 A standard definition of compactness of a topological space (X, τ) usually formulated as: Every open cover of X has a finite subcover. A proof of the following well-known result can be found, for example, in [67, Theorem 12.1.3]. Proposition 2.6. Let (X, d) be a metric space. Then the following conditions are equivalent for every A ⊆ X: (i) A is compact. (ii) Every infinite sequence of points of A contains a subsequence which converges to a point of A. (iii) A is complete and totally bounded. Let (X, τ) be a topological space and let S be a subset of X. If S ∩ A 6= ∅ holds for every nonempty A ∈ τ, then S is called a dense subset of X. For a metric space (X, d) a set A ⊆ X is dense if and only if for every a ∈ X there is a sequence (an)n∈N ⊆ A such that

a = lim an. n→∞ As usual we will say that a topological (or metric) space X is separable iﬀ X contains an at most countable dense subset. In particular, every ﬁnite topological space is separable. Recall that a point p of a metric space (X, d) is isolated if there is ε> 0 such that d(p, x) >ε for every x ∈ X \{p}. If p is not an isolated point of X, then p is called an accumulation point of X. We say that (X, d) is discrete if all points of X are isolated. It can be proved that for a given metric space (X, d) there exists exactly one (up to isometry) complete metric space (X, d) such that X contains a dense, isometric to (X, d), subset (see, for example, Theo- rem 4.3.19 [27]). The metric space (X, d) satisfying the above condition is called the completion of (X, d). e e e Proposition 2.7. The completion (X,e de) of a metric space (X, d) is compact if and only if (X, d) is totally bounded. e e For the proof see, for example, Corollary 4.3.30 in [27].

Lemma 2.8. The completion (X, d) is ultrametric for every ultrametric (X, d).

Proof. Since d is a metric on X,e toe prove that (X, d) is ultrametric it suﬃces to justify

e e d(x, y) 6 maxe e{d(x, z), d(z,y)} (2.1)

′ for all x, y, z ∈ X. By deﬁnition, the metrice space (X,ed) containse a dense subspace (X , d|X′×X′ ) which is ′ isometric to (X, d). For simplicity we may assume X = X and d|X′×X′ = d. e e e e 3 e Let (xn)n∈N, (yn)n∈N and (zn)n∈N be sequences of points of X such that

lim xn = x, lim yn = y and lim zn = z. (2.2) n→∞ n→∞ n→∞

Since d: X × X → R+ is a continuous mapping and d: X × X → R+ is an ultrametric, (2.2) implies

e e e d(x, y) = lim d(xn,yn) 6 max lim d(xn,zn), lim d(zn, xn) n→∞ n→∞ n→∞ n o e = max{d(x, z), d(z,y)}.

Inequality (2.1) follows. e e The following lemma will be useful for description of distance sets of totally bounded ultrametric spaces.

Lemma 2.9. For every compact metric space (X, d) the distance set D(X) is a compact subset of (R+,ρ) with respect to the standard Euclidean metric ρ. Proof. Let (X, d) be compact. The metric d: X ×X → R+ is a continuous function on the Cartesian product X × X. Any Cartesian product of compact spaces is compact by Tychonoff theorem ([27, Theorem 3.2.4]). Furthermore, every continuous image of compact Hausdorff space is compact ([27, Theorem 3.1.10]). To complete the proof it suffices to note that (X, d) evidently is a Hausdorff space and that Cartesian products of Hausdorff spaces are Hausdorff ([27, Theorem 2.3.11]). The next result can be found in [9] (see formula (12), Theorem 1.6).

Proposition 2.10. Let (X, d) be a nonempty ultrametric space and let A be a dense subset of X. Then the distance sets of (X, d) and of (A, d|A×A) are the same. Corollary 2.11. Let (X, τ) be an ultrametrizable separable topological space. Then the distance set D(X, d) is at most countable for every compatible with τ ultrametric d. Remark 2.12. In Theorem 4.12 of Section 4 of the paper we will prove that the converse to Corollary 2.11 is also true. Proposition 2.10 and Lemma 2.8 also imply the following well-known result: “No new values of the ultrametric after completion” (see, for example, [52, p. 4] or [9, Theorem 1.6, Statement (12)]).

Proposition 2.13. Let (X, d) be an ultrametric space and let (X, d) be the completion of (X, d). Then the equality D(X)= D(X) holds. e e 2.2. Totally orderede sets Recall that a reﬂexive and transitive binary relation 4 on a set Y is a partial order on Y if 4 has the antisymmetric property: x 4 y and y 4 x ⇒ (x = y) is valid for all x, y ∈ Y . Remark 2.14. Here and in what follows we use x 4 y instead of hx, yi ∈ 4.

Let 4Y be a partial order on a set Y . Then a pair (Y, 4Y ) is called to be a poset (a partially ordered set). 2 If Z is a subset of Y and 4Z is a partial order on Z given by 4Z = Z ∩ 4Y , then we say that Z = (Z, 4Z ) is a subposet of (Y, 4Y ). A poset (Y, 4Y ) is totally ordered if, for all points y1, y2 ∈ Y , we have

y1 4Y y2 or y2 4Y y1.

4 Example 2.15. Let Z− be the set of all negative integers,

Z− = {..., −3, −2, −1} and let Z− := Z− ∪ {−∞}. Let us deﬁne an order 4 on Z− by:

b −∞ 4 p and (q 4b l) ⇔ (q 6 l) for all p, q, l ∈ Z−, where 6 is the standard order on R. Then (Z−, 4) is totally ordered. Deﬁnition 2.16. 4 4 isotone Let (Q, Q) and (L, L) be posets. A mappingb f : Q → L is if, for all q1, q2 ∈ Q, we have (q1 4Q q2) ⇒ (f(q1) 4L f(q2)). If, in addition, the isotone mapping f : Q → L is bijective and the inverse mapping f −1 : L → Q is also isotone, then we say that f is an order isomorphism and that (Q, 4Q) and (L, 4L) have the same order type. Deﬁnition 2.17. A poset has the order type 1 + ω∗ if this poset is order isomorphic to (Z−, 4). Example 2.18. The subposets b 1 1 {0} ∪ : n ∈ N and {0} ∪ 1+ : n ∈ N n n of the poset (R, 6) have the order type 1 + ω∗. Remark 2.19. The symbol ω is the usual designation of the order type of (N, 6) and 1 denotes the order type of one-point posets. The symbol ω∗ is used for the backwards order type of ω. Thus, the subposet Z− of the poset (R, 6) has the order type ω∗. Moreover, by 1 + ω∗ we denote the sum of the order types 1 and ω∗ (see, for example, Section 1.4 in [64] for details).

We also remember that a point q0 ∈ Q is the called the smallest element of a poset (Q, 4) if the inequality q0 4 q holds for every q ∈ Q. The largest element of the poset can be deﬁned by duality. In particular, every totally ordered set having the order type 1 + ω∗ contains the largest element and the smallest one. Example 2.20. The point 0 is the smallest element of the distance set D(X, d) of every nonempty ultrametric space (X, d). The distance set D(X, d) contains the largest element iﬀ diam X ∈ D(X, d), where diam(X, d) = sup{d(x, y): x, y ∈ X}.

3. Distance sets of totally bounded ultrametric spaces

The following general result was obtained by Delhommé, Laflamme, Pouzet, and Sauer in paper [10]. Theorem 3.1. Let A be a subset of R+. Then A is the distance set of a nonempty ultrametric space if and only if 0 belongs to A. The proof of this theorem is based on the following simple construction. Example 3.2. Let X be a subset of R+ such that 0 ∈ X. Following [10] we define a mapping d: X×X → R+ by 0 if x = y, d(x, y)= (3.1) (max{x, y} if x 6= y. Then d is an ultrametric on X and the distance set of (X, d) coincides with X, D(X, d)= X. In Theorem 3.6 below we will refine Theorem 3.1 for the case of totally bounded ultrametric spaces. 5 Lemma 3.3. Let (X, d) be a totally bounded infinite ultrametric space. Then D(X) is countably infinite, |D(X)| = ℵ0. Proof. Every totally bounded metric space is separable. Consequently, D(X) is at most countable by Corollary 2.11. Suppose D(X) is finite, |D(X)| < ℵ0. Since X is infinite, the inequality |D(X)| > 2 holds. The double inequality 2 6 |D(X)| < ℵ0 implies that there is d1 ∈ D(X) such that

∗ 0 < d1 6 d (3.2)

∗ for every d ∈ D(X) \{0}. Let us consider the family F = Bd1/2(x): x ∈ X . Then F is a cover of X, X ⊆ B. B[∈F

Using (3.2) we obtain Bd1/2(x)∩Bd1/2(y)= ∅ for all distinct x, y ∈ X. Hence, the cover F does not contain any ﬁnite subcover. This contradicts Deﬁnition 2.3. The equality |D(X)| = ℵ0 follows. The following lemma is a special case of Lemma 1 from [21].

Lemma 3.4. Let (X,ρ) be an ultrametric space and let {a,b,c,d} be a four-point subset of X. Then the inequalities ρ(a,b) >ρ(a, d) and ρ(a,b) >ρ(b,c) imply the equality ρ(a,b)= ρ(c, d). Lemma 3.5. Let (X, d) be a compact ultrametric space with the distance set D(X) and let p> 0 belong to D(X). Then p is an isolated point of the metric space (D(X),ρ) with the standard metric ρ(x, y)= |x − y|. Proof. Suppose for every ε> 0 there is s ∈ D(X) satisfying

Then there is (sn)n∈N ⊆ D(X) such that

lim sn = p and p

d(xn,yn)= sn (3.5) for every n ∈ N and lim d(xn,yn)= p. (3.6) n→∞

By Proposition 2.6, there are points x, y ∈ X and subsequences (xnk )k∈N of (xn)n∈N and (ynk )k∈N of (yn)n∈N such that

lim d(x, xnk ) = lim d(y,ynk )=0. (3.7) k→∞ k→∞ Since d: X × X → R+ is continuous, from (3.4), (3.5) and (3.7) it follows that d(x, y)= p.

Let us consider the set {x, y, xnk ,ynk } ⊆ X. Equalities d(x, y) = p, (3.7) and Lemma 3.4 give us p = d(xnk ,ynk ) for all suﬃciently large k ∈ N, that contradicts (3.4). Thus, there is ε1 > 0 such that

(p,p + ε1) ∩ D(X)= ∅. (3.8)

Arguing similarly we can ﬁnd ε2 > 0 satisfying

(p − ε2,p) ∩ D(X)= ∅. (3.9)

From (3.8) and (3.9) it follows that p is an isolated point of D(X). 6 Theorem 3.6. Let A be a subset of R+. Then A is the distance set of an infinite totally bounded ultrametric space (X, d) if and only if the following conditions simultaneously hold: (i) A has the order type 1 + ω∗ as a subposet of (R+, 6). (ii) The point 0 is an accumulation point of A with respect to the standard Euclidean metric on R+. Proof. Suppose that there is an infinite totally bounded ultrametric space (X, d) such that A = D(X). Let us prove the validity of (i). The completion X of X is compact (see Proposition 2.7) and ultrametric (see Lemma 2.8). By Corollary 2.13, the equality D(X)= D(X) holds. Hence, without loss generality, we assume that X is compact. e By Lemma 3.3, we have |D(X)| = ℵ0. Consequently, it sufficese to show that the cardinality of

∗ Dd∗ := D(X) ∩ [d , diam X] is ﬁnite for every strictly positive d∗ ∈ D(X). By Lemma 2.9, the distance set D(X) is a compact subset of R. The classical Heine—Borel theorem (from Real Analysis) claims that a subset S of R is compact if and only if S is closed and bounded. Thus, Dd∗ is compact. By Lemma 3.5, every p ∈ Dd∗ is isolated. Hence, for every p ∈ Dd∗ , the one-point set {p} is an open ball in the metric subspace Dd∗ of R. It is clear that the open cover

{p}: p ∈ Dd∗ contains a finite subcover if and only if the cardinal number |Dd∗ | is finite. Now the inequality |Dd∗ | < ∞ follows from Borel—Lebesgue property (Definition 2.5). Condition (i) follows. Let us prove condition (ii). Using Proposition 2.13 and Lemma 2.9 we obtain that A is a compact subset of R+. If (ii) does not hold, then 0 is an isolated point of A. Consequently, by Lemma 3.5, all points of A are isolated. Since A is compact, it implies the finiteness of A, contrary to (i). Condition (ii) follows. Conversely, suppose that A satisfies conditions (i) and (ii). First of all we note that condition (ii) implies 0 ∈ A. Consequently, if we write X = A and define the ultrametric d: X × X → R+ as in Example 3.2. Then the equality A = D(X) holds. Using (i) and (ii), it is easy to prove that (X, d) is compact, ultrametric and infinite.

Remark 3.7. Considering the poset (R+, 6) and its subposet

1 {0} ∪ 1+ : n ∈ N n (Example 2.18), we see that conditions (i) and (ii) from Theorem 3.6 are independent of one another.

+ Example 3.8. Let (xn)n∈N ⊆ R be a strictly decreasing sequence such that

lim xn =0. n→∞

Let us deﬁne a set A ⊆ R+ by the rule

(x ∈ A) ⇔ (x =0 or ∃n ∈ N: x = xn).

Write X = A and let d: X × X → R+ be defined as in Example 3.2. Then (X, d) is a compact infinite ultrametric space satisfying the equality D(X, d)= A. Theorem 3.6 and Example 3.8 give us the following “constructive” description of distance sets of all infinite totally bounded ultrametric spaces. Corollary 3.9. The following statements are equivalent for every A ⊆ R+: (i) There is an infinite compact ultrametric space X such that A = D(X). 7 (ii) There is an infinite totally bounded ultrametric space X such that A = D(X). + (iii) There is a strictly decreasing sequence (xn)n∈N ⊆ R such that

lim xn =0 n→∞ holds and the equivalence (x ∈ A) ⇔ (x =0 or ∃n ∈ N: xn = x) is valid for every x ∈ R+.

4. From the distance sets to compactness and separability

Let (X, τ) be a topological space. Recall that a metric d: X × X → R+ is compatible with τ if the equality τ = τd holds, where we denote by τd a topology with the base BX = BX,d. The topological space (X, τ) is, by deﬁnition, ultrametrizable if there is a compatible with τ ultrametric on X. Before considering the set of all ultrametrics, which are compatible with an compact ultrametrizable topology, we will formulate some lemmas. The ﬁrst lemma is a reformulation of an elegant theorem that was obtained by Pongsriiam and Termwut- tipong [55].

Lemma 4.1. The following conditions are equivalent for every f : R+ → R+:

f (i) The mapping X × X −→d R+ −→ R+ is an ultrametric for every ultrametric space (X, d). (ii) f is increasing and the equality f(x)=0 holds if and only if x =0. Remark 4.2. The functions satisfying condition (i) of Lemma 4.1 are called ultrametric preserving. These functions together with the construction of ultrametrics given in [10] are played a critical role in the proofs of main results of the paper. Some future information on ultrametric preserving functions can be found in [3, 16, 38–41, 65, 69]

Lemma 4.3. Let (X,ρ) be an inﬁnite ultrametric space. Then there is an ultrametric d: X × X → R+ such ∗ that τd = τρ and the distance set D(X, d) has the order type 1 + ω .

Proof. Suppose ﬁrst that (X,ρ) is discrete. Let {Xn : n ∈ N} be a partitions of X, i.e.,

X = Xn and Xn1 6= ∅ = Xn1 ∩ Xn2 (4.1) N n[∈ + hold for all distinct n1, n2 ∈ N. Similarly to (3.1) we deﬁne a mapping d: X × X → R as

0 if x = y, d(x, y)= 1 1 (4.2) (max n , m if x 6= y and x ∈ Xn, y ∈ Xm.

It is clear that d is a symmetric nonnegative mapping and d(x, y)=0 iff x = y. We claim that the strong triangle inequality d(x, y) 6 max{d(x, z), d(z,y)} (4.3) holds for all x, y, z ∈ X. Indeed, it suffices to prove (4.3) when the points x, y and z are pairwise distinct (in opposite case inequality (4.3) is evidently valid). Let x, y, z be pairwise distinct and x ∈ Xp, y ∈ Xq, z ∈ Xl for some not necessarily different p, q, l ∈ N. Then using (4.2) we see that (4.3) is equivalent to the true inequality 1 1 1 1 1 1 max , 6 max max , , max , . (4.4) p q p l l q

8 Hence, d is an ultrametric on X. If x0 is an arbitrary point of X, then there is n ∈ N such that x0 ∈ Xn. Now > 1 (4.2) implies that the inequality d(x0, x) n holds for every x ∈ X. Hence, (X, d) is a discrete ultrametric space, that implies τd = τρ. From (4.1) and (4.2) follows the equality

1 D(X, d)= : n ∈ N ∪{0}. n Thus, D(X, d) has the order type 1 + ω∗ (see Example 2.18). If (X,ρ) contains an accumulation point p, then there is a sequence (xn)n∈N ⊆ X such that limn→∞ xn = p. Hence, (ρ(xn,p))n∈N ⊆ D(X, d) (4.5) and 0 is an accumulation point of the set D(X, d) with respect to the standard Euclidean metric on R+. Let us deﬁne a function f : R+ → R+ as

0 if t =0, f(t)= 1 if t > 1, (4.6)  1 1 6 1  n−1 if n−1

 ρ f Let us denote by d the mapping X × X −→ R+ −→ R+. Since f satisfies condition (ii) of Lemma 4.1, d is an ultrametric on X. Using (4.6) it is easy to see that, for every (yn)n∈N and every z ∈ X, we have limn→∞ d(yn,z)=0 if and only if limn→∞ ρ(yn,z)=0. It implies the equality τd = τρ. Moreover, from (4.5) 1 and definition of d it follows that (0, ∞) ∩ D(X, d) is an infinite subset of the set n : n ∈ N . All infinite subposets of 1 : n ∈ N has the order type ω∗. Hence, the poset D(X, d), n D(X, d)= {0} ∪ (0, ∞) ∩ D(X, d) , has the order type 1 + ω∗. In the next lemma we present a well-known property of open ultrametric balls.

Lemma 4.4. Let (X, d) be an ultrametric space. Then, for every Br(c) ∈ BX and every a ∈ Br(c), we have Br(c)= Br(a). For the proof of this lemma see, for example, Proposition 18.4 [66].

Corollary 4.5. Let (X, d) be an ultrametric space. Then, for all Br1 (c1) and Br2 (c2), we have

Br1 (c1) ⊆ Br2 (c2) (4.7)

whenever Br1 (c1) ∩ Br2 (c2) 6= ∅ and 0 < r1 6 r2 < ∞. In particular, for all Br(c1) and Br(c2) and every r> 0 we have Br(c1)= Br(c2) (4.8) whenever Br(c1) ∩ Br(c2) 6= ∅.

Proof. Let Br(c1) ∩ Br(c2) 6= ∅ hold and r> 0 be ﬁxed. Then, by Lemma 4.4, we obtain the equalities

Br(c1)= Br(a) and Br(c2)= Br(a) for every a ∈ Br(c1) ∩ Br(c2). Equality (4.8) follows.

If we have 0 < r1 6 r2 < ∞ and Br1 (c1)∩Br2 (c2) 6= ∅, then (4.8) implies the equality Br2 (c1)= Br2 (c2).

Now (4.7) follows from the last equality and the inclusion Br1 (c1) ⊆ Br2 (c1).

9 Lemma 4.6. Let (X, d) be an ultrametric space, let r > 0 be ﬁxed, and let C be a nonempty subset of X. Then there is C1 ⊆ C such that Br(c)= Br(c) (4.9) 1 c[∈C c∈[C 1 and Br(c1) ∩ Br(c2)= ∅ whenever c1 and c2 are distinct points of C .

Proof. Let us deﬁne a binary relation =r on the set C as

r (c1 = c2) ⇔ (Br(c1) ∩ Br(c2) 6= ∅).

r It is clear that the relation = is reﬂexive and symmetric,

c =r c, and (a =r b) ⇔ (b =r a) for all a, b, c ∈ C. Moreover, Corollary 4.5 implies the the transitivity of =r . Hence, =r is an equivalence relation on C. Using the well-known one-to-one correspondence between the equivalence relations and partitions (see, for example, [45, Chapter II, § 5]), we can find a partition C = {Ci : i ∈ I} of the set C generated by relation r r =. It means that, for all a, b ∈ C, we have a = b iff there is i ∈ I such that a, b ∈ Ci and e Ci = C, Ci ∩ Cj = ∅ and Ck 6= ∅ i[∈I whenever i, j, k ∈ I and i 6= j. Now if we define C1 ⊆ C as a system of distinct representatives for C,

1 C = {ci : i ∈ I and ci ∈ Ci}, e then, using Corollary 4.5, we obtain the equality

Br(ci)= Br(c) (4.10) c[∈Ci r for every i ∈ I. In addition, from the definition of = it follows that Br(ci1 ) and Br(ci2 ) are disjoint for all distinct i1, i2 ∈ I. To complete the proof it suffices to note that (4.9) follows from the definition of C and (4.10). e Theorem 4.7. Let (X, τ) be an ultrametrizable infinite topological space. Then the following conditions are equivalent: (i) The space (X, τ) is compact. (ii) The distance set D(X, d) of an ultrametric space (X, d) has the order type 1 + ω∗ whenever d is compatible with τ. (iii) If (X, d) and (X,ρ) are ultrametric spaces such that

τd = τ = τρ,

then the distance sets D(X, d) and D(X,ρ) are order isomorphic as subposets of (R+, 6). (iv) The distance set D(X, d) has the largest element whenever d is a compatible with τ ultrametric. Proof. (i) ⇒ (ii). This is valid by Theorem 3.6. (ii) ⇒ (iii). If (ii) holds, then we have (iii) by Definition 2.16. + (iii) ⇒ (iv). By Lemma 4.3, there is an ultrametric ρ: X × X → R such that τ = τρ and D(X,ρ) has the order type 1 + ω∗. If (iii) holds and an ultrametric d: X × X → R+ is compatible with τ, then D(X, d) also has the order type 1 + ω∗ by (iii) and, consequently, D(X, d) contains the largest element. 10 (iv) ⇒ (i). Let (iv) hold. Suppose that (X, τ) is not a compact. Using Lemma 4.3 we can find an + ∗ ultrametric ρ: X × X → R such that τ = τρ and 1 + ω is the order type of D(X,ρ). Hence, D(X,ρ) has the largest element diam(X,ρ). Thus, it suffices to find a compatible with τ ultrametric d: X × X → R+ such that D(X, d) does not have the largest element. We will construct a desirable ultrametric d: X×X → R+ by a modification of the ultrametric ρ. Consider first the case when (X,ρ) is not totally bounded. Then, by Definition 2.3, there exists r1 > 0 such that the open cover B B = {Br1 (x): x ∈ X, Br1 (x) ∈ X,ρ} of X does not have any finite subcover. Using Lemma 4.6, we find a set C ⊆ X such that e

Br1 (c)= X c[∈C and Br1 (c1) ∩ Br1 (c2)= ∅ whenever c1 and c2 are distinct points of C. Write B = {Br1 (c): c ∈ C}. Since the cover B has no finite subcovers, B has an infinite cardinality. Consequently, there is a surjective mapping + g : B → (r1, 2r1) ∩ Q. Let us define a mapping d: X × X → R as e ρ(x, y) if there is B ∈B such that x, y ∈ B, d(x, y)= 1 2 1 2 1 2 (4.11) (max{g(B ),g(B )} if x ∈ B and y ∈ B for distinct B , B ∈B.

We claim that d is an ultrametric on X. Indeed, d is symmetric and d(x, y)=0 holds iﬀ x = y. Thus, it suﬃces to prove the strong triangle inequality

d(x, y) 6 max {d(x, z), d(z,y)} (4.12) for all x, y, z ∈ X. Let us consider all possible cases of mutual arrangement of the points x, y and z. If there is B ∈B such that {x,y,z}⊆ B, then from (4.11) it follows that

d(x, y)= ρ(x, y), d(x, z)= ρ(x, z) and d(y,z)= ρ(y,z).

Hence, (4.12) follows because ρ is an ultrametric. If there are distinct B1, B2, B3 ∈ B such that x ∈ B1, y ∈ B2 and z ∈ B3, then using (4.11) we can rewrite (4.12) as the obviously correct inequality

max{g(B1),g(B2)} 6 max max{g(B1),g(B3)}, max{g(B3),g(B2)}

(cf. inequality (4.4) from the proof of Lemma 4.3). To complete the proof of (4.12) it suﬃces to consider the case when there exist some distinct B1, B2 ∈B such that {x,y,z}⊆ B1 ∪ B2 but

B \{x,y,z} 6= ∅ for every B ∈B. Without loss of generality, we suppose that

d(x, y) > d(x, z) > d(z,y) > 0 (4.13) and |B1 ∩{x,y,z}| =2, |B2 ∩{x,y,z}| =1. (4.14) Using Lemma 4.4 and (4.13), (4.14) we can prove the inclusions

{y,z}⊆ B1 and {x}⊆ B2.

Hence, d(z,y)= ρ(z,y), d(x, z)= d(x, y) = max{g(B1),g(B2)}

11 hold by (4.11). Lemma 4.4, the inclusion {y,z} ⊆ B1, and the equality d(z,y) = ρ(z,y) imply the strict inequality d(z,y) < r1. (4.15)

Now (4.12) follows from (4.15) and g(B) ⊂ (r1, 2r1). Let us prove the equality τd = τρ. The last equality holds iﬀ, for every (pn)n∈N ⊆ X and for every p ∈ X, we have the equivalence

lim ρ(pn,p)=0 ⇔ lim d(pn,p)=0 . (4.16) n→∞ n→∞ To prove (4.16), we first note that lim d(pn,p)=0 implies lim ρ(pn,p)=0 because d(x, y) > ρ(x, y) n→∞ n→∞ for all x, y ∈ X by (4.11). Conversely, let lim ρ(pn,p)=0 hold. The family B is a cover of X. Hence, we n→∞ can find B ∈B such that p ∈ B. Since B is an open ball in (X,ρ), the limit relation lim ρ(pn,p)=0 and the n→∞ membership p ∈ B imply that (pn)n∈N eventually belongs to B. Thus, by (4.11), we have ρ(pn,p)= d(pn,p) for all sufficiently large n, that implies the equality lim d(pn,p)=0. n→∞ Let us consider now the case when the ultrametric space (X,ρ) is totally bounded. By Corollary 3.9, + there is a strictly decreasing sequence (rn)n∈N ⊆ R such that lim rn =0 and n→∞

D(X,ρ)= {rn : n ∈ N}∪{0}.

Since (X,ρ) is totally bounded, we can ﬁnd a ﬁnite set C1 ⊆ X such that

X = Br1 (c). (4.17) 1 c∈[C 1 2 1 2 Using Lemma 4.6, we may also suppose that Br1 (c ) ∩ Br1 (c )= ∅ whenever c and c are different points of C1. Since (X,ρ) is totally bounded but not compact, Proposition 2.6 implies that there is a Cauchy sequence (xn)n∈N ⊆ X which does not have any limit in (X,ρ). Using the definition of Cauchy sequences we 1 can find c1 ∈ C such that xn ∈ Br1 (c1) for all sufficiently large n ∈ N. The ball Br1 (c1) is totally bounded as a subset of totally bounded space (X,ρ). Consequently, there is a finite C2 ⊆ X satisfying

Br1 (c1) ⊆ Br2 (c). (4.18) 2 c∈[C 2 Without loss of generality we suppose that Br1 (c) ∩ Br2 (c) 6= ∅ holds for every c ∈ C . Then, by Corol- 2 2 lary 4.5, for every c ∈ C we have the inclusion Br2 (c) ⊆ Br1 (c1) (because r2 < r1). Hence, C ⊆ Br1 (c1) holds and inclusion (4.18) comes to the equality

Br1 (c1)= Br2 (c). (4.19) 2 c∈[C By Lemma 4.6, there is C2,1 ⊆ C2 such that

Br2 (c)= Br2 (c) 2,1 2 c∈[C c∈[C and 1 2 Br2 (c ) ∩ Br2 (c )= ∅ (4.20) whenever c1 and c2 are distinct points of C2,1. Replacing, if necessary, the set C2 by C2,1, we can assume that (4.20) is satisﬁed whenever c1 and c2 are distinct points of C2. By deﬁnition of the Cauchy sequences, B there is a ball Br2 (x) ∈ X,ρ such that the sequence (xn)n∈N eventually belongs to Br2 (x). Since (xn)n∈N also eventually belongs to Br1 (c1), we have

Br2 (x) ∩ Br1 (c1) 6= ∅. (4.21) 12 By Corollary 4.5, the inequality r2 < r1 and (4.21) imply the inclusion

Br2 (x) ⊆ Br1 (c1).

2 Using (4.19) we also see that Corollary 4.5 implies the equality Br2 (x)= Br2 (c2) with an unique c2 ∈ C . n n Arguing similarly for every integer n > 3, we can ﬁnd a ﬁnite C ⊆ Brn−1 (cn−1) and a point cn ∈ C such that the sequence (xn)n∈N eventually belongs to Brn (cn), and

Brn−1 (cn−1)= Brn (c), (4.22) n c∈[C 1 2 1 2 n and Brn (c ) ∩ Brn (c )= ∅ whenever c and c are distinct points of C . n Let us consider B ⊆ BX,ρ deﬁned by the rule: B ∈ B iﬀ there are n ∈ N and c ∈ C \{cn} such that

B = Brn (c), where {cn} is the one point set consisting from the point cn. We claim that B is a partition of the set X. Indeed, it follows directly from the definition of the sets Cn, n = 1, 2, ..., that every two different B1, B2 ∈B are disjoint. Moreover, every B ∈B is nonempty. Consequently, B is a partition of X iff X = B (4.23) B[∈B holds. Suppose contrary that there is

p0 ∈ X \ B . ! B[∈B Then from (4.17) and

Br1 (c) ⊆ B 1 c∈C[\{c1} B[∈B it follows that p0 ∈ Br1 (c1). Similarly, using (4.19) we obtain that p0 ∈ Br2 (c2). Using (4.22), for arbitrary integer n > 3, we can prove the membership

p0 ∈ Brn (cn) (4.24)

by induction. Since, for every n ∈ N, the ball Brn (cn) eventually contains the sequence (xn)n∈N, membership (4.24) and the limit relation lim rn = 0 imply the equality lim ρ(xn,p0)=0. Thus, (xn)n∈N n→∞ n→∞ converges to the point p0 ∈ X, contrary to the definition. It should be also noted that the family B is countably infinite. Indeed, B is at most countable because n C is finite for every n ∈ N. If B is finite, then there is n0 ∈ N such that

Brn0 (cn0 )= Brn0+1 (cn0+1)= ....

Thus, (xn)n∈N is eventually constant and, consequently, convergent, which is impossible as was shown above. The equality |B| = ℵ0 implies that there is a bijection g : B → (r1, 2r1) ∩ Q. Now we can deﬁne a mapping d: X × X → R+ by (4.11). Since B is a cover of X, the mapping (x, y) 7→ d(x, y) is correctly deﬁned. Reasoning as above, we can show that d: X × X → R+ is an ultrametric. Now we can easily complete the proof of the truth of (iv) ⇒ (i) for both completely bounded and not completely bounded ρ. Indeed, (4.11) and the equality g(B) = (r1, 2r1) ∩ Q imply that

(r1, 2r1) ∩ Q ⊆ D(X, d) ⊆ [0, 2r1).

Hence, we have 2r1 = diam(X, d) ∈/ D(X, d). Consequently, the poset D(X, d) does not have the largest element (see Example 2.20).

13 Corollary 4.8. Let (X, τ) be an ultrametrizable infinite topological space. Then (X, τ) is compact if and only if every compatible with τ ultrametric is totally bounded. Proof. If (X, τ) is compact, then every compatible with τ ultrametric is also compact and, consequently, totally bounded. Conversely, suppose every compatible with τ ultrametric d: X × X → R+ is totally bounded. Then, by Theorem 3.6, all compatible with τ ultrametrics have the order isomorphic distance sets. Hence, (X, τ) is compact by Theorem 4.7. Remark 4.9. It is interesting to compare Corollary 4.8 with the following theorem: “A metrizable topological space (X, τ) is compact if and only if every compatible with τ metric is complete”. This result was proved by Niemytzki and Tychonoff in 1928 [51]. There is also an ultrametric modification of Niemytzki—Tychonoff theorem obtained by Yoshito Ishiki in 2020 [37]. Using the ultrametric version Niemytzky–Tychonoff theorem [37], we can expand Corollary 4.8 to the following result.

Proposition 4.10. Let (X, τ) be an ultrametrizable topological space. Then the following conditions are equivalent: (i) (X, τ) is compact. (ii) Every compatible with τ ultrametric is complete. (iii) Every compatible with τ ultrametric is totally bounded. To prove Proposition 4.10, we only note that statements (i), (ii) and (iii) are trivially true if X is a ﬁnite set. Let us now turn to the ultrametrizable separable topological spaces.

Lemma 4.11. Let (X,ρ) be an ultrametric space. Then the following conditions are equivalent: (i) (X,ρ) is not separable. (ii) There is r> 0 such that

X \ Br(a) 6= ∅ a[∈A for every at most countable A ⊆ X. Proof. It follows directly from the deﬁnition of separability that the implication

¬(i) ⇒ ¬(ii) is true, where ¬ is the negation symbol. Hence, it suﬃces to prove the validity of the implication (i) ⇒ (ii). Suppose that (i) holds but for every r> 0 there exists an at most countable set A(r) ⊂ X such that

X ⊆ Br(a). a∈[A(r) Write Q+ = (0, ∞) ∩ Q. Then the set A = A(r) + r∈[Q is at most countable and dense in (X,ρ). Hence, (X,ρ) is separable, contrary to (i). Theorem 4.12. Let (X, τ) be an ultrametrizable topological space. Then the following conditions are equivalent: (i) (X, τ) is separable.

14 (ii) The distance set D(X, d) of an ultrametric space (X, d) is at most countable whenever d: X ×X → R+ is compatible with τ. Proof. (i) ⇒ (ii). This implication is valid by Corollary 2.11. (ii) ⇒ (i). Suppose (ii) holds but (X, τ) is not separable. Let us consider a compatible with τ ultrametric + ρ: X × X → R . Then the ultrametric space (X,ρ) is not separable. By Lemma 4.11 there exists r0 > 0 such that the cover

B = {Br0 (c): c ∈ X} of X does not contain any countable subcover. Using Lemma 4.6, we can ﬁnd C ⊆ X such that e

B = {Br0 (c): c ∈ C} is a cover of X consisting disjoint balls only. Since B⊆ B holds, the set B is uncountable. Hence, there is a mapping g : B → (r0, 2r0) having the uncountable range set, e |g(B)| > ℵ0. (4.25) If we deﬁne d: X × X → R+ by formula (4.11), then, as was proved earlier, d is an ultrametric compatible with τ. Since the distance set D(X, d) contains the set g(B), inequality (4.25) implies that D(X, d) is uncountable, contrary to (ii).

5. Instead of conclusion The concept of ultrametrics admits a natural generalization for poset-valued mappings. Let (Γ, 4) be a partially ordered set with the smallest element γ0 and let X be a nonempty set. 2 Deﬁnition 5.1. A mapping dΓ : X → Γ is an ultrametric distance, if the following conditions hold for all x, y, z ∈ X and γ ∈ Γ.

(i) dΓ(x, y)= γ0 if and only if x = y. (ii) dΓ(x, y)= dΓ(y, x). (iii) If dΓ(x, y) 4 γ and dΓ(y,z) 4 γ, then dΓ(x, z) 4 γ. The ultrametric distances were introduced by Priess-Crampe and Ribenboim [56] and studied in [14, 57, 58, 62, 63]. This generalization of ultrametrics has some interesting applications to logic programming, computational logic and domain theory [44, 59, 68]. For the case of totally ordered (Γ, 4), condition (iii) from the above deﬁnition becomes the following modiﬁcation of the strong triangle inequality

dΓ(x, y) 4 max {dΓ(x, z), dΓ(z,y)} . (5.1) Now we can deﬁne the “open balls” as the sets

Bγ (c)= {x ∈ X : dΓ(c, x) ≺ γ}, c ∈ X, γ ∈ Γ \{γ0}, and show that the set B X,dΓ = {Bγ(c): γ ∈ Γ \{γ0}, c ∈ X} is a base for a topology τdΓ if |Γ| > 2. It seems to be interesting to describe, for ﬁxed (Γ, 4), the topological spaces (X, τ) which are Γ-ultrametrizable in the sense that τ = τdΓ holds with some ultrametric distance 2 dΓ : X → Γ. Conjecture 5.2 (prove or disprove). Let (Γ, 4) be a totally ordered set and let (X, τ) be a Γ-ultrametrizable topological space. The following conditions are equivalent: ∗ (i) (Γ, 4) is not a poset of order type 1 + ω but there is a subposet Γ0 of (Γ, 4) such that Γ0 has the order type 1 + ω∗.

(ii) (X, τ) is compact iﬀ {dΓ(x, y): x, y ∈ X} has the largest element whenever τ = τdΓ . 15 References

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