The axiom of choice in metric measure spaces and maximal $\delta$-separated sets

Abstract

We show that the Axiom of Countable Choice is necessary and sufficient to prove that the existence of a Borel measure on a pseudometric space such that the measure of open balls is positive and finite implies separability of the space. In this way a negative answer to an open problem formulated in Górka (Am Math Mon 128:84–86, 2020) is given. Moreover, we study existence of maximal $\delta$-separated sets in metric and pseudometric spaces from the point of view the Axiom of Choice and its weaker forms.

1 Introduction

Assuming the Axiom of Choice (AC) the following theorem holds

Theorem 1.1

Let (X, d) be a pseudometric space with a Borel measure $\mu$ such that the measure of every open ball is positive and finite. Then, the space X is separable.

We refer to [6, Theorem 4.1] for the proof of Theorem 1.1 in the case of metric spaces.¹ Let us stress that the condition that the distance between different points is positive is not a necessary assumption for the proof in [6] so it remains valid also for pseudometrics. In fact, the converse to the implication in Theorem 1.1 also holds, i.e. if a pseudometric space (X, d) is separable, then there exists a Borel measure on X such that the measure of every open ball is positive and finite (see [7, Theorem 2]) and it is theorem of ZF. It was shown in [7, Theorem 2] that the Axiom of Dependent Choice (DC) is sufficient to prove Theorem 1.1 for metric spaces. Furthermore, in [7] there was asked the question if Theorem 1.1 in the case of metric spaces is equivalent to DC.

The first aim of this paper is to give an answer to the above question. Namely, we shall show that the Axiom of Countable Choice (CC) (which is strictly weaker than DC) is sufficient to prove Theorem 1.1. Moreover, we shall prove that CC is necessary assumption for Theorem 1.1 to be true. In this manner, we show that Theorem 1.1 is not equivalent to DC.

Let us stress that the proof in [7, Theorem 2] uses the technique of maximal nonoverlapping balls. The idea behind it is similar to the concept of maximal $\delta$-separated sets but, especially in the terms of maximality, it considers the relation between points, not balls (see Remark 4.1 for further details). On the other hand, using maximal $\delta$-separated sets one can prove Theorem 1.1 (see² [5, Theorem 2.2.16]). Furthermore, the existence of maximal $\delta$-separated sets is widely used in analysis on metric spaces. Therefore, the second aim of the paper is focus on maximal $\delta$-separated sets and investigate them in terms of AC and weaker forms of choice.

The remainder of the paper is structured as follows. In Sect. 2, we introduce the notations and recall the definitions of various types of axioms of choice. Moreover, some basic objects from analysis on metric spaces are collected there. Section 3 is devoted to the proof of equivalence between the Axiom of Countable Choice and Theorem 1.1. We close this section with an open question. In the last section we focus on $\delta$-separated sets. The problem of existence of maximal $\delta$-separated sets is studied in many types of metric spaces, e.g. separable spaces, geometrically doubling spaces, spaces with doubling measure. We show the relations between the existence of maximal $\delta$-separtated sets and axioms of choice.

2 Preliminaries

2.1 Axioms of choice

Let us recall that AC states that every collection $\left\{ A_\alpha \right\} _{\alpha \in \Lambda }$ of nonempty sets has a choice function, i.e. $f \,:\, \Lambda \rightarrow \bigcup _{\alpha \in \Lambda } A_\alpha$ such that $f(\alpha ) \in A_\alpha$ for all $\alpha \in \Lambda$.

In similar manner CC states that every countable collection $\left\{ A_n\right\} _{n=1}^\infty$ of nonempty sets has a choice function, i.e. $f \,:\, {\mathbb {N}} \rightarrow \bigcup _{n=1}^\infty A_n$ such that $f(n) \in A_n$ for all $n \in {\mathbb {N}}$. In general, in this paper by countable set we mean its cardinality is less or equal to $\aleph _0$.

We will be applying the Axiom of Dependent Choice (DC) in some cases which is strictly weaker than AC and strictly stronger than CC (see [10, Theorem 2.12] for implications and [14, Sect. 8.2, Problem 5.26] for strictness). The DC states that for every nonempty set X and entire relation³R on X there exists a sequence $\left\{ x_i\right\} _{i=1}^\infty \subset X$ such that $x_i \, R \, x_{i+1}$ for every $i \in {\mathbb {N}}$. Informally DC guarantees that the procedure based on choice at each step (i.e. an element in the next step is not determined by the previous steps) and such that in general those choices are dependent on the previous ones can be continued up to (countable) infinity. In ZF (Zermelo-Fraenkel set theory without any kind of choice type axiom) we can make only finitely many of those steps.

In parentheses next to the Theorem, Lemma etc., we provide the appropriate assumption regarding choice type axiom—DC, CC or ZF.

We will denote the length of a finite sequence x as $|x|$ and its elements as $x = \left\{ x^i\right\} _{i=1}^{|x|}$.

Lemma 2.1

(DC) For every nonempty set X, $v \in X$ and entire relation R on X there exists a sequence $\left\{ x_i\right\} _{i=1}^\infty \subset X$ such that $x_i\,R\,x_{i+1}$ for all $i \in {\mathbb {N}}$ and $x_1 = v$.

The proof of the Lemma 2.1 is contained in [12, Note 101]. Nevertheless, for the sake of completeness we provide the proof.

Proof

Let ${\mathcal {S}}_{fin}^v$ be the family of all finite sequences $x \subset X$ such that $x_1 = v$ and $x_i\,R\,x_{i+1}$ for all $i = 1, \ldots |x| - 1$. Let ${\widetilde{R}}$ be a binary relation on ${\mathcal {S}}^v_{fin}$ defined as follows:

$$\begin{aligned} x\,{\widetilde{R}}\,y \Longleftrightarrow |y| = |x| + 1 \quad \text {and} \quad x^i = y^i \; \text { for all } \; i=1,\ldots ,|x|. \end{aligned}$$

Since R is entire, it is easy to see that ${\widetilde{R}}$ is entire too. Therefore, by DC there exists a sequence $\left\{ x_i\right\} _{i=1}^\infty \subset {\mathcal {S}}_{fin}^v$ such that $x_i\,{\widetilde{R}}\,x_{i+1}$ for all $i \in {\mathbb {N}}$. Let us denote $x_i = \left\{ x_i^j\right\} _{j=1}^{|x_i|}$. Thus, a sequence we are looking for is

$$\begin{aligned} v=x_1^1,\quad x_1^2,\quad x_1^3,\quad \ldots ,\quad x_1^{|x_1|},\quad x_2^{|x_2|},\quad x_3^{|x_3|},\quad \ldots \end{aligned}$$

$\square$

Let X be a set, R be a binary relation on X and x be a finite sequence in X. For the purposes of the next lemma we will say that an element $y \in X$ is a successor of x if $x^i\,R\,y$ for all $i=1,\ldots ,|x|$ and there is no $z \in X$ with this property such that $z\,R\,y$. In addition, we will say that x is maximal if x is a one-element sequence or $x^j$ is a successor of $\left\{ x^i\right\} _{i=1}^{j-1}$ for every $j=2,\ldots ,|x|$. Then, we will say that infinite sequence $\left\{ x^i\right\} _{i=1}^\infty \subset X$ is maximal if every finite subsequence $\left\{ x^i\right\} _{i=1}^N$ of $\left\{ x^i\right\} _{i=1}^\infty$ is maximal.

Lemma 2.2

(DC) Let X be a nonempty set, $v \in X$ and R be a binary relation on X. Let us suppose that for every maximal finite sequence x in X there exists successor $y \in X$ of x. Then, there exists a maximal sequence $\left\{ x_i\right\} _{i=1}^\infty$ in X with $x_1=v$.

Proof

We are reasoning in the similar way as before. Let ${\mathcal {S}}_{fin}^{max}$ be the family of maximal finite sequences in X. Obviously ${\mathcal {S}}_{fin}^{max}$ is nonempty since it contains one-element sequences. We define a binary relation ${\widetilde{R}}$ on ${\mathcal {S}}_{fin}^{max}$ in the same manner as before:

$$\begin{aligned} x\,{\widetilde{R}}\,y \Longleftrightarrow |y| = |x| + 1 \quad \text {and} \quad x^i = y^i \; \text { for all } \; i=1,\ldots ,|x|. \end{aligned}$$

Since every maximal finite sequence in X has successor, relation ${\widetilde{R}}$ is entire. By Lemma 2.1, there exists a sequence $\left\{ x_i\right\} _{i=1}^\infty$ in ${\mathcal {S}}_{fin}^{max}$ such that $x_i\,{\widetilde{R}}\,x_{i+1}$ for all $i \in {\mathbb {N}}$ and $x_1=\left\{ v\right\}$. It is easy to see the sequence $\left\{ x_i^{|x_i|}\right\} _{i=1}^\infty$ is a maximal sequence in X. $\square$

For more information regarding the Axiom of Choice and its weaker forms we refer to books [10, 14] or [13]. Some connections between choice type axioms and the topology of pseudometric spaces are done in [1, 4, 3].

If (X, d) is a pseudometric space we can define an equivalence relation $\sim$ on X as follows:

$$\begin{aligned} x \sim y \quad \Longleftrightarrow \quad d(x,y) = 0. \end{aligned}$$

Every pseudometric space (X, d) can be viewed as metric space $(X^*, d^*)$ where $X^* = X/{\sim }$ and $d^*$ is metric given by:

$$\begin{aligned} d^*\left( {[x], [y]}\right) = d(x,y), \end{aligned}$$

where [x] denotes the equivalence class of x. In general, properties of X and the induced metric space $X^*$ may vary depends on the choice principles we are dealing with. For more details and some connections between these spaces we refer to [11].

2.2 Basic tools from analysis on metric spaces

We shall recall basic definitions that play an important role in the analysis on metric spaces.

Let (X, d) be a metric space and $\delta > 0$. We say that set Y which consists of closed (resp. open) balls of radii $\delta$ in X is strict (resp. non-strict) $\delta$-cover in X if Y is covering of X.

We will focus only on strict $\delta$-covers and thus we will omit the word “strict”. We denote the closed ball ${\overline{B}}$ of center x and radius r as ${\overline{B}}(x, r)$ or ${\overline{B}}_r$ if we do not want to specify any center. Moreover, we say that metric space X is proper if every closed ball in X is compact.

We say that metric space X is geometrically doubling if there exists a constant $M \in {\mathbb {N}}$ such that for every $r>0$ every ball of radius r can be covered by at most M balls of radius r/2.

Remark 2.3

We do not distinguish open and closed balls in the above definition since one can easily show the space is geometrically doubling for closed balls if and only if it is geometrically doubling for open balls. However, in general, the best constant M does not have to be the same in both cases. For instance, in the case of ${\mathbb {R}}$ equipped with the Euclidean distance we have $M=2$ for closed balls and $M=3$ for open balls.

Remark 2.4

It is easy to see that if X is geometrically doubling, then every ball ${\overline{B}}_r$ can be covered by finitely many balls of arbitrarily small radius.

Let $(X, d, \mu )$ be a metric measure space equipped with a metric d and the Borel measure $\mu$. We say that the measure $\mu$ is doubling if the measure of every closed ball is finite and positive and there exists a constant⁴$C\ge 1$ such that for every $x\in X$ and $r>0$

$$\begin{aligned} \mu \left( {{\overline{B}}(x,2r)}\right) \le C\mu \left( {{\overline{B}}(x,r)}\right) . \end{aligned}$$

It is easy to see that measure is doubling for closed balls if and only if it is doubling for open balls and, in contrast to Remark 2.3, the best doubling constant in both cases is the same, i.e.

$$\begin{aligned} \sup _{x \in X, r>0} \frac{\mu \left( {B\left( {x, 2r}\right) }\right) }{\mu \left( {B\left( {x, r}\right) }\right) } = \sup _{x \in X, r>0} \frac{\mu \left( {{\overline{B}}\left( {x, 2r}\right) }\right) }{\mu \left( {{\overline{B}}\left( {x, r}\right) }\right) }. \end{aligned}$$

The same definitions applies to pseudometric spaces. One can read more about results concerning geometrically doubling spaces and doubling measures in the analysis on metric spaces, e.g. in [2, 8, 9] or [15]. It is not difficult to see that if $(X, d, \mu )$ is a metric-measure space with doubling measure $\mu$, then (X, d) is geometrically doubling (see Theorem 4.9 in Sect. 4.). Conversely, Luukkainen and Saksman in [15] prove that every complete⁵ geometrically doubling metric space carries a doubling measure.

3 Existence of nontrivial Borel measure vs separability of metric space

In this section we prove that Theorem 1.1 and CC are equivalent.

Theorem 3.1

The following statements are equivalent:

• (i) CC;

• (ii) Theorem 1.1.

Proof

(i) $\implies$ (ii): Let us fix $x_0 \in X$ and put $B_k = B(x_0, k)$ for all $k \in {\mathbb {N}}$. Let $l_0(n) = \frac{1}{\mu \left( {B\left( {x_0, 1/n}\right) }\right) } + 1$ for $n \in {\mathbb {N}}$. For $k,n,l \in {\mathbb {N}}$ such that $l \ge l_0(n)$ let ${\mathcal {A}}(k,n,l)$ be the set of families of nonoverlapping open balls contained in $B_k$ with radius smaller than 1/n and measure bigger than 1/l. Since $l\ge l_0(n)$, we have $\left\{ B(x_0, r_n)\right\} \in {\mathcal {A}}(k,n,l)$ for some⁶$r_n < 1/n$ so ${\mathcal {A}}(k,n,l)$ is nonempty.

For fixed $k,n,l \in {\mathbb {N}}$, $l\ge l_0(n)$, all families contained in ${\mathcal {A}}(k,n,l)$ are finite and their cardinalities are uniformly upper bounded by the constant value dependent on k, l. Indeed, assume there exists $\beta \in {\mathcal {A}}(k,n,l)$ such that $\#\beta > l\mu \left( {B_k}\right)$ (this set can be finite as well as infinite). Let ${\widetilde{\beta }} \subset \beta$ be finite and $\#{\widetilde{\beta }} > l\mu \left( {B_k}\right)$ (we do not need any kind of the axiom of choice to select a finite subset with an arbitrary big cardinality of any infinite set). Then,

$$\begin{aligned} \infty> \mu (B_k) \ge \mu \left( {\bigcup \beta }\right) \ge \mu \left( {\bigcup {\widetilde{\beta }}}\right) = \sum _{B \in {\widetilde{\beta }}} \mu (B) \ge \sum _{B \in {\widetilde{\beta }}} \frac{1}{l} = \frac{1}{l} \# {\widetilde{\beta }} > \mu (B_k). \end{aligned}$$

It gives us a contradiction, thus $\beta$ has to be a finite set and the cardinality of $\beta$ is no bigger than $\mu (B_k)l$.

Let $c_{k,n,l} = \max \left\{ {\#\beta \,:\, \beta \in {\mathcal {A}}(k,n,l)}\right\} < \infty$. Since the family $\left\{ {\mathcal {A}}(k,n,l)\right\} _{k,n,l}$ is countable, we can choose an element $\alpha _{k,n,l}$ from every set ${\mathcal {A}}(k,n,l)$ with cardinality equals $c_{k,n,l}$ and the choice is guaranteed by CC. The element $\alpha _{k,n,l}$ is a maximal element of ${\mathcal {A}}(k,n,l)$ under inclusion.

Next, for fixed k, n, l by finiteness of $\alpha _{k,n,l}$ there exists a set ${\mathcal {S}}(k,n,l)$ of exactly one center of each ball⁷ from $\alpha _{k,n,l}$. Therefore, by CC we can choose such a set ${\mathcal {S}}(k,n,l)$ from every $\alpha _{k,n,l}$. Finally, let us define the set

$$\begin{aligned} {\mathcal {S}} = \bigcup _{k=1}^\infty \bigcup _{n=1}^\infty \bigcup _{l=l_0(n)}^\infty {\mathcal {S}}(k,n,l), \end{aligned}$$

which is countable by CC as countable union of finite sets. We need to show that the set ${\mathcal {S}}$ is dense in X.

For this purpose we fix $x \in X$ and $\varepsilon >0$. Let K be such that $B(x, \varepsilon ) \subset B_{K}$, N such that $\frac{1}{2N} < \varepsilon /3$ and $L > l_0(N)$ such that $1/L < \mu \left( {B\left( {x, \frac{1}{2N}}\right) }\right)$. Then, $B\left( {x, \frac{1}{2N}}\right) \in {\mathcal {A}}(K,N,L)$, and therefore from the maximality of $\alpha _{K,N,L}$ there exists $y \in {\mathcal {S}}(K,N,L) \subset {\mathcal {S}}$ and $\delta < 1/N$ such that

$$\begin{aligned} B\left( {y, \delta }\right) \cap B\left( {x, \frac{1}{2N}}\right) \ne \emptyset . \end{aligned}$$

Therefore, $d(x, y)< \frac{3}{2N} < \varepsilon$ and it finishes the proof of implication (i) $\implies$ (ii).

(ii) $\implies$ (i): Let $\left\{ A_i\right\} _{i=1}^\infty$ be an arbitrary countable family of nonempty sets and let $X = \bigcup _{i=1}^\infty A_i$. Without loss of generality, we can assume that the sets $A_i$ are pairwise disjoint⁸. Let us define the function $d :X \times X \rightarrow \left\{ 0, 1\right\}$ as follows:

$$\begin{aligned} d(x, y) = {\left\{ \begin{array}{ll} 0, \quad \text {if } x,y \in A_i \text { for some } i \in {\mathbb {N}}; \\ 1, \quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Obviously, d is a pseudometric on X, and therefore (X, d) is a pseudometric space.

Open balls in this space are either sets $A_i$ or the whole space, thus the topology $\tau$ on X has the form

$$\begin{aligned} \tau = \left\{ \bigcup _{i \in I} A_i :I \subset {\mathbb {N}}\right\} \cup \left\{ \emptyset \right\} . \end{aligned}$$

It is easy to check that the topology $\tau$ is also $\sigma$-algebra on X. Thus, $\tau$ is the Borel $\sigma$-algebra ${\mathcal {B}}(X)$ on X.

Next, we define a Borel measure $\mu :{\mathcal {B}}(X) \rightarrow [0, 1]$. Since $A_i$ are pairwise disjoint and ${\mathcal {B}}(X) = \tau$, it is enough to define $\mu$ on every $A_i$. We put $\mu (A_i) = 1/2^i$. Then, $\mu (X) = 1$ and the measure of every open ball is finite and positive. This allows us to use Theorem 1.1 which gives the separability of X. Let $\left\{ d_i\right\} _{i=1}^\infty$ be a dense subset of X. We define function $f :{\mathbb {N}} \rightarrow X$ as follows: $f(i) = d_{j_i}$ where $j_i = \min \left\{ j :\, d_j \in A_i\right\}$. Those numbers $j_i$ are well-defined since for a fixed $i \in {\mathbb {N}}$ set $A_i \ne \emptyset$ is open, so there exists j such that $d_j \in A_i$. Therefore, f is a choice function for family $\left\{ A_i\right\} _{i=1}^\infty$. $\square$

From Theorem 3.1 we have the following corollary.

Theorem 3.2

(CC) Let (X, d) be a metric space with a Borel measure $\mu$ such that the measure of every open ball is positive and finite. Then, the space X is separable.

The above theorem is an essential improvement of [7, Theorem 2], where Theorem 3.2 has been proven assuming DC. In this way, we give an answer to the open problem from [7] which was mentioned in the introduction.

Problem 3.3

It is very natural to ask whether CC is equivalent to the statement in Theorem 3.2.

4 $\delta$-separated sets

In this section, we shall explore the idea of maximal $\delta$-separated sets mentioned in the introduction in terms of the choice type axioms. These sets play a crucial role in the analysis on metric spaces. Let us recall that a subset Y of a metric space (X, d) is called a strict (resp. non-strict) $\delta$-separated set if $d(x, y) > \delta$ (resp. $d(x, y) \ge \delta$) for all distinct points $x,y \in Y$.

The same definition applies to pseudometric or ultrametric spaces. In applications in the analysis on metric spaces there are essentially no differences between strict and non-strict $\delta$-separated sets so we will focus only on strict $\delta$-separated sets and thus we will omit the word “strict”. In most often cases, we need to be guaranteed the existence of maximal (under inclusion) $\delta$-separated sets for $\delta$’s of the form 1/n (in general $\delta < 1$).

Remark 4.1

As indicated in the introduction, the concepts of $\delta$-separated sets and pairwise disjoint balls are different in general, especially in the terms of maximality. It is easy to see that if Y is $\delta$-separated set in X, then $\left\{ {\overline{B}}\left( {y, \delta /2}\right) \right\} _{y\in Y}$ is a family of pairwise disjoint balls. The converse is not true. As an example one can take $X = \left\{ 0, 1\right\}$ with discrete metric. The family $\left\{ {\overline{B}}(0, 3/4), {\overline{B}}(1, 3/4)\right\}$ consists of pairwise disjoint balls with $\delta =3/2$ (and it is even maximal) but $d(0,1) = 1$ so X is not 3/2-separated set (and thus also not maximal). Looking at the same space, $\left\{ 0\right\}$ is maximal 3/2-separated set, while $\left\{ {\overline{B}}(0,3/4)\right\}$ is not maximal, since $\left\{ {\overline{B}}(0,3/4), {\overline{B}}(1,3/4)\right\}$ is bigger.

Our journey will begin with the following proposition.

Proposition 4.2

The following statements are equivalent:

• (i) AC;

• (ii) For every pseudometric space (X, d) and $\delta > 0$ there exists a maximal $\delta$-separated set;

• (iii) For every metric space (X, d) and $\delta > 0$ there exists a maximal $\delta$-separated set;

• (iv) For every ultrametric space (X, d) and $\delta > 0$ there exists a maximal $\delta$-separated set.

Proof

The implication (i) $\implies$ (ii) is clear by the Zorn Lemma, and the implications (ii) $\implies$ (iii) $\implies$ (iv) are obvious.

(iv) $\implies$ (i): Let $\left\{ A_\alpha \right\} _{\alpha \in \Lambda }$ be an arbitrary family of nonempty pairwise disjoint sets and let $X = \bigcup _{\alpha \in \Lambda } A_\alpha$. We define ultrametric d on X as follows:

$$\begin{aligned} d(x, y) = {\left\{ \begin{array}{ll} 0, \quad \quad \text {if } x=y; \\ 1/2, \quad \text {if } x\ne y \text { and } x,y \in A_\alpha \text { for some } \alpha \in \Lambda ; \\ 1, \quad \quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Let $Y \subset X$ be a maximal 3/4-separated set in (X, d). Then, Y contains exactly one element from each set $A_\alpha$. Indeed, Y contains at most one element from each set $A_\alpha$ since if $x,y \in Y \cap A_\alpha$ and $x \ne y$, then $d(x,y) = 1/2$ from the definition of d, but $d(x,y) > 3/4$ from the definition of Y. Moreover, Y has nonempty intersection with $A_\alpha$ for every $\alpha$. Indeed, let us suppose $Y \cap A_\alpha = \emptyset$ for some $\alpha$ and take $x \in A_\alpha$. If $y \in Y$, then $y \notin A_\alpha$ so $d(x,y) = 1 > 3/4$ for all $y \in Y$. This contradicts the maximality of Y. $\square$

As we can see the assumption of existence of a maximal $\delta$-separated set for every metric space is so strong that it implies AC. We will therefore look at the existence of a maximal $\delta$-separated set for metric or pseudometric spaces with certain properties which often appear in the analysis on metric spaces.

Theorem 4.3

The following two statements are true in ZF.

• (i) Let $\delta >0$ and (X, d) be a pseudometric space such that the cardinality of every $\delta$-separated set in X is finite and their cardinalities are uniformly upper bounded by some constant C. Then, there exists a maximal $\delta$-separated set;

• (ii) Let $\delta >0$ and (X, d) be a pseudometric space which contains a finite $\delta /2$-cover. Then, all $\delta$-separated sets in X are finite and their cardinalities are uniformly upper bounded. Hence, in particular, there exists a maximal $\delta$-separated set.

The following statements are equivalent with DC:

• (iii) Let $\delta >0$ and (X, d) be a pseudometric space which contains a countable $\delta /2$-cover. Then, there exists a maximal $\delta$-separated set;

• (iii)* Let $\delta >0$ and (X, d) be a metric space which contains a countable $\delta /2$-cover. Then, there exists a maximal $\delta$-separated set;

• (iv) Let $\delta >0$ and (X, d) be a pseudometric space such that all $\delta$-separated sets are finite. Then, there exists a maximal $\delta$-separated set;

• (iv)* Let $\delta >0$ and (X, d) be a metric space such that all $\delta$-separated sets are finite. Then, there exists a maximal $\delta$-separated set.

Let us pay an attention that (i) and (ii) are counterparts of (iv) and (iii), recpectively.

Proof

ZF $\implies$ (i): Let us take $x_1 \in X$. Then, we choose $x_2 \in X$ such that $d(x_2, x_1) > \delta$ (if any such $x_2$ exists). Next, we choose $x_3 \in X$ such that $d(x_3,x_2) > \delta$ and $d(x_3,x_1) > \delta$ (also if any such $x_3$ exists) and so on. After finite steps we will not be able to pick the next point such that selected points would create a $\delta$-separated set since the cardinality of any $\delta$-separated set in X is no bigger than C.

ZF $\implies$ (ii): By assumption we can write $X = \bigcup _{n=1}^N {\overline{B}}^{\,n}_{\delta /2}$. Since the distance between every two points of any ball ${\overline{B}}^{\,n}_{\delta /2}$ is at most $\delta$, every $\delta$-separated set has no more than N elements. By (i) there exists a maximal $\delta$-separated set.

DC $\implies$ (iii): We can assume there is no finite maximal $\delta$-separated set in X (otherwise, we can take one). Let $X = \bigcup _{n=1}^\infty {\overline{B}}^{\,n}_{\delta /2}$ and pick $v \in {\overline{B}}^{\,1}_{\delta /2}$. Let R be a binary relation on $X \times {\mathbb {N}}$ defined as follows:

$$\begin{aligned} (x,n)\,R\,(y,m) \Longleftrightarrow n<m, \;x \in {\overline{B}}^{\,n}_{\delta /2},\; y \in {\overline{B}}^{\,m}_{\delta /2} \quad \text {and} \quad d(x,y) > \delta . \end{aligned}$$

We will show that relation R satisfies the assumption of Lemma 2.2. Let $z = \left\{ (z_i, n_i)\right\} _{i=1}^N$ be a maximal sequence in $X \times {\mathbb {N}}$. Since $\left\{ z_i\right\} _{i=1}^N$ is a finite $\delta$-separated set, it cannot be maximal. Let us take ${\widetilde{z}} \in {\overline{B}}^{\,n}_{\delta /2}$ such that $d({\widetilde{z}},z_i) > \delta$ for all $i=1,\ldots ,N$, where n is the least $n > n_1$ for which such ${\widetilde{z}}$ exists. It has to be $n > \max \left\{ n_i:\, i=1,\ldots ,N\right\}$ since otherwise it would contradicts a maximality of z. Therefore, $\left( {z_i, n_i}\right) \,R\,\left( {{\widetilde{z}},n}\right)$ for all $i=1,\ldots ,N$ and $({\widetilde{z}},n)$ is a successor of z. By Lemma 2.2 there exists a maximal sequence $\left\{ \left( {z_i, n_i}\right) \right\} _{i=1}^\infty$ in $X\times {\mathbb {N}}$ and $(z_1,n_1)=(v,1)$. Thus, $\left\{ z_i\right\} _{i=1}^\infty$ is a maximal $\delta$-separated set in X.

(iii) $\implies$ (iii)*: Obvious.

(iii)* $\implies$ DC: Let us fix nonempty set X and entire relation R on X. We can assume X has more than one element. Let ${\mathcal {S}}_{fin}$ be the family of all finite sequences in X and denote by ${\mathcal {S}}_n$ the family of all finite sequences of length n in X. Clearly $\bigcup _{n=1}^\infty {\mathcal {S}}_n = {\mathcal {S}}_{fin}$. We define relation ${\widetilde{R}}$ on ${\mathcal {S}}_{fin}$ in the following manner:

$$\begin{aligned}&x\,{\widetilde{R}}\,y \Longleftrightarrow |x| < |y| \; \text { and } \; x^i = y^i \; \text { for all } \; i=1,\ldots ,|x| \\&\text { and } \;y^i\,R\,y^{i+1} \; \text { for all } \; i=|x|,\ldots ,|y|-1. \end{aligned}$$

It is easy to see that ${\widetilde{R}}$ is transitive. Next, we define metric⁹d on ${\mathcal {S}}_{fin}$:

We will show there exists a countable 5/4-cover in $({\mathcal {S}}_{fin}, d)$. To simplify the notation, for $z \in X$ let $nz \in {\mathcal {S}}_n$ be the sequence with all elements equal z. Let us take $x \in X$ and $y \in X$ such that $x\ne y$. Note that

$$\begin{aligned} {\mathcal {S}}_{n-1} \subset {\overline{B}}\left( {nx, 5/4}\right) \cup {\overline{B}}\left( {ny, 5/4}\right) \end{aligned}$$

for all $n \ge 2$. Indeed, let us fix $z \in {\mathcal {S}}_{n-1}$. If $z^1\ne x$, then $||z| - |nx|| = 1$ and (and obviously ). Hence, $d(z, nx) = 1 < 5/4$. If $z^1=x$, then $z^1\ne y$ and analogously $d(z, ny) = 1 < 5/4$. It proves the above inclusion, therefore, summing it over all n we get a countable 5/4-cover in ${\mathcal {S}}_{fin}$. By assumption there exists a maximal 5/2-separated set Y in ${\mathcal {S}}_{fin}$. Hence, for all $y_1,y_2 \in Y, y_1 \ne y_2$ we have $d(y_1,y_2) = 3$.

The set Y is infinite. Indeed, let us suppose Y is finite, i.e. $Y=\left\{ y_1,\ldots ,y_k\right\}$. Take $y_n$ with the greatest length among elements of Y and denote $N = |y_n|$. Since R is entire, there exist $x^1,x^2,x^3 \in X$ such that $y_n^N\,R\,x^1$, $x^1\,R\,x^2$, $x^2\,R\,x^3$. Let $x = \left( {y_n^1,\ldots ,y_n^N, x^1, x^2, x^3}\right)$. We get $||y_n|-|x|| = 3$ and $y_n\,{\widetilde{R}}\,x$ so $d(y_n, x) = 3$. Moreover, from the definition of $y_n$ we have $||y_i|-|x|| \ge 3$ for $i=1,\ldots ,k$ and since $d(y_i,y_n)=3$ for $i\ne n$, we have $y_i\,{\widetilde{R}}\,y_n$. Therefore, from transitivity of ${\widetilde{R}}$ we have $y_i\,{\widetilde{R}}\,x$ for all $i=1,\ldots ,k$. Thus, $d(y_i,x)=3$ for all $i=1,\ldots ,k$ and it contradicts the maximality of Y.

The set Y is countable. Let $f:\,Y \rightarrow {\mathbb {N}}$ be defined as $f(y) = |y|$. This function is injective. Indeed, if $f(y_1)=f(y_2)$, then $|y_1|=|y_2|$ so $y_1=y_2$ since $||y_1|-|y_2|| \ge 3$ for $y_1\ne y_2$.

We get that Y is countably infinite and we can enumerate elements of Y by the function f, i.e. $Y =\left\{ y_{n_i}\right\} _{i=1}^\infty$ where $|y_{n_i}|=n_i$ is increasing. Since $d(y_{n_i}, y_{n_{i+1}})=3$, we have $y_{n_i}\,{\widetilde{R}}\,y_{n_{i+1}}$ and $|y_{n_{i+1}}| \ge |y_{n_i}| + 3$. Finally, a required sequence we define as follows

$$\begin{aligned} y_{n_1}^{|y_{n_1}|},\quad y_{n_2}^{|y_{n_1}|+1},\quad \ldots , \quad y_{n_2}^{|y_{n_2}|},\quad y_{n_3}^{|y_{n_2}|+1}, \quad \ldots ,\quad y_{n_3}^{|y_{n_3}|},\quad \ldots \end{aligned}$$

(iv) $\implies$ (iv)*: Obvious.

(iv)* $\implies$ DC: Let us fix nonempty set X and an entire relation R on X. We define metric d on ${\mathcal {S}}_{fin}$ as in the proof of the previous implication. Let us suppose all 5/2-separated sets in $({\mathcal {S}}_{fin}, d)$ are finite. Then, by the assumption there exists a maximal and finite 5/2-separated set. Since all maximal 5/2-separated sets are infinite in that space, we get a contradiction. Thus, there exists an infinite 5/2-separated set and we can construct the required sequence in the same manner as in the last two paragraphs of the proof of the previous implication.

DC $\implies$ (iv): Let us suppose there is no maximal $\delta$-separated set in X. We will show there exists an infinite $\delta$-separated set in X. Let ${\mathcal {S}}_{fin}^{sep}$ be a family of all finite sequences of distinct points which are $\delta$-separated sets. Since all one-element sequences belong to ${\mathcal {S}}_{fin}^{sep}$, we have ${\mathcal {S}}_{fin}^{sep} \ne \emptyset$. Let us define relation R on ${\mathcal {S}}_{fin}^{sep}$ as follows.

$$\begin{aligned} x\,R\,y \Longleftrightarrow |y| = |x| + 1 \; \text { and } \; x^i = y^i \; \text { for all } \; i=1,\ldots ,|x|. \end{aligned}$$

The relation R is entire. Indeed, let us fix $x \in {\mathcal {S}}_{fin}^{sep}$. By an assumption x cannot be maximal $\delta$-separated set. Therefore, there exists ${\widetilde{x}} \in X$ such that ${\widetilde{x}} \ne x^i$ for all $i=1,\ldots |x|$ and $y = (x^1, \ldots x^{|x|}, {\widetilde{x}})$ creates $\delta$-separated set, so consequently $x\,R\,y$. By DC there exists a sequence $\left\{ x_i\right\} _{i=1}^\infty \subset {\mathcal {S}}_{fin}^{sep}$ such that $x_i\,R\,x_{i+1}$ for all $i \in {\mathbb {N}}$. Thus, the sequence $\left\{ x_i^{|x_i|}\right\} _{i=1}^\infty$ is an infinite $\delta$-separated set in X. $\square$

In view of Theorem 4.3 we immediately get two results.

Corollary 4.4

(ZF) For every totally bounded pseudometric space and every $\delta >0$ all $\delta$-separated sets are finite and their cardinalities are uniformly upper bounded. Hence there exists a maximal $\delta$-separated set for every $\delta >0$.

Proof

We can write the whole space as a finite sum of balls of radii $\delta /2$ and then apply Theorem 4.3 (ii). $\square$

Corollary 4.5

(DC) Let (X, d) be a separable pseudometric space and $\delta > 0$. Then, there exists a maximal $\delta$-separated set.

Proof

Let $\left\{ d_i\right\} _{i=1}^\infty$ be a dense subset of X. Then, $X = \bigcup _{i=1}^\infty {\overline{B}}(d_i, \delta /2)$, and we can apply Theorem 4.3 (iii). $\square$

Problem 4.6

What is the set-theoretic status of the existence of a maximal $\delta$-separated set in separable metric (or pseudometric) spaces in the sense of equivalence with choice type axioms?

Now, we shall consider proper spaces, geometrically doubling spaces and spaces which admit a doubling measure.

Theorem 4.7

(ZF) Let (X, d) be a separable pseudometric space such that one of the following conditions holds

• (i) X is proper,

• (ii) X is geometrically doubling,

• (iii) there exists a doubling measure $\mu$ on X.

Then, for every $\delta >0$ there exists a maximal $\delta$-separated set.

Proof

First of all we shall prove the following lemma.

Lemma 4.8

(ZF) Let (X, d) be a pseudometric space such that one of the following conditions holds

• (i) X is proper,

• (ii) X is geometrically doubling,

• (iii) there exists a doubling measure $\mu$ on X.

Then, for any $\delta , r>0$ all $\delta$-separated sets contained in ${\overline{B}}_r$ are finite and their cardinalities are uniformly upper bounded.

Proof

Obviously, we can assume that $r > \delta /2$.

• (i) : Since the set ${\overline{B}}_r$ is compact, the Lemma follows from Corollary 4.4.

• (ii) : Since ${\overline{B}}_r$ can be covered by finitely many balls of radius $\delta /2$, it is enough to apply Theorem 4.3 (ii).

• (iii) : Let $K \in {\mathbb {N}}$ be such that $2^K \delta \ge 2r$. Suppose there exists an infinite $\delta$-separated set Y in ${\overline{B}}_r$. Let us take a finite subset ${\widetilde{Y}}$ of Y such that $\#{\widetilde{Y}} > C^{K+2}$, where C is a doubling constant of $\mu$. Then, for every $y \in {\widetilde{Y}}$ we have ${\overline{B}}(y, r) \subset {\overline{B}}_{2r}$ and ${\overline{B}}_r \subset {\overline{B}}(y, 2r)$. Thus, we get the following string of inequalities

$$\begin{aligned} \infty&> \mu \left( {{\overline{B}}_{2r}}\right) \ge \mu \left( {\bigcup _{y \in {\widetilde{Y}}} {\overline{B}}(y, r)}\right) \ge \mu \left( {\bigcup _{y \in {\widetilde{Y}}} {\overline{B}}(y, \delta /2)}\right) = \sum _{y\in {\widetilde{Y}}} \mu \left( {{\overline{B}}(y, \delta /2)}\right) \\ {}&\ge \frac{1}{C^{K+1}} \sum _{y\in {\widetilde{Y}}} \mu \left( {{\overline{B}}(y, 2^K \delta )}\right) \ge \frac{1}{C^{K+1}} \sum _{y\in {\widetilde{Y}}} \mu \left( {{\overline{B}}(y, 2r)}\right) \ge \frac{1}{C^{K+1}} \sum _{y\in {\widetilde{Y}}} \mu \left( {{\overline{B}}_r}\right) \\ {}&= \frac{1}{C^{K+1}}\mu \left( {{\overline{B}}_r}\right) \#{\widetilde{Y}} > C\mu \left( {{\overline{B}}_{r}}\right) \ge \mu \left( {{\overline{B}}_{2r}}\right) , \end{aligned}$$

which is an obvious contradiction. Therefore, every $\delta$-separated set Y is finite and $\#Y \le C^{K+2}$. $\square$

If X is bounded, then by Lemma 4.8 all $\delta$-separated sets are finite and uniformly upper bounded, and thus the proof follows. Therefore, we can assume that X is an unbounded space. Let $\left\{ x_n\right\} _{n=1}^\infty$ be a dense subset of X. We define sequence $\left\{ n_k\right\} _{k=1}^\infty \subset {\mathbb {N}}$ as follows: $n_1=1$, $n_k$ is the smallest $n > n_{k-1}$ such that $d(x_n, x_{n_i}) > \delta$ for every $i = 1, \ldots , k-1$. Let us note that the construction goes to infinity since X is unbounded and the whole sequence is determined by $n_1$ so the procedure is possible in ZF. Clearly, $\left\{ x_{n_k}\right\} _{k=1}^\infty$ is a $\delta$-separated set. We claim that $\left\{ x_{n_k}\right\} _{k=1}^\infty$ is a maximal $\delta$-separated set.

Indeed, let us suppose there exists $x \in X$ such that $d(x, x_{n_k}) > \delta$ for every $k \in {\mathbb {N}}$. By Lemma 4.8, ball ${\overline{B}}(x, 2\delta )$ contains only finitely many elements of sequence $\left\{ x_{n_k}\right\} _{k=1}^\infty$. Hence, there exists $0 < \varepsilon \le \delta$ such that $d(x, x_{n_k}) > \delta + \varepsilon$ for every $k \in {\mathbb {N}}$. Let $m \in {\mathbb {N}}$ be such that $d(x, x_m) < \varepsilon$. Obviously $m \ne n_k$ for all $k \in {\mathbb {N}}$. Therefore, from the construction of the sequence $\left\{ n_k\right\} _{k=1}^\infty \subset {\mathbb {N}}$, there exists $k \in {\mathbb {N}}$ such that $d(x_m, x_{n_k}) \le \delta$. Thus,

$$\begin{aligned} \delta +\varepsilon< d(x, x_{n_k}) \le d(x, x_m) + d(x_m, x_{n_k}) < \delta + \varepsilon . \end{aligned}$$

This contradiction finishes the proof. $\square$

There is also a well-known result connecting geometrically doubling spaces with spaces which admit a doubling measure. We simply obtain it with the help of Lemma 4.8.

Theorem 4.9

(ZF, Coifmann-Weiss) Every pseudometric space which admits a doubling measure is geometrically doubling.

Proof

Let us fix $r > 0$. Then, by Lemma 4.8 (iii), there exists a maximal and finite r/2-separated set $\left\{ x_i\right\} _{i=1}^M$ in ${\overline{B}}_r$. From the proof of Lemma 4.8 (iii), M is upper bounded by a constant $C^{4}$, where C is a doubling constant. From maximality we have an inclusion ${\overline{B}}_r \subset \bigcup _{i=1}^M {\overline{B}}\left( {x_i, r/2}\right)$. This finishes the proof since every ball ${\overline{B}}_r$ can be covered by at most $C^{4}$ balls of radius r/2. $\square$

Theorem 4.10

The following statements are equivalent:

• (i) CC;

• (ii) For every $\delta >0$ and pseudometric space which admits a doubling measure there exists a maximal $\delta$-separated set;

• (iii) For every $\delta >0$ and geometrically doubling pseudometric space there exists a maximal $\delta$-separated set;

• (iv) Every geometrically doubling pseudometric space is separable.

Proof

(i) $\implies$ (iv): Let X be a geometrically doubling space and fix $x \in X$. For $n \in {\mathbb {N}}$ let ${\mathcal {A}}_n$ be the family that consists of finite sets of closed balls of radius $2^{-n}$ which cover ${\overline{B}}(x,n)$. Since X is geometrically doubling, ${\mathcal {A}}_n \ne \emptyset$. From CC we can choose one element ${\mathcal {B}}_n$ from every set ${\mathcal {A}}_n$. Let ${\mathcal {B}} = \bigcup _{n=1}^\infty {\mathcal {B}}_n$. The family ${\mathcal {B}}$ of closed balls is countable since every ${\mathcal {B}}_n$ is finite and we are assuming CC. Again, from CC we can choose one center $x_n$ from every ball in ${\mathcal {B}}$. We claim that $\left\{ x_n\right\} _{n=1}^\infty$ is dense in X.

Indeed, let us fix $y\in X$, $\varepsilon >0$ and let $N \in {\mathbb {N}}$ be such that $y \in {\overline{B}}(x, N)$ and $2^{-N} < \varepsilon$. From the construction there exists a subset $\left\{ x_{n_i}\right\} _{i=1}^M$ of $\left\{ x_n\right\} _{n=1}^\infty$ such that ${\overline{B}}(x,N) \subset \bigcup _{i=1}^M {\overline{B}}(x_{n_i}, 2^{-N})$. Hence, for some $j=1,\ldots ,M$ we have $y \in {\overline{B}}(x_{n_j}, 2^{-N}) \subset {\overline{B}}(x_{n_j}, \varepsilon )$ .

(iv) $\implies$ (iii): Let us fix $\delta > 0$. If a space is geometrically doubling, then by the assumption it is separable. Therefore, by Theorem 4.7 (true in ZF) there exists a maximal $\delta$-separated set.

(iii) $\implies$ (ii): It follows by the Coifmann-Weiss Theorem.

(ii) $\implies$ (i): Let $\left\{ A_i\right\} _{i \in {\mathbb {Z}}}$ be an arbitrary countable family of nonempty pairwise disjoint sets and let $X = \bigcup _{i \in {\mathbb {Z}}} A_i$. We define pseudometric on X in the following manner: $d(x,y) = |i-j|$ if and only if $x \in A_i$ and $y \in A_j$. Then, $\tau (X) = {\mathcal {B}}(X) = \left\{ \bigcup _{i \in I} A_i :I \subset {\mathbb {Z}}\right\} \cup \left\{ \emptyset \right\}$. We define measure $\mu$ on ${\mathcal {B}}(X)$ as follows: $\mu \left( {A_i}\right) = 1$ for all $i \in {\mathbb {Z}}$. Obviously, the measure of every ball is positive and finite. Moreover, $\mu$ is doubling since for every $r > 0$ and $x \in X$ we have

$$\begin{aligned} \frac{\mu \left( {{\overline{B}}\left( {x, 2r}\right) }\right) }{\mu \left( {{\overline{B}}\left( {x,r}\right) }\right) } = \frac{2{\lfloor 2r \rfloor }+1}{2{\lfloor r \rfloor }+1} \le 3. \end{aligned}$$

Therefore, for every $r>0$ we obtain $\mu \left( {{\overline{B}}_{2r}}\right) \le 3\mu \left( {{\overline{B}}_{r}}\right)$. Hence, from the assumption there exists a maximal 1/2-separated set in X. The rest of the proof goes in the same way as the proof of Proposition 4.2. $\square$