Proof. For each point $x \in X$ we can define a (unique) sequence $\hat {x}$ of elements of D (the countable dense set) by letting $\hat {x}_n$ be the first element of D which is in $B(x,1/(n+1))$ . It is clear that the function $x \mapsto \hat {x}$ is one to one so $|X| \le |D^{\aleph _0}| \le |\aleph _0^{\aleph _0}| = |\mathbb {R}|$ .

Proof. We give an outline of the proof for the case that $\boldsymbol {\Psi }$ is $\mathsf{PSE}$ and leave the remaining similar proofs for the interested reader.

Assume $\mathsf{PSE}(\delta _1)$ and let S be a strictly $\delta _2$ -separated set in the pseudometric space $(X,p)$ . Define a metric $p'$ on X by $p'(x,y) = \frac { \delta _1}{ \delta _2}p(x,y) $ . Then $(X,p')$ is a pseudometric space and S is a strictly $\delta _1$ -separated set in $(X,p')$ . Using $\mathsf{PSE}(\delta _1)$ there is a maximal strictly $\delta _1$ -separated set $S'$ in $(X,p')$ such that $S \subset S'$ and therefore $S'$ is a maximal strictly $\delta _2$ -separated set in $(X,p)$ .

Proof. Let us notice that the implications $\boldsymbol {\Phi }_1 \implies \boldsymbol {\Psi }_1$ and $\boldsymbol {\Phi }_2 \implies \boldsymbol {\Psi }_2$ are clear. We shall prove only the implication CC $+ \boldsymbol {\Psi }_1 \implies \boldsymbol {\Phi }_1$ since CC( $\mathbb {R}$ ) $+ \boldsymbol {\Psi }_2 \implies \boldsymbol {\Phi }_2$ follows in the similar manner using the fact that all separable metric spaces can be bijectively embedded in $\mathbb {R}$ (see Proposition 2.1). We prove the implication CC $+ \boldsymbol {\Psi }_1 \implies \boldsymbol {\Phi }_1$ for pair $(\boldsymbol {\Psi }_1, \boldsymbol {\Phi }_1) = (\mathsf{PN}, \mathsf{PNE})$ since the remaining case follows in the similar way.

Let us assume CC and PN hold. Let $(X, d)$ be a pseudometric separable space, $\delta> 0$ and $S \subset X$ be an arbitrary non-strictly $\delta $ -separated set. We define $N_\delta (S) = \bigcup _{x \in S} B\left ({x, \delta }\right )$ . Let us consider pseudometric space $\left ({X \setminus N_\delta (S), d}\right )$ . Under CC separability is hereditaryFootnote ³ (see [Reference Bentley and Herrlich1, Theorem 1.12]) so $X \setminus N_\delta (S)$ is separable. Thus by virtue of PN space $X \setminus N_\delta (S)$ contains a maximal non-strictly $\delta $ -separable set $S_0$ . We claim that $S' := S \cup S_0$ is the maximal non-strictly $\delta $ -separated set in X containing S.

Indeed, let $x, y \in S'$ , $x \neq y$ . If $x, y \in S$ or $x, y \in S_0$ , then $d(x, y) \ge \delta $ . If $x \in S$ and $y \in S_0$ , then $y \notin N_\delta (S)$ so $y \notin B(x, \delta )$ . It proves that $S'$ is non-strictly $\delta $ -separated set.

Let us suppose that $S'$ is not a maximal non-strictly $\delta $ -separated set, i.e., there exists $x_0 \notin S'$ such that $d(x_0, x) \ge \delta $ for all $x \in S'$ . In particular $d(x_0, x) \ge \delta $ for all $x \in S$ so $x_0 \notin N_\delta (S)$ . Since $x_0 \in X \setminus N_\delta (S)$ , $x_0 \notin S_0$ and $S_0$ is maximal non-strictly $\delta $ -separated set in $X \setminus N_\delta (S)$ , then $d(x_0, x) < \delta $ for some $x \in S_0$ and it gives a contradiction.

Proof. Let $(\boldsymbol {\Psi _1}, \boldsymbol {\Psi _2})$ be one of (PS, PSE), (MS, MSE), (PN, PNE), or (MN, MNE). By Remark 1.3 we have DC $\implies \boldsymbol {\Psi _1}$ . Thus by virtue of Proposition 2.3 we get $\mathsf{DC} \implies \mathsf{DC} + \boldsymbol {\Psi _1} \implies \mathsf{CC} + \boldsymbol {\Psi _1} \implies \boldsymbol {\Psi _2}$ .

Proof. By Theorem 2.2, we only have to prove the theorem for $\delta = 1$ . As in the proof of Theorem 2.4 this proof depends on ideas from [Reference Dybowski and Górka2]. Let $(X,p)$ be a separable pseudometric space with countable dense subset $D = \{d_i : i \in \mathbb {Z}^{+} \}$ and assume that $(X,p)$ has no finite maximal strictly $1$ -separated set.

Proof. The set $\Omega $ is open and by our assumption non-empty. Since D is dense in $(X,p)$ , there is an $i \in \mathbb {Z}^{+}$ such that $d_i \in \Omega $ . This completes the proof of the lemma.

Proof. By Theorem 2.2, it suffices to prove $\mathsf{PSE}(1) \implies \mathsf{CC}$ .

Assume that $\mathcal {Y} = \{Y_n : n \in \omega \}$ is a countable pairwise disjoint family of non-empty sets (and that the mapping $n \mapsto Y_n$ is one to one). Our plan is to apply $\mathsf{PSE}(1)$ to a certain separable pseudometric space $(X,d)$ and a strictly $1$ -separated set in $(X,d)$ to get a maximal strictly $1$ -separated set. From this we will be able to obtain a choice function for $\mathcal {Y}$ .

To construct X we assume without loss of generality that

(4) $$ \begin{align} \left(\mathbb{R}^{(\mathbb{Z}^+)} \times \omega\right) \cap \left(\bigcup \mathcal{Y}\right) = \emptyset. \end{align} $$

For $i \in \mathbb {Z}^+$ we define two sequences $a_i = (a_i(k))_{k\in \mathbb {Z}^+}$ and $a_i' = ( a_i'(k))_{k \in \mathbb {Z}^+}$ (both in $\mathbb {R}^{(\mathbb {Z}^+)}$ ) by

$$ \begin{align*} a_i(k) = \begin{cases}\frac{i+1}{i} & \text{ if } i = k\\ 0 & \text{ otherwise} \end{cases} , \,\,\,\, a_i'(k) = \begin{cases} \frac{1}{i} & \text{ if }i = k \\ 0 & \text{ otherwise}. \end{cases} \end{align*} $$

Let $\mathbf {z} \in \mathbb {R}^{(\mathbb {Z}^+)}$ be the zero sequence defined by $\mathbf {z}(k) = 0$ for all $k \in \mathbb {Z}^+$ and define A and $A_E$ , respectively, by

$$ \begin{align*} A = \{ a_i : i \in \mathbb{Z}^+ \} \cup \{ a_i' : i \in \mathbb{Z}^+ \}, \,\,\,\, A_E = \{ a_i : i \in \mathbb{Z}^+ \} \cup \{ a_i' : i \in \mathbb{Z}^+ \} \cup \{ \mathbf{z} \}. \end{align*} $$

Let $d_E$ be the usual Euclidean metric on $A_E$ defined by

$$\begin{align*}d_E(b,c) = \sqrt{\sum_{k=1}^\infty (c(k)-b(k))^2}. \end{align*}$$

For each $n \in \omega $ , let $X_n = (A \times \{ n \}) \cup Y_n$ and let $X = \bigcup _{n \in \omega } X_n$ . Finally, define a metric d on X by

$$\begin{align*}d(t,s) = \begin{cases} 3 & \text{ if } \exists m,n \in \omega \text{ with } m \ne n \land t \in X_n \land s \in X_m \\ 0 & \text{ if } \exists n \in \omega \text{ such that }t,s \in Y_n \\ d_E(x,y) & \text{ if } \exists n \in \omega \hskip.05in \exists x, y \in A \text{ such that } t = (x,n) \land s = (y,n) \\ d_E(x,\mathbf{z}) & \text{ if } \exists n \in \omega \hskip.05in \exists x \in A \text{ such that } (t= (x,n) \land s \in Y_n ) \text{ or }\\ & \hskip.2in(s = (x,n) \land t \in Y_n). \end{cases}. \end{align*}$$

Our assumption that the elements of $\mathcal {Y}$ are pairwise disjoint and the assumption (4) ensure that d is well defined. Furthermore, one can easily convince oneself that $(X,d)$ is a pseudometric space.

We now list some properties of $(X,d)$ which will be useful in proving Lemmas 3.5 through 3.7. The proofs are straightforward.

Proof. Since A is countable, $D = A \times \omega $ is countable. Furthermore, since $X = D \cup \left ( \bigcup _{n \in \omega } Y_n \right )$ , to show that D is dense in $(X,d)$ it suffices to prove that every $t \in \bigcup _{n \in \omega } Y_n$ is a limit point of D. By definition $\{ (a^{\prime }_i,n) : i \in \mathbb {Z}^+ \land n \in \omega \} \subseteq D$ . Therefore part (1) of Lemma 3.4 implies that t is a limit point of D.

Proof. Assume $(a_i,n)$ and $(a_j,m)$ are distinct elements in S. If $n \ne m$ then by the first clause of the definition of d, $d((a_i,n),(a_j,m)) = 3> 1$ . On the other hand, if $n = m$ then $i \ne j$ , so by part (2) of Lemma 3.4, $d((a_i,n),(a_j,m))> 1$ .

Proof. Assume $S'$ satisfies the hypotheses of the lemma and assume $n \in \omega $ . Clearly, $|S' \cap Y_n| \le 1$ . We prove $|S' \cap Y_n| \ge 1$ by contradiction. Suppose $S' \cap Y_n = \emptyset $ . Since $S'$ is maximal, for all $y \in Y_n$ there is a $t \in S' \setminus Y_n$ such that $d(t,y) \le 1$ . By part (4) of Lemma 3.4, $t = (a^{\prime }_i,n)$ for some $i \in \mathbb {Z}^+$ . But by part (3) in Lemma 3.4, $d((a^{\prime }_i,n), (a_i,n)) = 1$ . This is a contradiction since $(a_i,n) \in S \subseteq S'$ .

Proof. As in the proof of Theorem 3.3, it suffices to prove $\mathsf{PN}(1) \implies \mathsf{CC}$ . Assume $\mathsf{PN}(1)$ and let X be a countable family of non-empty, pairwise disjoint sets where the function $n \mapsto X_n$ is one to one from $\omega $ onto X. Also assume without loss of generality that $(\bigcup X) \cap \mathbb {R} = \emptyset $ . We will construct a choice function for X.

Proof of Lemma 4.2.

We begin by defining a certain subset $Y_0$ of the Euclidean plane $(\mathbb {R}^2, d)$ as follows. Let

$$ \begin{align*} A_0^0 &= \left\{ (x,y) \in \mathbb{R}^2: x^2 + (y-1)^2 = 1 \text{ and } 0 < x < \sqrt{3}/2 \text{ and } x \text{ is rational and } 0 < y < 1/2 \right\},\\ A_0^1 &= \left\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \text{ and } 0 < x < \sqrt{3}/2 \text{ and } x \text{ is rational and } 1/2 < y < 1 \right\}, \\ Y_0 &= A_0^0 \cup A_0^1. \end{align*} $$

Note that both $A_0^0$ and $A_0^1$ are countable. For every positive integer n we let $A_n^0 = \{ (a, b + 2n) : (a,b) \in A_0^0 \}$ , $A_n^1 = \{ (x, y + 2n) : (x,y) \in A_0^1 \}$ and $Y_n = A_n^0 \cup A_n^1$ . Since for every $(x,y) \in Y_0$ , we have $0 < y < 1$ we conclude that for every pair $(x,y) \in Y_n$ we have $2n < y < 2n + 1$ .

Sublemma 4.2.1. Let $n \in \omega $ , then for all $(a_1,b_1)$ and $(a_2, b_2)$ in $A_n^0$ and for all $(x_1,y_1)$ and $(x_2,y_2)$ in $A_n^1$ we have

(5) $$ \begin{align} d((a_1,b_1),(x_1, y_1)) < 1, & \end{align} $$

(6) $$ \begin{align} d((a_1,b_1), (a_2,b_2)) < 1 &\text{ and } d((x_1,y_1), (x_2,y_2)) < 1, \end{align} $$

(7) $$ \begin{align} d((a_1,b_1),(0,2n)) < 1 &\text{ and } d((x_1,y_1),(0,2n+1)) < 1, \end{align} $$

(8) $$ \begin{align} d(a_1,b_1), (0,2n+1)) = 1 &\text{ and } d((x_1,y_1),(0,2n)) = 1. \end{align} $$

Proof. It is enough to prove the Sublemma for $n=0$ . Parts (6)–(8) are clear from the picture. For part (5) an algebraic proof can be given by assuming that $(a_1,b_1) \in A_0^0$ and $(x_1,y_1) \in A_0^1$ and calculating the square of the distance from $(a_1,b_1)$ to $(x_1,y_1)$ .

Sublemma 4.2.2. Assume $z_1, z_2 \in A_n^0$ , $w_1, w_2 \in A_n^1$ , $u_1, u_2 \in X_n \times \{ 0 \}$ and $v_1, v_2 \in X_n \times \{ 1 \}$ , then

1. (1) $p(z_1, w_1) < 1$ ,

2. (2) $p(z_1, z_2) < 1$ and $p(w_1,w_2) < 1$ ,

3. (3) $p(z_1,u_1) < 1$ and $p(w_1,v_1) < 1$ ,

4. (4) $p(z_1, v_1) = 1$ and $p(w_1,u_1) = 1$ ,

5. (5) $p(u_1,v_1) = 1$ ,

6. (6) $p(u_1,u_2) = 0$ and $p(v_1,v_2) = 0$ .

Proof. Using the definitions of c and p, items (1) through (4) follow from equations (5) through (8) respectively and items (5) and (6) are clear from the definitions of c and p.

Proof. With the use of Proposition 2.3 statement (i) follows from Theorems 3.1 and 3.3 and statement (ii) follows from Theorem 4.1.

Proof. $\mathsf{MSE} \implies \mathsf{CC}(\mathbb {R})$ : Assume that $\mathcal {Y} = \left \{Y_n \colon n \in \omega \right \}$ is a countable family of pairwise disjoint non-empty subsets of $\mathbb {R}$ and that the mapping $n \mapsto Y_n$ is one-to-one. Since $\mathbb {R}$ is bijective with $(3n, 3n+1)$ for all $n \in \omega $ , we can assume that $Y_n \subset (3n, 3n+1)$ for all $n \in \omega $ . Let $Y = \bigcup \mathcal {Y}$ . Let $D_0$ be the set of dyadic numbers, i.e., $D_0 = \left \{m/ 2^n \colon \, m \in \mathbb {Z}, n \in \mathbb {Z}^+\right \}$ and $D_1 = D_0 + 1/3$ . Sets $D_0$ and $D_1$ are countable and dense in $\mathbb {R}$ and it is easy to check that $D_0 \cap D_1 = \emptyset $ . We can assumeFootnote ⁵ that $Y \cap \left ({D_0 \cup D_1}\right ) = \emptyset $ . Finally, we put $X = Y \cup D_0 \cup D_1$ and define metric d on X in the following manner:

$$\begin{align*}d(x, y) = \begin{cases} 0 & \text{if } x=y\\ 1 + |x - y| & \text{if } x\neq y \; \text{and} \; x,y \in D_1 \\ 1 + |y - x + 1/3| & \text{if } x \in D_1 \; \text{and} \; y \not\in D_1 \\ 1 + |x - y + 1/3| & \text{if } x \not\in D_1 \; \text{and} \; y \in D_1 \\ |x-y| & \text{otherwise}. \end{cases} \end{align*}$$

It is easy to check that d is indeed metric and $(X, d)$ is separable metric space with $D_0 \cup D_1$ as the countable and dense set in X. Since $D_1$ is strictly $1$ -separated set in X and $X \subset \mathbb {R}$ is separable, we can use the assumed statement so there exists a maximal strictly $1$ -separated set $S'$ in X such that $D_1 \subset S'$ . We shall show that $|Y_n \cap S'| = 1$ for all $n \in \omega $ .

Let us notice that for every $x \in D_0$ there exists $y \in D_1$ such that $d(x, y) = 1$ . Indeed, it is enough to take $y = x + 1/3 \in D_1$ and then $d(x, y) = 1 + |x - y + 1/3| = 1$ . Since $S'$ is strictly $1$ -separated set and $D_1 \subset S'$ , it means that $S' \cap D_0 = \emptyset $ .

Let us fix $n \in \omega $ . If $x, y \in Y_n \cap S' \subset Y_n \subset (3n, 3n + 1)$ , then $d(x, y) = |x - y| < 1$ . Thus $x = y$ since $x, y \in S'$ . Hence $|Y_n \cap S'| \le 1$ .

Let us suppose that $Y_n \cap S' = \emptyset $ for some $n \in \omega $ . Let us take $x \in Y_n$ and fix $y \in S'$ . Since $S' \cap D_0 = \emptyset $ , either $y \in D_1$ or $y \in Y$ . If $y \in D_1$ , then $y - 1/3 \in D_0$ and $Y_n \cap D_0 = \emptyset $ so $x \neq y - 1/3$ . Thus $d(x, y) = 1 + |x - y + 1/3|> 1$ . If $y \in Y$ , then $y \not \in Y_n$ since $Y_n \cap S' = \emptyset $ so $d(x, y)> 1$ . We have $d(x, y)> 1$ for all $y \in S'$ so it contradicts the maximality of $S'$ . Hence $Y_n \cap S' \neq \emptyset $ for all $n \in \omega $ .

$\mathsf{CC}(\mathbb {R}) \implies \mathsf{MSE}$ : By Proposition 2.3 we have that $\mathsf{CC}(\mathbb {R}) + \mathsf{MS} \implies \mathsf{MSE}$ but MS is true in ZFA by Theorem 3.1.

Proof. It follows immediately from Theorem 5.2.

Proof. Assume that $\mathcal {Y} = \left \{Y_n \colon n \in \omega \right \}$ is a countable family of pairwise disjoint non-empty subsets of $\mathbb {R}$ and that the mapping $n \mapsto Y_n$ is one-to-one. Let $S^1$ be unit circle in $\mathbb {R}^2$ , namely, $S^1 = \left \{(x, y) \in \mathbb {R}^2\colon \; x^2 + y^2 = 1\right \}$ . There exists a bijection $f\colon \mathbb {R} \to S^1$ . Let $\widehat {Y}_n = f(Y_n) + (4n, 0)$ for all $n \in \omega $ . Since f is bijective and $\mathbb {R}$ is linearly ordered, it is enough to prove that there exists set $S'$ such that $S' \cap \widehat {Y}_n$ is non-empty and finite for all $n \in \omega $ to complete the proof.

Let $D^1 = \left \{(x, y) \in \mathbb {Q}^2\colon \; x^2 + y^2 < 1\right \}$ and $D_n^1 = D^1 + (4n, 0)$ for all $n \in \omega $ . We consider metric space $(X, d)$ where $X = \bigcup _{n \in \omega } \left ({\widehat {Y}_n \cup D_n^1}\right )$ and d is the usual Euclidean metric on $\mathbb {R}^2$ . Obviously, X is separable with $\bigcup _{n \in \omega } D_n^1$ as a countable and dense set. Let $S = \left \{(4n, 0)\colon \; n \in \omega \right \}$ . It is non-strictly $1$ -separated set in X so by MNE there exists a maximal non-strictly $1$ -separated set $S'$ such that $S \subset S'$ .

Let us notice that for all $n \in \omega , x \in D_n^1$ , and $y \in \widehat {Y}_n$ we have $d\left ({(4n, 0), x}\right ) < 1$ and $d\left ({(4n, 0), y}\right ) = 1$ . Thus $S' \cap \bigcup _{n \in \omega } D_n^1 \setminus \{(4n,0)\} = \emptyset $ . Since $S'$ is maximal and $d(x, y) \ge 1$ for all $x \in D_n^1 \cup \widehat {Y}_n$ and $y \in D_m^1 \cup \widehat {Y}_m$ where $n \neq m$ , it is easy to see that $S' \cap \widehat {Y}_n \neq \emptyset $ for all $n \in \omega $ . Moreover for all $n \in \omega $ set $S' \cap \widehat {Y}_n$ is finite since every $1$ -separated set contained in $S^1 + (4n, 0)$ is finite.

Proof. It follows from Propositions 5.4 and 2.3.

Proof. For the implication “ $\Longleftarrow $ ” we note that $\mathsf{PN} \implies \mathsf{MN}$ is clear and from Theorem 4.1 we have $\mathsf{PN} \implies \mathsf{MN}$ .

For the other implication let $(X, p)$ be a separable pseudometric space and D be a countable and dense set. We convert this space into metric space in the standard way. We define an equivalence relation $\sim $ on X as follows: $x \sim y \Longleftrightarrow p(x,y) = 0$ for all $x, y \in X$ . Then $(X/{\sim }, d)$ is metric space where $d\left ({[x], [y]}\right ) = p(x,y)$ for all $x, y \in X$ and $[x]$ denotes the equivalence class of x. Obviously $D/{\sim }$ is a countable and dense set in $X/{\sim }$ so $(X/{\sim }, d)$ is separable.

Since we assumed MN, there exists a maximal non-strictly $1$ -separable set S in $(X/{\sim }, d)$ . Every non-strictly $1$ -separated in every separable metric space is at most countable so S is at most countable. Let $S = \left \{S_i\right \}_{i=1}^\infty $ . From CC there exists set $\widehat {S} = \left \{s_i\right \}_{i=1}^\infty \subset X$ such that $s_i \in S_i$ for every $i \in \mathbb {Z}^+$ . Then $[s_i] = S_i$ for every $i \in \mathbb {Z}^+$ and it is easy to see that $\widehat {S}$ is a maximal non-strictly $1$ -separated set in $(X, p)$ .

Proof. In [Reference Howard and Rubin3] a Fraenkel–Mostowski model is constructed in which CC is true and DC is false. By Remark 5.1 Part (1), MN is also true in this model. By Remark 5.1 Part (2), MN is injectively boundable. It follows from Theorem 2.5 that there is a model of ZF in which both CC and MN are true and DC is false. Therefore, by Theorem 5.6, in this model PN is true and DC is false.

Proof. First of all we shall prove the following lemma.

Proof. Let $\left \{x_i\right \}_{i=1}^\infty \subset X$ be a countable and dense subset of X. Let $F \in \mathcal {F}$ . We shall give the construction of $a_F \in F$ . We define sequence $\left \{n_k\right \}_{k=1}^\infty \subset \omega $ in the following manner:

• • $n_1 = \min \left \{i \in \mathbb {Z}^+\colon \; x_i \in \bigcup _{x \in F} B\left ({x, 1/2}\right )\right \}$ ;

• • $n_{k+1} = \min \left \{i \in \mathbb {Z}^+\colon \; x_i \in B\left ({x_{n_k}, 1/2^k}\right ) \cap \bigcup _{x \in F} B(x, 1/2^{k+1})\right \}$ for all $k \ge 1$ .

Obviously, set $\bigcup _{x \in F} B\left ({x, 1/2}\right )$ is open and non-empty so $n_1$ is well-defined. Let us assume that $n_k$ are defined for $k \le m$ . We will show that $n_{m+1}$ is well-defined. Since set $B\left ({x_{n_m}, 1/2^m}\right ) \cap \bigcup _{x \in F} B(x, 1/2^{m+1})$ is open, it is enough to show that it is non-empty. Let us notice that $x_{n_m} \in \bigcup _{x \in F} B\left ({x, 1/2^m}\right )$ so $B\left ({x_{n_m}, 1/2^m}\right ) \cap F \neq \emptyset $ . As a consequence $B\left ({x_{n_m}, 1/2^m}\right ) \cap \bigcup _{x \in F} B(x, 1/2^{m+1}) \neq \emptyset $ and it shows that $n_{m+1}$ is well-defined. The sequence $\left \{n_k\right \}_{k=1}^\infty $ has the following properties:

1. (1) $x_{n_{k+1}} \in B\left ({x_{n_k}, 1/2^{k}}\right )$ for all $k \ge 1$ ;

2. (2) $\text {dist}\left ({x_{n_k}, F}\right ) < 1/2^k$ for all $k \ge 1$ .

From $(1)$ we conclude that sequence $\left \{x_{n_k}\right \}_{k=1}^\infty $ is Cauchy sequence so it converges since X is complete. Let $\lim _{k\to \infty } x_{n_k} = x \in X$ . From $(2)$ , since $\text {dist}\left ({x, F}\right ) \le \text {dist}\left ({x_{n_k}, F}\right ) + d(x, x_{n_k})$ for every $k \ge 1$ , then $\text {dist}\left ({x, F}\right ) = 0$ . But F is closed so $x \in F$ . We put $a_F := x$ .

LetFootnote ⁶ $\left \{x_i\right \}_{i=1}^\infty \subset X$ be a countable and dense subset of X and let S be non-strictly $\delta $ -separated subset of X. By virtue of Lemma 5.9 let $\mathcal {F}$ be the family of all non-empty closed subsets of X and $\left \{a_F\right \}_{F \in \mathcal {F}} \subset X$ be a set such that $a_F \in F$ for all $F \in \mathcal {F}$ . Let us denote $N_\delta (S) = \bigcup _{x \in S} B\left ({x, \delta }\right )$ . If $N_\delta (S) = X$ , then S is maximal non-strictly $\delta $ -separated set so we assume $N_\delta (S) \neq X$ . Let $y \in X \setminus N_\delta (S)$ . We define sequences $\left \{y_n\right \}_{n=1}^\infty \subset X$ and $\left \{k_n\right \}_{n=1}^\infty \subset \omega $ as followsFootnote ⁷ :

• • $k_1 = 0, y_1 = y$ ;

• • $k_n = \min \left \{i> k_{n-1}\colon \; \overline {B}\left ({x_i, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{n-1} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right ) \neq \emptyset \right \}$ for all $n \ge 2$ ;

• • $y_n = a_F$ where $F = \overline {B}\left ({x_{k_n}, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{n-1} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right )$ for all $n \ge 2$ .

We shall show that $S' := S \cup \left \{y_n\right \}_{n=1}^\infty $ is a maximal non-strictly $\delta $ -separated set in X containing S. Let us fix $n \in \mathbb {Z}^+$ . Since $y_n \notin \left ({\bigcup _{j=1}^{n-1} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right )$ , then $d(y_n, y_j) \ge \delta $ for all $j < n$ and $d(y_n, s) \ge \delta $ for all $s \in S$ . But $n \in \mathbb {Z}^+$ was arbitrary so $d(y_n, y_j) \ge \delta $ for all $n, j \in \mathbb {Z}^+$ , $n \neq j$ . It proves that $S'$ is non-strictly $\delta $ -separated set.

Let us suppose that $S'$ is not maximal, i.e., there exists $x \in X$ such that $d(x, y_n) \ge \delta $ for all $n \in \mathbb {Z}^+$ and $d(x, s) \ge \delta $ for all $s \in S$ . There exists $i \in \mathbb {Z}^+$ such that $x \in \overline {B}\left ({x_i, \delta /3}\right )$ . Thus $x \in \overline {B}\left ({x_i, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{\infty } B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right )$ . Let us suppose that $i \neq k_n$ for any $n \in \mathbb {Z}^+$ . The sequence $\left \{k_n\right \}_{n=1}^\infty $ is increasing so there exists $n \in \mathbb {Z}^+$ such that $k_n < i < k_{n+1}$ . By the definition of $k_{n+1}$ we have $\overline {B}\left ({x_i, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{n} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right ) = \emptyset $ so $\overline {B}\left ({x_i, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{\infty } B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right ) = \emptyset $ which contradicts the fact that x is an element of this set. Hence $i = k_n$ for some $n \in \mathbb {Z}^+$ . Obviously $n \ge 2$ since $i \in \mathbb {Z}^+$ and $k_1 = 0 \notin \mathbb {Z}^+$ . It means that $x \in \overline {B}\left ({x_{k_n}, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{n-1} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right )$ . Since $y_n \in \overline {B}\left ({x_{k_n}, \delta /3}\right ) \setminus \left ({\bigcup _{j=1}^{n-1} B\left ({y_j, \delta }\right ) \cup N_\delta (S)}\right )$ either, we obtain $x, y_n \in \overline {B}\left ({x_{k_n}, \delta /3}\right )$ so $d(x, y_n) \le 2\delta /3$ . It gives a contradiction with the fact that $d(x, y_n) \ge \delta $ . Hence $S'$ is maximal non-strictly $\delta $ -separated set.