We start with studying the space (clos(X),{\rho}_{cl}).
Fix a point \theta \in X and set {R}^{\ast}\doteq {sup}_{x\in X}{\varrho}_{X}(\theta ,x). We are interested, above all, in unbounded metric spaces, i.e., in the situation when {R}^{\ast}=\mathrm{\infty}, but all the results obtained from now on hold also in the case of {R}^{\ast}<\mathrm{\infty}. Set
P\doteq \{\begin{array}{cc}[0,{R}^{\ast}],\hfill & \text{if}{R}^{\ast}\mathrm{\infty}\text{and}\mathrm{\exists}x\in X:{\varrho}_{X}(\theta ,x)={R}^{\ast},\hfill \\ [0,{R}^{\ast}),\hfill & \text{otherwise},\hfill \end{array}
and define the function D:P\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+} in the following way:
D(p,r)\doteq \{\begin{array}{cc}0,\hfill & \text{if}r\u2a7dp,\hfill \\ dist(\overline{{O}_{p}^{o}},\overline{{O}_{r}^{o}}),\hfill & \text{if}rp,r\in P,\hfill \\ {sup}_{x,u\in \overline{{O}_{p}^{o}}}{\varrho}_{X}(x,u),\hfill & \text{if}rp,r\notin P\hfill \end{array}
(the latter occurs only if {R}^{\ast}<\mathrm{\infty}).
We will need the following properties of this function: for every p\in P, the function D(p,\cdot ):{\mathbb{R}}_{+}\to {\mathbb{R}}_{+} is nondecreasing; for every r\in {\mathbb{R}}_{+}, the function D(\cdot ,r):P\to {\mathbb{R}}_{+} is nonincreasing.
Show that D(p,\cdot ) is nondecreasing. Let p<{r}_{1}<{r}_{2}. If {r}_{1},{r}_{2}\in P, then from \overline{{O}_{{r}_{2}}^{o}}\subset \overline{{O}_{{r}_{1}}^{o}}\subset \overline{{O}_{p}^{o}} it follows that D(p,r)=dist(\overline{{O}_{p}^{o}},\overline{{O}_{{r}_{1}}^{o}})\u2a7ddist(\overline{{O}_{p}^{o}},\overline{{O}_{{r}_{2}}^{o}})=D(p,{r}_{2}). In the case when {r}_{1}\in P, {r}_{2}\notin P (this situation appears only for {R}^{\ast}<\mathrm{\infty}), one gets
D(p,r)=dist(\overline{{O}_{p}^{o}},\overline{{O}_{{r}_{1}}^{o}})\u2a7d\underset{x\in \overline{{O}_{p}^{o}},\phantom{\rule{0.25em}{0ex}}u\in \overline{{O}_{{r}_{1}}^{o}}}{sup}{\varrho}_{X}(x,u)\u2a7d\underset{x,u\in \overline{{O}_{p}^{o}}}{sup}{\varrho}_{X}(x,u)=D(p,{r}_{2}).
Finally, if {r}_{1},{r}_{2}\notin P, {R}^{\ast}<\mathrm{\infty}, then D(p,{r}_{1})=D(p,{r}_{2}).
Let us illustrate the definition of the function D on some concrete metric spaces.
Example 1 For any linear normed space X with any choice of \theta \in X, the function D:{\mathbb{R}}_{+}^{2}\to {\mathbb{R}}_{+} is given by D(p,r)=rp, r>p.
If X=\mathbb{Z}, {\varrho}_{\mathbb{Z}}(n,m)=nm \mathrm{\forall}n,m\in \mathbb{Z}, then no matter what point is chosen as \theta \in \mathbb{Z}, for any r,p\in {\mathbb{R}}_{+}, r>p, the value of D(p,r) is equal to the number of integers belonging to the interval [p,r).
One gets the same values for D:{\mathbb{R}}_{+}^{2}\to {\mathbb{R}}_{+} if X=\mathbb{N}, {\varrho}_{\mathbb{N}}(n,m)=nm \mathrm{\forall}n,m\in \mathbb{N}, and \theta =1. If \theta \ne 1, the value D(p,r) coincides with the number of integers in the set [p,r) when \theta 1\u2a7er>p or r>p>\theta 1, and is equal to the number of integers in the set [1,\theta +r) when r>\theta 1\u2a7ep.
Next, consider X=[1,2], \theta =0, {\varrho}_{X}(x,u)=xu. Then {R}^{\ast}=2, P=[0,2], and for r>p, the function D:P\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+} is defined as follows:

(1)
if 0\u2a7dp\u2a7d1, then
D(p,r)=\{\begin{array}{cc}rp,\hfill & r\u2a7d1,\hfill \\ r+1,\hfill & 1<r\u2a7d2,\hfill \\ 3,\hfill & r>2;\hfill \end{array}

(2)
if 1\u2a7dp\u2a7d2, then
D(p,r)=\{\begin{array}{cc}rp,\hfill & r\u2a7d2,\hfill \\ 2p,\hfill & r>2.\hfill \end{array}
For arbitrary closed F,G\subset X and r\u2a7e0, we determine now dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G). Next, we show that this distance is finite and also give the properties of the function {\mathbb{R}}_{+}\ni r\mapsto dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\in {\mathbb{R}}_{+}; these properties will be used in the sequel.
Lemma 1 Let F,G\in clos(X) be given. Then dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)<\mathrm{\infty} for every r\u2a7e0 and the following statements hold true:

(1)
the function {\mathbb{R}}_{+}\ni r\mapsto dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\in {\mathbb{R}}_{+} is nondecreasing;

(2)
for every r\u2a7e0, the inequality
dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7ddist(F,G)
(6)
and the relation
\underset{r\to \mathrm{\infty}}{lim}dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)=dist(F,G)
(7)
hold;

(3)
denoted {r}_{\ast}\doteq min\{\varrho (\theta ,F),\varrho (\theta ,G)\}, the inequality dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7dD({r}_{\ast},r) holds for each r\u2a7e0;

(4)
denoted {r}^{\ast}\doteq max\{\varrho (\theta ,F),\varrho (\theta ,G)\}, the inequality {\rho}^{o}(F,G)\u2a7ddist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G) holds for each r\u2a7e{r}^{\ast}.
Proof First of all, note that for any sets A,B,C\in clos(X), one has
dist(A\cup C,B\cup C)\u2a7ddist(A,B).
(8)
To show this inequality, we estimate the deviation d(A\cup C,B\cup C). Since
\varrho (x,B\cup C)\u2a7d\varrho (x,B)\u2a7dd(A,B)
for every x\in A and
\varrho (x,B\cup C)=0\u2a7dd(A,B)
for every x\in C, we get the inequality d(A\cup C,B\cup C)\u2a7dd(A,B). Similarly, it can be checked that d(B\cup C,A\cup C)\u2a7dd(A,B). So, (8) is proved.
From inequality (8), it follows that dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)<\mathrm{\infty} for every r\u2a7e0. Indeed, this is obvious if {R}^{\ast}<\mathrm{\infty}; in the case when {R}^{\ast}=\mathrm{\infty}, one has {\mathfrak{S}}_{r}F=(F\cap {O}_{r})\cup \overline{{O}_{r}^{o}}, {\mathfrak{S}}_{r}G=(G\cap {O}_{r})\cup \overline{{O}_{r}^{o}} and hence,
dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7ddist(F\cap {O}_{r},G\cap {O}_{r})<\mathrm{\infty}.
Using (8) one gets relation (6)
dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)=dist(F\cup \overline{{O}_{r}^{o}},G\cup \overline{{O}_{r}^{o}})\u2a7ddist(F,G).
Next, for {r}_{2}\u2a7e{r}_{1}, {\mathfrak{S}}_{{r}_{1}}F={\mathfrak{S}}_{{r}_{2}}F\cup \overline{{O}_{{r}_{1}}^{o}}, {\mathfrak{S}}_{{r}_{1}}G={\mathfrak{S}}_{{r}_{2}}G\cup \overline{{O}_{{r}_{1}}^{o}} hold. So, as a consequence of (8), the following holds:
dist({\mathfrak{S}}_{{r}_{1}}F,{\mathfrak{S}}_{{r}_{1}}G)\u2a7ddist({\mathfrak{S}}_{{r}_{2}}F,{\mathfrak{S}}_{{r}_{2}}G).
Thus, the function r\mapsto dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G) is nondecreasing.
Now, we prove equality (7). If {R}^{\ast}<\mathrm{\infty}, then {\mathfrak{S}}_{r}F=F, {\mathfrak{S}}_{r}G=G for r>{R}^{\ast}, so (7) is true.
Let {R}^{\ast}=\mathrm{\infty}. First, suppose that dist(F,G)<\mathrm{\infty}. Since the function r\mapsto dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G) is nondecreasing and bounded, there exists \alpha \doteq {lim}_{r\to \mathrm{\infty}}dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G). Then for each r, the inequality \alpha \u2a7edist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G) takes place, and it follows that for any \epsilon >0, every x\in F, and r>{\varrho}_{X}(\theta ,x)+\alpha +2\epsilon, one has x\in {O}_{\alpha +\epsilon}^{o}({\mathfrak{S}}_{r}G) and x\notin \overline{{O}_{r\epsilon \alpha}^{o}}. Thus, taking into account \overline{{O}_{r\epsilon \alpha}^{o}}\supseteq {O}_{\alpha +\epsilon}^{o}(\overline{{O}_{r}^{o}}), one gets x\notin {O}_{\alpha +\epsilon}^{o}(\overline{{O}_{r}^{o}}), i.e., x\in {O}_{\alpha +\epsilon}^{o}(G). As above, for each point y\in G, y\in {O}_{\alpha +\epsilon}^{o}(F) holds. Hence, dist(F,G)\u2a7d\alpha +\epsilon for any \epsilon >0, which means that dist(F,G)\u2a7d\alpha. The inequality dist(F,G)\u2a7e\alpha is easily obtained by passing to the limit in (6).
Now, let dist(F,G)=\mathrm{\infty}. If (7) is not true, then there exists \alpha \doteq {lim}_{r\to \mathrm{\infty}}dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)<\mathrm{\infty}. So, arguing as before, one gets the inequality dist(F,G)\u2a7d\alpha <\mathrm{\infty}, which contradicts the initial assumption. Thus, equality (7) is proved.
In order to prove statement (3), note that for every r\u2a7d{r}_{\ast}, {\mathfrak{S}}_{r}F={\mathfrak{S}}_{r}G=\overline{{O}_{r}^{o}} holds and hence dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)=0. If r>{r}_{\ast}, r\in P, then, according to the inclusions
{\mathfrak{S}}_{r}F\subset \overline{{O}_{{r}_{\ast}}^{o}},\phantom{\rule{2em}{0ex}}\overline{{O}_{r}^{o}}\subset {\mathfrak{S}}_{r}G,
one gets
d({\mathfrak{S}}_{r}G,{\mathfrak{S}}_{r}F)\u2a7dd(\overline{{O}_{r}^{o}},\overline{{O}_{{r}_{\ast}}^{o}})\u2a7dD({r}_{\ast},r).
Similarly, it follows that d({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7dD({r}_{\ast},r). So, dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7dD({r}_{\ast},r). For every r\notin P, {\mathfrak{S}}_{r}G\subset \overline{{O}_{{r}_{\ast}}^{o}}, {\mathfrak{S}}_{r}F\subset \overline{{O}_{{r}_{\ast}}^{o}} holds and therefore,
dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)\u2a7d\underset{x,u\in \overline{{O}_{{r}_{\ast}}^{o}}}{sup}{\varrho}_{X}(x,u)=D({r}_{\ast},r).
Property (4) is a consequence of the wellknown estimate (see, e.g., [6])
\varrho (x,A)\varrho (x,B)\u2a7ddist(A,B),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in X,\phantom{\rule{0.25em}{0ex}}\mathrm{\forall}A,B\in clos(X).
(9)
Indeed, assuming A={\mathfrak{S}}_{r}F, B={\mathfrak{S}}_{r}G and taking into account that for every r\u2a7e{r}^{\ast}, {\rho}^{o}(F,G)={\rho}^{o}({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G) holds, one gets
{\rho}^{o}(F,G)=\varrho (\theta ,{\mathfrak{S}}_{r}F)\varrho (\theta ,{\mathfrak{S}}_{r}G)\u2a7ddist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G),\phantom{\rule{1em}{0ex}}\mathrm{\forall}r\u2a7e{r}^{\ast}.
□
Theorem 1 Equality (5) defines a metric in the space clos(X). If X is a complete metric space, then (clos(X),{\rho}_{cl}) is also complete.
Proof Show that {\rho}_{cl} satisfies all the axioms of a metric, i.e., that for any F,G,H\in clos(X), the following hold:

(1)
0\u2a7d{\rho}_{cl}(F,G)<\mathrm{\infty}, and {\rho}_{cl}(F,G)=0 if and only if F=G;

(2)
{\rho}_{cl}(F,G)={\rho}_{cl}(G,F);

(3)
{\rho}_{cl}(F,G)\u2a7d{\rho}_{cl}(F,H)+{\rho}_{cl}(H,G).
It is obvious that {\rho}_{cl}(F,G)\u2a7e0 and {\rho}^{o}(F,G)<\mathrm{\infty} for any F,G\in clos(X). According to Lemma 1, one has
\begin{array}{rcl}{\rho}^{\mathfrak{S}}(F,G)& \u2a7d& \underset{r>0}{sup}min\{D({r}_{\ast},r),\frac{1}{r}\}\u2a7dmax\{\underset{r\in [0,1]}{sup}D({r}_{\ast},r),\underset{r>1}{sup}\frac{1}{r}\}\\ =& max\{D({r}_{\ast},1),1\}\u2a7dmax\{D(0,1),1\}.\end{array}
Therefore, {\rho}_{cl}(F,G)<\mathrm{\infty}. Next,
\begin{array}{rcl}{\rho}_{cl}(F,G)=0\phantom{\rule{1em}{0ex}}& \iff & \phantom{\rule{1em}{0ex}}\{\begin{array}{c}{\rho}^{o}(F,G)=0,\hfill \\ dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}G)=0\phantom{\rule{1em}{0ex}}\mathrm{\forall}r>0\hfill \end{array}\\ \iff & \phantom{\rule{1em}{0ex}}\{F\cap {O}_{r}^{o}=G\cap {O}_{r}^{o}\phantom{\rule{0.25em}{0ex}}\mathrm{\forall}r>0\}\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}F=G.\end{array}
The symmetry of {\rho}_{cl} is straightforward; property (3) follows from the inequalities
Thus, (clos(X),{\rho}_{cl}) is a metric space.
Now, let X be complete; show that (clos(X),{\rho}_{cl}) is also a complete metric space.
Consider a fundamental sequence {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}}\subset clos(X). For any \epsilon >0, there is a number N(\epsilon ) such that for all i,j>N(\epsilon ), the following inequality takes place:
{\rho}_{cl}({F}^{i},{F}^{j})=\varrho (\theta ,{F}^{i})\varrho (\theta ,{F}^{j})+\underset{r>0}{sup}min\{dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}{F}^{j}),\frac{1}{r}\}\u2a7d\epsilon .
(10)
From (10), it follows that the number sequence {\{\varrho (\theta ,{F}^{i})\}}_{i=1}^{\mathrm{\infty}} is fundamental, so it is convergent; denote its limit by {r}_{0}; obviously, {r}_{0}\u2a7e0.
Take an arbitrary radius r>{r}_{0} and consider the sequence {\{{\mathfrak{S}}_{r}{F}^{i}\}}_{i=1}^{\mathrm{\infty}}\subset clos(X). It follows from (10) that for any \epsilon \in (0,1/r) and all i,j>N(\epsilon ), the inequality dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}{F}^{j})<\epsilon takes place. Show that for the sequence {\{{\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1}\}}_{i=1}^{\mathrm{\infty}}\subset clbd(X) and all i,j>N(\epsilon ), the similar inequality holds:
dist({\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1},{\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1})<\epsilon .
(11)
Without loss of generality, assume \epsilon <1. Pick an arbitrary x\in {\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1}. If x\in {O}_{r+1}\mathrm{\setminus}{O}_{r}^{o}, then \varrho (x,{\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1})=0 since {O}_{r+1}\mathrm{\setminus}{O}_{r}^{o}\subset {\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1}. In the case when x\in {\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r}, there exists a point y\in {\mathfrak{S}}_{r}{F}^{j} satisfying the estimate {\varrho}_{X}(x,y)<\epsilon. Held y\in \overline{{O}_{r+1}^{o}}, one gets {\varrho}_{X}(x,y)\u2a7e1, therefore y\in {\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1}. So,
d({\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1},{\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1})<\epsilon ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}i,j>N(\epsilon ).
Similarly, d({\mathfrak{S}}_{r}{F}^{j}\cap {O}_{r+1},{\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1})<\epsilon and relation (11) is proved.
The sequence {\{{\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1}\}}_{i=1}^{\mathrm{\infty}}\subset clbd(X) is fundamental with respect to the metric dist and since the space (clbd(X),dist) is complete (see [7]), this sequence converges to some ‘limit’ set {\mathfrak{F}}_{r}\in clbd(X). According to inequality (8), the following holds:
dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}{\mathfrak{F}}_{r})=dist(({\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1})\cup \overline{{O}_{r}^{o}},{\mathfrak{F}}_{r}\cup \overline{{O}_{r}^{o}})\u2a7ddist({\mathfrak{S}}_{r}{F}^{i}\cap {O}_{r+1},{\mathfrak{F}}_{r}),
and therefore {\mathfrak{S}}_{r}{F}^{i}\to {\mathfrak{S}}_{r}{\mathfrak{F}}_{r}. Moreover, \varrho (\theta ,{\mathfrak{F}}_{r})=\varrho (\theta ,{\mathfrak{S}}_{r}{\mathfrak{F}}_{r})={r}_{0} (this equality follows from statement (4) of Lemma 1).
For every r>{r}_{0}, the set {F}_{r}\doteq {\mathfrak{S}}_{r}{\mathfrak{F}}_{r}\cap {O}_{r}^{o} is not empty for \varrho (\theta ,{\mathfrak{S}}_{r}{\mathfrak{F}}_{r})={r}_{0}. Prove that for any \overline{r}>r, {\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}\cap {O}_{r}^{o}={F}_{r} holds. Let x\in {\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}\cap {O}_{r}^{o}. Then there exists a sequence {\{{x}_{i}\}}_{i=1}^{\mathrm{\infty}}, {x}_{i}\in {\mathfrak{S}}_{\overline{r}}{F}^{i}\subset {\mathfrak{S}}_{r}{F}^{i}, convergent to x. This means that x\in {\mathfrak{S}}_{r}{\mathfrak{F}}_{r}; moreover, x\in {O}_{r}^{o}. Conversely, let x\in {\mathfrak{S}}_{r}{\mathfrak{F}}_{r}\cap {O}_{r}^{o}, then x\in {\mathfrak{F}}_{r}, x\notin \overline{{O}_{r}^{o}}, and hence there is a sequence {\{{x}_{i}\}}_{i=1}^{\mathrm{\infty}}, {x}_{i}\in {\mathfrak{S}}_{r}{F}^{i}, {x}_{i}\to x. Starting with some index I, all the members of this sequence satisfy the condition {x}_{i}\notin \overline{{O}_{r}^{o}}, and therefore {x}_{i}\in {F}^{i}\subset {\mathfrak{S}}_{\overline{r}}{F}^{i} \mathrm{\forall}i>I. Since {\mathfrak{S}}_{\overline{r}}{F}^{i} converges to {\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}, it follows that x\in {\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}. So, x\in {\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}\cap {O}_{p}^{o}.
From what has been proved, one may conclude that as r increases, the sets {F}_{r} ‘expand’ in the following way: if \overline{r}\u2a7er, then {F}_{r}={\mathfrak{S}}_{r}{\mathfrak{F}}_{r}\cap {O}_{r}^{o}={\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}\cap {O}_{r}^{o}={\mathfrak{S}}_{\overline{r}}{\mathfrak{F}}_{\overline{r}}\cap {O}_{\overline{r}}^{o}\cap {O}_{r}^{o}={F}_{\overline{r}}\cap {O}_{r}^{o} (i.e., {F}_{r} is a subset of {F}_{\overline{r}} containing elements x such that {\varrho}_{X}(\theta ,x)<r.) Define now the set F\doteq {\bigcup}_{r>{r}_{0}}{F}_{r} and show it is closed. Note that for every r, F\cap {O}_{r}^{o}={F}_{r} holds and the set {F}_{r}\cup \overline{{O}_{r}^{o}}={\mathfrak{S}}_{r}F is closed. Consider a sequence {\{{y}_{i}\}}_{i=1}^{\mathrm{\infty}}\subset F convergent to y. This sequence is bounded, so starting with some index i, the inclusions {y}_{i}\in {O}_{p}^{o}, {y}_{i}\in {\mathfrak{S}}_{p}F, where p={\varrho}_{X}(\theta ,y)+1, hold. Thus, since {\mathfrak{S}}_{p}F is closed, it follows that y\in {\mathfrak{S}}_{p}F; on the other hand, y\notin \overline{{O}_{r}^{o}}. So, y\in {F}_{p}\subset F and F is closed.
Finally, show that the sequence {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}} converges to the set F with respect to the metric {\rho}_{cl}. For an arbitrary \epsilon >0, put \overline{r}(\epsilon )=2/\epsilon. Since the sequence {\{{\mathfrak{S}}_{\overline{r}(\epsilon )}{F}^{i}\}}_{i=1}^{\mathrm{\infty}} converges (in the Hausdorff metric) to the set {\mathfrak{S}}_{\overline{r}(\epsilon )}{\mathfrak{F}}_{\overline{r}(\epsilon )}={\mathfrak{S}}_{\overline{r}(\epsilon )}F, for every i, starting with some I, the following inequalities hold:
dist({\mathfrak{S}}_{\overline{r}(\epsilon )}{F}^{i},{\mathfrak{S}}_{\overline{r}(\epsilon )}F)<\epsilon /2,\phantom{\rule{2em}{0ex}}{\rho}^{o}({F}^{i},F)<\epsilon /2.
(12)
According to Lemma 1, the function r\mapsto dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F) is nondecreasing. Hence, from (12), it follows that dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)<\epsilon /2 for every r\u2a7d\overline{r}(\epsilon ). Next, for r>\overline{r}(\epsilon ), one has 1/r<\epsilon /2. Thus, {\rho}^{\mathfrak{S}}({F}^{i},F)\u2a7d\epsilon /2 for all i>I, and therefore {\rho}_{cl}({F}^{i},F)<\epsilon, i.e., {lim}_{i\to \mathrm{\infty}}{\rho}_{cl}({F}^{i},F)=0. □
Remark 1 The function {\rho}^{\mathfrak{S}}:clos(X)\times clos(X)\to {\mathbb{R}}_{+}, given by (4), also defines a metric in the space clos(X), but for a complete metric space X, the space (clos(X),{\rho}^{\mathfrak{S}}), unlike (clos(X),{\rho}_{cl}), may not be complete. For example, the sequence of the sets {F}^{i}\doteq \{i\} (each consisting of a natural number) in the space (clos(\mathbb{R}),{\rho}^{\mathfrak{S}}) is fundamental [4]:
\mathrm{\forall}j>i\phantom{\rule{1em}{0ex}}{\rho}^{\mathfrak{S}}(\{i\},\{j\})=\frac{8}{\sqrt{{i}^{2}+4}+i},
yet there is no nonempty set F\in clos(\mathbb{R}) such that
{\rho}^{\mathfrak{S}}({F}^{i},F)\to 0.
(13)
Let us show this. If such a set F\subset \mathbb{R} does exist, then taking any of its points {x}_{0}, one gets the following estimates: for every i\u2a7e{x}_{0}+1 and each r\in [{x}_{0},i],
dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\u2a7er{x}_{0}.
Thus, {\rho}^{\mathfrak{S}}({F}^{i},F)\u2a7e1/\overline{r}, where \overline{r} is a solution of the equation 1/r=r{x}_{0}. So, the inequality
{\rho}^{\mathfrak{S}}({F}^{i},F)\u2a7e\frac{2}{{x}_{0}+\sqrt{{x}_{0}^{2}+4}}
holds, which contradicts relation (13).
The Hausdorff distance between sets A,B\in clos(X) has a simple geometric interpretation:
dist(A,B)=inf\{\epsilon :\epsilon >0,A\subset {O}_{\epsilon}^{o}(B),B\subset {O}_{\epsilon}^{o}(A)\}.
Regarding the metric {\rho}_{cl}, the analogue of the εneighborhood of a set A is given by
{O}_{\epsilon}^{o}(A,cl)\doteq \bigcup _{{x}_{0}\in {\mathfrak{A}}_{\epsilon}}\{x\in X:{\varrho}_{X}(x,{x}_{0})<\epsilon \},\phantom{\rule{1em}{0ex}}{\mathfrak{A}}_{\epsilon}=A\cup \overline{{O}_{1/\epsilon}^{o}}
(i.e., by the εneighborhood of the set {\mathfrak{A}}_{\epsilon}). This means that the ‘main part’ {\rho}^{\mathfrak{S}} of the distance {\rho}_{cl} can be defined by the equality
{\rho}^{\mathfrak{S}}(A,B)=inf\{\epsilon :\epsilon >0,A\subset {O}_{\epsilon}^{o}(B,cl),B\subset {O}_{\epsilon}^{o}(A,cl)\}.
Let us now give a criterion for a sequence of closed sets to be convergent with respect to the metric {\rho}_{cl}.
Lemma 2 Given F,{F}_{i}\in clos(X), i=1,2,\dots , the convergence {\rho}_{cl}({F}_{i},F)\to 0 implies \varrho (\theta ,{F}^{i})\to \varrho (\theta ,F) and dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\to 0 for any r>0.
Conversely, let {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}}\subset clos(X). If there exist {\varrho}_{0}\doteq {lim}_{i\to \mathrm{\infty}}\varrho (\theta ,{F}^{i}) and {r}_{0}\u2a7e0 such that for every r>{r}_{0}, the sequence {\{{\mathfrak{S}}_{r}{F}^{i}\}}_{i=1}^{\mathrm{\infty}} converges in the metric dist to some set {F}_{r}\in clos(X), then for any r, \overline{r} such that \overline{r}>r>{r}_{0}, {F}_{\overline{r}}\cap {O}_{\overline{r}}^{o}\supset {F}_{r}\cap {O}_{r}^{o} holds, and in the space clos(X) the sequence {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}} converges (with respect to {\rho}_{cl}) to the set F\doteq {\bigcup}_{r>{r}_{0}}({F}_{r}\cap {O}_{r}^{o})\in clos(X); moreover, \varrho (\theta ,F)={\varrho}_{0}.
We omit the proof of this statement since it repeats that of the corresponding result for the finite dimensional case [4].
We conclude this section by proving a theorem on the connection between convergence in the metric {\rho}_{cl} and convergence in the metric dist. Note that if a sequence {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}}\subset clos(X) converges in the Hausdorff metric, then by statement (2) of Lemma 1, it converges (to the same limit) in the metric {\rho}^{\mathfrak{S}}; moreover, as it will be shown later, in the metric {\rho}_{cl}. The converse is not true. For example, in ℝ the sequence of sets {F}^{i}\doteq \{0,i\}, i=1,2,\dots , converges in the metric {\rho}_{cl} to the set F\doteq \{0\}. On the other hand, for any i, j, one has dist({F}^{i},{F}^{j})=ij, i.e., the sequence {\{{F}^{i}\}}_{i=1}^{\mathrm{\infty}} is divergent in the Hausdorff metric.
Theorem 2 Let F,{F}^{i}\in clos(X), i\in \mathbb{N}. Then the following statements hold:

(1)
if dist({F}^{i},F)\to 0, then {\rho}_{cl}({F}^{i},F)\to 0;

(2)
if {\rho}_{cl}({F}^{i},F)\to 0 and there is a p>0 such that {F}^{i}\subset {O}_{p}, i\in \mathbb{N}, then F\in {O}_{p} and dist({F}^{i},F)\to 0.
Proof (1) Suppose dist({F}^{i},F)\to 0, i\to \mathrm{\infty}. Then, for an arbitrary \epsilon >0, there is a number I=I(\epsilon ) such that dist({F}^{i},F)<\epsilon for every i>I. So, according to inequality (9), one gets \varrho (\theta ,{F}^{i})\varrho (\theta ,F)<\epsilon, and from (6) it follows that the inequality dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)<\epsilon holds for every r\u2a7e0. Thus, by Lemma 2, {\rho}_{cl}({F}^{i},F)\to 0.
(2) Now, let {\rho}_{cl}({F}^{i},F)\to 0. Then from Lemma 2, it follows that dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\to 0 for every r>0. Next, since for every i=1,2,\dots , {F}^{i}\subset {O}_{p} holds, then F\subset {O}_{p}. Indeed, if this inclusion fails, then there exists an f\in F such that \delta \doteq \varrho (f,{O}_{p})>0, so one gets the estimate
\begin{array}{rcl}dist({\mathfrak{S}}_{p+2\delta}F,{\mathfrak{S}}_{p+2\delta}{F}^{i})& \u2a7e& \varrho (f,{\mathfrak{S}}_{p+2\delta}{F}^{i})=\varrho (f,{F}^{i}\cup \overline{{O}_{p+2\delta}^{o}})\\ =& min\{\varrho (f,{F}^{i}),\varrho (f,\overline{{O}_{p+2\delta}^{o}})\}\u2a7emin\{\varrho (f,{O}_{p}),\delta \}=\delta \end{array}
that contradicts the relation dist({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}{F}^{i})\to 0. Hence F\subset {O}_{p}.
Let r\u2a7e3p. Since {F}^{i}\subset {\mathfrak{S}}_{r}{F}^{i}, for every i one has
d({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\u2a7ed({F}^{i},{\mathfrak{S}}_{r}F)=d({F}^{i},F\cup \overline{{O}_{r}^{o}}).
Next, for every f\in {F}^{i}, i\in \mathbb{N}, we have
\varrho (f,F)\u2a7d2p,\phantom{\rule{2em}{0ex}}\varrho (f,\overline{{O}_{r}^{o}})\u2a7e2p,
and therefore \varrho (f,F)\u2a7d\varrho (f,\overline{{O}_{r}^{o}}), i.e., \varrho (f,F)=\varrho (f,F\cup \overline{{O}_{r}^{o}}). Thus, d({F}^{i},F\cup \overline{{O}_{r}^{o}})=d({F}^{i},F) and
d({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\u2a7ed({F}^{i},F).
Similarly, d({\mathfrak{S}}_{r}F,{\mathfrak{S}}_{r}{F}^{i})\u2a7ed(F,{F}^{i}). So, it is proved that
dist({\mathfrak{S}}_{r}{F}^{i},{\mathfrak{S}}_{r}F)\u2a7edist({F}^{i},F),
hence dist({F}^{i},F)\to 0. □