In this section, we shall study the cone metric spaces over solid vector spaces. The theory of such cone metric spaces is very close to the theory of the usual metric spaces. We show that every cone metric space over a solid vector space is a metrizable topological space. Every cone normed space over a solid vector space is normable.
9.1 Topological structure of cone metric spaces
Definition 9.1 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). For a point {x}_{0}\in X and a vector r\in Y with r\succ 0, the set
U({x}_{0},r)=\{x\in X:d(x,{x}_{0})\prec r\}
is called an open ball with center {x}_{0} and radius r.
Theorem 9.1 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Then the collection
\mathcal{B}=\{U(x,r):x\in X,r\in Y,r\succ 0\}
of all open balls in X is a basis for a topology {\tau}_{d} on X.
Proof Suppose that U({x}_{1},{c}_{1}) and U({x}_{2},{c}_{2}) are two open balls in X and take
x\in U({x}_{1},{c}_{1})\cap U({x}_{2},{c}_{2}).
Then d(x,{x}_{i})\prec {c}_{i} for i=1,2. From (S3), we get {c}_{i}d(x,{x}_{i})\succ 0 for i=1,2. It follows from (S12) that there exists a vector c\in Y with c\succ 0 such that c\prec {c}_{i}d(x,{x}_{i}) for i=1,2. By (S3), we obtain d(x,{x}_{i})\prec {c}_{i}c for i=1,2. Now using the triangle inequality and (S10), it easy to show that
U(x,c)\subset U({x}_{1},{c}_{1})\cap U({x}_{2},{c}_{2}).
Therefore, the collection ℬ is a basis for a topology on X. □
Thanks to Theorem 9.1 we can give the following definition.
Definition 9.2 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). The topology {\tau}_{d} on X with basis formed by open balls in X is called the cone metric topology on X.
We shall always assume that a cone metric space (X,d) over a solid vector space Y is endowed with the cone metric topology {\tau}_{d}. Hence, every cone metric space is a topological space.
Definition 9.3 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Then:

(i)
A sequence ({x}_{n}) in X is called Cauchy if for every c?Y with c?0 there is N?\mathbb{N} such that d({x}_{n},{x}_{m})?c for all n,m>N.

(ii)
A cone metric space X is called complete if each Cauchy sequence in X is convergent.

(iii)
A complete cone normed space is called a cone Banach space.
In the following theorem, we show that each cone metric space (X,d) over a solid vector space is metrizable. Besides, if (X,d) is a complete cone metric space, then it is completely metrizable. Moreover, the cone metric is equivalent to a metric which preserve some inequalities.
Theorem 9.2 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Suppose \parallel \cdot \parallel :Y\to \mathbb{R} is the Minkowski functional of [b,b] for some b\in Y with b\succ 0. Define the metric ρ on X by \rho (x,y)=\parallel d(x,y)\parallel. Then:

(i)
The topology of (X,d) coincides with the topology of (X,?).

(ii)
(X,d) is complete if and only if (X,?) is complete.

(iii)
For x,{x}_{1},\dots ,{x}_{n}?X, y,{y}_{1},\dots ,{y}_{n}?X, a?Y and {?}_{1},\dots ,{?}_{n}?\mathbb{R},
d(x,y)?a+\underset{i=1}{\overset{n}{?}}{?}_{i}d({x}_{i},{y}_{i})\phantom{\rule{1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}?(x,y)=?a?+\underset{i=1}{\overset{n}{?}}{?}_{i}?({x}_{i},{y}_{i}).
Proof (i) It follows from Lemma 7.2(i) and Definition 8.1 that ρ is a metric on X. Denoting by B(x,\epsilon ) an open ball in the metric space (X,\rho ) and by U(x,c) an open ball in the cone metric space (X,d), we shall prove that each B(x,\epsilon ) contains some U(x,c) and vice versa. First, we shall show that
B(x,\epsilon )=U(x,\epsilon b)\phantom{\rule{1em}{0ex}}\text{for all}x\in X\text{and}\epsilon 0.
(9.1)
According to Lemma 7.2(ii), for all x,y\in X and \epsilon >0,
\parallel d(x,y)\parallel <\epsilon \phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}d(x,y)\prec \epsilon b,
that is,
\rho (x,y)<\epsilon \phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}d(x,y)\prec \epsilon b
(9.2)
which proves (9.1). Note that identity (9.1) means that every open ball in the metric space (X,\rho ) is an open ball in the cone metric space (X,d). Now let U(x,c) be an arbitrary open ball in the cone metric space (X,d). Choosing \epsilon >0 such that \epsilon b\prec c, we conclude by (9.1) that B(x,\epsilon )\subset U(x,c).
(ii) Let ({x}_{n}) be a sequence in X. We have to prove that ({x}_{n}) is dCauchy if and only if it is ρCauchy. First note that (9.2) implies that for each \epsilon >0 and all m,n\in \mathbb{N},
\rho ({x}_{n},{x}_{m})<\epsilon \phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}d({x}_{n},{x}_{m})\prec \epsilon b.
Let ({x}_{n}) be dCauchy and \epsilon >0 be fixed. Then there is an integer N such that d({x}_{n},{x}_{m})\prec \epsilon b for all m,n>N. Hence, \rho ({x}_{n},{x}_{m})<\epsilon for all m,n>N which means that ({x}_{n}) be ρCauchy.
Now, let ({x}_{n}) be ρCauchy and c\succ 0 be fixed. Choose \epsilon >0 such that \epsilon b\prec c. Then there is an integer N such that d({x}_{n},{x}_{m})<\epsilon for all m,n>N. Therefore, for these n and m we get d({x}_{n},{x}_{m})\prec \epsilon b\prec c, which means that ({x}_{n}) is dCauchy.
(iii) follows from the monotony of the norm \parallel \cdot \parallel and the definition of the metric ρ. □
As we have seen the identity (9.1) plays an important role in the proof of Theorem 9.2. It is easy to see that this identity holds also for closed balls in the spaces (X,\rho ) and (X,d). Namely, we have
\overline{B}(x,\epsilon )=\overline{U}(x,\epsilon b)\phantom{\rule{1em}{0ex}}\text{for all}x\in X\text{and}\epsilon 0.
(9.3)
The main idea of Theorem 9.2 can be formulated in the following theorem.
Theorem 9.3 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Then there exists a metric ρ on X such that the following statements hold true.

(i)
The topology of (X,d) coincides with the topology of (X,?).

(ii)
(X,d) is complete if and only if (X,?) is complete.

(iii)
For x,{x}_{1},\dots ,{x}_{n}?X, y,{y}_{1},\dots ,{y}_{n}?X and {?}_{1},\dots ,{?}_{n}?\mathbb{R},
d(x,y)?\underset{i=1}{\overset{n}{?}}{?}_{i}d({x}_{i},{y}_{i})\phantom{\rule{1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}?(x,y)=\underset{i=1}{\overset{n}{?}}{?}_{i}?({x}_{i},{y}_{i}).
Metrizable topological spaces inherit all topological properties from metric spaces. In particular, it follows from Theorem 9.3 that every cone metric space over a solid vector space is a Hausdorff paracompact space and firstcountable. Since every first countable space is sequential, we immediately get that every cone metric space is a sequential space. Hence, as a consequence of Theorem 9.3 we get the following corollary.
Corollary 9.1 Let (X,d) be a cone metric space over a solid vector space Y. Then the following statements hold true:

(i)
A subset of X is open if and only if it is sequentially open.

(ii)
A subset of X is closed if and only if it is sequentially closed.

(iii)
A function f:D?X?X is continuous if and only if it is sequentially continuous.
Lemma 9.1 Let (X,d) be a cone metric space over a solid vector space Y. Then every closed ball \overline{U}(a,r) in X is a closed set.
Proof According to Corollary 9.1 we have to prove that \overline{U}(a,r) is a sequentially closed set. Let ({x}_{n}) be a convergent sequence in \overline{U}(a,r) and let x\in Y be its limit. Let c\in Y with c\succ 0 be fixed. Since {x}_{n}\to x, then there exists n\in \mathbb{N} such that d({x}_{n},x)\prec c. Using the triangle inequality, we get d(x,a)\u2aafd({x}_{n},a)+d({x}_{n},x)\prec r+c. Hence, d(x,a)r\prec c for all c\in Y with c\succ 0. Then by (S11) we conclude that d(x,a)r\u2aaf0 which implies x\in \overline{U}(a,r). Therefore, \overline{U}(a,r) is a closed set in X. □
Remark 9.1 Theorem 9.3 plays an important role in the theory of cone metric spaces over a solid vector space. In particular, using this theorem one can prove that some fixed point theorems in cone metric spaces are equivalent to their versions in usual metric spaces. For example, the short version of the Banach contraction principle in complete cone metric spaces (see Theorem 11.2 below) follows directly from its short version in metric spaces. Du [34] was the first who showed that there are equivalence between some metric and cone metric results. He obtained his results using the socalled nonlinear scalarization function. One year later, Kadelburg, Radenović and Rakočević [37] showed that the same results can be obtained using Minkowski functional in topological vector spaces.
Remark 9.2 Theorem 9.2 generalizes and extends some recent results of Du [[34], Theorems 2.1 and 2.2], Kadelburg, Radenović and Rakočević [[37], Theorems 3.1 and 3.2], Çakalli, Sönmez and Genç [[40], Theorem 2.3], Simić [[39], Theorem 2.2], Abdeljawad and Rezapour [[43], Theorem 16] Arandelović and Kečkić [[38], Lemma 2]. All of these authors have studied cone metric spaces over a solid Hausdorff topological vector space. Note that the identity (9.1) was proved by Çakalli, Sönmez and Genç [[40], Theorem 2.2] provided that Y is a Hausdorff topological vector space.
Theorem 9.2 generalizes and extends also some recent results of AminiHarandi and Fakhar [[23], Lemma 2.1], Turkoglu and Abuloha [26], Khani and Pourmahdian [[29], Theorem 3.4], Sönmez [[24], Theorem 1], Asadi, Vaezpour and Soleimani [[30], Theorem 2.1], Feng and Mao [[65], Theorem 2.2]. These authors have studied cone metric spaces over a solid Banach space.
Note that Asadi, Rhoades and Soleimani [41] proved that the metrics of Feng and Mao [65] and Du [34] are equivalent.
Finally, let us note a work of Khamsi [27] in which he introduced a metrictype structure in cone metric spaces over a normal Banach space.
Definition 9.4 Let (X,\parallel \cdot \parallel ) be a cone normed space over a solid vector space Y. The cone metric topology {\tau}_{d} on X induced by the metric d(x,y)=\parallel xy\parallel is called the cone topology on X.
In the following theorem, we show that each cone normed space (X,\parallel \cdot \parallel ) over a solid vector space is normable. Moreover, if (X,\parallel \cdot \parallel ) is a cone Banach space, then it is completely normable.
Theorem 9.4 Suppose X is a vector space over a normed field (\mathbb{K},\cdot ). Let (X,\parallel \cdot \parallel ) be a cone normed space over a solid vector space (Y,\u2aaf,\prec ,\to ). Let \mu :Y\to \mathbb{R} be the Minkowski functional of [b,b] for some b\in Y with b\succ 0. Define the norm \u2980\cdot \u2980 by \u2980x\u2980=\mu (\parallel \cdot \parallel ). Then:

(i)
The topology of (X,?\xb7?) coincides with topology of (X,?\xb7?).

(ii)
(X,?\xb7?) is a cone Banach space if and only if (X,?\xb7?) is a Banach space.

(iii)
For x,{x}_{1},\dots ,{x}_{n}?X, a?Y and {?}_{1},\dots ,{?}_{n}?\mathbb{R},
?x??a+\underset{i=1}{\overset{n}{?}}{?}_{i}?{x}_{i}?\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}?x?=\mu (a)+\underset{i=1}{\overset{n}{?}}{?}_{i}?{x}_{i}?.
Proof The topology of (X,\parallel \cdot \parallel ) is induced by the cone metric d(x,y)=\parallel xy\parallel and the topology of (X,\u2980\cdot \u2980) is induced by the metric \rho (x,y)=\u2980xy\u2980. It is easy to see that \rho =\mu \circ d. Now the conclusions of the theorem follow from Theorem 9.2. □
Remark 9.3 Theorem 9.4(i) was recently proved by Çakalli, Sönmez and Genç [[40], Theorem 2.4] provided that \mathbb{K}=\mathbb{R} and Y is a Hausdorff topological vector space.
The following corollary is an immediate consequence of Theorem 9.4(i).
Corollary 9.2 Every cone normed space (X,\parallel \cdot \parallel ) over a solid vector space Y is a topological vector space.
9.2 Convergence in cone metric spaces
Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Let ({x}_{n}) be a sequence in X and x a point in X. We denote the convergence of ({x}_{n}) to x with respect to the cone metric topology, by {x}_{n}\stackrel{d}{\to}x or simply by {x}_{n}\to x. Obviously, {x}_{n}\stackrel{d}{\to}x if and only if for every vector c\in Y with c\succ 0, d({x}_{n},x)\prec c for all but finitely many n. This definition for the convergence in cone metric spaces over a solid Banach space can be found in the works of Chung [11, 12] published in the period from 1981 to 1982. The definition of complete cone metric space (Definition 9.3) in the case when Y is a solid Banach space also can be found in [11, 12].
Theorem 9.5 Let (X,d) be a cone metric space over a solid vector space Y. Then the convergence in X has the following properties:

(i)
Any convergent sequence has a unique limit.

(ii)
Any subsequence of a convergent sequence converges to the same limit.

(iii)
Any convergent sequence is bounded.

(iv)
The convergence and the limit of a sequence do not depend on finitely many of its terms.
Proof The properties (i), (ii) and (iv) are valid in any Hausdorff topological space. It remains to prove (iii). Let ({x}_{n}) be a sequence in X which converges to a point x\in X. Choose a vector {c}_{1}\in Y with {c}_{1}\succ 0. Then there exists N\in \mathbb{N} such that d({x}_{n},x)\prec {c}_{1} for all n\ge N. By (S13), there is a vector {c}_{2}\in Y such that d({x}_{n},x)\prec {c}_{2} for all n=1,\dots ,N. Again by (S13), we get that there is a vector c\in Y such that {c}_{i}\prec c for i=1,2. Then by the transitivity of ≺, we conclude that {x}_{n}\in U(x,c) for all n\in \mathbb{N}, which means that ({x}_{n}) is bounded. □
Applying Theorem 9.2, we shall prove a useful sufficient condition for convergence of a sequence in a cone metric space over a solid vector space.
Theorem 9.6 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Suppose ({x}_{n}) is a sequence in X satisfying
d({x}_{n},x)\u2aaf{b}_{n}+\alpha d({y}_{n},y)+\beta d({z}_{n},z)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}n,
(9.4)
where x is a point in X, ({b}_{n}) is a sequence in Y converging to 0, ({y}_{n}) is a sequence in X converging to y, ({z}_{n}) is a sequence in X converging to z, α and β are nonnegative numbers. Then the sequence ({x}_{n}) converges to x.
Proof Let \parallel \cdot \parallel be the Minkowski functional of [b,b] for some b\in Y with b\succ 0. Define the metric ρ on X as in Theorem 9.2. Then from (9.4), we get
\rho ({x}_{n},x)\u2aaf\parallel {b}_{n}\parallel +\alpha \rho ({y}_{n},y)+\beta \rho ({z}_{n},z)\phantom{\rule{1em}{0ex}}\text{for all}n.
(9.5)
According to Theorem 7.1(ii), {b}_{n}\to 0 implies \parallel {b}_{n}\parallel \to 0. Hence, the righthand side of (9.5) converges to 0 in ℝ. By the usual sandwich theorem, we conclude that {x}_{n}\stackrel{\rho}{\to}x which is equivalent to {x}_{n}\stackrel{d}{\to}x. □
Remark 9.4 A special case (\alpha =\beta =0) of Theorem 9.6 was given without proof by Kadelburg, Radenović and Rakočević [42] in the case when Y is a Banach space. This special case was proved by Şahin and Telsi [[22], Lemma 3.3].
It is easy to see that if ({x}_{n}) is a sequence in a cone metric space (X,d) over a solid vector space Y, then
d({x}_{n},x)\to 0\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}{x}_{n}\stackrel{d}{\to}x,
(9.6)
but the converse is not true (see Example 9.2(ii) below). Note also that in general case the cone metric is not (sequentially) continuous function (see Example 9.2(iii) below), that is, from {x}_{n}\to x and {y}_{n}\to y it need not follow that d({x}_{n},{y}_{n})\to d(x,y).
In the following theorem, we shall prove that the converse of (9.6) holds provided that Y is normal and solid.
Theorem 9.7 Suppose (X,d) is a cone metric space over a normal solid vector space (Y,\u2aaf,\prec ,\to ). Then
{x}_{n}\stackrel{d}{\to}x\phantom{\rule{1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\mathit{\text{if and only if}}\phantom{\rule{1em}{0ex}}d({x}_{n},x)\to 0.
(9.7)
Proof Let \parallel \cdot \parallel be the Minkowski functional of [b,b] for some b\in Y with b\succ 0. Define the metric ρ on X as in Theorem 9.2. By Theorem 9.2,
{x}_{n}\stackrel{d}{\to}x\phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}{x}_{n}\stackrel{\rho}{\to}x.
(9.8)
By Theorems 7.1 and 7.3, for each sequence ({u}_{n}) in Y
{u}_{n}\to 0\phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}\parallel {u}_{n}\parallel \to 0.
Applying this with {u}_{n}=d({x}_{n},x), we get
d({x}_{n},x)\to 0\phantom{\rule{1em}{0ex}}\text{if and only}\phantom{\rule{1em}{0ex}}\rho ({x}_{n},x)\to 0,
that is,
d({x}_{n},x)\to 0\phantom{\rule{1em}{0ex}}\text{if and only}\phantom{\rule{1em}{0ex}}{x}_{n}\stackrel{\rho}{\to}x.
(9.9)
Now (9.7) follows from (9.8) and (9.9). □
The following theorem follows immediately from Corollary 9.2. It can also be proved by Theorem 9.6.
Theorem 9.8 Let X be a vector space over a complete normed field (\mathbb{K},\cdot ), and let (X,\parallel \cdot \parallel ) be a cone normed space over a solid vector space (Y,\u2aaf,\prec ,\to ). Then the convergence in X satisfies the properties (i)(iv) of Theorem 9.5 and it satisfies also the following properties:

(v)
If {x}_{n}\to x and {y}_{n}\to y, then {x}_{n}+{y}_{n}\to x+y.

(vi)
If {\lambda}_{n}\to \lambda in \mathbb{K} and {x}_{n}\to x, then {\lambda}_{n}{x}_{n}\to \lambda x.
9.3 Complete cone metric spaces
Now we shall prove a useful sufficient condition for Cauchy sequence in cone metric spaces over a solid vector space. The second part of this result gives an error estimate for the limit of a convergent sequence in cone metric space. Also we shall prove a criterion for completeness of a cone metric space over a solid vector space.
Theorem 9.9 Let (X,d) be a cone metric space over a solid vector space (Y,\u2aaf,\prec ,\to ). Suppose ({x}_{n}) is a sequence in X satisfying
d({x}_{n},{x}_{m})\u2aaf{b}_{n}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}n,m\ge 0\phantom{\rule{0.1em}{0ex}}\mathit{\text{with}}\phantom{\rule{0.1em}{0ex}}m\ge n,
(9.10)
where ({b}_{n}) is a sequence in Y which converges to 0. Then:

(i)
The sequence ({x}_{n}) is a Cauchy sequence in X.

(ii)
If ({x}_{n}) converges to a point x?X, then
d({x}_{n},x)?{b}_{n}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}n=0.
(9.11)
Proof (i) Let c\in Y with c\succ 0 be fixed. According to (S5), {b}_{n}\to 0 implies that there exists N\in \mathbb{N} such that {b}_{n}\prec c for all n>N. It follows from (9.10) and (S2) that d({x}_{n},{x}_{m})\prec c for all m,n>N with m\ge n. Therefore, {x}_{n} is a Cauchy sequence in X.
(ii) Suppose {x}_{n}\to x. Let n\ge 0 be fixed. Choose an arbitrary c\in Y with c\succ 0. Since {x}_{n}\to x, then there exists m>n such that d({x}_{m},x)\prec c. By the triangle inequality, (9.10) and (S10), we get
d({x}_{n},x)\u2aafd({x}_{n},{x}_{m})+d({x}_{m},x)\prec {b}_{n}+c.
It follows from (S3) that d({x}_{n},x){b}_{n}\prec c holds for each c\succ 0, which according to (S11) means that d({x}_{n},x){b}_{n}\u2aaf0. Hence, d({x}_{n},x)\u2aaf{b}_{n}, which completes the proof. □
Remark 9.5 The part (i) of Theorem 9.9 was proved by Azam, Beg and Arshad [[36], Lemma 1.3] in the case when Y is a topological vector space. Note also that whenever the cone metric space (X,d) is complete, then the assumption of the second part of Theorem 9.9 is satisfied automatically.
A sequence of closed balls (\overline{U}({x}_{n},{r}_{n})) in a cone metric space X is called a nested sequence if
\overline{U}({x}_{1},{r}_{1})\supset \overline{U}({x}_{2},{r}_{2})\supset \cdots .
Now we shall prove a simple criterion for the completeness of a cone metric space over a solid vector space.
Theorem 9.10 (Nested ball theorem)
A cone metric space (X,d) over a solid vector space (Y,\u2aaf,\prec ,\to ) is complete if and only if every nested sequence (\overline{U}({x}_{n},{r}_{n})) of closed balls in X such that {r}_{n}\to 0 has a nonempty intersection.
Proof Let \parallel \cdot \parallel be the Minkowski functional of [b,b] for some b\in Y with b\succ 0. Define the metric ρ on X as in Theorem 9.2. By Theorem 9.2, (X,d) is complete if and only if (X,\rho ) is complete.
Necessity. If (\overline{U}({x}_{n},{r}_{n})) is a nested sequence of closed balls in (X,d) such that {r}_{n}\to 0, then according to Lemma 9.1 it is a nested sequence of closed sets in (X,\rho ) with the sequence of diameters ({\delta}_{n}) converging to zero. Indeed, it easy to see that \rho (x,y)=\parallel d(x,y)\parallel \u2aaf2\parallel {r}_{n}\parallel for all x,y\in \overline{U}({x}_{n},{r}_{n}). Hence, {\delta}_{n}\le 2\parallel {r}_{n}\parallel , which yields {\delta}_{n}\to 0. Applying Cantor’s intersection theorem to the metric space (X,\rho ), we conclude that the intersection of the sets \overline{U}({x}_{n},{r}_{n}) is nonempty.
Sufficiently. Assume that every nested sequence of closed balls in (X,d) with radii converging to zero has a nonempty intersection. We shall prove that each nested sequence (\overline{B}({x}_{n},{\epsilon}_{n})) of closed balls in (X,\rho ) such that {\epsilon}_{n}\to 0 has a nonempty intersection. By identity (9.3), we get
\overline{B}({x}_{n},{\epsilon}_{n})=\overline{U}({x}_{n},{r}_{n})\phantom{\rule{1em}{0ex}}\text{for all}n,
(9.12)
where {r}_{n}={\epsilon}_{n}b\to 0. Hence, according to the assumptions the balls \overline{B}({x}_{n},{\epsilon}_{n}) have a nonempty intersection. Applying the nested ball theorem to the metric space (X,\rho ), we conclude that it is complete and so (X,d) is also complete. □
9.4 Examples of complete cone metric spaces
We end this section with three examples of complete cone metric spaces. Some other examples on cone metric spaces can be found in [6].
Example 9.1 Let X be a nonempty set and let (Y,\u2aaf,\prec ,\to ) be a solid vector space. Suppose a is a vector in Y such that a\u2ab00 and a\ne 0. Define the cone metric d:X\times X\to Y by
d(x,y)=\{\begin{array}{cc}a\hfill & \text{if}x\ne y,\hfill \\ 0\hfill & \text{if}x=y.\hfill \end{array}
(9.13)
Then (X,d) is a complete cone metric space over Y. This space is called a discrete cone metric space.
Proof It is obvious that (X,d) is a cone metric space (even if Y is an arbitrary ordered vector space). We shall prove that every Cauchy sequence in X is stationary. Assume the contrary and choose a sequence ({x}_{n}) in X, which is Cauchy but not stationary. Then for every c\in Y with c\succ 0 there exist n,m\in \mathbb{N} such that d({x}_{n},{x}_{m})\prec c and {x}_{n}\ne {x}_{m}. Hence, a\prec c for each c\succ 0. Then by (S11) we conclude that a\u2aaf0 which together with a\u2ab00 leads to the contradiction a=0. Therefore, every Cauchy sequence in X is stationary and so convergent in X. □
Example 9.2 Let (Y,\u2aaf,\prec ,\to ) be a solid vector space, and let X be its positive cone. Define the cone metric d:X\times X\to Y as follows:
d(x,y)=\{\begin{array}{cc}x+y\hfill & \text{if}x\ne y,\hfill \\ 0\hfill & \text{if}x=y.\hfill \end{array}
(9.14)
Then the following statements hold true:

(i)
(X,d) is a complete cone metric space over Y.

(ii)
If Y is not normal, then there are sequences ({x}_{n}) in X such that {x}_{n}?0 but d({x}_{n},0)?0.

(iii)
If Y is not normal, then the cone metric d is not continuous.
Proof First we shall prove the following claim: A sequence ({x}_{n}) in X is Cauchy if and only if it satisfies one of the following two conditions:

(a)
The sequence ({x}_{n}) is stationary.

(b)
For every c?0 the inequality {x}_{n}?c holds for all but finitely many n.
Necessity. Suppose ({x}_{n}) is Cauchy but not stationary. Then for every c\in Y with c\succ 0 there exists N\in \mathbb{N} such that d({x}_{n},{x}_{m})\prec c for all n,m>N. Hence, for all n,m>N we have {x}_{n}+{x}_{m}\prec c whenever {x}_{n}\ne {x}_{m}. Let n>N be fixed. Since ({x}_{n}) is not stationary, there exists m>N such that {x}_{n}\ne {x}_{m}. Hence, {x}_{n}+{x}_{m}\prec c. From this taking into account that {x}_{m}\u2ab00, we get {x}_{n}\prec c and so ({x}_{n}) satisfies (b).
Sufficiently. Suppose that ({x}_{n}) satisfies (b). Then for every c\succ 0 there exists N\in \mathbb{N} such that for all n>N we have {x}_{n}\prec \frac{1}{2}c. Let n,m>N be fixed. Then
d({x}_{n},{x}_{m})\u2aaf{x}_{n}+{x}_{m}\prec c
which means that ({x}_{n}) is Cauchy.
Now we shall prove the statements of the example.
(i) Let ({x}_{n}) be a Cauchy sequence in X. If ({x}_{n}) satisfies (a), then it is convergent. Now suppose that ({x}_{n}) satisfies (b). Let c\succ 0 be fixed. Then d({x}_{n},0)\u2aaf{x}_{n}\prec c for all but finitely many n. This proves that {x}_{n}\to 0. Therefore, in both cases ({x}_{n}) is convergent.
(ii) Since Y is not normal, then there exist two sequences ({x}_{n}) and ({y}_{n}) in Y such that 0\u2aaf{x}_{n}\u2aaf{y}_{n} for all n, {y}_{n}\to 0 and {x}_{n}\nrightarrow 0. Let us consider ({x}_{n}) as a sequence in X. It follows from the definition of the cone metric d that d({x}_{n},0)={x}_{n} for all n. Hence, d({x}_{n},0)\u2aaf{y}_{n} for all n. Then by Theorem 9.6, we conclude that {x}_{n}\to 0. On the other hand, d({x}_{n},0)={x}_{n}\nrightarrow 0.
(iii) Assume that the cone metric d is a continuous. Let ({x}_{n}) be any sequence in X satisfying (ii). From {x}_{n}\to 0 and continuity of the cone metric d, we obtain d({x}_{n},0)\to d(0,0), i.e. {x}_{n}\to 0 in Y which is a contradiction. Hence, the cone metric d is not continuous. □
Example 9.3 Let X={\mathbb{K}}^{n} be ndimensional vector space over \mathbb{K}, where (\mathbb{K},\cdot ) is a complete normed field. Let Y={\mathbb{R}}^{n} be equipped with the coordinatewise convergence and the coordinatewise ordering (see Example 5.1). Define the cone norm \parallel \cdot \parallel :X\to Y by
\parallel x\parallel =({\alpha}_{1}{x}_{1},\dots ,{\alpha}_{n}{x}_{n}),
(9.15)
where x=({x}_{1},\dots ,{x}_{n}) and {\alpha}_{1},\dots ,{\alpha}_{n} are positive numbers. Then (X,\parallel \cdot \parallel ) is a cone Banach space over Y.