A unified theory of cone metric spaces and its applications to the fixed point theory

In this paper we develop a unified theory for cone metric spaces over a solid vector space. As an application of the new theory we present full statements of the iterated contraction principle and the Banach contraction principle in cone metric spaces over a solid vector space.

It is well known that cone metric spaces and cone normed spaces have deep applications in the numerical analysis and the fixed point theory. Some applications of cone metric spaces can be seen in Collatz [13] and Zabrejko [61]. Schröder [53,54] was the first who pointed out the important role of cone metric spaces in the numerical analysis. The famous Russian mathematician Kantorovich [32] was the first who showed the importance of cone normed spaces for the numerical analysis.
In Section 2 we introduce a simplified definition of a vector space with convergence which does not require an axiom for the uniqueness of the limit of a convergent sequence. Our axioms are enough to prove some fixed point theorems in cone metric spaces over solid vector spaces.
In Section 3 we present a criterion for the interior of a solid cone in a vector space with convergence (Theorem 3.3).
In Section 4 we introduce the definition of an ordered vector space and the well known theorem that the vector orderings and cones in a vector space with convergence are in one-to-one correspondence.
In Section 5 we introduce the new notion of a strict vector ordering on an ordered vector space. Then we show that an ordered vector space can be equipped with a strict vector ordering if and only if it is a solid vector space (Theorem 5.2). Moreover, if the positive cone of a vector space is solid, then there exists only one strict vector ordering on this space. Hence, the strict vector orderings and solid cones in an vector space with convergence are in one-to-one correspondence.
In Section 6, we show that every solid vector space can be endowed with an order topology τ and that x n → x implies x n τ → x (Theorems 6.2 and 6.5). As a consequence we show that every convergent sequence in a solid vector space has a unique limit (Theorem 6.6). In Section 7, using the Minkowski functional, we show that the order topology on every solid vector space is normable. We also show that every normal and solid vector space Y is normable in the sense that there exists a norm . on Y such that x n → x if and only if x n . → x (Theorem 7.7). Also we show that the convergence of sequences in a normal and solid vector space has the properties of the convergence in R (Theorem 7.10). This result shows that the Sandwich theorem plays an important role in solid vector spaces.
In Section 8 we introduce the definitions of cone metric spaces and cone normed spaces. Note that in our definition of a cone normed space (X, . ) we allow X to be a vector space over an arbitrary valued field K.
In Section 9 we study 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. For example, every cone metric space over a solid vector space is a metrizable topological space (Theorem 9.5) and in such spaces the nested ball theorem holds (Theorem 9.22). Among the other results in this section we prove that every cone normed space over a solid vector space is normable (Theorem 9.12). Also in this section we give some useful properties of cone metric spaces which allow us to establish convergence results for Picard iteration with a priori and a posteriori error estimates. Some of the results in this section generalize, extend and complement some results of Du [17], Kadelburg, Radenović and Rakočević [30,29], Çakalli, Sönmez and Genç [12], Simić [56], Abdeljawad and Rezapour [1], Arandelović and Kečkić [5], Amini-Harandi and Fakhar, [4], Khani and Pourmahdian [34], Sönmez [57], Asadi, Vaezpour and Soleimani [7], Şahin and Telsi [52]. Azam, Beg and Arshad [8].
In Section 10 we establish a full statement of the iterated contraction principle in cone metric spaces over a solid vector space. The main result of this section (Theorem 10.5) generalizes, extends and complements some results of Pathak and Shahzad [41], Wardowski [60], Ortega and Rheinboldt [40,Theorem 12.3.2] and others.
In Section 11 we establish a full statement of the Banach contraction principle in cone metric spaces over a solid vector space. The main result of this section (Theorem 11.1) generalizes, extends and complements some results of Rezapour and Hamlbarani [47], Du [17], Radenović and Kadelburg [46] and others.

Vector spaces with convergence
In this section we introduce a simplified definition for the notion of vector spaces with convergence. Our definition is different from those given in the monograph of Collatz [13] and in the survey paper of Zabrejko [61]. In particular, we do not need an axiom for the uniqueness of the limit of a convergence sequence.
Definition 2.1. Let Y be a real vector space and let S be the set of all infinite sequences in Y . A binary relation → between S and Y is called a convergence on Y if it satisfies the following axioms: (C1) If x n → x and y n → y, then x n + y n → x + y.
The pair (Y, →) is said to be a vector space with convergence. If x n → x, then (x n ) is said to be a convergent sequence in Y , and the vector x is said to be a limit of (x n ).
The following two properties of the convergence in a vector space (Y, →) follow immediately from the above axioms.
(C4) If x n = x for all n, then x n → x.
(C5) The convergence and the limits of a sequence do not depend on the change of finitely many of its terms.
x n ∈ A for all but finitely many n.
Remark 2.3. Let (Y, →) be a vector space with convergence. It is easy to prove that if a set A ⊂ Y is open, then Y \A is closed. Let us note that the converse holds true provided that each subsequence of a convergent sequence in Y is convergent with the same limits.
The following lemma follows immediately from the definition of an open set.   Remark 2.5. Lemma 2.4 shows that the family of all open subsets of (Y, →) defines a topology on Y . Note that in this paper we will never consider this topology on Y .
Lemma 2.6. Let (Y, →) be a vector space with convergence. Suppose U and V are nonempty subsets of Y . Then the following statements hold true.
Proof. (i) Let λ > 0 and U be an open subset of Y . Suppose (x n ) is a convergent sequence in Y with a limit x ∈ λU . Then there exists a vector a ∈ U such that x = λa. Consider the sequence (a n ) defined by a n = 1 λ x n . It follows from (C2) that a n → a since a = 1 λ x. Taking into account that U is open and a ∈ U, we conclude that a n ∈ U for all but finitely many n. Then x n ∈ λU for the same n since x n = λa n . Therefore, the set λU is open.
(ii) Let U be an arbitrary subset of Y and V be an open subset of Y . Suppose (x n ) is a convergent sequence in Y with limit x ∈ U + V . Then there exist a ∈ U and b ∈ V such that x = a + b. Consider the sequence (b n ) defined by b n = x n − a. It follows from (C1) and (C4) that b n → b since b = x − a. Taking into account that V is open and b ∈ V , we conclude that b n ∈ V for all but finitely many n. Then x n ∈ U + V for these n since x n = a + b n . Therefore, the set U + V is open.
Due to the first two statements of Lemma 2.4 we can give the following definition. The following lemma follows immediately from the definition of the notion of interior.
Lemma 2.8. Let A and B be two subsets of a vector space (Y, →). Then Example 2.9. Let (Y, τ ) be an arbitrary topological vector space and let τ → be the τ -convergence in Y . Obviously, (Y, τ →) is a vector space with convergence. It is well known that every τ -open subset of (Y, τ ) is sequentially open and every τ -closed set is sequentially closed. Recall also that a topological space is called a sequential space if it satisfies one of the following equivalent conditions: Let us note that according to a well known theorem of Franklin [22] every first countable topological vector space is a sequential space. For sequential topological spaces see a survey paper of Goreham [25].

Solid cones in vector spaces with convergence
In this section we establish a useful criterion for the interior of a solid cone. This criterion will play an important role in Section 5.
For more on cone theory, see the classical survey paper of Krein and Rutman [37], the classical monographs of Krasnoselskii [ A nontrivial cone K is said to be a solid cone if its interior is nonempty.
Then there is at most one nonempty open subset U of K satisfying the following conditions: (i) λU ⊂ U for any λ > 0; Proof. Let U be a nonempty open subsets of K satisfying conditions (i)-(iii). First we shall prove that every nonempty open subset V of K is a subset of U. Let a vector x ∈ V be fixed. Choose a vector a ∈ U with a = 0. This is possible since U is nonempty and 0 / ∈ U. Consider the sequence (x n ) in Y defined by x n = x − 1 n a. It follows from (C1) and (C4) that x n → x.
Since V is open and x ∈ V , then there exists n ∈ N such that x n ∈ V . Therefore, x n ∈ K since V ⊂ K. On the other hand it follows from (i) that 1 n a ∈ U since a ∈ U . Then from (ii) we conclude that x = x n + 1 n a ∈ U which proves that V ⊂ U. Now if U and V are two nonempty open subsets of K satisfying conditions (i)-(iii), then we have both V ⊂ U and U ⊂ V which means that U = V . Now we are ready to establish a criterion for the interior of a solid cone. Theorem 3.3. Let K be a solid cone in a vector space (Y, →). Then the interior K • of K has the following properties: Proof. First part. We shall prove that the interior K • of a solid cone K satisfies properties (i)-(iii).
(ii) By Lemma 2.6(ii) K + K • is an open set. It follows from K • ⊂ K and K + K ⊂ K that K + K • ⊂ K. Now from Lemma 2.8, we conclude that (iii) Assume that 0 ∈ K • . Since K is nonempty and nontrivial, then we can choose a vector a ∈ K with a = 0. By axiom (C3), − 1 n a → 0. Taking into account that K • is open, we conclude that there exists n ∈ N such that − 1 n a ∈ K • . Then it follows from (i) that −a ∈ K • . Since K • ⊂ K, we have both a ∈ K and −a ∈ K which implies a = 0. This is a contradiction which proves that 0 / ∈ K • . Second part. The second part of the theorem follows from Lemma 3.2. Indeed, suppose that K • is a nonempty open subset of K satisfying properties (i)-(iii). Then by Lemma 3.2 we conclude that K • is a unique nonempty open subset of K satisfying these properties. On the other hand, it follows from the first part of the theorem that the interior of K also satisfies properties (i)-(iii). Therefore, K • coincide with the interior of K.

Ordered Vector Spaces
Recall that a binary relation on a set Y is said to be an ordering on Y if it is reflexive, antisymmetric and transitive. (V2) If λ ≥ 0 and x y, then λx λy; (V3) If x n → x, y n → y, x n y n for all n, then x y.
A vector space (Y, →) equipped with a vector ordering is called an ordered vector space and is denoted by (Y, , →). If the convergence → on Y is produced by a vector topology τ , we sometimes write (Y, τ, ) instead of (Y, , →). Analogously, if the convergence → on Y is produced by a norm . , we sometimes write (Y, . , ).
Axiom (V3) is known as passage to the limit in inequalities. Obviously, it is equivalent to the following statement: (V3 ′ ) If x n → 0, x n 0 for all n, then x 0.
is called the positive cone of the ordering or positive cone of Y .
The following well known theorem shows that the positive cone is indeed a cone. It shows also that the vector orderings and cones in a vector space (Y, →) with convergence are in one-to-one correspondence. is a vector ordering on Y whose positive cone coincides with K.
Definition 4.4. Let (Y, , →) be an ordered vector space.  6. An ordered vector space (Y, , →) is called a normal vector space whenever for arbitrary sequences (x n ), (y n ), (z n ) in Y , x n y n z n for all n and x n → x and z n → x imply y n → x.
The statement (3) is known as sandwich theorem or rule of intermediate sequence. (a) Every bounded increasing sequence in Y is convergent.
(b) Every bounded decreasing sequence in Y is convergent.

Strict vector orderings and solid cones
In this section we introduce a notion of a strict vector ordering and prove that an ordered vector space can be equipped with a strict vector ordering if and only if it is a solid vector space.
Recall that a nonempty binary relation ≺ on a set Y is said to be a strict ordering on Y if it is irreflexive, asymmetric and transitive.
Definition 5.1. Let (Y, , →) be an ordered vector space. A strict ordering ≺ on Y is said to be a strict vector ordering if it is compatible with the vector ordering, the algebraic structure and the convergence structure on Y in the sense that the following are true: (S1) If x ≺ y , then x y; (S2) If x y and y ≺ z, then x ≺ z; (S3) If x ≺ y , then x + z ≺ y + z; (S4) If λ > 0 and x ≺ y, then λx ≺ λy; (S5) If x n → x, y n → y and x ≺ y, then x n ≺ y n for all but finitely many n.
An ordered vector space (Y, , →) equipped with a strict vector ordering ≺ is denoted by (Y, , ≺, →). It turns out that ordered vector spaces with strict vector ordering are just solid vector spaces (see Corollary 5.3 below). Axiom (S5) is known as converse property of passage to the limit in inequalities. It is equivalent to the following statement: (S5 ′ ) If x n → 0 and c ≻ 0, then x n ≺ c for all but finitely many n.
(S9) If x ≺ y and y z, then x ≺ z.
(S10) If x y and u ≺ v , then x + u ≺ y + v; (S11) If x ≺ c for each c ≻ 0, then x 0.
(S12) For every finite set A ⊂ Y consisting of strictly positive vectors, there exists a vector c ≻ 0 such that c ≺ x for all x ∈ A. Moreover, for every vector b ≻ 0, c always can be chosen in the form c = λ b for some λ > 0.
(S13) For every finite set A ⊂ Y , there is a vector c ≻ 0 such that −c ≺ x ≺ c for all x ∈ A. Moreover, for every vector b ≻ 0, c always can be chosen in the form c = λ b for some λ > 0.
Proof of (S11). Let x be a vector in Y such that x ≺ c for each c ≻ 0. Choose a vector b ∈ Y with b ≻ 0. It follows from (S4) that 1 n b ≻ 0 for each n ∈ N. Hence, x ≺ 1 n b for each n ∈ N. Passing to the limit in this inequality, we obtain x 0.
Proof of (S12). Let x be an arbitrary vector from A. Choose a vector b ∈ Y with b ≻ 0. Since 1 n b → 0 and 0 ≺ x, then from (S5) we deduce that 1 n b ≺ x for all but finitely many n. Taking into account that A is a finite set, we conclude that for sufficiently large n we have 1 n b ≺ x for all x ∈ A. Now every vector c = 1 n b with sufficiently large n satisfies c ≺ x for all x ∈ A. To complete the proof put λ = 1 n .
Proof of (S13). Let x be an arbitrary vector from A. Choose a vector b ∈ Y with b ≻ 0. Since 1 n x → 0 and − 1 n x → 0, then from (S5) we obtain that 1 n x ≺ b and − 1 n x ≺ b for all but finitely many n. From these inequalities, we conclude that −nb ≺ x ≺ nb. Taking into account that A is a finite set, we get that every vector c = nb with sufficiently large n satisfies −c ≺ x ≺ c for all x ∈ A. To complete the proof put λ = n.
The next theorem shows that an ordered vector space can be equipped with a strict vector ordering if and only if it is a solid vector space. Moreover, on every ordered vector space there is at most one strict vector ordering. In other words the solid cones and strict vector orderings on a vector space with convergence are in one-to-one correspondence.
Theorem 5.2. Let (Y, , →) be an ordered vector space and let K be its positive cone, i.e. K = {x ∈ Y : x 0}. If a relation ≺ is a strict vector ordering on Y , then K is a solid cone with the interior Conversely, if K is a solid cone with the interior K • , then the relation ≺ on Y defined by means of is a unique strict vector ordering on Y .
Proof. First part. Suppose a relation ≺ is a strict vector ordering on Y . We shall prove that the set K • defined by (4) is a nonempty open subset of K which satisfies conditions (i)-(iii) of Theorem 3.3. Then it follows from the second part of Theorem 3.3 that K • is the interior of K and that K is a solid cone. By the definition of strict ordering, it follows that the relation ≺ is nonempty. Therefore, there are at least two vectors a and b in Y such that we get x ≻ 0. Then by (S5), we conclude that x n ≻ 0 for all but finitely many n which means that K • is open. Conditions (i) and (ii) of Theorem 3.3 follow immediately from (S4) and (S10) respectively. It remains to prove that 0 / ∈ K • . Assume the contrary, that is 0 ∈ K • . By the definition of K • , we get 0 ≻ 0 which is a contradiction since the relation ≺ is irreflexive.
Second part. Let K be a solid cone and K • be its interior. Note that according to to the first part of Theorem 3.3, K • has the following properties: We have to prove that the relation ≺ defined by (5) is a strict vector ordering. First we shall show that ≺ is nonempty and irreflexive. Since K is solid, K • is nonempty and nontrivial. Hence, there exists a vector c ∈ K • such that c = 0. Now by the definition of ≺, we get 0 ≺ c which means that ≺ is nonempty. To prove that ≺ is irreflexive assume the contrary. Then there exists a vector x ∈ Y such that x ≺ x. Hence, 0 = x − x ∈ K • which is a contradiction since 0 / ∈ K • . Now we shall show that ≺ satisfies properties (S1)-(S5).
(S1) Let x≺ y. Using the definition (5), the inclusion K • ⊂ K, and the definition of the positive cone K, we have (S2) Let x y and y ≺ z. Using the definition of the positive cone K, the definition (5) and the inclusion K + K • ⊂ K • , we get (S3) follows immediately from the definition (5). (S4) Let x y and λ > 0. Using the definition (5) and the inclusion (S5) Let x n → x, y n → y and x ≺ y. This yields y n − x n → y − x and y − x ∈ K • . Since K • is open, we conclude that y n − x n ∈ K • for all but finitely many n. Hence, x n ≺ y n for all but finitely many n.
Uniqueness. Now we shall prove the uniqueness of the strict vector ordering on Y . Assume that ≺ and < are two vector orderings on Y . It follows from the first part of the theorem that From this and (S3), we get for all x, y ∈ Y , which means that relations ≺ and < are equal.
Note that property (S13) shows that every finite set in a solid vector space is bounded. Property (S12) shows that every finite set consisting of strictly positive vectors in a solid vector space is bounded below by a positive vector.
The following assertion is an immediate consequence of Theorem 5.2. (i) Y is a solid vector space.
(ii) Y can be equipped with a strict vector ordering.
Remark 5.4. The strict ordering ≺ defined by (5) was first introduced in 1948 by Krein and Rutman [37, p. 8] in the case when K is a solid cone in a Banach space Y . In this case they proved that ≺ satisfies axioms (S1)-(S4).
In conclusion of this section we present three examples of solid vector spaces We end the section with a remark which shows that axiom (S5) plays an important role in the definition of strict vector ordering.
Example 5.5. Let Y = R n with → the coordinate-wise convergence, and with coordinate-wise ordering defined by Then (Y, , ≺, →) is a solid vector space. This space is normal and regular.
Then (Y, . ∞ , , ≺) is a solid Banach space. This space is normal but nonregular. Consider, for example, the sequence ( Define the ordering and ≺ as in Example 5.6. Then (Y, . , , ≺) is is a solid Banach space. The space Y is not normal. Consider, for example, the sequences (x n ) and (y n ) in Y defined by x n (t) = t n n and y n (t) = 1 n . It is easy to see that 0 x n y n for all n, y n → 0 and x n → 0.
is a strict ordering on Y and it always satisfies axioms (S1)-(S4) and properties (S5)-(S10). However, it is not in general a strict vector ordering on Y . For example, from the uniqueness of strict vector ordering (Theorem 5.2) it follows that ≺ defined by (6) is not a strict vector ordering in the ordered vector spaces defined in Examples 5.5-5.7.

Order topology on solid vector spaces
In this section, we show that every solid vector space can be endowed with an order topology τ and that x n → x implies x n τ → x. As a consequence we show that every convergent sequence in a solid vector space has a unique limit.
It is easy to see that every open interval in Y is an infinite set. Indeed, one can prove that a + λ (b − a) ∈ (a, b) for all λ ∈ R with 0 < λ < 1. Proof. One has to prove that B satisfies the requirements for a basis. First, note that every vector x of Y lies in at least one element of B. Indeed, It remains to show that the topology τ is Hausdorff. We shall prove that for all x, y ∈ X with x = y there exists c ≻ 0 such that the intersection of the intervals (x − c, x + c) and (y − c, y + c) is empty. Assume the contrary. Then there exists x, y ∈ X with x = y such that for every Applying these inequalities to 1 2 c, we conclude that x − y ≺ c and y − x ≺ c for each c ≻ 0. Now it follows from (S11) that x y and y x which is a contradiction since x = y.
Thanks to Theorem 6.2 we can give the following definition.
is also a basis for the order topology τ on Y .
Theorem 6.5. Let (Y, , ≺, →) be a solid vector space and let τ be the order topology on Y . Then: Proof. The first claim follows from Remark 6.4. Let x n → x and (a, b) be a neighborhood of x. From a ≺ x ≺ b and (S5), we conclude that x n ∈ (a, b) for for all but finitely many n. Hence, x n τ → x which proves the second claim.
At the end of the next section we shall prove that the converse of the statement (ii) of Theorem 6.5 holds true if and only if Y is normal. Theorem 6.6. If (Y, , ≺, →) is a solid vector space, then the convergence on Y has the following properties.
(C6) Each convergent sequence in Y has a unique limit.
Assume that there are x, y ∈ Y such that x n → x and x n → y. It follows from Theorem 6.5 that x n τ → x and x n τ → y. According to Theorem 6.2 the topology τ is Hausdorff. Now by the uniqueness of the limit of a convergent sequence in the Hausdorff topological space (Y, τ ), we conclude that x = y.
(C7) Let (x n ) be a convergent sequence in Y and x n → x. By Theo-

Minkowski functional on solid vector spaces
In this section, using the Minkowski functional, we prove that the order topology on every solid vector space is normable. Also we show that every normal and solid vector space Y is normable in the sense that there exists a norm . on Y such that x n → x if and only if x n . → x. Finally, we give a criterion for a normal vector space and show that the convergence of a sequence in normal and solid vector space has the properties of the convergence in R. This last result shows that the Sandwich theorem plays an important role in solid vector spaces.
Definition 7.2. Let Y be a real vector space and A ⊂ Y an absorbing set. Then the functional . : Y → R defined by is called the Minkowski functional of A.
It is well known (see, e.g. [49,Theorem 1.35]) that the Minkowski functional of every absorbing, convex and balanced subset A of a vector space Y is a seminorm on Y . .
Definition 7.4. Let (Y, , →) be an ordered vector space, and let a, b ∈ Y be two vectors with a b. Then the set [a, b] = {x ∈ Y : a x b} is called a closed interval in Y .
Obviously, every closed interval [−b, b] in an ordered vector space Y is a convex, balanced, closed and bounded set. It follows from (S14) that is also an absorbing set provided that Y is a solid vector space and b ≻ 0.
. is a monotone norm on Y which can be defined by (ii) For x ∈ Y and ε > 0, Proof. (i) The claim with the exception of the monotonicity of the norm follows from Lemma 7.3. Let x and y be two vectors in Y such that 0 x y.
From (9), we get y y b. Hence, − y b x y b. Again from (9), we conclude that x y . Hence, . is a monotone norm.
The following theorem shows that the order topology on Y is normable.
Proof. (i) Denoting by B(x, ε) an open ball in the normed space (Y, . ), we shall prove that each B(x, ε) contains some interval (u, v) in Y and vice versa. First, we shall prove the following identity According to Lemma 7.6, for each x, y ∈ Y and ε > 0, which proves (11). Note that identity (11) means that every open ball in the normed space (Y, . ) is an open interval in Y . Now let (u, v) be an arbitrary open interval in Y and let x ∈ (u, v). Choose an interval of the type (x − c, x + c) which is a subset of (u, v), where c ∈ Y with c ≻ 0. Then choosing ε > 0 such that ε b ≺ c, we conclude by (11) that B(x, ε) ⊂ (u, v).
The main part of Theorem 7.7 can be formulated in the following theorem.
Theorem 7.8. Let (Y, , ≺, →) be a solid vector space. Then there exists a monotone norm . on Y such that the following statements hold true.
(i) The norm . generates the order topology on Y .
In the next theorem we shall give a criterion for a normal vector space. In particular, this theorem shows that every normal and solid vector space Y is normable in the sense that there exists a norm . on Y such that x n → x if and only if x n . → x. Analogous result for normability of normal topological vector space was proved by Vandergraft [59].  (iii) The convergence in Y is generated by the order topology on Y .
According to Lemma 7.6 the Minkowski functional of [−b, b] is a monotone norm on Y . We shall prove that the convergence in Y is generated by this norm. We have to prove that for a sequence (x n ) in Y , x n → x if and only if x n . → x. Without loss of generality we may assume that x = 0. Then we have to prove that x n → 0 if and only if x n → 0. Taking into account Theorem 7.7 we have only to prove that x n → 0 implies x n → 0. Let x n → 0. By Lemma 7.6, we get − x n b x n x n b for all n.
It follows from axiom (C3) that x n b → 0. Then by the Sandwich theorem we conclude that x n → 0.
(ii) → (iii). Suppose the convergence in Y is generated by a monotone norm . on Y , i.e. for a sequence (x n ) in Y , x n → x if and only if x n . → x. We shall prove that the convergence in Y is generated by the order topology τ on Y . According to Theorem 6.5 it is sufficient to prove that for a sequence (x n ) in Y , x n τ → x implies x n → x. Again without loss of generality we may assume that x = 0. Let x n τ → 0. Let ε > 0 be fixed. It follows from Theorem 6.5 that for every vector c ≻ 0, for all sufficiently large n. From (12), we obtain 0 ≺ c − x n ≺ 2 c. By monotonicity of the norm, we conclude that c − x n ≤ 2 c which implies that x n ≤ 3 c . Now choosing a vector c ≻ 0 such that c < ε/3, we obtain x n < ε for all sufficiently large n. Hence, x n → 0 which equivalent x n → x.
(iii) → (i). Suppose the convergence in Y is generated by the order topology on Y . We shall prove that Y is normal. Obviously, condition (3) in Definition 4.6 is equivalent to the following 0 x n y n for all n and y n → 0 imply x n → 0.
Let (x n ) and (y n ) be two sequences in Y such that 0 x n y n for all n and y n → 0. We have to prove that x n → 0. Let c ≻ 0 be fixed. It follows from y n → 0 and (S5) that y n ≺ c for all but finitely many n. From this and 0 x n y n , we conclude that −c ≺ x n ≺ c for all sufficiently large n. Now it follows from Theorem 6.5 that x n τ → x which is equivalent to x n → 0.
Note that Theorem 7.9 remains true if we replace in it "monotone norm" by "semimonotone norm".
The following theorem shows that the convergence in a normal and solid vector space has the properties of the convergence in R.
Theorem 7.10. If (Y, , ≺, →) is a normal and solid vector space, then the convergence on Y has the following additional properties.
(C8) Each subsequence of a convergent sequence converges to the same limit.
(C9) The convergence of a sequence and its limit do not depend on finitely many of its terms.
(C10) If λ n → λ in R and x n → x, then λ n x n → λ x.
(C11) If λ n → 0 in R and (x n ) is a bounded sequence in Y , then λ n x n → 0.
(C12) If (λ n ) is a bounded sequence in R and x n → 0, then λ n x n → 0.
Proof. Let . be a norm on Y that generates the convergence in Y . The existence of such norm follows from Theorem 7.9. The properties C8)-(C10) are valid in any normed space. Property (C13) follows from Theorems 6.5 and 7.9. The proofs of (C11) and (C12) are similar. We will prove only (C11). Since (x n ) is bounded, there exist a, b ∈ Y such that a x n b for all n. This implies |λ n | a |λ n | x n |λ n | b By axiom (C3), we get |λ n | a → 0 and |λ n | b → 0. Applying the Sandwich theorem to the inequalities (14), we conclude that |λ n | x n → 0. Then by Theorem 7.9, we obtain |λ n | x n → 0, that is, λ n x n → 0. Again by Theorem 7.9, we conclude that λ n x n → 0.

Cone metric spaces and cone normed spaces
In this section we introduce the notions of cone metric spaces and cone normed spaces. Cone metric spaces were first introduced in 1934 by Kurepa [38]. Cone normed spaces were first introduced in 1936 by Kantorovich [31,32]. For more on these abstract metric spaces, see the monograph of Collatz [13] and the survey paper of Zabrejko [61]. (ii) d(x, y) = d(x, y) for all x, y ∈ X; (iii) d(x, y) d(x, z) + d(z, y) for all x, y, z ∈ X.
The pair (X, d) is called a cone metric space over Y . The elements of a cone metric space X are called points.
Obviously, every metric space is a cone metric space over R. In Section 9 we show that the theory of cone metric spaces over solid vector spaces is very close to the theory of the metric spaces. (ii) | x y| = | x| .| y| for all x, y ∈ X; An absolute value is called trivial if | x| = 1 for x = 0. A field (K, | . |) equipped with a nontrivial absolute value is called a valued field.
Note that finite fields and their extensions only have the trivial absolute value. A valued field is always assumed to carry the topology induced by the metric ρ(x, y) = | x − y|, with respect to which it is a topological field. An absolute value is also called a multiplicative valuation or a norm. For more on valuation theory, see Engler and Prestel [20].
One of the most important class of cone metric spaces is the class of cone normed spaces. (ii) λ x = | λ| y for all λ ∈ K and x ∈ X; (iii) x + y x + y for all x, y ∈ X.
The pair (X, . ) is said to be a cone normed space over Y .
It is easy to show that every cone normed space (X, . ) over an ordered vector space Y is a cone metric space over Y with the cone metric defined by d(x, y) = x − y .
We end this section with the definitions of closed balls and bounded sets in cone metric spaces. Definition 8.4. Let (X, d) be a cone metric space over an ordered vector space (Y, , →). For a point x 0 ∈ X and a vector r ∈ Y with r 0, the set is called a closed ball with center x 0 and radius r.
Definition 8.5. Let X be a cone metric space.
(a) A set A ⊂ X is called bounded if it is contained in some closed ball.
(b) A sequence (x n ) in X is called bounded if the set of its terms is bounded.
Let (X, d) be a cone metric space over an ordered vector space (Y, , →). It is easy to show that a nonempty set A ⊂ X is bounded if and only if there exists a vector b ∈ Y such that d(x, y) b for all x, y ∈ A.
Analogously, if (X, . ) is a cone normed space over an ordered vector space (Y, , →), then a nonempty set A ⊂ X is bounded if and only if there exists a vector b ∈ Y such that x b for all x ∈ A.

Cone metric spaces over solid vector spaces
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.

Topological structure of cone metric spaces
Definition 9.1. Let (X, d) be a cone metric space over a solid vector space (Y, , ≺, →). For a point x 0 ∈ X and a vector r ∈ Y with r ≻ 0, the set is called an open ball with center x 0 and radius r.
It follows from (S12) that there exists a vector c ∈ Y with c ≻ 0 such that c ≺ c i − d(x, x i ) for i = 1, 2. By (S3), we obtain d(x, x i ) ≺ c i − c for i = 1, 2. Now using the triangle inequality and (S10), it easy to show that U(x, c) ⊂ U(x 1 , c 1 ) ∩ U(x 2 , c 2 ). Therefore, the collection B is a basis for a topology on X.
Thanks to Theorem 9.2 we can give the following definition. We shall always assume that a cone metric space (X, d) over a solid vector space Y is endowed with the cone metric topology τ d . Hence, every cone metric space is a topological space. (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. Moreover, if (X, d) is a complete cone metric space, then it is completely metrizable.
Theorem 9.5. Let (X, d) be a cone metric space over a solid vector space (i) The metric ρ : X × X → R defined by ρ(x, y) = d(x, y) generates the cone metric topology on X.
(ii) The cone metric space (X, d) is complete if and only if the metric space (X, ρ) is complete.
According to Lemma 7.6(ii), for all x, y ∈ X and ε > 0, which proves (15). Note that identity (15)  Let (x n ) be d-Cauchy and ε > 0 be fixed. Then there is an integer N such that d(x n , x m ) ≺ εb for all m, n > N . Hence, ρ(x n , x m ) < ε for all m, n > N which means that (x n ) be ρ-Cauchy . Now, let (x n ) be ρ-Cauchy and c ≻ 0 be fixed. Choose ε > 0 such that εb ≺ c. Then there is an integer N such that d(x n , x m ) < ε for all m, n > N . Therefore, for these n and m we get d(x n , x m ) ≺ εb ≺ c which means that (x n ) is d-Cauchy.
(iii) follows from the monotony of the norm . and the definition of the metric ρ.
As we have seen the identity (15) plays an important role in the proof of Theorem 9.5. It is easy to see that this identity holds also for closed balls in the spaces (X, ρ) and (X, d). Namely, we have The main idea of Theorem 9.5 can be formulated in the following theorem.
Theorem 9.6. Let (X, d) be a cone metric space over a solid vector space (Y, , ≺, →). Then there exists a metric ρ on X such that the following statements hold true.
(i) The metric ρ generates the cone metric topology on X.
(ii) The cone metric space (X, d) is complete if and only if the metric space (X, ρ) is complete.
(iii) For x i , y i ∈ X (i = 0, 1, . . . , n) and λ i ∈ R (i = 1, . . . , n), Metrizable topological spaces inherit all topological properties from metric spaces. In particular, it follows from Theorem 9.6 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 9.6 we get the following corollary.    Proof. According to Corollary 9.7 we have to prove that U(a, r) is a sequentially closed set. Let (x n ) be a convergent sequence in U (a, r) and let x ∈ Y be its limit. Let c ∈ Y with c ≻ 0 be fixed. Since x n → x, then there exists n ∈ N such that d(x n , x) ≺ c. Using the triangle inequality, we get Then by (S11) we conclude that d(x, a) − r 0 which implies x ∈ U (a, r). Therefore, U (a, r) is a closed set in X.
Remark 9.9. Theorem 9.6 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 [17] was the first who showed that there are equivalence between some metric and cone metric results. He obtained his results using the so-called nonlinear scalarization function. One year later, Kadelburg, Radenović and Rakočević [30] showed that the same results can be obtained using Minkowski functional in topological vector spaces. Note that Asadi and Soleimani [6] proved that the metrics of Feng and Mao [21] and Du [17] are equivalent.
Finally, let us note a work of Khamsi [33] in which he introduced a metric type structure in cone metric spaces over a normal Banach space. Definition 9.11. Let (X, . ) be a cone normed space over a solid vector space (Y, , ≺, →). The cone metric topology τ d on X induced by the metric d(x, y) = x − y is called the cone topology on X.
In the following theorem we show that each cone normed space (X, . ) over a solid vector space is normable. Moreover, if (X, . ) is a cone Banach space, then it is completely normable. (iii) For x i ∈ X and λ i ∈ R (i = 0, 1, . . . , n), Proof. The cone topology on the space (X, . ) is induced by the cone metric d(x, y) = x − y and the topology on (X, ||| . |||) is induced by the metric ρ(x, y) = |||x − y|||. It is easy to see that ρ = µ • d. Now the conclusions of the theorem follow from Theorem 9.5.
Remark 9.13. Theorem 9.12(i) was recently proved by Çakalli, Sönmez and Genç [12,Theorem 2.4] provided that K = R and Y is a Hausdorff topological vector space.
The following corollary is an immediate consequence of Theorem 9.12(i).
Corollary 9.14. Every cone normed space (X, . ) over a solid vector space Y is a topological vector space.

Convergence in cone metric spaces
Let (X, d) be a cone metric space over a solid vector space (Y, , ≺, →). 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 d → x or simply by x n → x. Obviously, x n d → x if and only if for every vector c ∈ Y with c ≻ 0, d(x n , x) ≺ 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 [14,15] published in the period from 1981 to 1982. The definition of complete cone metric space (Definition 9.4) in the case when Y is a solid Banach space also can be found in [14,15].
Theorem 9.15. 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 ∈ X. Choose a vector c 1 ∈ Y with c 1 ≻ 0. Then there exists N ∈ N such that d(x n , x) ≺ c 1 for all n ≥ N. By (S13), there is a vector c 2 ∈ Y such that d(x n , x) ≺ c 2 for all n = 1, . . . , N. Again by (S13), we get that there is a vector c ∈ Y such that c i ≺ c for i = 1, 2. Then by the transitivity of ≺, we conclude that x n ∈ U(x, c) for all n ∈ N which means that (x n ) is bounded.
Applying Theorem 9.5, we shall prove a useful sufficient condition for convergence of a sequence in a cone metric space over a solid vector space. Theorem 9.16. Let (X, d) be a cone metric space over a solid vector space (Y, , ≺, →). Suppose (x n ) is a sequence in X satisfying d(x n , x) b n + α d(y n , y) + β d(z n , z) for all n, 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 real numbers. Then the sequence (x n ) converges to x.
Define the metric ρ on X as in Theorem 9.5. Then from (18), we get ρ(x n , x) b n + α ρ(y n , y) + β ρ(z n , z) for all n, According to Theorem 7.7(ii), b n → 0 implies b n → 0. Hence, the righthand side of (19) converges to 0 in R. By usual Sandwich theorem, we conclude that x n ρ → x which is equivalent to x n d → x.
Remark 9.17. A special case (α = β = 0) of Theorem 9.16 was given without proof by Kadelburg, Radenović and Rakočević [29] in the case when Y is a Banach space. This special case was proved by Şahin and Telsi [52, 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 but the converse is not true (see Example 9.24(ii) below). Note also that in general case the cone metric is not (sequentially) continuous function (see Example 9.24(iii) below), that is, from x n → x and y n → y it need not follow that d(x n , y n ) → d(x, y).
In the following theorem we shall prove that the converse of (20) holds provided that Y is normal and solid. Applying this with u n = d(x n , x), we get Now (21) follows from (22) and (23).
The following theorem follows immediately from Corollary 9.14. It can also be proved by Theorem 9.16.
Theorem 9.19. Suppose X is a vector space over a valued field (K, | . | ). Let (X, . ) be a cone normed space over a solid vector space (Y, , ≺, →). Then the convergence in X satisfies the properties (i )-(iv ) of Theorem 9.15 and it satisfies also the following properties.
(v) If x n → x and y n → y, then x n + y n → x + y.
(vi) If λ n → λ in K and x n → x, then λ n x n → λ x.

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 9.20. Let (X, d) be a cone metric space over a solid vector space 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 for all n ≥ 0.
Proof. (i) Let c ∈ Y with c ≻ 0 be fixed. According to (S5), b n → 0 implies that there is N ∈ N such that b n ≺ c for all n > N . It follows from (24) and (S2) that d(x n , x m ) ≺ c for all m, n > N with m ≥ n. Therefore, x n is a Cauchy sequence in X.
(ii) Suppose x n → x. Let n ≥ 0 be fixed. Choose an arbitrary c ∈ Y with c ≻ 0. Since x n → x, then there exists m > n such that d(x m , x) ≺ c. By the triangle inequality, (24) and (S10), we get It follows from (S3) that d(x n , x) − b n ≺ c holds for each c ≻ 0. Which according to (S11) means that d(x n , x) − b n 0. Hence, d(x n , x) b n which completes the proof.
Remark 9.21. Theorem 9.20(i) was proved by Azam, Beg and Arshad [8,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 9.20 is satisfied automatically.
A sequence of closed balls (U(x n , r n )) in a cone metric space X is called a nested sequence if U (x 1 , r 1 ) ⊃ U (x 2 , r 2 ) ⊃ . . . Now we shall prove a simple criterion for the completeness of a cone metric space over a solid vector space.
Theorem 9.22 (Nested ball theorem). A cone metric space (X, d) over a solid vector space (Y, , ≺, →) is complete if and only if every nested sequence (U(x n , r n )) of closed balls in X such that r n → 0 has a nonempty intersection.
Define the metric ρ on X as in Theorem 9.5. By Theorem 9.5, (X, d) is complete if and only if (X, ρ) is complete.
Necessity. If (U(x n , r n )) is a nested sequence of closed balls in (X, d) such that r n → 0, then according to Lemma 9.8 it is a nested sequence of closed sets in (X, ρ) with the sequence of diameters (δ n ) converging to zero. Indeed, it easy to see that ρ(x, y) = d(x, y) 2 r n for all x, y ∈ U(x n , r n ). Hence, δ n ≤ 2 r n which yields δ n → 0. Applying Cantor's intersection theorem to the metric space (X, ρ), we conclude that the intersection of the sets 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 (B(x n , ε n )) of closed balls in (X, ρ) such that ε n → 0 has a nonempty intersection. By identity (17), we get where r n = ε n b → 0. Hence, according to the assumptions the balls B(x n , ε n ) have a nonempty intersection. Applying the nested ball theorem to the metric space (X, ρ), we conclude that it is complete and so (X, d) is also complete.

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 [61].
Example 9.23. Let X be a nonempty set and let (Y, , ≺, →) be a solid vector space. Suppose a is a vector in Y such that a 0 and a = 0. Define the cone metric d : 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 ∈ Y with c ≻ 0 there exist n, m ∈ N such that d(x n , x m ) ≺ c and x n = x m . Hence, a ≺ c for each c ≻ 0. Then by (S11) we conclude that a 0 which together with a 0 leads to the contradiction a = 0. Therefore, every Cauchy sequence in X is stationary and so convergent in X.
Example 9.24. Let (Y, , ≺, →) be a solid vector space, and let X be its positive cone. Define the cone metric d : X × X → Y as follows 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 ∈ Y with c ≻ 0 there exists N ∈ N such that d(x n , x m ) ≺ c for all n, m > N . Hence, for all n, m > N we have x n + x m ≺ c whenever x n = x m . Let n > N be fixed. Since (x n ) is not stationary, there exists m > N such that x n = x m .
Hence, x n + x m ≺ c. From this taking into account that x m 0, we get x n ≺ c and so (x n ) satisfies (b). Sufficiently. Suppose that (x n ) satisfies (b). Then for every c ≻ 0 there exists N ∈ N such that for all n > N we have x n ≺ 1 2 c. Let n, m > N be fixed. Then d(x n , x m ) x n + x m ≺ 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 ≻ 0 be fixed. Then d(x n , 0) x n ≺ c for all but finitely many n. This proves that x n → 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 x n y n for all n, y n → 0 and x n → 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) y n for all n. Then by Theorem 9.16, we conclude that x n → 0. On the other hand d(x n , 0) = x n → 0.
(iii) Assume that the cone metric d is a continuous. Let (x n ) be any sequence in X satisfying (ii). By x n → 0 and continuity of d, we obtain d(x n , 0) → d(0, 0), i.e. x n → 0 in Y which is a contradiction. Hence, the cone metric d is not continuous.
Example 9.25. Let X = K n be n-dimensional vector space over K, where (K, | . | ) is a valued field. Let Y = R n be n-dimensional real vector space with the coordinate-wise convergence and the coordinate-wise ordering (see Example 5.5). Define the cone norm . : X → Y by where x = (x 1 , . . . , x n ) and α 1 , . . . , α n are positive real numbers. Then (X, . ) is a cone Banach space over Y .

Iterated contractions in cone metric spaces
The iterated contraction principle in usual metric spaces was first mentioned in 1968 by Rheinboldt [48] as a special case of a more general theorem. Two years later, an explicit formulation of this principle (with a posteriori error estimates) was given in the monograph of Ortega and Rheinboldt [40,Theorem 12.3.2]. Great contribution to the iterated contraction principle in metric spaces and its applications to the fixed point theory was also given by Hicks and Rhoades [27], Park [42] and others (see Proinov [45,Section 6]). In this section we shall establish a full statement of the iterated contraction principle in cone metric spaces. We shall formulate the result for nonself mappings since the case of selfmappings is a special case of this one.
Let (X, d) be a cone metric space over a solid vector space (Y, , ≺, →), and let T : D ⊂ X → X be an arbitrary mapping in X. Then starting from a point x 0 ∈ D we can build up the Picard iterative sequence associated to the mapping T . We say that the iteration (30) is well defined if x n ∈ D for all n = 0, 1, 2, . . . The main problems which arise for the Picard iteration are the following: (i) Convergence problem. To find initial conditions for x 0 ∈ D which guarantee that the Picard iteration (30) is well defined and converging to a point ξ ∈ D.
(ii) Existence problem. To find conditions which guarantee that ξ is a fixed points of T .
(iii) Uniqueness problem. To find a subset of D in which ξ is a unique fixed point of T .
(iv) Error estimates problem. To find a priory and a posteriori estimates for the cone distance d(x n , ξ).
In our opinion, the solving of problem (i) for the convergence of the Picard iteration plays an important role for the solving of problem (ii) for existence of fixed points of T . It turns out that in many cases the convergence of the Picard iteration to a point ξ ∈ D implies that ξ is a fixed point of T . For example, such situation can be seen in the next proposition. Proof. If (F1) or (F2) is satisfied, then the conclusion follows from Theorem 9.15 and definition (30) of the Picard iteration. Suppose that condition (F3) holds. Since the norm . is semimonotone, there exists a constant K > 0 such that x ≤ K y whenever 0 x y. First we shall prove that x n → ξ implies d(x n , x n+1 ) → 0. We claim that for every ε > 0 there exists a vector c ∈ Y such that c ≻ 0 and c < ε. To prove this take a vector b ∈ Y with b ≻ 0. We have 1 n b = 1 n b → 0. Hence, every vector c = 1 n b with sufficiently large n satisfies c < ε. Now let ε > 0 be fixed. Choose a vector c ∈ Y such that c ≻ 0 and c < ε/K. From the triangle inequality, we get d(x n , x n+1 ) d(x n , ξ) + d(x n+1 , ξ). Now it follows from x n → ξ that d(x n , x n+1 ) ≺ c for all but finitely many n. Hence, d(x n , x n+1 ) K c < ε for these n. Therefore, d(x n , x n+1 ) → 0. Now taking into account that G is lower semicontinuous at ξ we conclude that Suppose that condition (F4) is satisfied. By substituting x = x n , we get From this, taking into account that x n → ξ, we conclude that d(ξ, T ξ) ≺ c for each c ∈ Y with c ≻ 0. According to (S11), this implies d(ξ, T ξ) 0. Therefore, d(ξ, T ξ) = 0 which means that ξ is a fixed point of T .
Remark 10.2. Obviously, if the space (X, d) in Proposition 10.1 is a metric space, then the function G in (F4) can be defined by G(x) = d(x, T x). In a metric space setting this is a classical result (see [27]). Let us note also that if the space Y in Proposition 10.1 is a normal and solid normed space with norm . , then one can choose in (F4) just this norm (see [60]).
Throughout this and next section for convenience we assume in R that 0 0 = 1 by definition.
where 0≤ λ < 1. Then (x n ) is a Cauchy sequence in X and lies in the closed ball U (x 0 , r) with radius Moreover, if (x n ) converges to a point ξ in X, then the following estimates hold: Proof. From (31) by induction on n ≥ 0, we get where b n = λ n 1−λ d(x 0 , x 1 ). Indeed, for all n, m ≥ 0 with m ≥ n, we have It follows from axiom (C3) that b n → 0 in Y . Then by Theorem 9.20(i) we conclude that (x n ) is a Cauchy sequence in X. Putting n = 0 in (35) we obtain that d(x m , x 0 ) b 0 for every m ≥ 0. Hence, the sequence (x n ) lies in the ball U (x 0 , r) since r = b 0 . Now suppose that (x n ) converges to a point ξ ∈ X. Then it follows from Theorem 9.20(ii) that (x n ) satisfies the inequality d(x n , ξ) b n (for every n ≥ 0) which proves (32). Applying (32) with n = 0, we conclude that the first two terms of the sequence (x n ) satisfy the inequality Note that for every n ≥ 0 the sequence (x n , x n+1 , x n+2 , . . .) also satisfies (31) and converges to ξ. Therefore, applying the last inequality to the first two terms of this sequence we get (33). The inequality (34) follows from (33)  Theorem 10.5 (Iterated contraction principle). Let (X, d) be a complete cone metric space over a solid vector space (Y, , ≺, →). Suppose T : D ⊂ X → X be a mapping satisfying the following conditions: . Then the following hold true: (i) Convergence of the iterative method. The Picard iteration (30) starting from x 0 is well defined, remains in the closed ball U (x 0 , r) and converges to a point ξ ∈ U(x 0 , r).
(ii) A priori error estimate. The following estimate holds: (iii) A posteriori error estimates. The following estimates hold: (iv) Existence of fixed points. If at least one of the conditions (F1 )-(F2 ) is satisfied, then ξ is a fixed point of T .
It follows from ρ(x 0 ) = r and (b) that the set U is not empty. We shall prove that T (U) ⊂ U. Let x be a given point in U. It follows from the definition of ρ that d(x, T x) ρ(x) which means that T x ∈ U (x, ρ(x)) ⊂ D. Therefore, T x ∈ D. Further, we shall show that Indeed, suppose that y ∈ U(T x, ρ(T x)). Then which means that y ∈ U (x, ρ(x)). Hence, U (T x, ρ(T x)) ⊂ D and so T x ∈ U . This proves that T (U) ⊂ U which means that Picard iteration (x n ) is well defined. From (a), we deduce that it satisfies (31). Now conclusions (i)-(iii) follow from Proposition 10.3. Conclusion (iv) follows from Proposition 10.1.
Remark 10.6. Obviously, whenever T is a selfmapping of X, condition (b) of Theorem 10.5 is satisfied automatically for every x 0 ∈ X and so it can be omitted. If D is closed and T (D) ⊂ D, then condition (b) also can be dropped. In a cone metric space setting full statement of the Banach contraction principle for a nonself mapping is given by the following theorem.
Theorem 11.1 (Banach contraction principle). Let (X, d) be a complete cone metric space over a solid vector space (Y, , ≺, →). Let T : D ⊂ X → X be a mapping satisfying the following conditions: There is x 0 ∈ D such that U (x 0 , r) ⊂ D, where r = 1 1−λ d(x 0 , T x 0 ). Then the following hold true: (i) Existence and uniqueness. T has a unique fixed point ξ in D.
(ii) Convergence of the iterative method. The Picard iteration (30) starting from x 0 is well defined, remains in the closed ball U (x 0 , r) and converges to ξ.
(v) Rate of convergence. The rate of convergence of the Picard iteration is given by d(x n+1 , ξ) λ d(x n , ξ) for all n ≥ 1; d(x n , ξ) λ n d(x 0 , ξ) for all n ≥ 0.
Proof. Using the triangle inequality and the contraction condition (a), one can see that condition (F4) holds with α = λ and β = 1. Conclusions (i)-(iv) with the exception of the uniqueness of the fixed point follow immediately from Theorem 10.5 since every contraction mapping is an iterated contraction mapping. Suppose T has two fixed points x, y ∈ D. Then it follows from (a) that d(x, y) λ d(x, y) which leads to (1 − λ)d(x, y) 0 and so d(x, y) 0. Hence, d(x, y) = 0 which means that x = y. Therefore, ξ is a unique fixed point of T in D. Conclusion (v) follows from (a) and (i) by putting x = x n and y = ξ.
Kirk in his paper [35] wrote for the Banach contraction principle in usual metric spaces the following "The great significance of Banach's principle, and the reason it is one of the most frequently cited fixed point theorems in all of analysis, lies in the fact that (i)-(v) contain elements of fundamental importance to the theoretical and practical treatment of mathematical equations". We would add that in general cone metrics give finer estimates than usual metrics.
Recall that a selfmapping T of a cone metric space (X, d) is called contraction on X if there exists 0 ≤ λ < 1 such that d(T x, T y) λ d(x, y) for all x, y ∈ X. The following short version of the Banach contraction principle for selfmappings in cone metric spaces follows immediately from Theorem 11.1. Note that the short version of Banach's principle follows also from the short version of Banach's principle in metric spaces and Theorem 9.5.
Theorem 11.2. Each contraction T on a cone metric space (X, d) over a solid vector space Y has a unique fixed point and for each x 0 ∈ X the Picard iteration (30) converges to the fixed point.

Conclusion
In the first part of this paper (Sections 2-7) we develop a unified theory for solid vector spaces. A real vector space Y with convergence (→) is called a solid vector space if it is equipped with a vector ordering ( ) and a strict vector ordering (≺). It turns out that every convergent sequence in a solid vector space has a unique limit. Every solid vector space Y can be endowed with an order topology τ such that x n → x implies x n τ → x. It turns out that the converse of this implication holds if and only if the space Y is normal, i.e. the Sandwich theorem holds in Y . Using the Minkowski functional, we show that the order topology on every solid vector space is normable with a monotone norm. Among the other results in this part of the paper, we show that an ordered vector space can be equipped with a strict vector ordering if and only if it has a solid positive cone. Moreover, if the positive cone of the vector ordering is solid, then there exists a unique strict vector ordering on this space.
In the second part of the paper (Sections 8-9) we develop a unified theory for cone metric spaces and cone normed spaces over a solid vector space. We show that every (complete) cone metric space (X, d) over a solid vector space Y is a (completely) metrizable topological space. Moreover, there exists an equivalent metric ρ on X that preserve some inequalities. In particular, an inequality of the type implies the inequality Using this result 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 a cone metric space is equivalent to its version in a metric space because the Banach contractive condition d(T x, T y) λ d(x, y) is of the type (44). Let us note that the above mentioned result cannot be applied to many contractive conditions in a cone metric space. That is why we need further properties of cone metric spaces. Further, we give some useful properties of cone metric spaces which allow us to prove convergence results for Picard iteration with a priori and a posteriori error estimates. Among the other results in this part of the paper we prove that every cone normed space over a solid vector space is normable.
In the third part of the paper (Sections 8-9) applying the cone metric theory we present full statements of the iterated contraction principle and the Banach contraction principle in cone metric spaces over a solid vector space.
Let us note that some of the results of the paper (Theorems 9.5, 9.12, 9.16 and 9.20; Propositions 10.3 and 10.1) give a method for obtaining convergence theorems (with error estimates) for Picard iteration and fixed point theorems in a cone metric space over a solid vector space.
Finally, let us note that we have come to the idea of a general theory of cone metric spaces (over a solid vector spaces) dealing with convergence problems of some iterative methods for finding all zeros of a polynomial f simultaneously (i.e., as a vector in C n , where n is the degree of f ). In our next papers we will continue studying the cone metric space theory and its applications. For instance, we shall show that almost all results given in Proinov [44,45] can be extended in cone metric spaces over a solid vector space. Also we shall present new convergence theorems for some iterative methods for finding zeros of a polynomial simultaneously. These results generalize, improve and complement a lot of of results given in the monographs of Sendov, Andreev, Kyurkchiev [55] and Petkovic [43]. In particular, it turns out that the cone norms in C n give better a priori and a posteriori error estimates for iterative methods in C n than usual norms.