A Cardinal Function on the Category of Metric Spaces 1

International Journal of Contemporary Mathematical Sciences Vol. 9, 2014, no. 15, 703 - 713 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijcms.2014.4442

A Cardinal Function on the Category of Metric Spaces

Fatemah Ayatollah Zadeh Shirazi

Faculty of Mathematics, Statistics and Computer Science, College of Science University of Tehran, Enghelab Ave., Tehran, Iran

Zakieh Farabi Khanghahi

Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran, Babolsar, Iran

Copyright c 2014 Fatemah Ayatollah Zadeh Shirazi and Zakieh Farabi Khanghahi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In the following text in the metric spaces category we introduce a topologically invariant cardinal function, Θ (clearly Θ is a tool to classify metric spaces). For this aim in metric space X we consider cone metrics (X, H, P, ρ) such that H is a Hilbert space, image of ρ has nonempty interior, and this cone metric induces the original metric topology on X. We prove that for all suﬃciently large cardinal numbers α, there exists a metric space (X, d) with Θ(X, d) = α. Mathematics Subject Classiﬁcation: 54E35, 46B99

Keywords: cone, cone metric order, Fp−cone metric order, Hilbert space, Hilbert-cone metric order, metric space, solid cone

1 Introduction

As it has been mentioned in several texts, cone metric spaces in the following form has been introduced for the ﬁrst time in [9] as a generalization of metric 704 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi spaces (e.g. see [8]). Several papers has been published regarding the matter since 2007. A large number of these papers deal with ﬁxed point theorems (e.g. see [9], [1], [6]), however there are texts deal with the other properties (e.g., [2]) amongst these texts some authors try to study metrizability or interaction between metric spaces and cone metric spaces ([5], [3], [4]). In this text R denotes the set of all real numbers, and N = {1, 2,...} denotes the set of all natural numbers. Note. All vector spaces assumed in this text are nonzero real vector spaces. In (real) norm vector space (E, k k) we say P(⊆ E) is a cone if: • P is a closed nonempty subset of E,

• for all x, y ∈ P and λ, µ ≥ 0 we have λx + µy ∈ P,

• P ∩ −P = {0}, and in addition it is a solid cone in E, if P◦ 6= ∅. Suppose P is a cone in norm vector space E. For all x, y ∈ E we say x ≤P y if y − x ∈ P. Obviously in this way (E, ≤P) is a partial ordered set. For all x, y ∈ E we say x

• d(a, b) = d(b, a),

• d(a, b) ≤P d(a, c) + d(c, b). In the cone metric space (X, E, P, d) for ε 0 and x ∈ E let Bd(x, ε) or simply B(x, ε) is {z ∈ E : d(z, x) ε}. It is easy to see that {B(x, ε): x ∈ E, ε 0} is a topological basis on X. We consider cone metric space (X, E, P, d) under topology generated by the above basis on X. It is well-known that for every (real) Hilbert space H there exists a nonzero cardinal number α such that H and `2(α) are isomorphic (as Hilbert spaces), where for nonempty set Γ we have:   2  Γ X 2  ` (Γ) = (xλ)λ∈Γ ∈ R : |xλ| < ∞  λ∈Γ 

X equipped with inner product < (xλ)λ∈Γ, (yλ)λ∈Γ >:= xλyλ and therefore λ∈Γ 1   2 X 2 2 norm k(xλ)λ∈Γk =  |xλ|  (for (xλ)λ∈Γ, (yλ)λ∈Γ ∈ ` (Γ)). λ∈Γ A cardinal function on the category of metric spaces 705

We recall that for all nonzero cardinal number α, we have α = {β ∈ ON : β < α} where ON is the class of all ordinal numbers (by CN we mean the class of all cardinal numbers which is a proper subclass of ON). We denote the least inﬁnite cardinal number with ℵ0 or ω and the cardinality of R with c. Also it is well-known that `2(n) can be considered as Rn with Euclidean norm for 1 ≤ n < ω.

2 First steps

In this section we get ready to deﬁne Hilbert-cone metric order of a metric space (X, d).

Lemma 2.1 In norm vector space (F, k k) if V is a nonvoid open subset of F, then card(V ) = card(F) = max(c, α), where α = dimR(F) (α is the cardinality of a Hamel basis of F over R).

Proof. For r > 0 let Ur = {x ∈ F : kxk < r}. For s > 0, card(Ur) = card(Us), s S since ϕ : Ur → Us with ϕ(x) = r x is bijective. On the other hand F = {Un : n ∈ N}, which leads to (since for all n ∈ N we have card(U1) = card(Un)):

card(U1) ≤ card(F) ≤ ℵ0card(U1) .

Using ℵ0 < card(R) ≤ card(F) ≤ ℵ0card(U1) we have ℵ0card(U1) = card(U1). By card(U1) ≤ card(F) ≤ ℵ0card(U1) = card(U1) we have:

card(F) = card(U1) .

Suppose V is a nonvoid open subset of F, there exist z ∈ V and r > 0 such that z + Ur ⊆ V . Since η : U1 → V with η(x) = rz + x is injective, we have:

card(F) ≥ card(V ) ≥ card(U1) = card(F) .

Therefore card(V ) = card(F) = max(c, α).

Theorem 2.2 In cone metric space (X, E, P, d) if d(X × X)◦ 6= ∅, then:

card(X) ≥ card(E) .

Proof. Using Lemma 2.1, card(d(X × X)◦ = card(E). Moreover

card(X × X) ≥ card(d(X × X)) ≥ card(d(X × X)◦) = card(E) and X is inﬁnite, therefore card(X) = card(X × X) ≥ card(E). Using Theorem 2.2, in cone metric space (X, E, P, d) if card(X) < card(E), then d(X × X)◦ = ∅. 706 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi

Lemma 2.3 For α ≥ c we have card(`2(α)) = α.

2 Proof. Consider α ≥ c, for all (xθ)θ∈α ∈ ` (α) there exists a countable subset 2 D of α such that for all θ ∈ α \ D we have xθ = 0. Therefore card(` (α)) ≤ card({(D, (xθ)θ∈D): D is a countable subset of α and xθ ∈ R for all α ∈ D}), 2 which leads to card(` (α)) ≤ card({(f, (xn)n∈N): f : N → α is an injection and xn ∈ R for all n ∈ N}) and:

card(`2(α)) ≤ card(αN × RN) = αℵ0 cℵ0 = α therefore card(`2(α)) = α.

Deﬁnition 2.4 In metric space (X, d) let

Θ(X, d) := sup({0} ∪ {α ∈ CN : there exists cone metric space (X, `2(α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(X × X)◦ 6= ∅}).

Using Theorem 2.2 and Lemma 2.3, Θ(X, d)(≤ card(X)) exists. We call Θ(X, d), cone metric order of (X, d) with respect to Hilbert spaces, or simply Hilbert-cone metric order of (X, d).

Note. Naturally Θ gives a classiﬁcation of metric spaces. Instead of classifying metric spaces regarding collection {`2(α): α ∈ CN \{0}}, one may consider p classiﬁcation regarding Fp = {` (α): α ∈ CN \{0}} for 1 ≤ p ≤ +∞, so let

Θp(X, d) := sup({0} ∪ {α ∈ CN : there exists cone metric space (X, `p(α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(X × X)◦ 6= ∅}).

And call Θp(X, d), cone metric order of (X, d) with respect to Fp, or simply Fp- cone metric order of (X, d). Now one may be interested to study the relation between Θp(X, d) and Θq(X, d) for p, q ≥ 1.

3 Towards main theorem: A useful example

The main aim of this section is to prove that for β ≥ c there exists metric space (Z, ψ) with Θ(Z, ψ) = β. In this section suppose α ∈ CN is a nonzero cardinal number and G = (Gθ)θ∈α where G0 = 1 and Gθ = 0 for θ 6= 0. Moreover set 2 P = {(xλ)λ∈α ∈ ` (α): x0 ≥ 0 ∧ (∀β ∈ α (|xβ| ≤ x0))} , Q = {(xθ)θ∈α ∈ P : sup{|xθ| : θ 6= 0} < min(x0, 1)} , A cardinal function on the category of metric spaces 707

X = `2(Q) \{0} . v v v Also for v ∈ Q, suppose uv = (δθ )θ∈Q with δv = 1 and δθ = 0 for θ 6= v. Now deﬁne d : X × X → R with:  min(|λ − µ|, 1) v ∈ Q, λ, µ ∈ R, x = λu , y = µu  v v d(x, y) = 0 x = y  4 otherwise and ρ : X × X → P with:  d(x, y)v v ∈ Q, x, y ∈ Ru  v ρ(x, y) = 0 x = y  4G otherwise

Lemma 3.1 The set P is a solid cone in `2(α). Proof. We prove the lemma step by step:

2 2 • For β ∈ α the map ϕβ : ` (α) → R with ϕβ((xλ)λ∈α) = (x0, xβ) is continuous, since |xβ| ≤ k(xλ)λ∈αk and |x0| ≤ k(xλ)λ∈αk. Therefore

\ −1 2 P = {ϕβ ({(x, y) ∈ R : |y| ≤ x}): β ∈ α} and P is close. Moreover clearly P ∩ −P = {0}.

• Suppose x = (xλ)λ∈α, y = (yλ)λ∈α ∈ P and r, s > 0. We have:

x, y ∈ P ⇒ (x0 ≥ 0 ∧ y0 ≥ 0 ∧ (∀β ∈ α (|xβ| ≤ x0 ∧ |yβ| ≤ y0)))

⇒ (rx0 + sy0 ≥ 0 ∧ (∀β ∈ α |rxβ + syβ| ≤ rx0 + sy0)) ⇒ rx + sy ∈ P .

2 • If z = 2G = (zλ)λ∈α, then z ∈ P and for x = (xλ)λ∈α ∈ ` (α) we have:

kz − xk < 1 ⇒ |z0 − x0| < 1 ∧ (∀β ∈ α |zβ − xβ| < 1)

⇒ |2 − x0| < 1 ∧ (∀β ∈ α \{0} |xβ| < 1)

⇒ x0 > 1 ∧ (∀β ∈ α \{0} |xβ| < 1)

⇒ x0 ≥ 0 ∧ (∀β ∈ α \{0} |xβ| < x0)

⇒ x0 ≥ 0 ∧ (∀β ∈ α |xβ| ≤ x0) ⇒ x ∈ P

Therefore {x ∈ `2(α): kz − xk < 1} ⊆ P and z ∈ P◦.

2 Note 3.2 For each (xθ)θ∈α ∈ ` (α) with sup{|xθ| : θ ∈ α} < 1 we have (xθ)θ∈α

Lemma 3.3 (X, `2(α), P, ρ) is a cone metric space. Proof. It is clear that (X, d) is a metric space. Suppose x, y, z ∈ X. If x 6= y and for all v ∈ Q, x∈ / Ruv or y∈ / Ruv, then ρ(x, y) = 4G ∈ P, on the other hand if there exists v ∈ Q, such that x, y ∈ Ruv, then d(x, y) ≥ 0 and v ∈ P, thus ρ(x, y) = d(x, y)v ∈ P. Moreover (note to the fact that 0 ∈/ Q):

ρ(x, y) = 0 ⇔ x = y ∨ (∃v ∈ Q(x, y ∈ Ruv ∧ 0 = ρ(x, y) = d(x, y)v)) ⇔ x = y ∨ d(x, y) = 0 ⇔ x = y It is clear that ρ(x, y) = ρ(y, x). We complete the proof, using the following cases:

• For v ∈ Q if x, y, z ∈ Ruv, then d(x, y) + d(y, z) − d(x, z) ≥ 0, thus: ρ(x, y) + ρ(y, z) − ρ(x, z) = (d(x, y) + d(y, z) − d(x, z))v ∈ P .

• For v ∈ Q if x, y ∈ Ruv and z∈ / Ruv, then: ρ(x, y) + ρ(y, z) − ρ(x, z) = ρ(x, y) + 4G − 4G = ρ(x, y) ∈ P .

• For v ∈ Q if x, z ∈ Ruv and y∈ / Ruv, then using Note 3.2, 2G−d(x, z)v ∈ P, since d(x, z) ≤ 1. Thus ρ(x, y) + ρ(y, z) − ρ(x, z) = 8G − ρ(x, z) = 6G + (2G − d(x, z)v) ∈ P + P ⊆ P.

• For v ∈ Q if y, z ∈ Ruv and x∈ / Ruv, then ρ(x, y) + ρ(y, z) − ρ(x, z) = 4G − ρ(y, z) − 4G = ρ(y, z) ∈ P.

◦ • If x, y, z∈ / Ruv for all v ∈ P , and x, y, z are pairwise distinct, then ρ(x, y) + ρ(y, z) − ρ(x, z) = 4G + 4G − 4G = 4G ∈ P. • If x = y or y = z, then ρ(x, y) + ρ(y, z) − ρ(x, z) = 0 ∈ P. If x = z, then ρ(x, y) + ρ(y, z) − ρ(x, z) = ρ(x, y) + ρ(y, z) ∈ P + P ⊆ P.

Therefore ρ(x, y) + ρ(y, z) − ρ(x, z) ∈ P and ρ(x, z) ≤P ρ(x, y) + ρ(y, z). Lemma 3.4 ρ(X × X)◦ 6= ∅. 1 v 1 Proof. For all v ∈ Q we have ρ(uv, 2 uv) = 2 , thus 2 Q ⊆ ρ(X × X). Moreover 2 for y = (yθ)θ∈α ∈ ` (α) we have: ky − G/2k < 1/4 ⇒ k2y − Gk < 1/2

⇒ ∀θ ∈ α |2yθ − Gθ| < 1/2

⇒ |2y0 − 1| < 1/2 ∧ (∀θ ∈ α \{0} |yθ| < 1/4)

⇒ 1/4 < y0 < 3/4 ∧ sup{|yθ| : θ 6= 0} ≤ 1/4

⇒ y0 > 0 ∧ sup{|yθ| : θ 6= 0} < y0 = min(y0, 1) ⇒ y ∈ Q A cardinal function on the category of metric spaces 709

n o 2 G 1 Using y ∈ ` (α): y − 2 < 4 ⊆ Q, we have:

2 G 1 1 y ∈ ` (α): y − < ⊆ Q ⊆ ρ(X × X) 4 8 2 which leads to the desired result. Lemma 3.5 Induced topology of metric d on X and induced topology of cone metric ρ on X are the same. S Proof. For all x ∈ X \ {Ruv : v ∈ Q}, we have {y ∈ X : ρ(x, y)

0 since v ∈ Q). For all y ∈ X \{x}, if d(x, y) < r, then y ∈ Ruv \{0} and ρ(x, y) = d(x, y)v(6= 0). Moreover v 6t0G, thus

ρ(x, y) = d(x, y)v 6t0d(x, y)G 6t0rG ε .

Hence:

∀ε 0 ∃r > 0 {y ∈ X : d(x, y) < r} ⊆ {y ∈ X : ρ(x, y) ε} .

s Conversely, if r > 0, using t0 > 0, there exists s > 0 such that < r and t0 s < 4. We have ε := sG 0. For all y ∈ X if ρ(x, y) P sG

ρ(x, y) sG ⇒ sG − d(x, y)v ∈ P◦ ⊆ P

⇒ s − d(x, y)t0 ≥ 0 s ⇒ d(x, y) ≤ < r . t0 Hence:

∀r > 0 ∃ε 0 {y ∈ X : ρ(x, y) ε} ⊆ {y ∈ X : d(x, y) < r} , which completes the proof. Corollary 3.6 (X, d) is a disconnected space with Θ(X, d) ≥ α. Proof. Use Lemma 3.3, Lemma 3.4 and Lemma 3.5. Lemma 3.7 If α ≥ c, then card(X) = α. Proof. Using the proof of Lemma 3.4 we have Q◦ 6= ∅. So by Lemma 2.1, card(Q◦) = card(`2(α)), which leads to card(Q) = card(`2(α)) = α. Therefore card(X) = card(`2(Q)) = card(`2(α)) = α. 710 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi

Lemma 3.8 If α ≥ c, then Θ(X, d) = α. Proof. Using Lemma 3.7, Corollary 3.6 and Θ(X, d) ≤ card(X), we have Θ(X, d) = α. Theorem 3.9 (Main Theorem) For β ≥ c there exists (disconnected) metric space (Z, ψ) with Θ(Z, ψ) = β. Proof. Use Lemma 3.8.

4 Products of metric spaces and Θ function

In this section for two metric spaces (X, d) and (Y, k), consider X × Y under metric σ(d,k), where σ(d,k)((x1, y1), (x2, y2)) = d(x1, x2) + k(y1, y2) for all (x1, y1), (x2, y2) ∈ X × Y . Also in two norm vector spaces (E, ρ) and (F, µ), consider norm vector space (E × F, σ(ρ,µ)), where σ(ρ,µ)(x, y) = ρ(x) + µ(y) for all (x, y) ∈ E × F.

4.1 Product of two cone metric space For f : A → B and g : C → D deﬁne f × g : A × C → B × D with (f × g)(x, y) = (f(x), g(y)). In the following we will prove Θ(X × Y, σ(d,k)) ≥ Θ(X, d) + Θ(Y, k) for two metric spaces (X, d) and (Y, k). Lemma 4.1 If P is a cone in norm vector space E and Q is a cone in norm vector space F, then P × Q is a cone in norm vector space E × F. Moreover P and Q are solid if and only if P × Q is solid. Proof. It is clear that −(P × Q) ∩ (P × Q) = (−P ∩ P) × (−Q ∩ Q) = ∅ × ∅ = ∅. On the other hand for (x, y), (z, w) ∈ P × Q and λ, µ ≥ 0 we have λx + µz ∈ P and λy + µw ∈ Q, therefore λ(x, y) + µ(z, w) = (λx + µz, λy + µw) ∈ P × Q. P × Q is a closed nonempty subset of E × F, since P is a nonempty closed subset of E and Q is a nonempty closed subset of F. Thus P × Q is a cone in E × F. Note to the fact that (A × B)◦ = A◦ × B◦, to complete the proof. Lemma 4.2 If (X, E, P, d) and (Y, F, Q, k) are cone metric spaces, then (X×Y, E×F, P×Q, d×k) is a cone metric space, where (d×k)((x1, y1), (x2, y2)) = (d(x1, x2), k(y1, y2)) for all (x1, y1), (x2, y2) ∈ X × Y . Proof. Use Lemma 4.1. Theorem 4.3 (Product of cone metric spaces) Consider two cone metric spaces (X, E, P, d) and (Y, F, Q, k). Cone metric topology on (X × Y, E × F, P × Q, d × k) is product topology on X × Y when X is considered under cone metric topology (X, E, P, d) and Y considered under cone metric topology (Y, F, Q, k). A cardinal function on the category of metric spaces 711

Proof. Note to the fact that for (%1,%2) ∈ E × F we have (0, 0) P×Q (%1,%2) if and only if 0 P %1 and 0 Q %2. Thus if 0 P ε1 and 0 Q ε2 and (x, y) ∈ E × F, then we have: {(z, w) ∈ E × F :(d × k)((z, w), (x, y)) P×Q (ε1, ε2)} = {z ∈ E : d(z, x) P ε1} × {w ∈ F : k(w, y) Q ε2}, which leads to the desired result.

Remark 4.4 For nonzero cardinal numbers α, β ∈ CN, `2(α) × `2(β) and `2(α + β) are isomorphic.

Corollary 4.5 If (X, d) and (Y, k) are metric spaces and Θ(X, d) ≥ α and Θ(Y, k) ≥ β, then Θ(X × Y, σ(d,k)) ≥ α + β.

Proof. Use Theorem 4.3 and Remark 4.4.

Theorem 4.6 If (X, d) and (Y, k) are metric spaces, then:

Θ(X × Y, σ(d,k)) ≥ Θ(X, d) + Θ(Y, k) .

Proof. Use Corollary 4.5.

Note 4.7 If (X, E, P, d) and (Y, E, P, k) are cone metric spaces and:

σ(d,k)((x1, y1), (x2, y2)) = d(x1, x2) + k(y1, y2) ((x1, y1), (x2, y2) ∈ X × Y ) , then (X × Y, E, P, σ(d,k)) is a cone metric space.

4.2 More details on Rn

n For n ∈ N suppose En is usual metric on R induced from its Euclidean norm. n+m n+m Using Theorem 4.6 two metric spaces (R , σ(En,Em)) and (R ,En+m) are n homeomorph, hence we have Θ(R ,En) ≥ nΘ(R,E1) ≥ 1. Here we want to n prove Θ(R ,En) ≥ n directly.

Note 4.8 For nonzero n ∈ ω let:

n P = {(x1, ..., xn) ∈ R : x1 ≥ 0, ..., xn ≥ 0} , then P is a solid cone in `2(n) = Rn with Euclidean norm.

Theorem 4.9 For nonzero n ∈ ω, consider Rn under its usual metric in- n n duced from Euclidean norm on R , we denote this metric on R with En. n Then Θ(R ,En) ≥ n. 712 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi

n Proof. Let P = {(x1, ..., xn) ∈ R : x1 ≥ 0, ..., xn ≥ 0} as in Note 4.8. Deﬁne n n d : R × R → P with d((x1, ..., xn), (y1, ..., yn)) = (|x1 − y1|, ..., |xn − yn|). In cone metric space (Rn, Rn, P, d) we have d(Rn × {x}) = P and d(Rn × ◦ ◦ n {x}) = P 6= ∅ for all x ∈ R . Moreover for all ε = (ε1, ..., εn) 0 and n x = (x1, ..., xn) ∈ R we have ε1, ..., εn > 0 and:

n n {y ∈ R : d(x, y) ε} = {(y1, ..., yn) ∈ R :(ε1 − |x1 − y1|, ..., εn − |xn − yn|) 0} n = {(y1, ..., yn) ∈ R : |x1 − y1| < ε1, ..., |xn − yn| < εn}

= (x1 − ε1, x1 + ε1) × · · · × (xn − εn, xn + εn)

d n Therefore B (x, ε) is an open subset of (R ,En). Moreover {(x1 −µ, x1 +µ)× d · · · × (xn − µ, xn + µ): x1, ..., xn ∈ R, µ > 0} (= {B (x, (µ, ..., µ)) : µ > 0, x ∈ n n R }) is a topological basis for (R ,En). Therefore induced topology from met- n n n n 2 ric space (R ,En) and cone metric space (R , R , P, d) (= (R , ` (n), P, d)) n are coincide. Hence Θ(R ,En) ≥ n.

5 Some Arising Problems

For each nonzero cardinal number α, suppose Tα is the class of all metric spaces (X, d) such that Θ(X, d) ≥ α. Using Theorem 3.9, Tα 6= ∅ for each nonzero α ∈ CN. Also for α ≥ c, T2α is a proper subclass of Tα. Also it is well-known [7] that for all connected metric space (X, d) with at least two distinct elements a, b we have [0, d(a, b)) ⊆ d(X × {a}), therefore considering cone metric space (X, R, [0, +∞), d) leads us to Θ(X, d) ≥ 1. In other words: The class of all connected metric spaces with at least two elements is a subclass of T1. However in Theorem 3.9, we prove that for α ≥ c there exists metric space (X, d) with Θ(X, d) = α. Now we have the following problems: Problem 5.1 For nonzero α < c, ﬁnd a metric space (X, d) with Θ(X, d) = α.

Problem 5.2 For nonzero α ∈ CN ﬁnd a connected metric space (X, d) with Θ(X, d) = α. In particular {{(X, d):(X, d) is a connected metric space with Θ(X, d) = α} : α is a nonzero cardinal number} is a meaningful partition of the class of all connected metric spaces?

Problem 5.3 For α ≥ ω in metric space (X, d) suppose Θ(X, d) = α. Is there any cone metric space (X, `p(α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(X × X)◦ 6= ∅? In other words can we replace “sup” in Deﬁnition 2.4 with “max”? A cardinal function on the category of metric spaces 713

Acknowledgement

The authors are grateful to the research division of the University of Tehran, for the grant which supported this research under the ref. no. 6103027/1/06.

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Received: April 15, 2014; Published: November 27, 2014