A Cardinal Function on the Category of Metric Spaces

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1 International Journal of Contemporary Mathematical Sciences Vol. 9, 2014, no. 15, HIKARI Ltd, 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, F p 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

2 704 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi spaces (e.g. see [8]). Several papers has been published regarding the matter since 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, ) 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. 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 < P y if x P y and x y, moreover we say x P y or simply x y if y x P. We say (X, E, P, d) is a cone metric space if P is a cone in E and d : X X P for all a, b, c X satisfies the following conditions: d(a, b) = 0 if and only if a = b, 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 B d (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 l 2 (α) are isomorphic (as Hilbert spaces), where for nonempty set Γ we have: l 2 (Γ) = (x λ) λ Γ R Γ : x λ 2 < λ Γ equipped with inner product < (x λ ) λ Γ, (y λ ) λ Γ >:= λ Γ x λ y λ and therefore norm (x λ ) λ Γ = x λ 2 λ Γ 1 2 (for (x λ ) λ Γ, (y λ ) λ Γ l 2 (Γ)).

3 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 ℵ 0 or ω and the cardinality of R with c. Also it is well-known that l 2 (n) can be considered as R n 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, ) if V is a nonvoid open subset of F, then card(v ) = card(f) = max(c, α), where α = dim R (F) (α is the cardinality of a Hamel basis of F over R). Proof. For r > 0 let U r = {x F : x < r}. For s > 0, card(u r ) = card(u s ), since ϕ : U r U s with ϕ(x) = s r x is bijective. On the other hand F = {U n : n N}, which leads to (since for all n N we have card(u 1 ) = card(u n )): card(u 1 ) card(f) ℵ 0 card(u 1 ). Using ℵ 0 < card(r) card(f) ℵ 0 card(u 1 ) we have ℵ 0 card(u 1 ) = card(u 1 ). By card(u 1 ) card(f) ℵ 0 card(u 1 ) = card(u 1 ) we have: card(f) = card(u 1 ). Suppose V is a nonvoid open subset of F, there exist z V and r > 0 such that z + U r V. Since η : U 1 V with η(x) = rz + x is injective, we have: card(f) card(v ) card(u 1 ) = 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), 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) =.

4 706 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi Lemma 2.3 For α c we have card(l 2 (α)) = α. Proof. Consider α c, for all (x θ ) θ α l 2 (α) there exists a countable subset D of α such that for all θ α \ D we have x θ = 0. Therefore card(l 2 (α)) card({(d, (x θ ) θ D ) : D is a countable subset of α and x θ R for all α D}), which leads to card(l 2 (α)) card({(f, (x n ) n N ) : f : N α is an injection and x n R for all n N}) and: therefore card(l 2 (α)) = α. card(l 2 (α)) card(α N R N ) = α ℵ 0 c ℵ 0 = α Definition 2.4 In metric space (X, d) let Θ(X, d) := sup({0} {α CN : there exists cone metric space (X, l 2 (α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(x X) }). 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 {l 2 (α) : α CN \ {0}}, one may consider classification regarding F p = {l p (α) : α CN \ {0}} for 1 p +, so let Θ p (X, d) := sup({0} {α CN : there exists cone metric space (X, l p (α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(x X) }). And call Θ p (X, d), cone metric order of (X, d) with respect to F p, or simply F p - cone metric order of (X, d). Now one may be interested to study the relation between Θ p (X, d) and Θ q (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 = (G θ ) θ α where G 0 = 1 and G θ = 0 for θ 0. Moreover set P = {(x λ ) λ α l 2 (α) : x 0 0 ( β α ( x β x 0 ))}, Q = {(x θ ) θ α P : sup{ x θ : θ 0} < min(x 0, 1)},

5 A cardinal function on the category of metric spaces 707 X = l 2 (Q) \ {0}. Also for v Q, suppose u v = (δθ) v θ Q with δv v = 1 and δθ v = 0 for θ v. Now define d : X X R with: min( λ µ, 1) v Q, λ, µ R, x = λu v, y = µu 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 l 2 (α). Proof. We prove the lemma step by step: For β α the map ϕ β : l 2 (α) R 2 with ϕ β ((x λ ) λ α ) = (x 0, x β ) is continuous, since x β (x λ ) λ α and x 0 (x λ ) λ α. Therefore P = {ϕ 1 β ({(x, y) R2 : y x}) : β α} and P is close. Moreover clearly P P = {0}. Suppose x = (x λ ) λ α, y = (y λ ) λ α P and r, s > 0. We have: x, y P (x 0 0 y 0 0 ( β α ( x β x 0 y β y 0 ))) (rx 0 + sy 0 0 ( β α rx β + sy β rx 0 + sy 0 )) rx + sy P. If z = 2G = (z λ ) λ α, then z P and for x = (x λ ) λ α l 2 (α) we have: z x < 1 z 0 x 0 < 1 ( β α z β x β < 1) 2 x 0 < 1 ( β α \ {0} x β < 1) x 0 > 1 ( β α \ {0} x β < 1) x 0 0 ( β α \ {0} x β < x 0 ) x 0 0 ( β α x β x 0 ) x P Therefore {x l 2 (α) : z x < 1} P and z P. Note 3.2 For each (x θ ) θ α l 2 (α) with sup{ x θ : θ α} < 1 we have (x θ ) θ α < P 2G, i.e. 2G (x θ ) θ α P. Moreover G 0 and (x θ ) θ α < P 2G leads us to 3G (x θ ) θ α.

6 708 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi Lemma 3.3 (X, l 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 y and for all v Q, x / Ru v or y / Ru v, then ρ(x, y) = 4G P, on the other hand if there exists v Q, such that x, y Ru v, 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 Ru v 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 Ru v, 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 Ru v and z / Ru v, 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 Ru v and y / Ru v, 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 Ru v and x / Ru v, then ρ(x, y) + ρ(y, z) ρ(x, z) = 4G ρ(y, z) 4G = ρ(y, z) P. If x, y, z / Ru v 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). Proof. For all v Q we have ρ(u v, 1u 2 v) = v, thus 1 Q ρ(x X). Moreover 2 2 for y = (y θ ) θ α l 2 (α) we have: y G/2 < 1/4 2y G < 1/2 θ α 2y θ G θ < 1/2 2y 0 1 < 1/2 ( θ α \ {0} y θ < 1/4) 1/4 < y 0 < 3/4 sup{ y θ : θ 0} 1/4 y 0 > 0 sup{ y θ : θ 0} < y 0 = min(y 0, 1) y Q

7 A cardinal function on the category of metric spaces 709 Using { y l 2 (α) : y G < 4} 1 Q, we have: 2 { y l 2 (α) : which leads to the desired result. y G < 1 } Q ρ(x X) 2 Lemma 3.5 Induced topology of metric d on X and induced topology of cone metric ρ on X are the same. Proof. For all x X \ {Ru v : v Q}, we have {y X : ρ(x, y) < P 4G} = {x}. Therefore {y X : ρ(x, y) 4G} = {y X : d(x, y) < 1} = {x}. Now suppose x = λu v X for λ R \ {0} and v = (t θ ) θ α Q. If ε 0, then there exists real number r < min(t 0 sup{ t θ : θ 0}, 1) such that 0 < r and 0 6rt 0 G ε (t 0 > 0 since v Q). For all y X \ {x}, if d(x, y) < r, then y Ru v \ {0} and ρ(x, y) = d(x, y)v( 0). Moreover v 6t 0 G, thus Hence: ρ(x, y) = d(x, y)v 6t 0 d(x, y)g 6t 0 rg ε. ε 0 r > 0 {y X : d(x, y) < r} {y X : ρ(x, y) ε}. Conversely, if r > 0, using t 0 > 0, there exists s > 0 such that s t 0 < r and s < 4. We have ε := sg 0. For all y X if ρ(x, y) P sg < P 4G, then y Ru v \ {0} and we have: Hence: ρ(x, y) sg sg d(x, y)v P P s d(x, y)t 0 0 d(x, y) s t 0 < r. 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. So by Lemma 2.1, card(q ) = card(l 2 (α)), which leads to card(q) = card(l 2 (α)) = α. Therefore card(x) = card(l 2 (Q)) = card(l 2 (α)) = α.

8 710 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 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) ((x 1, y 1 ), (x 2, y 2 )) = d(x 1, x 2 ) + k(y 1, y 2 ) for all (x 1, y 1 ), (x 2, y 2 ) 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)((x 1, y 1 ), (x 2, y 2 )) = (d(x 1, x 2 ), k(y 1, y 2 )) for all (x 1, y 1 ), (x 2, y 2 ) 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).

9 A cardinal function on the category of metric spaces 711 Proof. Note to the fact that for (ϱ 1, ϱ 2 ) E F we have (0, 0) P Q (ϱ 1, ϱ 2 ) if and only if 0 P ϱ 1 and 0 Q ϱ 2. Thus if 0 P ε 1 and 0 Q ε 2 and (x, y) E F, then we have: {(z, w) E F : (d k)((z, w), (x, y)) P Q (ε 1, ε 2 )} = {z E : d(z, x) P ε 1 } {w F : k(w, y) Q ε 2 }, which leads to the desired result. Remark 4.4 For nonzero cardinal numbers α, β CN, l 2 (α) l 2 (β) and l 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: Proof. Use Corollary 4.5. Θ(X Y, σ (d,k) ) Θ(X, d) + Θ(Y, k). Note 4.7 If (X, E, P, d) and (Y, E, P, k) are cone metric spaces and: σ (d,k) ((x 1, y 1 ), (x 2, y 2 )) = d(x 1, x 2 ) + k(y 1, y 2 ) ((x 1, y 1 ), (x 2, y 2 ) X Y ), then (X Y, E, P, σ (d,k) ) is a cone metric space. 4.2 More details on R n For n N suppose E n is usual metric on R n induced from its Euclidean norm. Using Theorem 4.6 two metric spaces (R n+m, σ (En,E m)) and (R n+m, E n+m ) are homeomorph, hence we have Θ(R n, E n ) nθ(r, E 1 ) 1. Here we want to prove Θ(R n, E n ) n directly. Note 4.8 For nonzero n ω let: P = {(x 1,..., x n ) R n : x 1 0,..., x n 0}, then P is a solid cone in l 2 (n) = R n with Euclidean norm. Theorem 4.9 For nonzero n ω, consider R n under its usual metric induced from Euclidean norm on R n, we denote this metric on R n with E n. Then Θ(R n, E n ) n.

10 712 F. Ayatollah Zadeh Shirazi and Z. Farabi Khanghahi Proof. Let P = {(x 1,..., x n ) R n : x 1 0,..., x n 0} as in Note 4.8. Define d : R n R n P with d((x 1,..., x n ), (y 1,..., y n )) = ( x 1 y 1,..., x n y n ). In cone metric space (R n, R n, P, d) we have d(r n {x}) = P and d(r n {x}) = P for all x R n. Moreover for all ε = (ε 1,..., ε n ) 0 and x = (x 1,..., x n ) R n we have ε 1,..., ε n > 0 and: {y R n : d(x, y) ε} = {(y 1,..., y n ) R n : (ε 1 x 1 y 1,..., ε n x n y n ) 0} = {(y 1,..., y n ) R n : x 1 y 1 < ε 1,..., x n y n < ε n } = (x 1 ε 1, x 1 + ε 1 ) (x n ε n, x n + ε n ) Therefore B d (x, ε) is an open subset of (R n, E n ). Moreover {(x 1 µ, x 1 +µ) (x n µ, x n + µ) : x 1,..., x n R, µ > 0} (= {B d (x, (µ,..., µ)) : µ > 0, x R n }) is a topological basis for (R n, E n ). Therefore induced topology from metric space (R n, E n ) and cone metric space (R n, R n, P, d) (= (R n, l 2 (n), P, d)) are coincide. Hence Θ(R n, E n ) n. 5 Some Arising Problems For each nonzero cardinal number α, suppose T α is the class of all metric spaces (X, d) such that Θ(X, d) α. Using Theorem 3.9, T α for each nonzero α CN. Also for α c, T 2 α is a proper subclass of T α. 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 T 1. 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, l p (α), P, ρ) which induces the same topology as metric topology on (X, d) and ρ(x X)? In other words can we replace sup in Definition 2.4 with max?

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