# On Random Topological Structures

Authors: Ravi P. Agarwal; Yeol Je Cho; Reza Saadati
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We present some topics about random spaces. The main purpose of this paper is to study topological structure of random normed
spaces and random functional analysis. These subjects are important to the study of nonlinear analysis in random normed spaces.


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## Body

## 1. Introduction

The purpose of this paper is to give a comprehensive text to the study of nonlinear random analysis such as the study of fixed point theory, approximation theory and stability of functional equations in random normed spaces. The notion of random normed space goes back to Šherstnev [1] and Hadžić [2–4] who were dulled from Menger [5], and Schweizer and Sklar [4] works. Some authors [6–11] considered some properties of probabilistic normed and metric spaces also fuzzy metric and normed spaces [12–21]. Fixed point theory [3, 22–26], approximation theory [27–31], and stability of functional equations [32–38] are studied at random normed space and its depended space that is, fuzzy normed space.This paper is introduced as a survey of the latest and new results on the following topics.(i)
Basic theory of triangular norms.(ii)
Topological structure of random normed spaces.(iii)
Random functional analysis.(iv)
Relationship between random normed spaces and fuzzy normed spaces.


## 2. Triangular Norms

Triangular norms first appeared in the framework of probabilistic metric spaces in the work Menger [5]. It also turns out that this is a crucial operation in several fields. Triangular norms are an indispensable tool for the interpretation of the conjunction in fuzzy logics [39] and, subsequently, for the intersection of fuzzy sets [40]. They are, however, interesting mathematical objects themselves. We refer to some papers and books for further details (see [2, 4, 41–44]).Definition 2.1.
Atriangular norm (shorter t-norm) is a binary operation on the unit interval [0,1], that is, a function T:[0,1]×[0,1]→[0,1] such that for all a,b,c∈[0,1] the following four axioms are satisfied:
(T1)T(a,b)=T(b,a) (commutativity);
(T2)T(a,(T(b,c)))=T(T(a,b),c) (associativity);
(T3)T(a,1)=a  (boundary condition);
(T4)T(a,b)≤T(a,c) whenever b≤c (monotonicity).The commutativity of (T1), boundary condition (T3), and the the monotonicity (T4) imply that for eacht-norm T and for each x∈[0,1] the following boundary conditions are also satisfied:
(2.1)T(x,1)=T(1,x)=x,T(x,0)=T(0,x)=0,
and therefore all t-norms coincide on the boundary of the unit square [0,1]2.The monotonicity of at-norm T in its second component (T3) is, together with the commutativity (T1), equivalent to the (joint) monotonicity in both components, that is, to
(2.2)T(x1,y1)≤T(x2,y2)wheneverx1≤x2,y1≤y2.
Basic examples are the Łukasiewicz t-norm TL:
(2.3)TL(a,b)=max⁡(a+b-1,0),∀a,b∈[0,1]
and the t-norms TP,TM,TD,  where
(2.4)TP(a,b):=ab,TM(a,b):=min⁡{a,b},TD(a,b):={min⁡(a,b),ifmax⁡(a,b)=1,0,otherwise.If, for twot-norms T1 and T2, the inequality T1(x,y)≤T2(x,y) holds for all (x,y)∈[0,1]2, then we say that T1 is weaker than T2 or, equivalently, that T2 is stronger than T2.As a result of (2.2) we obtain, for each (x,y)∈[0,1]2,
(2.5)T(x,y)≤T(x,1)=x,T(x,y)≤T(1,y)=y.
Since, for all (x,y)∈(0,1)2, trivially T(x,y)≥0=TD(x,y), we get for an arbitrary t-norm T,
(2.6)TD≤T≤TM.
That is, TD is weaker and TM is stronger than any other t-norms. Also since TL<TP, we obtain the following ordering for four basic t-norms
(2.7)TD<TL<TP<TM.Proposition 2.2 (see [2]).
(i) The minimumTM is the only t-norm satisfying T(x,x)=x for all x∈(0,1).
(ii) The weakestt-norm TD is the only t-norm satisfying T(x,x)=0for all x∈(0,1).Proposition 2.3 (see [2]).
At-norm T is continuous if and only if it is continuous in its first component, that is, if for each y∈[0,1] the one-place function
(2.8)T(⋅,y):[0,1]⟶[0,1],x⟼T(x,y),
is continuous.For example, the minimumTM and Łukasiewicz t-norm TL are continuous but the t-norm TΔ defined by
(2.9)TΔ(x,y):={xy2,ifmax⁡(x,y)<1,xy,otherwise,
for x,y∈[0,1], is not continuous.Definition 2.4.
(i) At-norm T is said to be strictly monotone if
(2.10)T(x,y)<T(x,z)wheneverx∈(0,1),y<z.(ii) A t-norm T is said to be strict if it is continuous and strictly monotone.For example, thet-norm TΔ is strictly monotone but the minimum TM and Łukasiewicz t-norm TL are not.Proposition 2.5 (see [2]).
At-norm T is strictly monotone if and only if
(2.11)T(x,y)=T(x,z),x>0⟹y=z.IfT is a t-norm, then xT(n) is defined for every x∈[0,1] and n∈N∪{0} by 1, if n=0 and T(xT(n-1),x), if n≥1.Definition 2.6.
At-norm T is said to be Archimedean if for all (x,y)∈(0,1)2 there is an integer n∈ℕ such that
(2.12)xT(n)<y.Proposition 2.7 (see [2]).
At-norm T is Archimedean if and only if for each x∈(0,1) one has
(2.13)lim⁡n→∞xT(n)=0.Proposition 2.8 (see [2]).
Ift-norm T is Archimedean, then for each x∈(0,1) one has
(2.14)T(x,x)<x.For example, the productTP, Łukasiewicz t-norm TL, and weakest t-norm TD are all Archimedean but the minimum TM is not an Archimedean t-norm.At-norm T is said to be of Hadžić-type (we denote by T∈ℋ) if the family (xT(n))n∈N is equicontinuous at x=1, that is,
(2.15)∀ɛ∈(0,1)∃δ∈(0,1),x>1-δ⟹xT(n)>1-ɛ,∀n≥1.TM is a trivial example of a t-norm of Hadžić type, but TP is not of Hadžić type.Proposition 2.9 (see [2]).
If a continuoust-norm T is Archimedean, then it cannot be a t-norm of Hadžić type.Other important triangular norms are [see [45]]:(i)
theSugeno-Weber family{TλSW}λ∈[-1,∞] is defined by T-1SW=TD,T∞SW=TP and
(2.16)TλSW(x,y)=max⁡(0,x+y-1+λxy1+λ)if λ∈(-1,∞);(ii)
theDomby family{TλD}λ∈[0,∞], defined by TD, if λ=0,TM, if λ=∞ and
(2.17)TλD(x,y)=11+(((1-x)/x)λ+((1-y)/y)λ)1/λif λ∈(0,∞);(iii)
theAczel-Alsina family{TλAA}λ∈[0,∞], defined by TD, if λ=0,TM, if λ=∞ and
(2.18)TλAA(x,y)=e-(|log⁡x|λ+|log⁡y|λ)1/λifλ∈(0,∞).At-norm T can be extended (by associativity) in a unique way to an n-array operation taking for any (x1,…,xn)∈[0,1]n the value T(x1,…,xn) defined by
(2.19)Ti=10xi=1,Ti=1nxi=T(Ti=1n-1xi,xn)=T(x1,…,xn).T can also be extended to a countable operation taking for any sequence (xn)n∈N in [0,1] the value
(2.20)Ti=1∞xi=lim⁡n→∞Ti=1nxi.
The limit on the right side of (2.20) exists since the sequence (Ti=1nxi)n∈ℕ is nonincreasing and bounded from below.Proposition 2.10 (see [45]).
(i) ForT≥TL, the following implication holds:
(2.21)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.
(ii) IfT is of Hadžić-type, then
(2.22)lim⁡n→∞Ti=1∞xn+i=1
for every sequence (xn)n∈N in [0,1] such that limn→∞xn=1.
(iii) IfT∈{TλAA}λ∈(0,∞)∪{TλD}λ∈(0,∞), then
(2.23)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)λ<∞.
(iv) IfT∈{TλSW}λ∈[-1,∞), then
(2.24)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.Definition 2.11.
LetT and T′ be two continuous t-norms. Then T′dominatesT, denoted by T′≫T, if for all x1,x2,y1,y2∈[0,1],
(2.25)T[T′(x1,x2),T′(y1,y2)]≤T′[T(x1,y1),T(x2,y2)].Now, we extended definitions and results on the triangular norm on lattices. Letℒ=(L,≤L) be a complete lattice, that is, a partially ordered set in which every nonempty subset admits supremum and infimum, and 0ℒ=infL, 1ℒ=supL.Definition 2.12 (see [46]).
At-norm on L is a mapping 𝒯:(L)2→L satisfying the following conditions:
(a)(∀x∈L)(𝒯(x,1ℒ)=x) (boundary condition);
(b)(∀(x,y)∈(L)2)(𝒯(x,y)=𝒯(y,x))   (commutativity);
(c)(∀(x,y,z)∈(L)3)(𝒯(x,𝒯(y,z))=𝒯(𝒯(x,y),z))   (associativity);
(d)(∀(x,x′,y,y′)∈(L)4)(x≤Lx′ and y≤Ly′⇒𝒯(x,y)≤L𝒯(x′,y′))   (monotonicity).Let{xn} be a sequence in L converges to x∈L (equipped order topology). The t-norm 𝒯 is said to be a continuoust-norm if
(2.26)lim⁡n→∞T(xn,y)=T(x,y),
for each y∈L.Note that we put𝒯=T whenever L=[0,1].Definition 2.13 (see [46]).
A continuoust-norm  𝒯 on L=[0,1]2 is said to be continuoust-representable if there exist a continuous t-norm * and a continuous t-conorm ⋄ on [0,1] such that, for all x=(x1,x2),y=(y1,y2)∈L,
(2.27)T(x,y)=(x1*y1,x2⋄y2).For example,(2.28)T(a,b)=(a1b1,min⁡{a2+b2,1}),M(a,b)=(min⁡{a1,b1},max⁡{a2,b2})
for all a=(a1,a2),b=(b1,b2)∈[0,1]2 are continuous t–representable.Define the mapping𝒯∧ from L2 to L by:
(2.29)T∧(x,y)={x,ify≥Lx,y,ifx≥Ly.Anegation on ℒ is any decreasing mapping 𝒩:L→L satisfying 𝒩(0ℒ)=1ℒ and 𝒩(1ℒ)=0ℒ. If 𝒩(𝒩(x))=x for all x∈L, then 𝒩 is called an involutive negation. In the following ℒ is endowed with a (fixed) negation 𝒩.


## 3. Distribution Functions and Fuzzy Sets

We denote,Δ+, the space of all distribution functions, that is, the space of all mappings F:ℝ∪{-∞,+∞}→[0,1], such that F is left-continuous and nondecreasing on ℝ, F(0)=0 and F(+∞)=1. D+ is a subset of Δ+ consisting of all functions F∈Δ+ for which l-F(+∞)=1, where l-f(x) denotes the left limit of the function f at the point x, that is, l-f(x)=limt→x-f(t). The space Δ+ is partially ordered by the usual point-wise ordering of functions, that is, F≤G if and only if F(t)≤G(t) for all t in ℝ. The maximal element for Δ+ in this order is the distribution function ɛ0 given by
(3.1)ɛ0(t)={0,ift≤0,1,ift>0,Example 3.1.
The functionG(t), defined by
(3.2)G(t)={0,ift≤0,1-e-t,ift>0,
is a distribution function. Since limt→∞G(t)=1, then G∈D+. Note that G(t+s)≥TP(G(t),G(s)) for each t,s>0.Example 3.2.
The functionF(t), defined by
(3.3)F(t)={0ift≤0,tif0≤t≤1,1if1≤t,
is a distribution function. Since limt→∞F(t)=1, then F∈D+. Note that F(t+s)≥TM(F(t),F(s)) for each t,s>0.Example 3.3 (see [6]).
The functionGp(t), p>0, defined by
(3.4)Gp(t)={0ift≤0,exp⁡(-|p|1/2)if0<t<+∞,1ift=+∞,
is a distribution function. Since limt→∞Gp(t)≠1, then G∈Δ+∖D+. Note that Gp(t+s)≥TM(Gp(t),Gp(s)) for each t,s>0.Definition 3.4.
IfW is a collection of objects denoted generically by w, then a fuzzy setA in W is a set of ordered pairs:
(3.5)A={(w,λA(w)):w∈W}.λA(w) is called the membership function or grade of membership of w in A which maps W to the membership space M. Note that, when M contains only the two points 0 and 1, A is nonfuzzy and λA(w) is identical to the characteristic function of a non-fuzzy set. The range of the membership function is [0,1] or a complete lattice.Example 3.5.
Consider the fuzzy setA which is real numbers considerably larger than 10,
(3.6)A={(w,λA(w)):w∈W},
where
(3.7)λA(w)={0,ifw<10,11+(w-10)-2,ifw≥10.Example 3.6.
Consider the fuzzy setA which is real numbers close to 10, like (3.6) where
(3.8)λA(w)=11+(w-10)2.
Note that, in this paper, in short, we apply membership functions instead of fuzzy sets.Definition 3.7 (see [47]).
Letℒ=(L,≤L) be a complete lattice and let U be a nonempty set called the universe. An ℒ-fuzzy set in U is defined as a mapping 𝒜:U→L. For each u in U, 𝒜(u) represents the degree (in L) to which u is an element of 𝒜.Lemma 3.8 (see [46]).
Consider the setL* and operation ≤L* defined by
(3.9)L*={(x1,x2):(x1,x2)∈[0,1]2andx1+x2≤1},(x1,x2)≤L*(y1,y2)⟺x1≤y1,x2≥y2
for all (x1,x2),(y1,y2)∈L*. Then (L*,≤L*) is a complete lattice.Definition 3.9 (see [48]).
Anintuitionistic fuzzy set𝒜ζ,η in the universe U is an object 𝒜ζ,η={(u,ζ𝒜(u),η𝒜(u)):u∈U}, where ζ𝒜(u)∈[0,1] and η𝒜(u)∈[0,1] for all u∈U are called the membership degree and the nonmembership degree, respectively, of u in 𝒜ζ,η and, furthermore, satisfy ζ𝒜(u)+η𝒜(u)≤1.Example 3.10.
Consider the intuitionistic fuzzy set𝒜ζ,η which is real numbers considerably larger than 10 for first place and real numbers close to 10 in second place,
(3.10)Aζ,η={(w,ζA(w),ηA(w)):w∈W},
where
(3.11)(ζA(w),ηA(w))={(0,11+(w-10)2),ifw<10,(11+(w-10)-2,11+(w-10)2),ifw≥10.
As we said above, we apply𝒜ζ,η(w)=(ζA(w),ηA(w)).


## 4. Random Normed Spaces

Random (probabilistic) normed spaces were introduced by Šerstnev in 1962 [1] by means of a definition that was closely modelled on the theory of (classical) normed spaces and used to study the problem of best approximation in statistics. In the sequel, we will adopt usual terminology, notation, and conventions of the theory of random normed spaces, as in [4, 6–8].Definition 4.1.
AMenger probabilistic metric space (or random metric spaces) is a triple (X,ℱ,T), where X is a nonempty set, T is a continuous t-norm, and ℱ is a mapping from X×X into D+ such that, if Fx,y denotes the value of ℱ at the pair (x,y), the following conditions hold:
(PM1)Fx,y(t)=ɛ0(t) for all t>0 if and only if x=y;
(PM2)Fx,y(t)=Fy,x(t);
(PM3)Fx,z(t+s)≥T(Fx,y(t),Fy,z(s)) for all x,y,z∈X and t,s≥0.Definition 4.2 (see [1]).
Arandom normed space (briefly, RN-space) is a triple (X,μ,T), where X is a vector space, T is a continuous t-norm, and μ is a mapping from X into D+ such that the following conditions hold:
(RN1)μx(t)=ɛ0(t) for all t>0 if and only if x=0;
(RN2)μαx(t)=μx(t/|α|) for all x∈X, α≠0;
(RN3)μx+y(t+s)≥T(μx(t),μy(s)) for all x,y∈X and t,s≥0.Example 4.3.
Let(X,∥·∥) be a linear normed space. Define
(4.1)μx(t)={0,ift≤0,tt+‖x‖,ift>0.
Then (X,μ,TP) is a random normed space. (RN1) and (RN2) are obvious and we show (RN3). 							(4.2)TP(μx(t),μy(s))=tt+‖x‖⋅ss+‖y‖=11+(‖x‖/t)⋅11+(‖y‖/s)≤11+(‖x‖/(t+s))⋅11+(‖y‖/(t+s))≤11+((‖x‖+‖y‖)/(t+s))≤11+((‖x+y‖)/(t+s))=t+st+s+‖x+y‖=μx+y(t+s)
for all x,y∈X and t,s≥0. Also, (X,μ,TM) is a random normed space.Example 4.4.
Let(X,∥·∥) be a linear normed spaces. Define
(4.3)μx(t)={0,ift≤0,e-(∥x∥/t),ift>0.
Then (X,μ,TP) is a random normed space. (RN1) and (RN2) are obvious and we show (RN3).
(4.4)TP(μx(t),μy(s))=e-(‖x‖/t)⋅e-(‖y‖/s)≤e-(‖x‖/(t+s))⋅e-(‖y‖/(t+s))=e-((‖x‖+‖y‖)/(t+s))≤e-((‖x+y‖)/(t+s))=μx+y(t+s)
for all x,y∈X and t,s≥0. Also, (X,μ,TM) is a random normed space.Example 4.5 (see [36]).
Let(X,∥·∥) be a linear normed space. Define
(4.5)μx(t)={max⁡{1-‖x‖t,0},ift>0,0,ift≤0.
Then (X,μ,TL) is an RN-space (this was essentially proved by Mušthari in [49], see also [50]). Indeed, μx(t)=1 for all t>0⇒∥x∥/t=0 for all t>0⇒x=0 and obviously μλx(t)=μx(t/λ)forallx∈X and t>0. Next, for every x,y∈X and t,s>0 we have
(4.6)μx+y(t+s)=max⁡{1-‖x+y‖t+s,0}=max⁡{1-‖x+yt+s‖,0}=max⁡{1-‖xt+s+yt+s‖,0}≥max⁡{1-‖xt‖-‖ys‖,0}=TL(μx(t),μy(s)).Letφ be a function defined on the real field ℝ into itself with the following properties:(a)
φ(-t)=φ(t) for every t∈ℝ;(b)
φ(1)=1;(c)
φ is strictly increasing and continuous on [0,∞), φ(0)=0 and limα→∞φ(α)=∞.Examples of such functions areφ(t)=|t|; φ(t)=|t|p, p∈(0,∞); φ(t)=2t2n/(|t|+1), n∈ℕ.Definition 4.6 (see [51]).
Arandomφ-normed space is a triple (X,ν,T), where X is a real vector space, T is a continuous t–norm, and ν is a mapping from X into D+ such that the following conditions hold:
(φ-RN1) νx(t)=ɛ0(t) for all t>0 if and only if x=0;
(φ-RN2) ναx(t)=νx(t/φ(α)) for all x in X, α≠0 and t>0;
(φ-RN3) νx+y(t+s)≥T(νx(t),νy(s)) for all x,y∈X and t,s≥0.Example 4.7 (see [37]).
An important example is the space(X,ν,TM), where (X,∥·∥p) is a p-normed space and
(4.7)νx(t)={0,ift≤0,tt+‖x‖p,p∈(0,1],ift>0.
(φ-RN1) and (φ-RN2) are obvious and we show (φ-RN3). Let νx(t)≤νy(s), then for each x,y∈X we have
(4.8)‖y‖ps≤‖x‖pt.
Now, if x=y, then we have t≤s. Now, since we have
(4.9)‖x‖pt+‖x‖pt≥‖x‖pt+‖y‖ps≥2‖x‖pt+s+2‖y‖pt+s≥2‖x+y‖pt+s,
then it follows that
(4.10)1+∥x∥pt≥1+∥x+y∥pt+s,
which implies that νx(t)≤νx+y(t+s). Hence νx+y(t+s)≥TM(νx(t),νy(s)) for all x,y∈X and t,s≥0.By anon-Archimedean field we mean a field 𝒦 equipped with a function (valuation) |·| from 𝒦 into [0,∞) such that |r|=0 if and only if r=0, |rs|=|r||s|, and |r+s|≤max{|r|,|s|} for all r,s∈𝒦. Clearly |1|=|-1|=1 and |n|≤1 for all n∈ℕ. By the trivial valuation we mean the mapping |·| taking everything but 0 into 1 and |0|=0.  Let X be a vector space over a field 𝒦 with a non-Archimedean non-trivial valuation |·|, that is, that there is an a0∈𝒦 such that |a0| is not in {0,1}.The most important examples of non-Archimedean spaces arep-adic numbers. In 1897, Hensel [52] discovered the p-adic numbers as a number theoretical analogue of power series in complex analysis. Fix a prime number p. For any nonzero rational number x, there exists a unique integer nx∈ℤ such that x=(a/b)pnx, where a and b are integers not divisible by p. Then |x|p:=p-nx defines a non-Archimedean norm on ℚ. The completion of ℚ with respect to the metric d(x,y)=|x-y|p is denoted by ℚp, which is called the p-adic number field.A function∥·∥:X→[0,∞) is called a non-Archimedean norm if it satisfies the following conditions:(i)
∥x∥=0 if and only if x=0;(ii)
for anyr∈𝒦,x∈X, ∥rx∥=|r|∥x∥;(iii)
the strong triangle inequality (ultrametric); namely,(4.11)‖x+y‖≤max⁡{‖x‖,‖y‖},∀x,y∈X.Then(X,∥·∥) is called a non-Archimedean normed space. Due to the fact that
(4.12)‖xn-xm‖≤max⁡{‖xj+1-xj‖:m≤j≤n-1},∀n,m∈N(n>m),
a sequence {xn} is a Cauchy sequence if and only if {xn+1-xn} converges to zero in a non-Archimedean normed space. By a complete non-Archimedean normed space we mean one in which every Cauchy sequence is convergent.Definition 4.8.
Anon-Archimedean random normed space (briefly, non-Archimedean RN-space) is a triple (X,μ,T), where X is a linear space over a non-Archimedean field 𝒦, T is a continuous t-norm, and μ is a mapping from X into D+ such that the following conditions hold:
(NA-RN1)μx(t)=ɛ0(t) for all t>0 if and only if x=0;
(NA-RN2)μαx(t)=μx(t/|α|) for all x∈X, t>0,α≠0;
(NA-RN3)μx+y(max{t,s})≥T(μx(t),μy(s)) for all x,y,z∈X and t,s≥0.
It is easy to see that, if (NA-RN3) holds then so will (RN3)μx+y(t+s)≥T(μx(t),μy(s)).Example 4.9.
As a classical example, if(X,∥·∥) is a non-Archimedean normed linear space, then the triple (X,μ,TM), where
(4.13)μx(t)={0t≤‖x‖,1t>‖x‖,
is a non-Archimedean RN-space.Example 4.10.
Let(X,∥·∥) be a non-Archimedean normed linear space. Define
(4.14)μx(t)=tt+‖x‖,∀x∈X,t>0.
Then (X,μ,TM) is a non-Archimedean RN-space.Definition 4.11.
Let(X,μ,T) be a non-Archimedean RN-space. Let {xn} be a sequence in X. Then {xn} is said to be convergent if there exists x∈X such that
(4.15)lim⁡n→∞μxn-x(t)=1
for all t>0. In that case, x is called the limit of the sequence {xn}.
A sequence{xn} in X is called a Cauchy sequence if for each ɛ>0 and t>0 there exists n0 such that for all n≥n0 and p>0 we have μxn+p-xn(t)>1-ɛ.
If each Cauchy sequence is convergent, then the random normed space is said to becomplete and the non-Archimedean RN-space is called a non-Archimedean  random Banach  space.Remark 4.12 (see [53]).
Let(X,μ,TM) be a non-Archimedean RN-space, then
(4.16)μxn+p-xn(t)≥min⁡{μxn+j+1-xn+j(t):j=0,1,2,…,p-1}.
So, the sequence {xn} is a Cauchy sequence if for each ɛ>0 and t>0 there exists n0 such that for all n≥n0 we have
(4.17)μxn+1-xn(t)>1-ɛ.


## 5. Topological Structure of Random Normed Spaces

Definition 5.1.
Let(X,μ,T) be an RN-space. We define the open ballBx(r,t) and the closed ballBx[r,t] with center x∈X and radius 0<r<1, t>0 as follows:
(5.1)Bx(r,t)={y∈X:μx-y(t)>1-r},Bx[r,t]={y∈X:μx-y(t)≥1-r}.Theorem 5.2.
Let(X,μ,T) be an RN-space. Every open ball Bx(r,t) is an open set.Proof.
LetBx(r,t)be an open ball with center x and radius r with respect t. Let y∈Bx(r,t). Then μx-y(t)>1-r. Since μx-y(t)>1-r, there exists t0∈(0,t) such that μx-y(t0)>1-r. Put r0=μx,y(t0). Since r0>1-r, there exists s∈(0,1) such that r0>1-s>1-r. Now for given r0 and s such that r0>1-s, there exists r1∈(0,1) such that T(r0,r1)>1-s. Consider the open ball By(1-r1,t-t0). We claim By(1-r1,t-t0)⊂Bx(r,t). Now, let z∈By(1-r1,t-t0). Then μy-z(t-t0)>r1. Therefore, we have
(5.2)μx-z(t)≥T(μx-y(t0),μy-z(t-t0))≥T(r0,r1)≥1-s>1-r.
Thus z∈Bx(r,t) and hence By(1-r1,t-t0)⊂Bx(r,t).Different kinds of topologies can be introduced in a random normed space [4]. The (r,t)-topology is introduced by a family of neighborhoods
(5.3){Bx(r,t)}x∈X,t>0,r∈(0,1).
In fact, every random norm μ on X generates a topology ((r,t)-topology) on X which has as a base the family of open sets of the form
(5.4){Bx(r,t)}x∈X,t>0,r∈(0,1).Remark 5.3.
Since{Bx(1/n,1/n):n=1,2,3,…} is a local base at x, the (r,t)-topology is first countable.Theorem 5.4.
Every RN-space(X,μ,T) is a Hausdorff space.Proof.
Let(X,μ,T) be an RN-space. Let x and y be two distinct points in X and t>0. Then 0<μx-y(t)<1. Put r=μx-y(t). For each r0∈(r,1), there exists r1 such that T(r1,r1)≥r0. Consider the open balls Bx(1-r1,t/2) and By(1-r1,t/2). Then clearly Bx(1-r1,t/2)∩By(1-r1,t/2)=∅. For, if there exists
(5.5)z∈Bx(1-r1,t2)∩By(1-r1,t2),
then we have
(5.6)r=μx-y(t)≥T(μx-z(t2),μy-z(t2))≥T(r1,r1)≥r0>r,
which is a contradiction. Hence (X,μ,T) is a Hausdorff space.Definition 5.5.
Let(X,μ,T) be an RN-space. A subset A of X is said to be R-bounded if there exists t>0 and r∈(0,1) such that μx-y(t)>1-r for all x,y∈A.Theorem 5.6.
Every compact subsetA of an RN-space (X,μ,T) is R-bounded.Proof.
LetA be a compact subset of an RN-space (X,μ,T). Fix t>0 and 0<r<1. Consider an open cover {Bx(r,t):x∈A}. Since A is compact, there exist x1,x2,…,xn∈A such that
(5.7)A⊆⋃i=1nBxi(r,t).
Let x,y∈A. Then x∈Bxi(r,t) and y∈Bxj(r,t) for some i,j. Thus we have μx-xi(t)>1-r and μy-xj(t)>1-r. Now, let
(5.8)α=min⁡{μxi,xj(t):1≤i,j≤n}.
Then α>0. Now, we have
(5.9)μx-y(3t)≥T2(μx-xi(t),μxi,xj(t),μy-xj(t))≥T2(1-r,1-r,α)>1-s.
Taking t′=3t, we have μx-y(t′)>1-s for all x,y∈A. Hence A is R-bounded.Remark 5.7.
In an RN-space(X,μ,T) every compact set is closed and R-bounded.Theorem 5.8 (see [4]).
If(X,μ,T) is an RN-space and {xn} is a sequence such that xn→x, then limn→∞μxn(t)=μx(t) almost everywhere.Theorem 5.9.
Let(X,μ,T) be an RN-space such that every Cauchy sequence in X has a convergent subsequence. Then (X,μ,T) is complete.Proof.
Let{xn} be a Cauchy sequence and let {xin} be a subsequence of {xn} that converges to x. We prove that xn→x. Let t>0 and ϵ∈(0,1) such that
(5.10)T(1-r,1-r)≥1-ϵ.
Since {xn} is a Cauchy sequence, there is n0∈ℕ such that
(5.11)μxm-xn(t)>1-r
for all m,n≥n0. Since xin→x, there is positive integer ip such that ip>n0 and
(5.12)μxip-x(t2)>1-r.
Then, if n≥n0, then we have
(5.13)μxn-x(t)≥T(μxn-xip(t2),μxip-x(t2))>T(1-r,1-r)≥1-ϵ.
Therefore, xn→x and hence (X,μ,T) is complete.Lemma 5.10.
Let(X,μ,T) be an RN-space. If one defines
(5.14)Fx,y(t)=μx-y(t),
then F is a random (probabilistic) metric on X, which is called the random (probabilistic) metric induced by the random norm μ.Lemma 5.11.
A random (probabilistic) metricF which is induced by a random norm on a RN-space (X,μ,T) has the following properties for all x,y,z∈X and every scalar α≠0:
(i)Fx+z,y+z(t)=Fx,y(t),
(ii)Fαx,αy(t)=Fx,y(t/|α|).Proof.
We have(5.15)Fx+z,y+z(t)=μ(x+z)-(y+z)(t)=μx-y(t)=Fx,y(t).
Also, we have
(5.16)Fαx,αy(t)=μαx-αy(t)=μx-y(t|α|)=Fx,y(t|α|).Lemma 5.12.
If(X,μ,T) is an RN-space, then
(i) the function(x,y)→x+y is continuous,
(ii) the function(α,x)→αx is continuous.Proof.
Ifxn→x and yn→y, then as n→∞(5.17)μ(xn+yn)-(x+y)(t)≥T(μxn-x(t2),μyn-y(t2))⟶1.
This proves (i).
Now, ifxn→x, αn→α and αn≠0, then
(5.18)μαnxn-αx(t)=μαn(xn-x)+x(αn-α)(t)≥T(μαn(xn-x)(t2)μx(αn-α)(t2))=T(μxn-x(t2αn),μx(t2(αn-α)))⟶1
as n→∞ and this proves (ii).Definition 5.13.
The RN-space(X,μ,T) is said to be a random Banach space whenever X is complete with respect to the random metric induced by random norm.Lemma 5.14.
Let(X,μ,T) be an RN-space and define
(5.19)Eλ,μ:X→R+∪{0}
by
(5.20)Eλ,μ(x)=inf⁡{t>0:μx(t)>1-λ}
for each λ∈(0,1) and x∈X. Then we have
(i)Eλ,μ(αx)=|α|Eλ,μ(x) for every x∈X and α∈ℝ;
(ii) ifT satisfies (2.15), then for any γ∈(0,1), there is λ∈(0,1) such that
(5.21)Eγ,μ(x1+⋯+xn)≤Eλ,μ(x1)+⋯+Eλ,μ(xn)
for any x,y∈X;
(iii) the sequence{xn}n∈ℕ is convergent with respect to a random norm μ if and only if Eλ,μ(xn-x)→0. Also the sequence {xn} is a Cauchy sequence with respect to a random norm μ if and only if it is a Cauchy sequence with Eλ,μ.Proof.
For (i), we find(5.22)Eλ,μ(αx)=inf⁡{t>0:μαx(t)>1-λ}=inf⁡{t>0:μx(t|α|)>1-λ}=|α|inf⁡{t>0:μx(t)>1-λ}=|α|Eλ,μ(x).
For (ii), by (2.15), for every α∈(0,1) we can find λ∈(0,1) such that
(5.23)Tn-1(1-λ,…,1-λ)≥1-α.
Thus we have
(5.24)μx1+⋯+xn(Eλ,μ(x1)+⋯+Eλ,μ(xn)+nδ)≥LTn-1(μx1(Eλ,M(x1)+δ),…,μxn(Eλ,P(xn)+δ))≥T(1-λ,…,1-λ)≥1-α
for every δ>0, which implies that
(5.25)Eα,μ(x1+⋯+xn)≤Eλ,μ(x1)+⋯+Eλ,μ(xn)+nδ.
Since δ>0 is arbitrary, we have
(5.26)Eα,μ(x1+⋯+xn)≤Eλ,μ(x1)+⋯+Eλ,μ(xn).
For (iii), note that sinceμ is continuous, Eλ,μ(x) is not an element of the set {t>0:μx(t)>1-λ} as soon as x≠0. Hence we have
(5.27)μxn-x(η)>1-λ⟺Eλ,μ(xn-x)<η
for every η>0. This completes the proof.Definition 5.15.
A functionf from an RN-space (X,μ,T) to an RN-space (Y,ν,T′) is said to be uniformly continuous if for given r∈(0,1) and t>0, there exist r0∈(0,1) and t0>0 such that
(5.28)μx-y(t0)>1-r0
implies
(5.29)νf(x),f(y)(t)>1-r.Theorem 5.16 (uniform continuity theorem).
Iff is continuous function from a compact RN-space (X,μ,T) to an RN-space (Y,ν,T′), then f is uniformly continuous.Proof.
Lets∈(0,1) and t>0 be given. Then we can find r∈(0,1) such that
(5.30)T′(1-r,1-r)>1-s.
Since f:X→Y is continuous, for each x∈X, we can find rx∈(0,1) and tx>0 such that
(5.31)μx-y(tx)>1-rx
implies
(5.32)νf(x)-f(y)(t2)>1-r.
But rx∈(0,1) and then we can find sx<rx such that
(5.33)T(1-sx,1-sx)>1-rx.
Since X is compact and
(5.34){Bx(sx,tx2):x∈X}
is an open covering of X, there exist x1,x2,…,xk in X such that
(5.35)X=⋃i=1kBxi(sxi,txi2).
Put s0=minsxi and t0=min(txi/2), i=1,2,…,k. For any x,y∈X, if
(5.36)μx-y(t0)>1-s0,
then
(5.37)μx-y(txi2)>1-sxi.
Since x∈X, there exists xi such that
(5.38)μx-xi(txi2)>1-sxi.
Hence we have
(5.39)νf(x),f(xi)(t2)>1-r.
Now, since we have
(5.40)μy-xi(txi)≥T(μx-y(txi2),μx-xi(txi2))≥T(1-sxi,1-sxi)>1-rxi,
it follows that
(5.41)νf(y)-f(xi)(t2)>1-r.
Now, we have
(5.42)νf(x)-f(y)(t)≥T(νf(x)-f(xi)(t2),νf(y)-f(xi)(t2))≥T(1-r,1-r)>1-s.
Hence f is uniformly continuous.Remark 5.17.
Letf be a uniformly continuous function from RN-space (X,μ,T) to RN-space (Y,ν,T′). If {xn} is a Cauchy sequence in X, then {f(xn)} is also a Cauchy sequence in Y.Theorem 5.18.
Every compact RN-space is separable.Proof.
Let(X,μ,T) be the given compact RN-space. Let r∈(0,1) and t>0. Since X is compact, there exist x1,x2,…,xn in X such that
(5.43)X=⋃i=1nBxi(r,t).
In particular, for each n∈ℕ, we can choose a finite subset An such that
(5.44)X=⋃a∈AnBa(rn,1n),
in which rn∈(0,1). Let
(5.45)A=⋃n∈NAn.
Then A is countable. We claim that X⊂A¯. Let x∈X. Then for each n∈ℕ, there exists an∈An such that x∈Ban(rn,1/n). Thus an converges to x. But since an∈A for all n, x∈A¯. Hence A is dense in X and thus X is separable.Definition 5.19.
LetX be any nonempty set and (Y,ν,T′) be an RN-space. Then a sequence {fn} of functions from X to Y is said to be uniformly  convergent to a function f from X to Y if for given r∈(0,1) and t>0, there exists n0∈ℕ such that
(5.46)νfn(x)-f(x)(t)>1-r
for all n≥n0 and x∈X.Definition 5.20.
A familyℱ of functions from an RN-space (X,μ,T) to a complete RN-space (Y,ν,T′) is said to be equicontinuous if for any r∈(0,1) and t>0, there exist r0∈(0,1) and t0>0 such that
(5.47)μx-y(t0)>1-r0⟹νf(x)-f(y)(t)>1-r
for all f∈ℱ.Lemma 5.21.
Let{fn} be an equicontinuous sequence of functions from an RN-space (X,μ,T) to a complete RN-space (Y,ν,T′). If {fn} converges for each point of a dense subset D of X, then {fn} converges for each point of X and the limit function is continuous.Proof.
Lets∈(0,1) and t>0 be given. Then we can find r∈(0,1) such that
(5.48)T′2(1-r,1-r,1-r)>1-s.
Since ℱ={fn} is equicontinuous family, for given r∈(0,1) and t>0, there exist r1∈(0,1) and t1>1 such that, for each x,y∈X,
(5.49)μx-y(t1)>1-r1⟹νfn(x)-fn(y)(t3)>1-r
for all fn∈ℱ. Since D is dense in X, there exists
(5.50)y∈Bx(r1,t1)∩D,
and {fn(y)} converges for that y. Since {fn(y)} is a Cauchy sequence, for any r∈(0,1) and t>0, there exists n0∈ℕ such that
(5.51)νfn(y)-fm(y)(t3)>1-r
for all m,n≥n0. Now, for any x∈X, we have
(5.52)νfn(x)-fm(x)(t)≥T′2(νfn(x)-fn(y)(t3),νfn(y)-fm(y)(t3),νfm(x)-fm(y)(t3))≥T′2(1-r,1-r,1-r)>1-s.
Hence {fn(x)} is a Cauchy sequence in Y. Since Y is complete, fn(x) converges. Let f(x)=limfn(x). We claim that f is continuous. Let so∈1-r and t0>0 be given. Then we can find r0∈1-r such that
(5.53)T′2(1-r0,1-r0,1-r0)>1-s0.
Since ℱ is equicontinuous, for given r0∈(0,1) and t0>0, there exist r2∈(0,1) and t2>0 such that
(5.54)μx-y(t2)>1-r2⟹νfn(x)-fn(y)(t03)>1-r0
for all fn∈ℱ. Since fn(x) converges to f(x), for given r0∈(0,1) and t0>0, there exists n1∈ℕ such that
(5.55)νfn(x)-f(x)(t03)>1-r0.
Also since fn(y) converges to f(y), for given r0∈(0,1) and t0>0, there exists n2∈ℕ such that
(5.56)νfn(y)-f(y)(t03)>1-r0
for all n≥n2. Now, for all n≥max{n1,n2}, we have
(5.57)νf(x)-f(y)(t0)≥T′2(νf(x)-fn(x)(t03),νfn(x)-fn(y)(t03),νfn(y)-f(y)(t03))≥T′2(1-r0,1-r0,1-r0)>1-s0.
Hence f is continuous.Theorem 5.22 (Ascoli-Arzela Theorem).
Let(X,μ,T) be a compact RN-space and (Y,ν,T′) be a complete RN-space. Let ℱ be an equicontinuous family of functions from X to Y. If {fn}n∈ℕ is a sequence in ℱ such that
(5.58){fn(x):n∈N}¯
is a compact subset of Y for each x∈X, then there exists a continuous function f from X to Y and a subsequence {gn} of {fn} such that gn converges uniformly to f on X.Proof.
Since(X,μ,T) be a compact RN-space, by Theorem 5.18, X is separable. Let
(5.59)D={xi:i=1,2,…}
be a countable dense subset of X. By hypothesis, for each i,
(5.60){fn(xi):n∈N}¯
is compact subset of Y. Since every RN-space is first countable space, every compact subset of Y is sequentially compact. Thus by standard argument, we have a subsequence {gn} of {fn} such that {gn(xi)} converges for each i=1,2,…. By Lemma 5.21, there exists a continuous function f from X to Y such that {gn(x)} converges to f(x) for all x∈X.
Now we claim that{gn} converges uniformly to f on X. Let s∈(0,1) and t>0 be given. Then we can find r∈(0,1) such that
(5.61)T′2(1-r,1-r,1-r)>1-s.
Since ℱ is equicontinuous, there exist r1∈(0,1) and t1>0 such that
(5.62)μx-y(t1)>1-r1⟹νgn(x),gn(y)(t3)>1-r
for all n∈ℕ. Since X is compact, by Theorem 5.16, f is uniformly continuous. Hence for given r∈(0,1) and t>0, there exist r2∈(0,1) and t2>0 such that
(5.63)μx-y(t2)>1-r2⟹νf(x)-f(y)(t3)>1-r
for all x,y∈X. Let r0=min{r1,r2} and t0=min{t1,t2}. Since X is compact and D is dense in X,
(5.64)X=⋃i=1kBxi(r0,t0)
for some k. Thus for each x∈X, there exists i, i≤i≤k, such that
(5.65)μx-xi(t0)>1-r0.
But since r0=min{r1,r2} and t0=min{t1,t2}, we have, by the equicontinuity of ℱ,
(5.66)νgn(x)-gn(xi)(t3)>1-r
and we also have, by the uniform continuity of f,
(5.67)νf(x)-f(xi)(t3)>1-r.
Since {gn(xj)} converges to f(xj), for any r∈(0,1) and t>0, there exists n0∈ℕ such that
(5.68)νgn(xj)-f(xj)(t3)>1-r
for all j=1,2,…,n.
Now, for eachx∈X, we have
(5.69)νgn(x)-f(x)(t)≥T′2(νgn(x)-gn(xi)(t3),νgn(xi)-f(xi)(t3),νf(xi)-f(x)(t3))≥T′2(1-r,1-r,1-r)>1-s.
Hence {gn} converges uniformly to f on X.Lemma 5.23.
A subsetA of ℝ is R-bounded in (ℝ,μ,T) if and only if it is bounded in ℝ.Proof.
LetA be a subset in ℝ which is R-bounded in (ℝ,μ,T), then there exist t0>0 and r0∈(0,1) such that μx(t0)>1-r0 for each x∈A and therefore we have
(5.70)t0≥Er0,μ(x)=|x|Er0,μ(1).
Now, Er0,μ(1)≠0. If we put k=t0/(Er0,μ(1)), then we have |x|≤k for each x∈A, that is, A is bounded in ℝ. The converse is easy to see.Lemma 5.24.
A sequence{βn}n∈ℕ is convergent in the RN-space (ℝ,μ,T) if and only if it is convergent in (ℝ,|·|).Proof.
Letβn→β in ℝ, by Lemma 5.14 (i), we have
(5.71)Eλ,μ(βn-β)=|βn-β|Eλ,μ(1)⟶0.
Then, by Lemma 5.14 (iii), βn→μβ.
Conversely, letβn→μβ, then, by Lemma 5.14, we have
(5.72)lim⁡n→+∞|βn-β|Eλ,μ(1)=lim⁡n→+∞Eλ,μ(βn-β)=0.
Now, Eλ,μ(1)≠0 and so βn→β in ℝ.Corollary 5.25.
If the real sequence{βn}n∈ℕ is R-bounded, then it has at least one limit point.Definition 5.26.
The 3-tuple(ℝn,Φ,T) is called a random Euclidean normed space if T is a continuous t-norm and Φx(t) is a random Euclidean norm defined by
(5.73)Φx(t)=∏j=1nμxj(t),
where ∏j=1naj=T′n-1(a1,…,an), T′≫T, x=(x1,…,xn), t>0, and μ is a random norm.
For example, letΦx(t)=exp(∥x∥/t)-1, μxj(t)=exp(|xj|/t)-1, and T=min. Then we have Φx(t)=minjμxj(t) or equivalently ∥x∥=maxj|xj|.Lemma 5.27.
Suppose that the hypotheses of Definition5.26 are satisfied. Then (ℝn,Φ,T) is an RN-space.Proof .
The properties of (RN1) and (RN2) follow immediately from the definition. For the triangle inequality (RN3), suppose thatx,y∈X and t,s>0. Then
(5.74)T(Φx(t),Φy(s))=T(∏j=1nPxj(t),∏j=1nPyj(s))=T(T′n-1(Px1(t),…,Pxn(t)),T′n-1(Py1(t),…,Pyn(t)))≤T′n-1(T(Px1(t),Py1(t)),…,T(Pxn(t),Pyn(t)))≤T′n-1(Px1+y1(t+s),…,Pxn+yn(t+s))=∏j=1nPxj+yj(t+s)=Φx+y(t+s).Lemma 5.28.
Suppose that(ℝn,Φ,T) is a random Euclidean normed space and A is an infinite and R-bounded subset of ℝn. Then A has at least one limit point.Proof.
Let{x(m)}m∈ℕ⊆A be an infinite sequence. Since A is R-bounded and so is {x(m)}m∈ℕ. Therefore there exist t0>0 and r0∈(0,1) such that 1-r0<Φx(t0) for each x∈A, which implies that Er0,Φ(x)≤t0. However, we have
(5.75)Er0,Φ(x)=inf⁡{t>0:1-r0<Φx(t)}=inf⁡{t>0:1-r0<∏j=1nμxj(t)}≥inf⁡{t>0:1-r0<μxj(t)}=Er0,μ(xj)
for each 1≤j≤n. Therefore |xj|≤k in which k=t0/(Er0,μ(1)), that is, the real sequences {xj(m)} for all j∈{1,…,n} are bounded. Hence there exists a subsequence {x1(mk1)} which converges to x1 in A w.r.t. the random norm μ. The corresponding sequence {x2(mk1)} is bounded and so there exists a subsequence {x2(mk2)} of {x2(mk1)} which converges to x2 with respect to the random norm μ. Continuing like this, we find a subsequence {x(mk)} converging to x=(x1,…,xn)∈ℝn.Lemma 5.29.
Let(ℝn,Φ,T) be a random Euclidean normed space. Let {Q1,Q2,…} be a countable collection of nonempty subsets in ℝn such that Qk+1⊆Qk, each Qk is closed and Q1 is R-bounded. Then ∩k=1∞Qk is nonempty and closed.Proof.
Using the above lemma, the proof proceeds as in the classical case [see Theorem 3.25 in [54]].We call ann-dimensional ball Bx(r,t) a rational ball if x∈ℚn, r0∈(0,1), and t∈ℚ+.Theorem 5.30.
Let(ℝn,Φ,T) be a random Euclidean normed space in which T satisfies (2.15). Let G={A1,A2,…} be a countable collection of n-dimensional rational open balls. If x∈ℝn and S is an open subset of ℝn containing x, then there exists a Ak∈G such that x∈Ak⊆S, for some k≥1.Proof.
Sincex∈S and S is an open, there exist r∈(0,1) and t>0 such that Bx(r,t)⊆S. By (2.15), we can find η∈(0,1) such that 1-r<T(1-η,1-η). Let {ξk}k=1n be a finite sequence such that 1-η<∏k=1n(1-ξk) and x=(x1,…,xn) and then we can find y=(y1,…,yn)∈ℚn such that 1-ξk)<μxk-yk(t/2). Therefore, we have
(5.76)1-η<∏k=1n(1-ξk)≤Φx-y(t2)=∏k=1nμxk-yk(t2)
and so x∈By(η,t/2). Now we prove that By(η,t/2)⊆Bx(r,t). Let z∈By(η,t/2). Then Φy-z(t/2)>1-η and hence
(5.77)1-r<T(1-η,1-η)≤T(Φx-y(t2),Φy-z(t2))≤Φx-z(t).
On the other hand, there exists t0∈ℚ such that t0<t/2 and x∈By(η,t0)⊆By(η,t/2)⊆Bx(r,t)⊆S. Now By(η,t0)∈G and the proof is complete.Corollary 5.31.
In a random Euclidean normed space(ℝn,Φ,T) in which T satisfies (2.15), every closed and R-bounded set is compact.Proof.
The proof is similar to the proof of Theorem 3.29 in [54].Corollary 5.32.
Let(ℝn,Φ,T) be a random Euclidean normed space in which T satisfies (2.15) and S⊆ℝn. Then S is compact set if and only if it is R-bounded and closed.Corollary 5.33.
The random Euclidean normed space(ℝn,Φ,T) is complete.Proof.
Let{xm} be a Cauchy sequence in the random Euclidean normed space (ℝn,Φ,T). Since
(5.78)Eλ,Φ(xn-xm)=inf⁡{t>0:Φxn-xm(t)>1-λ}=inf⁡{t>0:∏j=1nPxm,j-xn,j(t)>1-λ}≥inf⁡{t>0:Pxm,j-xn,j(t)>1-λ}=Eλ,P(xm,j-xn,j)=|xm,j-xn,j|Eλ,P(1),
the sequence {xm,j} in which j=1,…,n is a Cauchy sequence in ℝ and convergent to xj∈ℝ then, and by Lemma 5.14, the sequence {xm,j} is convergent in RN-space (ℝ,μ,T). We prove that {xm} is convergent to x=(x1,…,xn) and
(5.79)lim⁡m→∞Φxm-x(t)=lim⁡m→∞∏j=1nPxm,j-xj(t)=T′n-1(1,…,1)=1.


## 6. Random Functional Analysis

Theorem 6.1.
Let{x1,…,xn} be a linearly independent set of vectors in vector space X and (X,μ,T) be an RN-space. Then there is c≠0 and an RN-space (ℝ,μ′,T) such that for every choice of the n real scalars α1,…,αn we have
(6.1)μα1x1+⋯+αnxn(t)≤μc′∑j=1n|αj|(t).Proof.
Puts=|α1|+⋯+|αn|. If s=0, all αjs must be zero and so (6.1) holds for any c. Let s>0. Then (6.1) is equivalent to the inequality which we obtain from (6.1) by dividing by s and putting βj=(αj)/s, that is,
(6.2)μβ1x1+⋯+βnxn(t′)≤μc′(t′),
where t′=ts and ∑j=1n|βj|=1. Hence it suffices to prove the existence of a c≠0 and random norm μ′ such that (6.2) holds. Suppose that this is not true. Then there exists a sequence {ym} of vectors,
(6.3)ym=β1,mx1+⋯+βn,mxn,
where ∑j=1n|βj,m|=1, such that
(6.4)μym(t)⟶1
as m→∞ for every t>0. Since ∑j=1n|βj,m|=1, we have |βj,m|≤1 and then, by Lemma 5.23, the sequence of {βj,m} is R-bounded. According the Corollary 5.25, {β1,m} has a convergent subsequence. Let β1 denote the limit of that subsequence, and let {y1,m} denote the corresponding subsequence of {ym}. By the same argument, {y1,m} has a subsequence {y2,m} for which the corresponding of real scalars β2(m) convergence; let β2 denote the limit. Continuing this process, after n steps, we obtain a subsequence {yn,m}m of {ym} such that
(6.5)yn,m=∑j=1nγj,mxj,
where ∑j=1n|γj,m|=1, and γj,m→βj as m→∞. By the Lemma 5.14 (ii) for any α∈(0,1) there exists λ∈(0,1) such that
(6.6)Eα,μ(yn,m-∑j=1nβjxj)=Eα,μ(∑j=1n(γj,m-βj)xj)≤∑j=1n|γj,m-βj|Eλ,μ(xj)⟶0
as m→∞. By the Lemma 5.14 (iii), we conclude
(6.7)lim⁡m→∞yn,m=∑j=1nβjxj,
where ∑j=1n|βj|=1, so that not all βj can be zero. Put y=∑j=1nβjxj. Since {x1,…,xn} is a linearly independent set, we thus have y≠0. Since
(6.8)μym(t)⟶1,
by assumption, we have
(6.9)μyn,m(t)⟶1.
Hence it follows that
(6.10)μy(t)=μ(y-yn,m)+yn,m(t)≥T((μy-yn,mt2),μyn,m(t2))⟶1
and so y=0, which is a contradiction.Definition 6.2.
Let(X,μ,T) and (X,ν,T′) be two RN-spaces. Then two random norms μ and ν are said to be equivalent whenever xn→μx in (X,μ,T) if and only if xn→νx in (X,ν,T′).Theorem 6.3.
On a finite dimensional vector spaceX, every two random norms μ and ν are equivalent.Proof.
LetdimX=n and {v1,…,vn} be a basis for X. Then every x∈X has a unique representation x=∑j=1nαjvj. Let xm→μx in (X,μ,T), but, for each m∈ℕ, xm has a unique representation, that is,
(6.11)xm=α1,mv1+⋯+αn,mvn.
By Theorem 6.1, there is c≠0 and a random norm μ′ such that (6.1) holds. So
(6.12)μxm-x(t)≤μc∑j=1n|αj,m-αj|′(t)≤μc|αj,m-αj|′(t).
Now, if m→∞, then
(6.13)μxm-x(t)⟶1
for every t>0 and hence |αj,m-αj|→0 in ℝ. On the other hand, by Lemma 5.14 (ii), for any α∈(0,1), there exists λ∈(0,1) such that
(6.14)Eα,ν(xm-x)≤∑j=1n|αj,m-αj|Eλ,ν(vj).
Since |αj,m-αj|→0, then we have xm→νx in (X,ν,T′). With the same argument xm→x in (X,ν,T′) imply xm→x in (X,μ,T).Definition 6.4.
A linear operatorΛ:(X,μ,T)→(Y,ν,T′) is said to be random bounded if there exists a constant h∈ℝ-{0} such that for every x∈X and t>0(6.15)νΛx(t)≥μhx(t).Note that, by Lemma5.14 and last definition, we have
(6.16)Eλ,ν(Λx)=inf⁡{t>0:νΛx(t)>1-λ}≤inf⁡{t>0:μx(t|h|)>1-λ}=|h|inf⁡{t>0:μx(t)>1-λ}=|h|Eλ,μ(x).Theorem 6.5.
Every linear operatorΛ:(X,μ,T)→(Y,ν,T′) is  random bounded if and only if it is continuous.Proof.
By (6.15) every random bounded linear operator is continuous. Now, we prove that the converse it. Let the linear operator Λ be continuous but not random bounded. Then, for each n∈ℕ there is xn in X such that Eλ,ν(Λxn)≥nEλ,μ(pn). If we let
(6.17)yn=xnnEλ,μ(xn),
then it is easy to see yn→0 but Λyn do not tend to 0.Definition 6.6.
A linear operatorΛ:(X,μ,T)→(Y,ν,T′) is called a random topological isomorphism if Λ is one-to-one and onto and both Λ and Λ-1 are continuous. RN-spaces (X,μ,T) and (Y,ν,T′) for which such a Λ exists are said to be random topologically isomorphic.Lemma 6.7.
A linear operatorΛ:(X,μ,T)→(Y,ν,T′) is random topological isomorphism if Λ is onto and there exists constants a,b≠0 such that μax(t)≤νΛx(t)≤μbx(t).Proof.
By hypothesisΛ is random bounded and, by last theorem, is continuous and, since Λx=0, we have
(6.18)1=νΛx(t)≤μx(t|b|)
and consequently x=0. So Λ is one-to-one. Thus Λ-1 exists and, since
(6.19)νΛx(t)≤μbx(t)
is equivalent to
(6.20)νy(t)≤μbΛ-1y(t)=μΛ-1y(t|b|)
or
(6.21)ν(1b)y(t)≤μΛ-1y(t),
where y=Λx, we see that Λ-1 is random bounded and, by last theorem, is continuous. Hence Λ is an random topological isomorphism.Corollary 6.8.
Random topologically isomorphism preserves completeness.Theorem 6.9.
Every linear operatorΛ:(X,μ,T)→(Y,ν,T′) where dimX<∞ but other, not necessarily finite dimensional, is continuous.Proof.
If we define(6.22)ηx(t)=T′(μx(t),νΛx(t)),
where T′≫T, then (X,η,T) is an RN-space because (RN1) and (RN2) are immediate from definition, for the triangle inequality (RN3),
(6.23)T(ηx(t),ηz(s))=T[T′(μx(t),νΛx(t)),T′(μz(s),νΛz(s))]≤T′[T(μx(t),μz(s))T(νΛx(t),νΛz(s))]≤T′(μx+z(t+s),νΛ(x+z)(t+s))=ηx+z(t+s).
Now, let xn→μx, then, by Theorem 6.3, xn→ηx but since, by (6.15),
(6.24)νΛx(t)≥ηx(t),
then Λxn→νΛx. Hence Λ is continuous.Corollary 6.10.
Every linear isomorphism between finite dimensional RN-spaces is topological isomorphism.Corollary 6.11.
Every finite dimensional RN-space(X,μ,T) is complete.Proof.
By Corollary6.10, (X,μ,T) and (ℝn,Φ,T) are random topologically isomorphic. Since (ℝn,Φ,T) is complete and random topological isomorphism preserves completeness, (X,μ,T) is complete.Definition 6.12.
Let(V,μ,T) be an RN-space, W be a linear manifold in V and Q:V→V/W be the natural mapping, Qx=x+W. For any t>0, we define
(6.25)μ¯(x+W,t)=sup⁡{μx+y(t):y∈W}.Theorem 6.13.
LetW be a closed subspace of an RN-space (V,μ,T). If x∈V and ϵ>0, then there is x′ in V such that x′+W=x+W, Eλ,μ(x′)<Eλ,μ¯(x+W)+ϵ.Proof.
By the properties ofsup, there always exists y∈W such that Eλ,𝒫(x+y)<Eλ,μ¯(x+W)+ϵ. Now, it is enough to put x′=x+y.Theorem 6.14.
LetW be a closed subspace of RN-space (V,μ,T) and μ¯ be given in the above definition. Then
(1)μ¯ is an RN-space on V/W,
(2)μ¯Qx(t)≥μx(t),
(3)   if(V,μ,T) is a random Banach space, then so is (V/W,μ¯,T).Proof.
It is clear thatμ¯x+W(t)>0. Let μ¯x+W(t)=1. By definition, there is a sequence {xn} in W such that μx+xn(t)→1. Thus x+xn→0 or, equivalently, xn→(-x) and since W is closed, x∈W and x+W=W, the zero element of V/W. Now, we have
(6.26)μ¯(x+W)+(y+W)(t)=μ¯(x+y)+W(t)≥μ(x+m)+(y+n)(t)≥T(μx+m(t1),μy+n(t2))
for m,n∈W, x,y∈V, and t1+t2=t. Now, if we take the sup, then we have
(6.27)μ¯(x+W)+(y+W)(t)≥T(μ¯x+W(t1),μ¯y+W(t2)).
Therefore, μ¯ is random norm on V/W.
(2) By Definition6.12, we have
(6.28)μ¯Qx(t)=μ¯x+W(t)=sup⁡{μx+y(t):y∈W}≥μx(t).
Note that, by Lemma 5.14,
(6.29)Eλ,μ¯(Qx)=inf⁡{t>0:μ¯Qx(t)>1-λ}≤inf⁡{t>0:μx(t)>1-λ}=Eλ,μ(x).
(3) Let{xn+W} be a Cauchy sequence in V/W. Then there exists n0∈ℕ such that, for every n≥n0, Eλ,μ¯((xn+W)-(xn+1+W))≤2-n. Let y1=0. Choose y2∈W such that
(6.30)Eλ,μ(x1-(x2-y2),t)≤Eλ,μ¯((x1-x2)+W)+12.
However, Eλ,μ¯((x1-x2)+W)≤1/2 and so Eλ,μ(x1-(x2-y2))≤1/22.
Now, suppose thatyn-1 has been chosen and so choose yn∈W such that
(6.31)Eλ,μ((xn-1+yn-1)-(xn+yn))≤Eλ,μ¯((xn-1-xn)+W)+2-n+1.
Hence we have
(6.32)Eλ,μ((xn-1+yn-1)-(xn+yn))≤2-n+2.
However, for every positive integerm>n and λ∈(0,1), by Lemma 5.14, there exists γ∈(0,1) such that
(6.33)Eλ,μ((xm+ym)-(xn+yn))≤Eγ,μ((xn+1+yn+1)-(xn+yn))+⋯+Eγ,μ((xm+ym)-(xm-1+ym-1))≤∑i=nm2-i.
By Lemma 5.14, {xn+yn} is a Cauchy sequence in V. Since V is complete, there is x0 in V such that xn+yn→x0 in V. On the other hand,
(6.34)xn+W=Q(xn+yn)⟶Q(x0)=x0+W.
Therefore, every Cauchy sequence {xn+W} is convergent in V/W and so V/W is complete. Thus (V/W,μ¯,T) is a random Banach space.Theorem 6.15.
LetW be a closed subspace of an RN-space (V,μ,T). If two of the spaces V, W, and V/W are complete, then so is the third one.Proof.
IfV is a random Banach space, then so are V/W and W. Hence all that needs to be checked is that V is complete whenever both W and V/W are complete. Suppose that W and V/W are random Banach spaces and let {xn} be a Cauchy sequence in V. Since
(6.35)Eλ,μ¯((xn-xm)+W)≤Eλ,μ(xn-xm)
for each m,n∈ℕ, the sequence {xn+W} is a Cauchy sequence in V/W and so  it converges to y+W for some y∈W. Thus there is n0∈ℕ such that, for every n≥n0,
(6.36)Eλ,μ¯((xn-y)+W)<2-n.
Now, by the last theorem, there exist a sequence {yn} in V such that yn+W=(xn-y)+W and
(6.37)Eλ,μ(yn)<Eλ,μ¯((xn-y)+W)+2-n.
Thus we have
(6.38)lim⁡nEλ,μ(yn)≤0
and, by Lemma 5.14, μyn(t)→1 for every t>0, that is, limnyn=0. Therefore, {xn-yn-y} is a Cauchy sequence in W and thus is convergent to a point z∈W. This implies that {xn} converges to z+y and hence V is complete.Theorem 6.16 (open mapping theorem).
IfT is a random bounded linear operator from an RN-space (V,μ,T) onto an RN-space (V′,ν,T), then T is an open mapping.Proof.
The theorem will be proved in several steps.
Step 1.
LetE be a neighborhood of the 0 in V. We show 0∈(T(E)¯)o. Let W be a balanced neighborhood of 0 such that W+W⊂E. Since T(V)=V′ and W is absorbing, it follows that V′=∩nT(nW) and so there exists an n0∈ℕ such that T(n0W)¯ has nonempty interior. Therefore, 0∈(T(W)¯)o-(T(W)¯)o. On the other hand,
(6.39)(T(W)¯)o-(T(W)¯)o⊂T(W)¯-T(W)¯=T(W)+T(W)¯⊂T(E)¯.
So, the set T(E)¯ includes the neighborhood (T(W)¯)o-(T(W)¯)o of 0.
Step 2.
We show0∈(T(E))o. Since 0∈E and E is an open set, there exist 0<α<1 and t0∈(0,∞) such that B0(α,t0)⊂E. However, 0<α<1 and so a sequence {ϵn} can be found such that
(6.40)Tm-n(1-ϵn+1,…,1-ϵm)⟶1,1-α<lim⁡nTn-1(1-ϵ1,1-ϵn),
in which m>n. On the other hand, 0∈T(B0(ϵn,tn′))¯, where tn′=(1/2n)t0 and so, by Step 1, there exist 0<σn<1 and tn>0 such that
(6.41)B0(σn,tn)⊂T(B0(ϵn,tn′))¯.
Since the set {B0(r,1/n)} is a countable local base at zero and tn′→0 as n→∞, so tn and σn can be chosen such that tn→0 and σn→0 as n→∞.
Now, we show that(6.42)B0(σ1,t1)⊂(T(E))o.
Suppose y0∈B0(,σ1,t1). Then y0∈T(B0(ϵ1,t1′))¯ and so for 0<σ2 and t2>0 the ball By0(σ2,t2) intersects T(B0(ϵ1,t1′)). Therefore, there exists x1∈B0(ϵ1,t1′) such that Tx1∈By0(σ2,t2), that is,
(6.43)νy0-Tx1(t2)>1-σ2
or equivalently
(6.44)y0-Tx1∈B0(σ2,t2)⊂T(B0(ϵ1,t1′))¯.
By the similar argument, there exist x2 in B0(ϵ2,t2′) such that
(6.45)νy0-(Tx1+Tx2)(t3)=ν(y0-Tx1)-Tx2(t3)>1-σ3.
If this process is continued, then it leads to a sequence {xn} such that xn∈B0(ϵn,tn′) and
(6.46)νy0-∑j=1n-1Txj(tn)>1-σn.
Now, if n,m∈ℕ and m>n, then we have
(6.47)μ∑j=1nxj-∑j=n+1mxj(t)=μ∑j=n+1mxj(t)≥Tm-n(μxn+1(tn+1),μxm(tm)),
where tn+1+tn+2+⋯+tm=t. Put t0′=min{tn+1,tn+2,…,tm}. Since tn′→0, there exists n0∈ℕ such that 0<tn′≤t0′ for n>n0. Therefore, for m>n we have
(6.48)Tm-n(μxn+1(t0′),μxm(t0′))≥Tm-n(μxn+1(tn+1′),μxm(tm′))≥Tm-n(1-ϵn+1,1-ϵm).
Hence it follows that
(6.49)lim⁡n→∞μ∑j=n+1mxj(t)≥lim⁡n→∞Tm-n(1-ϵn+1,1-ϵm)=1.
That is,
(6.50)μ∑j=n+1mxj(t)⟶1
for all t>0. Thus the sequence {∑j=1nxj} is a Cauchy sequence and consequently the series {∑j=1∞xj} converges to some point x0∈V because V is a complete space.
By fixingt>0, there exists n0∈ℕ such that t>tn for n>n0 because tn→0. Thus
(6.51)νy0-T(∑j=1n-1xj)(t)≥νy0-T(∑j=1n-1xj)(tn)≥1-σn
and thus
(6.52)νy0-T(∑j=1n-1xj)(t)⟶1.
Therefore, we have
(6.53)y0=lim⁡nT(∑j=1n-1xj)=T(lim⁡n∑j=1n-1xj)=Tx0.
But it follows that
(6.54)μx0(t0)=lim⁡nμ∑j=1nxj(t0)≥Tn(lim⁡n(μx1(t1′),μxn(tn′)))≥lim⁡nTn-1(1-ϵ1,…,1-ϵn)>1-α.
Hence x0∈B0(α,t0).
Step 3.
LetG be an open subset of V and x∈G. Then we have
(6.55)T(G)=Tx+T(-x+G)⊃Tx+(T(-x+G))o.
Hence T(G) is open, because it includes a neighborhood of each of its point.Corollary 6.17.
Every one-to-one random bounded linear operator from a random Banach space onto random Banach space has a random bounded inverse.Theorem 6.18 (closed graph theorem).
LetT be a linear operator from the random Banach space (V,μ,T) into the random Banach space (V′,ν,T). Suppose that, for every sequence {xn} in V such that xn→x and Txn→y for some elements x∈V and y∈V′, it follows that Tx=y. Then T is random bounded.Proof.
For anyt>0, x∈X and y∈V′, define
(6.56)Φ(x,y)(t)=T′(μx(t),νy(t)),
where T′≫T.
First we show that(V×V′,Φ,T) is a complete RN-space. The properties of (RN1) and (RN2) are immediate from the definition. For the triangle inequality (RN3), suppose that x,z∈V, y,u∈V′, and t,s>0, then
(6.57)T(Φ(x,y)(t),Φ(z,u)(s))=T[T′(μx(t),νy(t)),T′(μz(s),νu(s))]≤T′[T(μx(t),μz(s)),T(νy(t),νu(s))]≤T′(μx+z(t+s),νy+u(t+s))=Φ(x+z,y+u)(t+s).
Now, if{(xn,yn)} is a Cauchy sequence in V×V′, then for every ϵ>0 and t>0 there exists n0∈ℕ such that
(6.58)Φ(xn,yn)-(xm,ym)(t)>1-ϵ
for m,n>n0. Thus for m,n>n0,
(6.59)T′(μxn-xm(t),νyn-ym(t))=Φ(xn-xm,yn-ym)(t)=Φ(xn,yn)-(xm,ym)(t)>1-ϵ.
Therefore, {xn} and {yn} are Cauchy sequences in V and V′, respectively, and there exist x∈V and y∈V′ such that xn→x and yn→y and consequently (xn,yn)→(x,y). Hence (V×V′,Φ,T) is a complete RN-space. The remainder of the proof is the same as the classical case.


## 7. Fuzzy Normed Spaces

Now, we define the fuzzy normed spaces and give an example of these spaces. Here thet-norms notation is showed by *.Definition 7.1.
The triple(X,M,*) is said to be a fuzzy metric space if X is an arbitrary set, * is a continuous t-norm, and M is a fuzzy set on X2×(0,∞) satisfying the following conditions: for all x,y,z∈X and t,s>0,
(FM1)M(x,y,0)>0;
(FM2)M(x,y,t)=1 for all t>0 if and only if x=y;
(FM3)M(x,y,t)=M(y,x,t);
(FM4)M(x,y,t)*M(y,z,s)≤M(x,z,t+s) for all t,s>0;
(FM5)M(x,y,·):(0,∞)→[0,1] is continuous.Definition 7.2.
The triple(X,N,*) is said to be a fuzzy normed space if X is a vector space, * is a continuous t-norm, and N is a fuzzy set on X×(0,∞) satisfying the following conditions: for every x,y∈X and t,s>0;
(FN1)N(x,t)>0;
(FN2)N(x,t)=1 if and only if x=0;
(FN3)N(αx,t)=N(x,t/|α|), for all α≠0;
(FN4)N(x,t)*N(y,s)≤N(x+y,t+s);
(FN5)N(x,·):(0,∞)→[0,1] is continuous;
(FN6)limt→∞N(x,t)=1.Lemma 7.3.
LetN be a fuzzy norm, then we have
(i)N(x,t) is nondecreasing with respect to t for each x∈X,
(ii)N(x-y,t)=N(y-x,t).Proof.
Lett<s then k=s-t>0 and we have
(7.1)N(x,t)=N(x,t)*1=N(x,t)*N(0,k)≤N(x,s)
and this proves (i).
To prove (ii) we have(7.2)N(x-y,t)=N((-1)(y-x),t)=N(y-x,t|-1|)=N(y-x,t).Example 7.4.
Let(x,∥·∥) be a normed space, define a*b=ab or a*b=min(a,b) and
(7.3)N(x,t)=ktnktn+m‖x‖
for all k,m,n∈ℝ+. Then (X,N,*) is a fuzzy normed space. In particular, if k=n=m=1, then we have
(7.4)N(x,t)=tt+‖x‖,
which is called the standard fuzzy norm induced by norm ∥·∥.Lemma 7.5.
Let(X,N,*) be a fuzzy normed space. If one defines
(7.5)M(x,y,t)=N(x-y,t),
then M is a fuzzy metric on X, which is called the fuzzy metric induced by the fuzzy norm N.We can see both definition and properties fuzzy normed spaces are very similar to random normed spaces. ThenX equipped with μx(t)=N(x,t) and T=* can be regarded as an RN-space. Now, we extend the definition of fuzzy metric space. In fact we extend the range of fuzzy sets to arbitrary lattice.Definition 7.6.
The triple(X,𝒫,𝒯) is said to be an  ℒ-fuzzy normed space (briefly, ℒF-normed space) if X is a vector space, 𝒯 is a continuous t-norm on ℒ, and 𝒫 is an ℒ-fuzzy set on X×]0,+∞[ satisfying the following conditions: for every x,y in X and t,s in ]0,+∞[;
(a)𝒫(x,t)>L0ℒ;
(b)𝒫(x,t)=1ℒ if and only if x=0;
(c)𝒫(αx,t)=𝒫(x,t/|α|) for each α≠0;
(d)𝒯(𝒫(x,t),𝒫(y,s))≤L𝒫(x+y,t+s);
(e)𝒫(x,·):]0,∞[→L is continuous;
(f)limt→∞𝒫(x,t)=1ℒ.In this case,𝒫 is called an  ℒ-fuzzy norm (briefly, ℒF-norm). If 𝒫=𝒫μ,ν is an intuitionistic fuzzy set and the t-norm 𝒯 is t-representable then the triple (X,𝒫μ,ν,𝒯) is said to be an intuitionistic fuzzy normed space (briefly, IF-normed space).Example 7.7.
Let(X,∥·∥) be a normed space. Denote 𝒯(a,b)=(a1b1,min(a2+b2,1)) for all a=(a1,a2),b=(b1,b2)∈L* and let M and N be fuzzy sets on X×(0,∞) defined as follows:
(7.6)PM,N(x,t)=(htnhtn+m‖x‖,m‖x‖htn+m‖x‖)
for all t,h,m,n∈ℝ+. Then (X,𝒫M,N,𝒯) is an IF-normed space.



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*Source: 762361-2011-06-22.xml*