FUZZY TOPOLOGICAL VECTOR SPACE

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1 CHAPTER 6 FUZZY TOPOLOGICAL VECTOR SPACE AND TOPOLOGICAL GROUP 6.1 Introduction: Fuzzy topological vector spaces are introduced by A. K. Katsaras and D. B. Liu [K; L]. In 1981, Katsaras [kat], changed the definition of fuzzy topological vector spaces. After a considerable period of time, J. J. Buckley and Aimien Yan [B; A] opened the way towards the development of fuzzy topological vector space by introducing the notion of fuzzy vector space in a new manner. In this chapter, we establish a relation between fuzzy topological vector spaces and topological group. Also some properties of fuzzy topological vector spaces are discussed. 6.2 Fuzzy topological vector space and topological group Theorem Let X be a fuzzy vector space and let z be a topology on F. Then z is compatible with the vector space structure of x, that is, addition and scalar multiplication are continuous if and only if

2 (i) T is compatible with the additive group structure of (ii) for each as R, the mapping [u] -+ [au], ([u] E X ) is continuous at (iii) for each [u] E X, the mapping a+ [au], (a R) is continuous at (iv) the mapping (a,[u])+[au], (as R, [ u]~ T) is continuous Assume that T is con~patible with vector space structure. Then conditions (ii)-(iv) follows trivially fi-om the definition of a fuzzy topological vector space over R and condition (i) results from the continuity of ([u], [v]) + [LI] + [v] and [u] -+ (-l)[u] = [-u]. Conversely, let E R, [uo]~ X, the problem is to show that (a,[u]) 4 [au] is continuous at (ao, [u "1). Let U be a neighborhood

3 of 6, then [quo] + U is a typical neighborhood of [quo]. We seek neighborhood A, W of 0, o such that Let V be a neighborhood of 6 such that v+v+v cu. By (iii), there exists a neighbourhood Al of 0 such that Al[uo]c V. By (ii), there exists a neighbourhood W, of G such that a,,wi c V. By (iv), there exists a neighbourhood A2 of 0 and a neighbourhood W2 of O such that A2Wz c V. Let A = A) naz and W - WI n W2, then evidently A[uo] c V, aow c V and AW c V. Hence by distributivity, Hence the result.

4 Definition If X is fuzzy vector space, a homothetic mapping is a mapping of the form [u] + [au] + [v], ([VIE 2). Where a E R and [v] E.F are fixed. If a #O, such mapping is bijective with a homothetic inverse mapping namely, [u] -t [a-'u] - [d'v]. Note In a topological group, the translation mapping plays an important role; in a fuzzy topological vector space; the mapping that play the analogous role are homotheties. Theorem In a fuzzy topological vector space every non-constant homothety is a homeomorphism. Define the homothety, f([u]) = [au] + [v], where a + 0. Since the inverse of a homothety is also a homothety, it is enough to show that f is continuous and this is evident from the factorization [u] + [au] + [au] + [v].

5 Note One consequence of this proposition is that every topology z on a fuzzy topological vector space is translation invariant. A set E c 2 is open if and only if each of its translates [u] + E is open. Then T is completely determined by any local base. A local base of a fuzzy topological vector space is thus a neighborhoods of 0 such that every neighborhood of o contains a member of (B. The open sets of F are then precisely those that are unions of translates of members Theorem If X is a fuzzy topological vector space over R and M fuzzy vector sub space over R then M, with the relative topology is also a fuzzy topological vector space over R. The restriction of the continuous mappings ([u], [v]) + [u] + [v] - - and (a,[u]) + [au], to M x M and Rx M respectively are continuous for the relative topology of M.

6 Theorem In a fuzzy topological vector space, the closure of a fuzzy vector subspace is a fuzzy vector subspace. If M is a fuzzy vector subspace of a fuzzy topological vector space over R, then the continuous mapping ([u], [v])-t [u] + [v], which maps x a -+ M also maps clo(m x M ) = clom x clog + clom, then [u]~clom whenever [u], [v] E CIOM. Similarly the continuous mapping (a, [u])+[au], which maps Rx M + M also maps clo(rx ii?) = RxcloM -+ clom. 1.e.; [au]~cloe, kfa~ R and [u] E clom. Theorem In a fuzzy topological vector space 2 over R - (a) every neighborhood of o contains a balanced neighborhood of 0 and (b) every convex neighborhood of 8 contains a balanced convex neighborhood of 0.

7 Proof :(a) Suppose U is a neighborhood of Si in F. Since scalar multiplication is continuous, there is a 6>0 and there is a neighborhood V of 6 in 2 such that avc U whenever 1 a 1 < 6. Let W be the union of all these sets av. Then W is a neighborhood of Si, W is balanced and Wc u. Proof :(b) Suppose U is a convex neighborhood of in F.Let A = n au, where -1 < a i 1. Choose W as in part (a). Since W is balanced, a-'w = W, when ( a ( = 1. Hence W c au. Thus W c A, which implies that the interior of A is a neighborhood of 6. Clearly, interior of A is a subset of U. Being an intersection, with convex sets, A is convex. Hence, so is inta. To prove that inta is a neighbourhood with desired property, we have to show that inta is balanced; for this it is enough to prove that A is balanced. Choose c such that 0 i c < 1, then Since au is a convex set that contains G, we have

8 Thus ca c A. Wl~ich completes the proof. Theorem Suppose V is a neighborhood of 8 in a fuzzy topological vector space over R. (a)if O<r, <r2<:... and r,+ m as n-+ m,then (b) Every compact subset K of 3 is bounded. (c)if 61>82>... and 6,--0 as n+ m andif Vis bounded, then the collection (6,V, n=1,2,3,...) is a local base for X. (a) Fix [u] E 2. Since a + [au] is a continuous mapping of R+ F, then the set of all a with [au] E V is open, contains 0, hence 1 1 contains - for all large n. Thus - [u] E V or [u] E r, V, for rn r,, large n. (b) Then by (a) Let W be a balanced neighbourhood of 6 such that W c V.

9 Since K is compact, there are integers n, < n2 <...< n, such that K c nlwun2wu... un,w=n,w. The equality holds since W is balanced. If t > n,, it follows that K c tw c tv. (c) Let U be a neighborhood of 0 in Y. If V is bounded, there exists 6 > 0 such that V c: tu, V t z s. If n is so large that 6,) < I, it follows that V c i:i 1 - U. Hence U actually contains all but finitely many of the sets 6,,V Note This property is often described as neighborhoods of 0 are absorbing. Theorem Let Y and F be fuzzy topological vector spaces over R. If A : T + F is linear and continuous at 6, then A is continuous. In fact, A is uniformly continuous in the following sense: To each

10 neighborhood W of o in F there corresponds a neighborhood V of o in F such that [v] - [u] E V 3 A [v] - A [u] E W. Once W is chosen, the continuity of A at O shows that A V c W for some neighborhood V of 6. If [v] - [u] E V, the linearity of A shows that A [v] - A [u] = ~([v] - [u]) E W. Thus A maps the neighborhood [u] + V of [u] into the neighborhood A [u] + W of A [u]. Which says that /\ is continuous at [u]. Theorem Let F be a fuzzy topological vector space over R. Let A : F + R is linear. Assume A [u:l # 0, for some [u] E 2. Then each of the following four properties implies the other three. (a) A is continuous. (b) The null space N(A) is closed. (c) N(A) is not dense in T. (d) A is bounded in some neighborhood V of 6. Since N(A) = 12-'((0)) and (0) is the closed subset of R,

11 (a) 3 (b). By hypothesis, N(A) # F. Hence (b) = (c). Assume (c) holds. That is assume that the complement of N(A) has non-empty interiors. ([ul + V ) n N(A) = 4, (16) for some [u] E X and some balanced neighborhood V of 6. Then AV is a balanced subset of R. Thus either AV is bounded, in which case (d) holds, or A\.' = R. In the later case there exists [v] E V such that A[v] = -A[u] and so [v] + [u] E N(A) in contradiction to (16). Then (c) a (d). Finally if (d) holds, then 1 A[u] 1 < M, V [U]E V and forsome M< GO. If r>oand if W= - V, then - I A[u] I < r, V [u] E: W. Hence A is continuous at 0, by the theorem , and this implies (a). Definition A fuzzy topological vector space over R is said to be metrizable, if it is metrizable as a topological space. In view of (2.3.17), a separated fuzzy topological vector space E is metrizable if and only if there exists a hndamental sequence of neighbourhoods of - 0, in which case there exists a metric d, compatible with the topology of E, which is additively invariant.

12 Lemma Let E be a fuzzy vector space, with a metrizable topology such that (i) (ii) the topology is compatible with the additive group structure for each [u] E E, the mapping a + [au] is continuous at a = 0 and (iii) for each a E R, the mapping [u] + [au] is continuous at [u] = 0; then the topology is compatible with the vector space structure. That is E is a metrizable fuzzy topological vector space over R. The point is that in the presence of metrizability, condition (iv) of [6.2.1] is superfluous. Thus in view of [6.2.1], it will suffice to verify that (a, [u]) -+ [au] is continuous at (0, G ). We observe first that the mapping (a,[u]) -+ [au] is separately continuous in a and in [u]; this follows from (i) - (iii) and the formulas [auo]- [aouo]= [(a -- ao)uo] and [sou] - [aouo]= [ao(u -- uo)]. Let d be any metric that generates the topology. Assuming a,+o and d([u,], G) -+ 0, we wish to show that d([a,u,], G) + 0. Suppose to the contrary that there exists an E > 0 such that

13 d([anu,], 6) t E, for infinitely many n. We can suppose that d([a,,un], 6) 2 E, Vn. Let U be a neighourhood of o such that d([u] + [v], 6) < E, V [U:/,[V] E U (this is possible by (i)). We can suppose U to be closed and symmetric. For each n, let Since each of the sets {a: I:aui] U} is closed (it is the inverse image of the closed set U under the continuous mapping a + [aui]), it follows that the C, are closed. More over, For, given any a E R, one has [aui] -+ 6, by (iii); consequently there exists an index n such that taui] E U, Vi 2 n. i.e. a E C,. According to the relation (i), R is the union of a sequence of closed sets C,. It is a classical property of R is that at least one of the sets C, must have an interior point. Let % be an interior point of Ck. For a suitable 6 > 0, the open set B= {a: la-aoi<6)cck. Let A= {b: lb) ~6);then A is a symmetric neighbourhood of 0, and

14 ao+a=bcck. (17) Choose an index n so large that a, E A, Vi 2 n (possible, since a,-+o), [sou] E U, Vi 2 n (possible by (5)). Let m = maxfk,n}. Since m 2 n, we have a, E A (by the definition of n), and therefore % + am E Ck, by (17). Since m 2 k, it follows from the definition of Ck that (% + a,)[u,] E U. Also m 2 n 3 % [u,, ] E U (by the definition of n) and therefore [-sou,] E -U = U. Taking [u] = [-sou,,] and [v] = (ao + a,,,)[um]; we have [u], [v] EU and [u] + [v] = [a,,~,,,]; then d([a,u,], d([a,u,], 6) = d([u] + [v], 0 ) < E, by the definition of U, where as 6) 2 E, by the choice of E. Which is a contradiction. Hence the proof. Defmition A sub set M of a fuzzy topological vector space over R is said to be bounded, if to every neighborhood V of 0 in F corresponds a number 6 > 0 such that set M (= fv, Vt > 6. Theorem (a) If d is a translation invariant metric on a fuzzy vector space 2 over R, then

15 d([nu], G ) 5 nd([u], 0 ), V [u] E X and for n = 1,2,3,... (b) If {[u,]} is a sequence in a metrizable fuzzy topological vector space Fover R and if [u,] -+ 6 as n + m, then there are positive numbers r, such that r,, + m) and [r,u,] + 0. Statement (a) follows from d([nu], 6) I xd(k[~~],(k - l)[u]), by triangle inequality h=i = n d([u],o), by translation invariant. To prove (b), let d be a metric as in (a), compatible with the topology of F. Since d([u,],o) + 0, there is an increasing sequence of positive integers nk, such that For such n, d([u,], 8 ) < k-', if n 2 nk. Put r,= L, if n<nl and r,= k; if nklnlnk+,. d([r,u,], 0) = d([ku,],0) 5 kd ([u,], G) < k-i. Hence [r,u,] -+ G as n + m. Theorem Let X be a fuzzy topological vector space over R and E c F. Then the following two properties are equivalent.

16 (a) E is bounded (b) If {[u,]) is a sequence in E and r } is a sequence of real numbers such that r, -+ 0 as n -+ m, then [r,,u,]-+ G Suppose E is bounded. Let V be a balanced neighborhood of - 0 in X. Then Ec tv, for some t. If [un] E E and r,, + 0, there exists N such that rnt < 1, if n > N. Since t-'e c V and V is balanced, [rnun], V n > N. Thus [r,u,] -+ ii. Conversely, if B is not bounded, there is a neighborhood V of 0 and a sequence r + such that no r,v contains E. Choose [u,,] E E such that [u,] G r,v. Then no [r;' u,] is in V, so that {[rilun] $ does not converge to a.