ON SOME TOPOLOGICAL PROPERTIES OF NEW TYPE OF DIFFERENCE SEQUENCE SPACES

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1 Acta Universitatis Apulensis ISSN: No. 32/2012 pp ON SOME TOPOLOGICAL PROPERTIES OF NEW TYPE OF DIFFERENCE SEQUENCE SPACES Çiğdem A. BEKTAŞ and Gülcan ATICİ Abstract. In this paper, we define the sequence spaces l ( m,v ), c ( m,v ) and c 0 ( m,v ) (m N). Also we give some topological properties and inclusion relations of these sequence spaces. Keywords: Difference sequence spaces, Solid space, Symmetric space, Completeness Mathematics Subject Classification: 40C05, 40A05, 46A45. 1.Introduction Let l, c and c 0 be the linear spaces of bounded, convergent and null sequences x = (x ) with complex terms, respectively, normed by x = sup x where N = {1, 2,...}, the set of positive integers. The difference sequence spaces X( ) = {x = (x ) : x X} first defined by Kızmaz [1], where x = ( x ) = (x x +1 ) and X is any of the sets l, c and c 0, and showed that these are Banach spaces with norm x = x 1 + sup x. Then Çola [2] defined the sequence spaces v (X) = {x = (x ) : v x X}, where v x = ( v x ) = (v x v +1 x +1 ) and is any sequence space, and investigated some topological properties of this spaces. 61

2 A. Betaş and G. Atici - On Some Topological Properties of New Type of... Tripathy and Esi [3] defined the new type of difference sequence spaces Z( m ) = {x = (x ) : m x Z} for Z = l, c and c 0, where m N be fixed, m x = ( m x ) = (x x +m ) for all N and showed that these are Banach spaces with norm x m = x r + sup m x. Definition 1.1.[4] Let X be a sequence space. Then X is called: (i) Solid (or normal), if (α x ) X whenever (x ) X for all sequences (α ) of scalar with α 1. (ii) Monotone provided X contains the canonical preimages of all its stepspace. (iii) Symmetric if (x ) X implies ( x π() ) X, where π () is a permutation of N. (iv) A sequence algebra if (x ), (y ) X implies (x y ) X. (v) Convergence free if (y ) X whenever (x ) X and y = θ whenever x = θ. (vi) For r > 0, nonempty subset V of linear space is said to be absolutely r- convex if x, y V and λ r + µ r 1 together imply that λx + µx V. A linear topological space X is said to be r-convex if every neighborhood of θ X contains as absolutely r-convex neighborhood of θ X (see for instance [5]). 2. Main Results Let v = (v ) be any fixed sequence of nonzero complex numbers. Now we define l ( m,v ) = {x = (x ) : m,v x l }, c( m,v ) = {x = (x ) : m,v x c}, c 0 ( m,v ) = {x = (x ) : m,v x c 0 } where m N be fixed, m,v x = ( m,v x ) = (v x v +m x +m ) for all N. If we tae (v ) = (1, 1,...), then we obtain l ( m ), c ( m ) and c 0 ( m ). Theorem 2.1. The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are normed linear spaces, normed by x = v r x r + sup m,v x. (2.1) 62

3 A. Betaş and G. Atici - On Some Topological Properties of New Type of... Proof. We shall prove only for l ( m,v ). The other cases can be proved similarly. Let α, β be scalars and (x ), (y ) l ( m,v ). Then Hence sup sup m,v x < and sup m,v y <. (2.2) m,v (αx + βy ) α sup m,v x + β sup m,v y < by (2.2). Hence l ( m,v ) is a linear space. Next for x = θ, we have θ = 0. Conversely, let x = 0. Then x = v r x r + sup m,v x = 0. Since v r 0 for r N, we have x r = 0 for r = 1, 2,..., m and m,v x = 0 for all N. Consider = 1 i.e. m,v x 1 = 0 v 1 x 1 v 1+m x 1+m = 0 x 1+m = 0 (v 1+m 0), since x 1 = 0 (v 1 0). Proceeding in this way we have x = 0, for all N. After then we write x + y = Finally v r (x r + y r ) + sup m,v (x + y ) ( m = x + y. v r x r + sup m,v x λx = = λ x. ) + ( m λv r x r + sup m,v (λx ) v r y r + sup m,v y Hence. is a norm on the sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ). This completes the proof. Theorem 2.2. The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are Banach spaces under the norm (2.1). Proof. Let (x s ) be a Cauchy sequence in l ( m,v ), where x s = (x s i ) = (xs 1, xs 2,...) l ( m,v ), for each s N. Then x s x t = v r (x s r x t r) + sup m,v x s m,vx t 0 ) 63

4 A. Betaş and G. Atici - On Some Topological Properties of New Type of... as s, t. Hence for given ε > 0, there exists n 0 N such that x s x t = v r (x s r x t r) + sup m,v x s m,vx t < ε (2.3) for all s, t n 0. Hence we obtain v (x s xt ) < ε and since v 0 for all N we have x s xt < ε for all s, t n 0 and = 1, 2,..., m. Therefore (x t ) is a Cauchy sequence in C for each N. Since C is a complete space, (x t ) is a convergent in C for = 1, 2,..., m. Let lim x t = x say for each N. From (2.3) we have m,v x s m,vx t < ε for all s, t n 0 and all N. Hence ( m,v x t ) is a Cauchy sequence in C for all N. Thus ( m,v x t ) is a convergent in C and let lim m,vx t = y say for each N. Then we have lim v r (x s r x t r) = v r (x s r x r ) < ε s n 0, and lim v x s v x t (v +mx s +m v +mx t +m ) = v x s v x (v +m x s +m v +mx +m ) < ε for all N and s n 0. Hence for all s n 0, we have Thus we obtain by sup m,v x s m,vx < ε. v r (x s r x r ) + sup m,v x s m,vx < 2ε and (x s x) l ( m,v ) for all s n 0. Since l ( m,v ) is a linear space, we have x = x s (x s x) l ( m,v ), for all s n 0. Therefore l ( m,v ) is complete. It can be shown that c( m,v ) and c 0 ( m,v ) are closed subspaces of l ( m,v ). Therefore these sequence spaces are Banach spaces with norm (2.1). Theorem 2.3. The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are BKspaces with the same norm as in (2.1). Proof. These sequence spaces showed to be Banach space in Theorem 2.2. Now let x n x 0 64

5 A. Betaş and G. Atici - On Some Topological Properties of New Type of... as n. Then for m and x n x 0 (n ) m,v (x n x ) 0 (n ) for all N. Here also we obtain x n x 0 (n ) for all N. Hence sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are BK-spaces. Theorem 2.4. (i) X( ) X( m,v ), for X = l, c, c 0 and the inclusions are strict. (ii) c 0 ( m,v ) c( m,v ) l ( m,v ) and the inclusions are strict. Proof. (i) The proof is obtain for m = 1 and v = 1 for all N. To show the inclusions are strict consider the following example. Example 1. Let m = 2, v = 1 for all N and consider the sequence (x ) defined by x = 1 for odd and x = 0 for even. Then the sequence (x ) belongs to c 0 ( m,v ) but does not belong to c 0 ( ). Let m = 1, v = 1 for all N and x = (2 ). Then the sequence (x ) belongs to c( m,v ) l ( m,v ) but does not belong to c( ) l ( ). (ii) The inclusion c 0 ( m,v ) c( m,v ) is obvious. Now let x c( m,v ). Since m,v (x ) c l, we obtain x l ( m,v ). Thus c( m,v ) l ( m,v ). To show the inclusions are strict consider the following example. Example 2. Let m = 1, v = 1 for all N and x =. Then the sequence (x ) belongs to c( m,v ) but does not belong to c 0 ( m,v ). Let m = 1, v = 1 for all N and consider the sequence (x ) defined by x = 1 for odd and x = 0 for even. Then the sequence (x ) belongs to l ( m,v ) but does not belong to c( m,v ). Theorem 2.5. The sequence spaces c( m,v ) and c 0 ( m,v ) are closed subsets in l ( m,v ). Proof. Since c l, then c( m,v ) l ( m,v ) by Theorem 2.4 (ii). Now we show that c( m,v ) = c( m,v ), where c( m,v ), the closure of c( m,v ) and c, the closure of c. Let x c( m,v ), then there exists a sequence (x n ) in c( m,v ) such that in c( m,v ), and so x n x 0 (n ) v r (x n x r ) + sup m,v x n m,vx (n ) 65

6 A. Betaş and G. Atici - On Some Topological Properties of New Type of... in c. Thus m,v x c. Hence x c( m,v ). Conversely if x c( m,v ), then m,v (x) c. Since c is closed, x c( m,v ) c( m,v ). Hence x c( m,v ). This completes the proof. The proof of c 0 ( m,v ) is similar to that of c( m,v ). Theorem 2.6. The sequence spaces c( m,v ) and c 0 ( m,v ) are separable spaces. Proof. The proof is similar to that of Theorem 2.5. Theorem 2.7. The sequence spaces c( m,v ) and c 0 ( m,v ) are nowhere dense subsets of l ( m,v ). Proof. Suppose that c o o =, but c( m,v ). Then c contains no neighborhood and B(a) c( m,v ), where B(a) is a neighborhood of center a and radius r. Hence This implies that m,v (a) c. So a B(a) c( m,v ) = c( m,v ). B( m,v (a)) c. This contradicts to c o o =. Hence c( m,v ) =. The proof of c 0 ( m,v ) is similar to that of c( m,v ). The proofs of the following theorems are obtained by using the same technique of Tripathy and Esi [3], therefore we give it without proof. Theorem 2.8. The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are not solid, not monotone and not convergence free. Theorem 2.9. The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are not symmetric for m > 1. Theorem The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are not sequence algebra. Proof. The proof follows from the following examples. Example 3. Let m = 1, v = 1 for all N, x = and y = for all N. Then x, y c( ) l ( ), but (x.y) / c( ) l ( ). 66

7 A. Betaş and G. Atici - On Some Topological Properties of New Type of... Let m = 1, v = 1 for all N, x = and y = for all N. Then x, y c 0 ( m,v ), but (x.y) / c 0 ( m,v ). Theorem The sequence spaces l ( m,v ), c( m,v ) and c 0 ( m,v ) are 1- convex. Proof. If 0 < δ < 1, then V = {x = (x ) : x δ} is an absolutely 1-convex set, for x, y V and λ + µ 1, then λx + µy = This completes the proof. v r (λx r + µy r ) + sup m,v (λx + µy ) λ x + µ y δ( λ + µ ) δ. References [1] H. Kızmaz, On certain sequence spaces. Canad. Math.Bull., 24, (1981), [2] R. Çola, On Some Generalized Sequence Spaces, Commun. Fac. Sci. Uni. An. Series A 1, 38, (1989), [3] B.C. Tripathy, and A. Esi, A New Type of Difference Sequence spaces, Int. J. Sci. & Tech., 1(1), (2006), [4] P. K. Kamthan, and M. Gupta, Sequence spaces and series, Lecture Notes in Pure and Applied Mathematics, 65, Marcel Deer Incorporated, New Yor, [5] I.J. Maddox and J.W. Roles, Absolute convexity in certain topological linear spaces, Proc.Camb.Phil.Soc., 66, (1969), [6] I.J. Maddox, Elements of Functional Analysis, Cambridge University Press, Cambridge, London and New Yor, Çiğdem A. BEKTAŞ Department of Mathematics Firat University, Elazig, 23119, TURKEY. cigdemas78@hotmail.com Gülcan ATICİ Department of Mathematics Muş Alparslan University, Muş, 49100, TURKEY. gatici23@hotmail.com 67