In this paper, denotes a complete, non-trivially valued non-archimedean field. In the present paper, statistical convergence of sequences and statistically Cauchy sequences are defined and a few theorems on statistically convergent sequences are proved in such field

Krishnamurthy, Suja

Vaithinathasamy, Srinivasan

English

KHALSA PUBLICATIONS

ISSN 2347-1921 1038 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            On Statistically Convergent and Statistically Cauchy Sequences in Non-Archimedean Fields Suja Krishnamurthy, Srinivasan Vaithinathasamy Department of Mathematics Assistant Professor SRM University, India hari_se04@rediffmail.com Department of Mathematics Professor SRM University, India drvsrinivas.5@gmail.com  ABSTRACT In this paper, 𝐾 denotes a complete, non-trivially valued non-archimedean field. In the present paper, statistical convergence of sequences and statistically Cauchy sequences are defined and a few theorems on statistically convergent sequences are proved in such fields 𝐾. Keywords: Statistically convergent sequences; Statistically Cauchy sequences; Statistical limit superior; Statistical limit inferior. Academic Discipline And Sub-Disciplines: Mathematics (Analysis) Mathematics Subject  Classification: 40A05, 40A35, 46S10 Type (Method/Approach): Literary Analysis                       Council for Innovative Research Peer Review Research Publishing System Journal: Journal of Advances in Mathematics Vol 6, No. 3 editor@cirworld.com www.cirworld.com, member.cirworld.com  ISSN 2347-1921 1039 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            INTRODUCTION  The concept of statistical convergence was introduced by Fast in 1951. It was further studied, in detail by Kolk, Maddox, Bulut and Huseyin Cakalli [5]. The purpose of this paper is to give characterizations of statistical convergence of sequences and statistical Cauchy sequences in Non-Archimedean fields, which are analogous to the work of D. Rath and B.C. Tripathy [7], in classical case. Definition 1.  Let 𝐾 be a complete, non-trivially valued non-archimedean field.  A sequence 𝑥 =  𝑥𝑘  in 𝐾 is said to be statistically convergent to a limit ‘𝑙’ if for any  ϵ>o ,                             lim𝑛→∞1𝑛  𝑘 ≤ 𝑛:  𝑥𝑛 −  𝑙 ≥ є   = 0 Symbolically we write stat- lim𝑛→∞𝑥𝑘 = 𝑙       (or)       𝑥𝑘𝑠𝑡𝑎𝑡   𝑙.  Definition 2.  𝑥 =  𝑥𝑘  is a statistically Cauchy sequence if for any  ϵ > 𝑜 ,    there exists an  𝑛 ∈  ℕ 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡                               lim𝑛→∞1𝑛  𝑘 ≤ 𝑛:  𝑥𝑛+1 −  𝑥𝑛 ≥ є   =0                                                        Theorem 1. A sequence {𝑥𝑛 } is statistically convergent if and only if the following condition is satisfied:                           lim𝑛→∞1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 −  𝑥𝑘 ′(𝑟) ≥ є  = 0  𝑤ℎ𝑒𝑟𝑒  𝑥𝑘 ′ 𝑟   𝑖𝑠 𝑎 𝑠𝑢𝑏𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 𝑜𝑓  𝑥𝑛   𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡      lim⁡        𝑟→∞𝑥𝑘 ′  𝑟 = 𝑙 Proof: Let a sequence {𝑥𝑛 } be  statistically convergent . To prove   lim𝑛→∞1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 −  𝑥𝑘 ′(𝑟) ≥ є   =0                                                                                                                                                                    ….     (1) is satisfied. By  definition of statistical convergence of a sequence {𝑥𝑛 } to limit 𝑙 ,we have,                             lim𝑛→∞1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 −  𝑙 ≥ є   = 0                                                                                     ….     (2) Now,               lim𝑛→∞1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟  ≥ є                                =  lim𝑛→∞ 1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟 + 𝑙 − 𝑙 ≥ є                             =   lim           𝑛→∞ 1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  (𝑥𝑘 − 𝑙) + (𝑙 − 𝑥𝑘 ′ 𝑟 ) ≥ є                             ≤   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є   +   lim𝑛→∞ 1𝑛  𝑘′ ≤ 𝑛:  𝑙 − 𝑥𝑘 ′ 𝑟  ≥ є                             ≤ 𝑚𝑎𝑥    lim𝑛→∞ 1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є  ,   lim𝑛→∞ 1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′(𝑟) − 𝑙 ≥ є                              ≤ 𝑚𝑎𝑥  0,   lim𝑛→∞ 1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′(𝑟) − 𝑙 ≥ є                                                                           …. (3)  (using (2))  It is given that          lim𝑟→∞𝑥𝑘 ′(𝑟) = 𝑙. Since it is convergent, it is also statistically convergent. Therefore we can write               lim   𝑛→∞  1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′ 𝑟 − 𝑙 ≥ є       =0                                                                                               …. (4)                                                                                                  In view of (3) & (4) ∴          lim   𝑛→∞  1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟  ≥ є   =0 ISSN 2347-1921 1040 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            Conversly, let              lim   𝑛→∞   1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟  ≥ є   =0                                                                                    …. (5)            To prove that the sequence {𝑥𝑛} is statistically convergent. To this end, consider                                lim   𝑛→∞  1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є                                       =       lim   𝑛→∞  1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟 + 𝑥𝑘 ′(𝑟) − 𝑙 ≥ є                                     ≤     lim   𝑛→∞   1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟  ≥ є                                                             +    lim   𝑛→∞     1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′ 𝑟 − 𝑙 ≥ є                                     ≤   max⁡[  lim                 𝑛→∞  1𝑛  𝑘 ≤ 𝑛, 𝑘′ ≤ 𝑛:  𝑥𝑘 − 𝑥𝑘 ′ 𝑟  ≥ є   ,                                                              lim   𝑛→∞     1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′ 𝑟 − 𝑙 ≥ є  ]                                  ≤ max [0 ,  lim   𝑛→∞1𝑛  𝑘′ ≤ 𝑛:  𝑥𝑘 ′ 𝑟 − 𝑙 ≥ є   ]  , ( using (5)) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡    lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є   = 0,  (using (4)) This implies that  the sequence  {𝑥𝑛} is statistically convergent. Theorem 2. If          lim   𝑘→∞    𝑥𝑘 = 𝑙 and    𝑠𝑡𝑎𝑡 −   lim   𝑘→∞𝑦𝑘 = 0, then                                         𝑠𝑡𝑎𝑡 −   lim   𝑘→∞ 𝑥𝑘 + 𝑦𝑘 =   lim   𝑘→∞𝑥𝑘  . Proof. Given           lim   𝑘→∞    𝑥𝑘 = 𝑙                      (i.e)  𝑥𝑘 − 𝑙 = 0, as 𝑘 → ∞                                                                  …. (6) Also ,          𝑠𝑡𝑎𝑡 −    lim   𝑘→∞  𝑦𝑘 = 0    That is   lim   𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑦𝑛 − 0 ≥ 𝜖  = 0                                                    …. (7) Now,  Let                           𝑠𝑡𝑎𝑡 −   lim   𝑘→∞ 𝑥𝑘 + 𝑦𝑘 = 𝑙’                                                                                                  …. (8) Therefore, 𝑠𝑡𝑎𝑡 −   lim   𝑘→∞ 𝑥𝑘 + 𝑦𝑘 =   lim   𝑘→∞1𝑘  𝑛 ≤ 𝑘:   𝑥𝑛 + 𝑦𝑛 − 𝑙′ ≥∈     = 0                                                                ….(9) This implies that            lim𝑘→∞ 𝑥𝑘 − 𝑙′ + lim𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑦𝑛 − 0   ≥∈   = 0 That is     𝑚𝑎𝑥    lim   𝑘→∞  𝑥𝑘 − 𝑙′ ,          lim   𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑦𝑛 − 0 ≥ 𝜖       = 0 That is  𝑚𝑎𝑥    lim   𝑘→∞  𝑥𝑘 − 𝑙′ ,   0     = 0 ,      ( using (2)) which implies that  lim                                                    𝑘→∞  𝑥𝑘 − 𝑙′  = 0 𝑇ℎ𝑎𝑡 𝑖𝑠   lim𝑘→∞𝑥𝑘 = 𝑙′       . But   lim𝑘→∞𝑥𝑘 = 𝑙.  This ⇒ 𝑙′ = 𝑙                                                                            ….(10) From (8) & (10) , it is proved that        𝑠𝑡𝑎𝑡 −   lim   𝑘→∞ 𝑥𝑘 + 𝑦𝑘 = lim𝑘→∞ 𝑥𝑘  . Theorem  3. If a sequence 𝑥 =  𝑥𝑘  is statistically convergent to 𝑙, then there are  sequences  ISSN 2347-1921 1041 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            𝑦 =  𝑦𝑘  and 𝑧 =  𝑧𝑘  such that  lim   𝑘→∞  𝑦𝑘   = 𝑙, 𝑥 = 𝑦 + 𝑧  and   lim   𝑛→∞1𝑛  𝑘 ≤ 𝑛: 𝑥𝑘 ≠ 𝑦𝑘   = 0 and  𝑧 =  𝑧𝑘  is a statistically null sequence. Proof. Given a sequence 𝑥 =  𝑥𝑘  is statistically convergent to 𝑙  , 𝑇ℎ𝑎𝑡 𝑖𝑠    lim   𝑛→∞      1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є   = 0                                                                                            …. (11)      To prove (i) there exist sequences    𝑦 =  𝑦𝑘  and 𝑧 =  𝑧𝑘     such that                    𝑦𝑘 − 𝑙 → 0, 𝑘 → ∞  where   𝑥 = 𝑦 + 𝑧; 𝑥 =  𝑥𝑘  (ii)  lim   𝑛→∞1𝑛  𝑘 ≤ 𝑛: 𝑥𝑘 ≠ 𝑦𝑘   = 0   and (iii)   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛:  𝑧𝑘 − 0 ≥ є   = 0         (i)                  lim       𝑛→∞1𝑛  𝑘 ≤ 𝑛:  𝑥𝑘 − 𝑙 ≥ є   = 0      which we write as                                   𝑠𝑡𝑎𝑡 −   lim   𝑘→∞𝑥𝑘 = 𝑙                                                                                                            …. (12) Now, 𝑠𝑡𝑎𝑡 −     lim𝑥𝑘   𝑘→∞       means that     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑙 ≥ є   = 0 That is     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑙 + 𝑦𝑛 − 𝑦𝑛 ≥ є      = 0 That is     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  (𝑥𝑛 − 𝑦𝑛 ) + (𝑦𝑛 − 𝑙) ≥ є     =0 Therefore     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑦𝑛 ≥ є   +     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑦𝑛 − 𝑙 ≥ є  = 0                                                                                                                                                                 ….  (13) implies that max    lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑦𝑛 ≥ є   ,    lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑦𝑛 − 𝑙 ≥ є    =0 That is  max     lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑦𝑛 ≥ є    ,0   = 0    (𝑠𝑖𝑛𝑐𝑒  lim   𝑛→∞  𝑦𝑛   = 𝑙) Hence    lim       𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑦𝑛 ≥ є    = 0, which implies that         lim             𝑘→∞1𝑘  𝑛 ≤ 𝑘:  𝑥𝑛 − 𝑦𝑛 ↛ 0    = 0 That is  lim   𝑘→∞1𝑘  𝑛 ≤ 𝑘: 𝑥𝑛 ≠ 𝑦𝑛   = 0 (or)   lim   𝑛→∞1𝑛  𝑘 ≤ 𝑛: 𝑥𝑘 ≠ 𝑦𝑘  = 0 Since  lim   𝑘→∞  𝑦𝑘   = 𝑙 and since 𝑥 = 𝑦 + 𝑧   and    lim   𝑛→∞1𝑛  𝑘 ≤ 𝑛: 𝑥𝑘 ≠ 𝑦𝑘  = 0  , in view of  previous theorem, we have that  𝑠𝑡𝑎𝑡 −  lim   𝑘→∞ 𝑦𝑘 + 𝑧𝑘  =   lim   𝑘→∞⁡    𝑦𝑘     (=𝑙) implies that  𝑠𝑡𝑎𝑡 −     lim   𝑘→∞𝑧𝑘   must be  =0  𝑖𝑒   𝑧 =  𝑧𝑘  is a statistically null sequence. Definition3. For a sequence  𝑥 =  𝑥𝑘   , let 𝐵𝑥  denote the set                 𝐵𝑥  =  𝑏 ∈ 𝐾/ 𝑥𝑘 > 𝑏    ISSN 2347-1921 1042 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            Similarly   ,                        𝐴𝑥  =  𝑎 ∈ 𝐾/ 𝑥𝑘 < 𝑎 . Definition 4. If 𝑥 =  𝑥𝑘  is a sequence, then statistical limit superior of x is given by 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 =  𝑠𝑢𝑝𝐵𝑥          ,   𝑖𝑓  𝐵𝑥 ≠ ∅                                     −∞            , 𝑖𝑓  𝐵𝑥 = ∅                          Definition 5. If 𝑥 =  𝑥𝑘  is a sequence, then statistical limit superior of 𝑥 is given by                                    𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 =  𝑠𝑢𝑝𝐴𝑥         ,   𝑖𝑓  𝐴𝑥 ≠ ∅                              +∞           , 𝑖𝑓  𝐴𝑥 = ∅ Theorem 4. If 𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 is finite, then for every 𝜖 > 0,                 lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛽 + 𝜖 < 𝑥𝑘 < 𝛽 − 𝜖  = 0                                                                                            ….(14) Conversly, if (14) holds, then      𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥.          Proof. Given  𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥   is finite                                                                                                            ….  (15) To prove                     lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛽 + 𝜖 < 𝑥𝑘 < 𝛽 − 𝜖  = 0 Let us consider the case of statistical limit superior as                         𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 = −∞ ,   if 𝐵𝑥 = ∅  𝑖. 𝑒          𝑖𝑓                    𝐵𝑥   =   𝑏 ∈ 𝑘: 𝑥𝑘 > 𝑏  = 0                                                                                          …. 16  ⇒             lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛽 + 𝜖 < 𝑥𝑘 < 𝛽 − 𝜖  = 0                                                                                          … .  17  Conversly, let us suppose that                   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛽 + 𝜖 < 𝑥𝑘 < 𝛽 − 𝜖  = 0  To prove                      𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡  𝐵𝑥   =   𝑏 ∈ 𝑘: 𝑥𝑘 > 𝑏  = 0,  by the definition  That is,  𝐵𝑥 = ∅,  𝑊ℎ𝑖𝑐ℎ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡,          𝑙𝑢𝑏𝐵𝑥 = −∞  𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 In view of (17) we have that                  𝑥𝑘 < 𝛽 − 𝜖 ≠ 0 𝑇ℎ𝑎𝑡 𝑖𝑠, 𝐵𝑥 ≠ ∅  𝑊ℎ𝑖𝑐ℎ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡, 𝛽 = 𝑙𝑢𝑏𝐵𝑥 = 𝑠𝑢𝑝𝐵𝑥             𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 In any case,        𝛽 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑠𝑢𝑝𝑥 . Theorem 5. If 𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 is finite, then for every ∈> 0,                   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛼 + 𝜖 < 𝑥𝑘 < 𝛼 − 𝜖  = 0                                                                                    ….(18) Conversly, if (18) holds, then      𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥.          ISSN 2347-1921 1043 | P a g e                                                F e b r u a r y  2 0 ,  2 0 1 4            Proof. Given  𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥   is finite                                                                                                        … . (19) To prove                     lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛼 + 𝜖 < 𝑥𝑘 < 𝛼 − 𝜖  = 0 Let us consider the case of statistical limit inferior as                         𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 = +∞ ,   if 𝐴𝑥 = 𝜙 That is,         𝑖𝑓                    𝐴𝑥  =   𝑎 ∈ 𝑘: 𝑥𝑘 < 𝑎  = 0                                                                             ….  20  ⇒                   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛼 + 𝜖 < 𝑥𝑘 < 𝛼 − 𝜖  = 0                                                                              … .  21          Conversly, let us suppose that                   lim   𝑛→∞ 1𝑛  𝑘 ≤ 𝑛: 𝛼 + 𝜖 < 𝑥𝑘 < 𝛼 − 𝜖  = 0  To prove                      𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 We know that    𝐴𝑥   =   𝑎 ∈ 𝑘: 𝑥𝑘 < 𝑎  = 0,  by the definition                That is   𝐴𝑥 = ∅,  𝑤ℎ𝑖𝑐ℎ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡, 𝛼 = 𝑙𝑢𝑏𝐴𝑥 = +∞  Therefore,                     𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 In view of (21) we have that  𝑥𝑘 > 𝛼 + 𝜖 ≠ 0 That is, 𝐴𝑥 ≠ ∅ which implies that,   𝑙𝑢𝑏𝐴𝑥 = 𝑖𝑛𝑓𝑥  . Therefore,                       𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 In any case,                    𝛼 = 𝑠𝑡𝑎𝑡 − 𝑙𝑖𝑚𝑖𝑛𝑓𝑥 . REFERENCES [1] Fridy, J.A. (1985). On Statistical Convergence. Analysis. 5, 301-313. [2] Fridy, J.A. (1993). Statistical Limit Points. Proc. Amer. Math. Soc. 118(4), 1187-1192. [3] Fridy, J.A. and Orhan. C. (1997). Statistical Limit Superior And Limit Inferior, Proc. Amer. Math. Soc. 125 (97), 3625-3631. [4] Hardy, G.H. (1949). Divergent Series. Oxford Univ. Press, London. MR 11:25a. [5] Huseyin Cakalli (2009). A Study On Statistical Convergence, Functional Analysis, Approximation and Computation. 1(2), 19-24. [6] Lawrence Narici, Edward Beckenstein and George Bachman. (1971). Functional Analysis and Valuation Theory. Marcel Dekker Inc. New York. [7] Rath. D. and Tripathy. B.C. (1994). On Statistically Convergent and Statiscally Cauchy Sequences. Indian J. Pure Appl. Math. 25(4), 381-386.  

On Statistically Convergent and Statistically Cauchy Sequences in Non-Archimedean fields

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On Statistically Convergent and Statistically Cauchy Sequences in Non-Archimedean fields

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