Selçuk J. Appl. Math. Selçuk Journal of Vol. 9. No. 2, pp. 9 –18 , 2008 Applied Mathematics
Some I-related properties of triple sequences
Ahmet Sahiner1, Binod Chandra Tripathy2
1Department of Mathematics, Suleyman Demirel University, Cunur Campus, 32260,
Isparta, Turkey
e-mail: sahiner@fef.sdu.edu.tr
2
Mathematical Sciences Division, Institute of Advanced Study in Science and Technology, Paschim Boragaon,Garcuk; Guwahati-781 035, India e-mail: tripathybc@yahoo.com
Abstract. Ideal convergence was presented by Kostyrko et al. in 2001. This concept was extended to the double sequences by Tripathy et al. in 2006. In this paper we introduce the notions of Iconvergence, Ibounded,
I inferior and Isuperior for triple sequences. We also investigate some further properties of Ilimit superior and Ilimit inferior of triple sequences.
Key words: Double sequence; triple sequence; statistical convergence; I-convergence; double natural density; triple natural density
2000 Mathematics Subject Classification. 40A05, 26A03.
1. Preliminaries
In this article we aimed to extend the notion of statistically convergent triple sequences to I convergent triple sequences. Now we recall some definitions and notions introduced in [14].
Definition 1. A function x : NNNR
C is called a real (complex) triple sequence.Definition 2. A triple sequence
xnkl is said to be convergent to Lin Pringsheim's sense if for every 0, there exists n0
N such that xnklL whenever n,k,ln0 .Definition 3. A triple sequence
xnkl is said to be Cauchy sequence if for every 0, there exists n0
N such that
nkl
pqr x
x whenever pnn0,qk n0,rln0.
Definition 4. A triple sequence
xnkl is said to be bounded if there exists M 0 such that xnkl M for all n ,,k lN .We denote the set of all bounded triple sequences by 3. It can easily be shown that 3 is a normed space by ,3sup, , nkl . l k n x x
The notion of statistically convergent double sequences was introduced by Tripathy [12]. Recall that a subset E of NN is said to have density
E if
E E
n k q k p n pq q plim , 1 ,
exists, where E
n,k is the characteristic function of the set E. Thus a double sequence
xnk is said to be statistically convergent to L in Pringsheim's sense iffor every 0,
, :
0 n k N N xnk L
[13]. The notion of statistically convergent double sequences was extended to I convergent double sequences by Tripathy in [13]. The notion of statistically convergent triple sequences was introduced by Sahiner [14]. Recall that a subset A of NNN is said to have density
A if
n k l
pqr A A r l q k p n r q p , , 1 lim , ,
exists. For example if A
n3,k3,l3
: n,k,lN
then
lim
, ,
lim 0 3 3 3 , , , , pqr r q p pqr r q p K A r q p r q p where
p q r
n k l
p n q k r l
K , , , , NNN : , , and K
p,q,r
is the cardinality of K
p,q,r
. Thus a triple sequence
xnkl is said to be statistically convergent to L in Pringsheim's sense iffor every 0 ,
, , :
0. n k l N N N xnkl L
Let X be a non-empty set, then a non-void class I2X (power set of X ) is called an ideal if I is additive
i.e.A,BIABI
and hereditary
i.e.AIandBABI
. An ideal I2X is said to be non-trival if I2X.A non-trival ideal I is said to be admissible if I contains every finite subset of N. A non-trival ideal I is said to be maximal if there does not exist any non trival ideal JI containing I as a subset.
In this article we aimed to introduce and examine I related interesting properties of triple sequences.
We denote the ideals of 2 by I ; the ideals of N 2NN by I and the 2
ideals of 2NNN by I . 3
2. Ideal convergence of triple sequences
Definition 5. Let I be an ideal of 3 2NNN, then a triple sequence
xnkl is said to be I convergent to L in Pringsheim's sense if forevery 0,
n,k,l NNN : xnklL
I3.If
xnkl is I convergent to L we write I - 3 limxnklL .Now we give some examples of ideals and corresponding
I convergences.
(I) Let I3
f be the family of all finite subsets of NNN . Then
f3
I is an admissible ideal in NNN and I3
f convergence coincides with the convergence of triple sequences in Pringsheim's sense.(II) Let ANNN be a three dimensional set of positive integers and let A
p,q,r
be the cardinality of
n ,,k l
in A such thatp
n , kq , lr. In case of the sequence
Appqrqr
r q p , , , ,
lim has a limit in Pringsheim's sense then we say that A has a triple natural density
and we denote this by lim
, ,
. , , pqr A r q p A r q p Put
:
0
. 3 ANNN A I Then I3
is an admissible ideal in NNN and I3
convergence coincides with the statistical convergence in Pringsheim's sense [14].Example 1. Let II3
. Define the triple sequence
xnkl by otherwise. , 5 squares are and , if , 1 n k l xnkl Then for every 0
, , : 5
lim 0. , , pqr r q p x l k n r q p nkl N N NThis implies that I limxnkl5 in Pringsheim's sense. But, the sequence
xnkl is not convergent to 5 in Pringsheim's sense.Remark 1. If I is admissible and 3
xnkl converges to L inPringsheim's sense, then
xnkl is I convergent to L in Pringsheim'ssense.
3. I- limit superior and I- limit inferior for triple sequences
Definition 6. Let I be an ideal of 3 N N N
2 . A number is said to be an I3limit point of the triple sequence
xnkl provided that thereexists a set M
n1n2...
k1k2...
l1l2...
NNN such that MI3 and m j ikl
n
x
P lim for all i,j,m1,2,... .
Definition 7. A number is called to be an I cluster point of the triple sequence
xnkl if for each 0 ,
n,k,l NNN: xnkl
I3.Definition 8. A real triple sequence
xnkl is said to be bounded if there is a K 0 such that
n,k,l
NNN: xnkl K
I3.Now let
xnkl be a triple sequence and tR . Then we set the following sets to be able to give the definitions of Iliminf x andx sup lim I of
xnkl .
n,k,l : x t
, M
n,k,l
: x t
. Mt nkl t nklDefinition 9. (a) If there is a tR such that MtI3, we put
:
.sup sup
lim I3
I x tR Mt
If MtI3 holds for each tR then we put Ilimsupx.
(b )If there is a tR such that tI3,
M we put
:
. inf inf lim I3 I t M t x R If tI3M holds for each tR then we put Iliminfx.
Example 2. If we define
xnkl by nonsquare even an is , 0 nonsquare odd an is , 1 square even an is , 2 square odd an is , n n n n n xnkl or nonsquare even an is , 0 nonsquare odd an is , 1 square even an is , 2 square odd an is , k k k k k xnkl or nonsquare even an is , 0 nonsquare odd an is , 1 square even an is , 2 square odd an is , l l l l l xnklthen, in each case,
xnkl is not bounded above but it is I bounded. Also,
tR : MtI3
,1
,
tR : MtI3
0, and thus Ilimsupx1 , Iliminfx0 . On the other hand
xnkl can not be I convergent in Pringsheim's sense and the set of I cluster points in Pringsheim's sense is
0,1. So we have the following.Theorem 1. (i) Ilimsupx if and only if for each 0 ,
n,k,l NNN: xnkl
I3and
n,k,l NNN: xnkl
I3(ii) Iliminfx if and only if for each 0 ,
n,k,l NNN: xnkl
I3and
n,k,l NNN: xnkl
I3.Proof. (i) We prove necessity first. Let 0 be given. Since ,
t : MtI3
and
n,k,l NNN: xnkl
I3. Similarly, since , there exists some t such that t and t
t : MtI3
. Thus
n,k,l
NNN: xnkl t
I3 and
n,k,l NNN: xnkl
I3.Now we prove sufficiency. If 0 then
t : MtI3
and. sup
lim
3 xnkl
I On the other hand we already have
limsupxnkl
3
I and this means I3limsupxnkl as
desired.
(ii) can be proved analogously.
Theorem 2. For every real triple sequence
xnkl ,nkl nkl x x limsup inf lim 3 3 I I .
Proof. If
xnkl is any real triple sequence we have three possibilities: (1) The case I3limsupxnkl is clear.(2) If I3limsupxnkl. Then we have . and 3 3 I I t t M M t R
Thus, I3liminf inf
: tI3
infRnkl t M x and nkl nkl x x limsup inf lim 3 3 I I .
(3) If I3limsupxnkl and if I3limsupxnkl then for any tR, . and 3 3 I I t t M M t
But this means I3liminf inf
: I3
.t
nkl t M
x
Theorem 3. For any I bounded real triple sequence
xnkl
wehave the following inequalities.
. sup lim sup lim inf lim inf lim xnkl 3 xnkl 3 xnkl P xnkl P I I
Proof. The case Plimsupxnkl is straightforward. Let
x L
P limsup nkl . Then for any t L, MtI3. So
: I3
t Mt
t implies I3limsupxnklsup
t : MtI3
t and .sup lim
3 xnklL
I This proves the last inequality. For the first one, if Pliminfxnkl then clearly the inequality holds. Let
x T
P liminf nkl . Then for any tT, I3. t M So
: I3
t M tt implies I3liminf xnklsup
t : MtI3
t and .sup lim
3 xnklT
I
Remark 2. If I3 limxnkl exists then
xnkl is I bounded. Remark 3. Note that ideal boundedness of triple sequences implies that I3limsup and I3liminf are finite.
Recall that the core of a bounded double sequence xnk, that is,
xnk ,core
P is the interval
Pliminfxnk,Plimsupxnk
Pcore
xnk ; I core of bounded double sequence x is the interval nk
I2liminf xnk,I2limsupxnk
. Analogously we give the definitions of Pcore and Icore of a bounded triple sequence xnkl.Definition 10. We define the Pcore of bounded real triple sequence
nkl
x by
Definition 11. If
xnkl is any I bounded real triple sequence then 3 we define its Icore by
I3liminf xnkl,I3limsupxnkl
. We use I3core
xnkl to denote I core of 3 xnkl .Corollary 1. If
xnkl is any real triple sequence then we have
.3core xnkl Pcore xnkl
I
Theorem 4. A real triple sequence
xnkl is I3 convergent if and only if I3liminf xnklI3limsupxnkl .Proof. Let LI3limxnkl. Then
n,k,l NNN: xnklL
I3 and
n,k,l NNN: xnklL
I3. Then for any tL and ,
L
t the sets M and t Mt are in I Hence 3.
t : Mt 3
Lsup I and inf
t : MtI3
L . To prove sufficiency let 0 and LI3liminf xnklI3limsupxnkl. Then since
L x l k n L x l k n L x l k n nkl nkl nkl : , , : , , : , , N N N N N N N N Nwe conclude that LI3limxnkl .
Note that if
xnkl is a bounded real triple sequence then we denote the set of all I3 cluster points of
xnkl by I3
x .Theorem 5. Suppose
xnkl is a bounded real triple sequence then
x nklx
3
3 limsup maxI
I and
. min inf lim 3 3 xnkl I x I Proof. Let
3
3 limsup sup : , , : I
I xnkl L t n k l NNN xnklt . If L
n,k,l N: xnklL
I3 and this means there exists some0 such that
n,k,l NNN: xnklL
I3, that is, L
x .Now, we show L is really an I3 cluster point of
xnkl . Clearly, for each 0 there exists some t
L,L
such that
n,k,l NNN: xnkl t
I3 and this implies
n,k,l NNN: xnklL
I3.4. Further properties
In this section we prove some further results on Ilimsup and inf
lim
I of a triple sequence.
Theorem 6. Let I be an ideal of 3 2 . N N N
If x
xnkl ,y
ynkl aretwo I bounded triple sequences in Pringsheim's sense, then (i) Ilimsup
xy
IlimsupxIlimsupy(ii) Iliminf
x y
Iliminf xIliminf y.Proof. Since the proof of
ii is analogous we prove only
i Let . ,sup lim
1I x
2 Ilimsupy and 0 be given. We know that both and 1 are finite. Let 2
: , , : I3
. c n k l x y c
A R nkl nkl
We can also assume that A is not empty. Now since
2 1 2 1 : , , 2 : , , 2 : , , nkl nkl nkl nkl y x l k n y l k n x l k n we have
. 2 : , , 2 : , , : , , 2 1 2 1 nkl nkl nkl nkl y l k n x l k n y x l k n
n,k,l : xnklynkl 12
I3.If cA, then
n,k,l
: xnklynkl c
I3. We claim that .2
1
c For, otherwise we would have
n,k,l : xnklynklc
n,k,l
: xnklynkl12
which means
n,k,l
: xnklynklc
I3, a contradiction. Hence 1 2 c and we deduce
sup . sup lim 1 2 3 xnklynkl A ISince 0 is arbitrary, this completes the proof.
We need the following definition for the subsequent theorem.
Definition 12. Let I be an ideal of 3 2 . N N N
A sequence x
xnklis said to be I convergent to in Pringsheim's sense
or
if for every real number G0,
n,k,l
: xnklG
I3
or n,k,l : xnkl G I3
.Theorem 7. If I is an admissible ideal and 3 Ilimsupx, then
there exists a subsequence of
xnkl that is I convergent to inPringsheim's sense.
Proof. Since I3 and I is admissible, we can assume that 3
xnkl is a non-constant triple sequence having infinite number ofdistinct elements.
Case (i) : If . Then from definition,
tR : MtI
. Hence, if K 0, then
n,k,l
: xnkl2K
I3. Since
n,k,l : xnklK
n,k,l
: xnkl 2K
, we have
n,k,l
: xnklK
I3 and so Ilimx.Case (ii) : If then
tR : MtI
R. So for any tR,
n,k,l : xnkl t
I3. Let xn1k1l1 be an arbitrary term of
xnkl andlet
, ,
: 1
. 1 1 1 1 1 1kl nkl nkl n n k l x x A Since I3 , 1 1 1kl n A is not empty and also 3.1 1 1kl I
n
A We claim that there is at least one
n,k,l
An1k1l1 such that nn11,kk11, ll11. For, otherwise
1,1,1, 2,2,2,..., 1, 1, 1, 1 1, 1 1, 1 1
, 1 1 1 n n n n n n Ankl which is amember of I (since 3 I is admissible) and so 3 3,
1 1 1kl I
n
A a
contradiction. We call this
n ,,k l
as
n2,k2,l2
. Thus . 1 1 1 1 2 2 2kl nkl n xx Proceeding in this way we obtain a subsequence
xnikili of
xnkl with xnikili xni1ki1li1 1 for all .i Since for any, 0 K
n k l
x K
i i ikl n i ii, , : is a finite set, it must belong to I 3,
because I is admissible and so 3 3lim .
kl
n
x
I
Case (iii) : If . By Theorem 1(i),
n,k,l : xnkl 1
I3 so that
n,k,l
: xnkl1
. Weobserve that there is at least one element, say n1k1l1, in this set for
which 21, 1 1 1kl n x for otherwise
2
3 1 : , , 1 : , ,k l xnkl n k l xnkl I n which is acontradiction. Hence we have
. 1 2 1 1 1 1 1 xnkl
Next we proceed to choose an element
2 2 2kl n x from
xnkl , n2 n1 , 1 2 kk and l2 l1 such that
2 1 2 2 2kl n x , for otherwise
2
1 1 1
1 , , ,..., 2 , 2 , 2 , 1 , 1 , 1 : , ,k l x n n n n nkl is a member of 3I which contradicts
i of Theorem 1. Hence
( ) . 2 1 and , , : , , 1 1 1 1 1 1 say E x l l k k n n l k n nkl nkl Now if
1 1 1 , ,k l Enkl n always implies 21 nkl x then
. 4 1 : , , 2 1 : , , 1 1 1 nkl nkl l k n n k l x n k l x EBy
i of Theorem 1, the right-hand set belongs to I and so 3 . 3 1 1 1kl I n E Since I is admissible, 3
1,1,1, 2,2,2
,..., n1,n1,n1
I3 and thus
1,1,1
, 2,2,2
,..., , ,
. 2 1 : , , 1 1 1 1 1 1 nkl nkl n n n E x l k n So
2
3 1 : , ,k l xnkl I n , a contradiction to Theorem 1.and l2 l1 such that 21. 2 1 2 2 2
xn kl Proceeding in this way we
obtain a subsequence
i i ikl n x of
xnkl , ni ni1,ki ki1 and 1 i i l l such that i xnikili i 1 1 for each .i The subsequence
xnikili therefore converges to in Pringsheim's sense and is thus
I convergent to in Pringsheim's sense by Remark 1. This proves the theorem.
The reasonings made up to now give the folowing results.
Theorem 8. If Iliminfx, then there is a subsequence of
xnklwhich is I convergent to in Pringsheim's sense.
Theorem 9. Every I bounded triple sequence
xnkl inPringsheim's sense has a subsequence which is I convergent to a finite real number in Pringsheim's sense.
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