# OnMλm,n-Statistical Convergence

Authors: Nazlım Deniz Aral; Şerife Günal
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we introduce a new type of statistical convergence method for double sequences by using theM,λm,n-method of summability which was defined by Natarajan. We also obtain some inclusion relations between statistical convergence and Mλm,n-statistical convergence for double sequences.


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## Body

## 1. Introduction

The subject of statistical convergence has been studied by many researchers since the emergence of the idea of statistical convergence in 1935. Statistical convergence was introduced by Fast [2] and Steinhaus [3] independently in the same year 1951 as a generalization of ordinary convergence and was later reintroduced by Schoenberg [4]. Quite a few researchers have generalized or extended this concept and applied different fields of mathematics such as Erdös and Tenenbaum [5], Miller [6], Zygmund [7], Freedmanet al. [8], Connor [9], Salat [10], Duman and Orhan [11], Et et al. [12], Çakallı [13, 14], Çakallı and Savaş [15], Edely et al. [16], Mursaleen et al. [17, 18], Natarajan [19], Tok and Başarır [20], Aral and Küçükaslan [21–23], and Taylan [24].Letx=xm,n be a double sequence. Then, x=xm,nm,n=0∞ is said to be convergent to L in the Pringsheim sense if for every ε>0, there exists N∈ℕ such that xm,n−L<ε, whenever m,n>N. In this case, we write P−limx=L [25].Also, a double sequencex=xm,nm,n=0∞ is said to be bounded if there exists a positive number M such that xm,n<M for all m,n∈ℕ×ℕ. By ℓ∞2, we denote the set of all bounded double sequences.LetK⊆ℕ×ℕ and Km,n=m,n:k≤m,j≤n. Then, the double natural density of K is given by(1)δK≔P−limm,n⟶∞1mn∑k,j=1m,nχKm,nk,jif the limit exists. A double sequence x=xm,n is said to be statistically convergent to L provided that, for every ε>0, the set(2)m,n:k≤mandj≤n:xm,n−L≥εhas double natural density zero [26]. In this case, we write st2−limx=L. By st2, we denote the set of all statistically convergent double sequences. Later, a lot of works have been done on the statistical convergence of double sequences (see [27–33]).The following definition which is required for our study is given by Natarajan.Definition 1 (see [1]).
Letλ=λm,n be a double sequence such that ∑m,n=0∞,∞λm,n<∞. Then, the M,λm,n-method is defined by the 4-dimensional infinite matrix am,n,k,j, where(3)am,n,k,j≔λm−k,n−j,ifk≤m,j≤n,0,otherwise.
It is well known from Theorem 3.4 of [1] that the M,λm,n-method is regular if and only if ∑m,n=0∞,∞λm,n=1.
The purpose of this paper is to give a new statistical convergence definition using the above definition for double sequences and some relations between statistical convergence andMλm,n-statistical convergence. In addition, we have used it to prove a Korovkin-type approximation theorem as an application of our method.
We acknowledge that the definition ofMλm,n-statistical convergence for double sequences was presented at the International Conference on Multidisciplinary, Engineering, Science, Education and Technology in 2017 [34].
Letx=xm,n be a given real-valued sequence. A sequence tm,n of Mλm,n-mean of xm,n is defined by(4)tm,n=∑k,j=0m,nλm−k,n−jxk,j,for all m,n∈ℕ.Definition 2.
The sequencex=xm,n is said to be Mλm,n-summable to L∈ℝ if tm,n⟶L and is denoted by(5)xm,n⟶LMλm,n.Definition 3.
A double sequencex=xm,n is said to be strongly Mλm,n-summable to L∈ℝ if xm,n−L is Mλm,n-summable to zero. In this case, we write xm,n⟶LMλm,n. The set of all strongly Mλm,n-summable sequences is denoted by Mλm,n as(6)Mλm,n=x=xk,j:limm,n⟶∞,∞∑k,j=0m,nλm−k,n−jxm,n−L=0.
By considering the matrixMλm,n in (3) for any λ=λm,n, natural density and statistical convergence can be defined as follows.Definition 4 (Mλm,n-density).
LetK be a subset of ℕ×ℕ. Then, Mλm,n-density of K is denoted by δλm,nK and defined by(7)limm,n⟶∞∑k,j=0m,nλm−k,n−jχKk,jif the limit exists.Definition 5 (Mλm,n-statistical convergence).
A double sequencex=xm,n is said to be Mλm,n-statistically convergent to L if for every ε>0, Mλm,n-density of the set Kε≔m,n:xm,n−L≥ε is zero, i.e.,(8)limm,n⟶∞∑k,j=0m,nλm−k,n−jχKεk,j=0.
It is denoted byxm,n⟶LMλm,n−st. The set of all Mλm,n-statistically convergent sequences is denoted by Mλm,nst, i.e.,(9)Mλm,nst≔x=xm,n:∃L∈ℝsuch thatxm,n⟶LMλm,n−st.
Letp=pn and q=qn be sequences of positive natural numbers and Pm≔∑k=0mpk⟶∞ and Qn≔∑k=0nqk⟶∞. Take λ=λm,n, where λm,n=λm,n,k,j with(10)λm,n,k,j≔pjqkPmQn,if0≤j≤m,0≤k≤n,0,otherwise.
If we considerλ=λm,n as in (10), then it is clear that Mλm,n-statistical convergence coincides weighted statistical convergence (or SN2¯) which was defined and studied by Çınar and Et in [29].


## 2. Main Results and Proofs

In this section, we will give some properties ofMλm,n-statistical convergence and comparison with strong Mλm,n-summability. Moreover, inclusion results for Mλm,nst are given.Theorem 1.
Letx=xm,n and y=ym,n be two double sequences. Then,(i)
Ifxm,n⟶L1Mλm,n−st and ym,n⟶L2Mλm,n−st, then xm,n+ym,n⟶L1+L2Mλm,n−st(ii)
Ifxm,n⟶L1Mλm,n−st and c∈ℂ, then cxm,n⟶cL1Mλm,n−stProof.
Omitted.
We define each of the following sets:(11)Λ≔λ=λm,n:∑m,n=0∞,∞λm,n<∞,Λ0≔λ∈Λ:∑m,n=0∞,∞λm,n=1.Theorem 2.
Letλ=λm,n∈Λ0. If xm,n⟶L1Mλm,n−st and xm,n⟶L2Mλm,n−st, then L1=L2.Proof.
Assume thatxm,n⟶L1Mλm,n−st, xm,n⟶L2Mλm,n−st, and L1≠L2. Take any ε<1/2L1−L2, and denote the sets(12)AεL1=m,n:xm,n−L1≥ε,AεL2=m,n:xm,n−L2≥ε.
Sincexm,n⟶L1Mλm,n−st and xm,n⟶L2Mλm,n−st, we can write δλm,nAεL1=0 and δλm,nAεL2=0. It is clear that inclusion(13)m,n:xm,n−L2<ε⊂AεL1holds. Therefore, the inequality(14)δλm,nm,n:xm,n−L2<ε≤δλm,nAεL1=0holds. Also, the sets m,n:xm,n−L2<ε and AεL2 are disjoint and(15)ℕ×ℕ=m,n:xm,n−L2<ε∪AεL2.
From (15), we obtain(16)δλm,nℕ×ℕ=δλm,nm,n:xm,n−L2<ε+δλm,nAεL2.
The last equality gives thatδλm,nℕ×ℕ=0, but this is a contradiction to δλm,nℕ×ℕ=1. So, Mλm,n-statistical limit of x=xm,n is unique.Theorem 3.
Letλ=λm,n be a sequence from Λ0. Then, the following statements hold true:(i)
Ifxm,n⟶LMλm,n, then xm,n⟶LMλm,n−st and Mλm,n⊊Mλm,nst(ii)
Ifx=xm,n∈ℓ∞2 and xm,n⟶LMλm,n−st, then xm,n⟶LMλm,nProof.
(i) Letε>0 and xm,n⟶LMλm,n. In this case, we obtain(17)∑k,j=0m,nλm−k,n−jxk,j−L=∑k,j=0,k,j∈Kεm,nλm−k,n−jxk,j−L+∑k,j=0,k,j∉Kεm,nλm−k,n−jxk,j−L≥∑k,j=0,k,j∈Kεm,nλm−k,n−jxk,j−L≥ε∑k,j=0m,nλm−k,n−jχKεk,j.(ii)
After taking the limit whenm,n⟶∞ in the above inequality, we obtain xm,n⟶LMλm,n−st.(iii)
Assume thatxm,n⟶LMλm,n−st and x∈ℓ∞2. Then, there exists a positive constant H>0 such that(18)xm,n−L≤H,for all m,n∈ℕ.
For a givenε>0, the following inequality holds:(19)∑k,j=0m,nλm−k,n−jxkj−L=∑k,j=0,k,j∈Kεm,nλm−k,n−jxk,j−L+∑k,j=0,k,j∉Kεm,nλm−k,n−jxk,j−L≤H∑k,j=0m,nλm−k,n−jχKεk,j+ε∑k,j=0,k,j∉Kεm,nλm−k,n−j.
If we take the limit asm,n⟶∞, then we obtain that xk,j⟶LMλm,n.Remark 1.
The converse of (i) in Theorem3 does not hold.
Let us considerλ=λm,n=1/mn+12 and x=xm,n as follows:(20)xm,n=s3r3,m=r2,n=s2,0,otherwise.
Therefore, we have(21)limm,n⟶∞∑k,j=0m,n1m−kn−j+12⋅χKεk,j=limm,n⟶∞∑r,s=0m,n1m−rn−s+12≤0,but the following inequality holds:(22)limm,n⟶∞∑k,j=0m,n1m−kn−j+12⋅xkj=limm,n⟶∞∑r,s=0m,n1m−rn−s+12s3r3≥limm,n⟶∞1mn+12⋅∑r,s=0m,ns3r3=∞.
This gives thatMλm,n⊂Mλm,nst is strict.Corollary 1.
Ifxm,n⟶Lm,n⟶∞, then xm,n⟶LMλm,n−st for any λ∈Λ0.Proof.
Sincexm,n⟶Lm,n⟶∞ and Mλm,n is regular for any λ∈Λ0, then xm,n⟶LMλm,n. Therefore, Theorem 3 (i) gives that xm,n⟶LMλm,n−st.
Forλ,μ∈Λ0, let us consider the series(23)λx,y=∑m,n=0∞,∞λm,nxmyn,μx,y=∑m,n=0∞,∞μm,nxmyn,(24)μx,yλx,y=kx,y=∑m,n=0∞,∞km,nxmyn,λx,yμx,y=lx,y=∑m,n=0∞,∞lm,nxmyn.
The seriesλx,y and μx,y are convergent for all x<1,y<1.Theorem 4.
For anyλ,μ∈Λ, there exists a double sequence γ=γm,n∈Λ such that Mλm,nst⊂Mγm,nst and Mμm,nst⊂Mγm,nst.Proof.
Forλ=λm,n∈Λ0 and μ=μm,n∈Λ0, let us consider the sequence γ=γm,n as(25)γm,n≔λm,nμ0,0+λm−1,n−1μ1,1+⋯+λ0,0μm,n,for m,n∈ℕ. Let us show that γ∈Λ. Since λ,μ∈Λ, we have(26)∑m,n=0∞,∞γm,n=∑m,n=0∞,∞λm,nμ00+λm−1,n−1μ11+⋯+λ00μm,n≤∑m,n=0∞,∞λm,nμ0,0+λm−1,n−1μ1,1+⋯+λ0,0μm,n=∑m,n=0∞,∞λm,n∑m,n=0∞,∞μm,n<∞.
Lettm,nλ and tm,nμ be the Mλm,n−st and Mμm,n−st transformations of x=xm,n, respectively.
tm,nγ is also the Mγm,n-st transformation of xm,n, where(27)tm,nγ≔∑k,j=0m,nγm−k,n−jχKεk,j=γm,nχKε0,0+⋯+γ0,0χKεm,n.
From (25), we have the following equality:(28)tm,nγ=λm,nμ0,0+λm−1,n−1μ1,1+⋯+λ0,0μm,nχKε0,0+⋯+λ0,0μ0,0χKεm,n=λ0,0μ0,0χKεm,n+⋯+λm,nμ0,0+⋯+λ0,0μm,nχKε0,0=λ1,1μ0,0χKεm,n+⋯+μm,nχKε0,0+⋯+λm,nμm,nχKε0,0=λ1,1tm,nμ+λ2,2tm−1,n−1μ+⋯+λm,nt1,1μ.
Sincetm,nμ⟶0, as m,n⟶∞, we obtain limm,n⟶∞tm,nγ=0. This implies that xm,n⟶LMγm,n−st.
By using the same argument, we can getMλm,nst⊂Mγm,nst.Remark 2.
Theorem4 is valid for any λ,μ∈Λ0.Theorem 5.
For any twoλ,μ∈Λ, the methods Mλm,nst and Mμm,nst are consistent.Proof.
Letxm,n be a sequence such that xm,n⟶sMλm,n−st and xm,n⟶tMμm,n−st. If we consider the sequence γ=γm,n∈Λ as (25) in Theorem 4, we obtain xm,n⟶sMγm,n−st and xm,n⟶tMγm,n−st. From the uniqueness of the limit, we obtain that s=t.Theorem 6.
Letλ,μ∈Λ0. Then, Mλm,nst⊂Mμm,nst if and only if ∑m,n=0∞km,n<∞ and ∑m,n=0∞,∞km,n=1.Proof.
Lettm,nλ and tm,nμ be the Mλm,nst- and Mμm,nst-transforms of the sequence x=xm,n and y=ym,n, respectively.
Ifx<1 and y<1, we obtain(29)∑m,n=0∞,∞tm,nμxm,nym,n=∑m,n=0∞μm,nχKε0,0+⋯+μ0,0χKεm,nxmnymn=∑m,n=0∞,∞μm,nxmnymn∑m,n=0∞,∞χKεm,nxmnymn.
Similarly, ifx<1 and y<1,(30)∑m,n=0∞,∞tm,nλxmnymn=∑m,n=0∞,∞λm,nxmnymn∑m,n=0∞,∞χKεm,nxmnymn.
Then, it is clear from (24) that(31)μx,y=λx,ykx,y,for all x<1 and y<1.
It follows from the above equality and (23) and (24) that(32)kx,yλx,y∑m,n=0∞,∞χKεm,nxmnymn=μx,y∑m,n=0∞,∞χKεm,nxmnymn.
So, we obtain(33)kx,y∑m,n=0∞,∞tm,nλxm,nym,n=∑m,n=0∞,∞tm,nμxm,nym,n.
Then,(34)tm,nμ≔k0,0tm,nλ+k1,1tm−1,n−1λ+⋯+km,nt0,0λ=∑i,j=0∞,∞am,n,i,jti,jλ,where(35)am,n,i,j=km−i,n−j,ifi≤m,j≤n,0,otherwise.
Then, for the last discussion, we haveMλst⊂Mμst if and only if the matrix am,n,i,j is regular. So, we obtain(36)supm,n≥0∑i,j=0∞,∞am,n,i,j=supm,n≥0∑i,j=0∞,∞km−i,n−j<∞.
That is,∑m,n=0∞,∞km,n<∞. Also, we have(37)limm,n⟶∞∑i,j=0∞,∞am,n,i,j=limm,n⟶∞∑i,j=0m,nkm−i,n−j=1,i.e., ∑m,n=0∞,∞km,n=1.
The proof is completed.Corollary 2.
Letλ,μ∈Λ0. Then, Mλm,nst=Mμm,nst if and only if ∑m,n=0∞,∞km,n<∞, ∑m,n=0∞,∞hm,n<∞, and ∑m,n=0∞,∞km,n=∑m,n=0∞,∞hm,n=1.Definition 6 (see [25]).
Two sequencesλ=λm,n and μ=μm,n in Λ are said to be equivalent if(38)limm,n⟶∞λm,nμm,n=1,and it is denoted by λ∼μ.Theorem 7.
Letλ and μ be sequences in Λ such that λ∼μ. Then, Mλst=Mμst.Proof.
Letx=xm,n∈Mλm,nst be an arbitrary sequence such that(39)limm,n⟶∞∑k,j=0m,nλm−k,n−jχKεk,j=0,for any ε>0. Therefore,(40)limm,n⟶∞∑k,j=0m,nμm−k,n−jχKεk,j=limm,n⟶∞∑k,j=0m,nμm−k,n−jλm−k,n−jλm−k,n−jχKεk,j.
Sinceλ∼μ, for any fixed ε0>0, there exists n0≡n0ε0∈ℕ and m0≡m0ε0∈ℕ such that(41)1−ε0<μm−k,n−jλm−k,n−j<1+ε0holds for all n≥n0 and n≥m0. Hence, the following inequality holds:(42)∑k,j=0m,nμm−k,n−jχKεk,j=∑k,j=0m0ε0,n0ε0μm−k,n−jλm−k,n−jλm−k,n−jχKεk,j+∑m0ε0,n0ε0m,nμm−k,n−jλm−k,n−jλm−k,n−jχKεk,j≤∑k,j=0m0ε0,n0ε0μm−k,n−jλm−k,n−jλm−k,n−jχKεk,j+1+ε0∑m0ε0,n0ε0m,nλm−k,n−jχKεk,j≤c∑k,j=0m,nλm−k,n−jχKεk,jwhen(43)c=max1+ε,μmnλmn,μm−1,n−1λm−1,n−1,…,μm−m0,n−n0λm−m0,n−n0.
Now, by taking the limit asm,n⟶∞, we obtain that xm,n⟶LMμm,n−st. We conclude that Mλm,nst⊂Mμm,nst. Converse of this inclusion result can be obtained by the same way. Hence, the proof is completed.Corollary 3.
LetE=λm,n:m,n∈ℕ and F=μm,n:m,n∈ℕ be associated sets of λ=λm,n,μ=μm,n∈Λ. If E\ForF\E is finite, then Mλm,nst=Mμm,nst.Theorem 8.
Letx=xm,n be a sequence and λ∈Λ. Then, xm,n⟶Lst implies xm,n⟶LMλm,n−st if and only if mnλm−k,n−j=O1 for 0≤k≤n and 0≤j≤m.Proof.
LetKε=m,n:xm,n−L≥ε for any ε>0. Since mnλm−k,n−j=O1, we have(44)∑k,j=0m,nλm−k,n−jχKεk,j=1mn∑k,j=0m,nmnλm−k,n−jχKεk,j≤O11mn∑k,j=0m,nχKεk,j.
If we take the limit asm,n⟶∞, the desired result is obtained if and only if mnλm−k,n−j=O1.Theorem 9.
Letx=xm,n be a sequence and λ∈Λ. Then, xm,n⟶LMλ−st implies xm,n⟶Lst if and only if the sequence 1/mnλm−k,n−j=O1 for k≤m and j≤n.Proof.
For anyε>0, let Kε=m,n:xm,n−L≥ε. Then, the inequality(45)limm,n⟶∞1mn∑k,j=0m,nχKεk,j=limm,n⟶∞1mn∑k,j=0m,nλm−k,n−jλm−k,n−jχKεk,j=limm,n⟶∞∑k,j=0m,n1mnλm−k,n−jλm−k,n−jχKεk,j≤O1limm,n⟶∞∑k,j=0m,nλm−k,n−jχKεk,jholds. This gives the proof.
On the contrary, let us recall thatCD is the space of all continuous real-valued functions on any compact subset of the real two-dimensional space. We know that CD is a Banach space with norm(46)f∞≔supx,y∈Dfx,y,f∈CD.
Suppose thatL is a linear operator from CD into CD. It is clear that if f≥0 implies Lf≥0, then the linear operator L is positive on CD. Also, we denote the value of Lf at a point x,y by Lfu,v;x,y or only Lf;x,y. The classical Korovkin approximation theorem (see [35]) was extended from single sequences to double sequences [36].
Now, we give the following theorem to prove the approximation theorem.Theorem 10.
LetLm,n be a double sequence of positive linear operators from CD into CD. Then,(47)limm,nLm,nf−f∞⟶0Mλm,n−stfor all f∈CD if and only if(48)limm,nLm,nfi−fi∞⟶0,Mλm,n−st,where f0x,y=1, f1x,y=x, f2x,y=y, f3x,y=x2+y2, and i=0,1,2,3.Proof.
Since eachfi∈CDi=0,1,2,3, assertion (48) follows immediately from assertion (47). Assume that (48) holds. Since f∈CD, we have fx,y≤M, where M=f∞. Using the continuity of f on D for every ε>0, there is δ>0 such that fv,z−fx,y<ε for all v,z∈D satisfying v−x<δ and z−y<δ. Then, we get(49)fv,z−fx,y<ε+2Mδ2v−x2+z−y2.
Also, by the linearity and positivity of the operatorsLm,n and from (49), we obtain(50)Lm,nf;x,y−fx,y=Lm,nfv,z−fx,y;x,y−fx,yLm,nf0;x,y−f0x,y≤Lm,nfv,z−fx,y;x,y+MLm,nf0;x,y−f0x,y≤Lm,nε+2Mδ2v−x2+z−y2;x,y+MLm,nf0;x,y−f0x,y≤ε+M+2Mδ2A2+B2Lm,nf0;x,y−f0x,y+4Mδ2ALm,nf1;x,y−f1x,y+4Mδ2BLm,nf2;x,y−f2x,y+2Mδ2Lm,nf3;x,y−f3x,y+ε,where A≔maxx and B≔maxy. Then, taking supx,y∈D, we obtain(51)Lm,nf−f∞≤RLm,nf0−f0∞+Lm,nf1−f1∞+Lm,nf2−f2∞+Lm,nf3−f3∞+ε,where(52)R=maxε+M+2Mδ2E2+F2,4Mδ2E,4Mδ2F,2Mδ2.
Similarly, we obtain(53)Lm,nf;x,yλm−k,n−j−fx,y∞≤ε+R∑i=03Lm,nfi;x,yλm−k,n−j−fix,y∞.
We now replaceLm,n.;x,yλm−k,n−j by(54)Tm,n.;x,y=∑k,lLm,n.;x,yλm−k,n−j.
We chooseε′>0 such that ε′<r for a given r>0. Then, define the sets(55)S≔m,n∈ℕ×ℕ:Tm,nf−fCK≥r,Si=m,n∈ℕ×ℕ:Tm,nfi−fiCK≥r−ε′4R,i=0,1,2,3. It is clear that S⊂∪i=03Si, and so, δS≤δS0+δS1+δS2+δS3. Therefore, using condition (53), we obtain(56)limm,nLm,nf−f∞⟶0Mλm,n−st.
This completes the proof.Corollary 4.
LetLm,n be a double sequence of positive linear operators from CD into CD. Then,(57)P−limm,nLm,nf−f∞=0for all f∈CD if and only if(58)P−limm,nLm,nfi−fi∞=0,where f0x,y=1, f1x,y=x, f2x,y=y, f3x,y=x2+y2, and i=0,1,2,3.Remark 3.
We now show in the following an example of a sequence of positive linear operators of two variables satisfying the conditions of Theorem10 but does not satisfy the conditions of the Korovkin theorem. Consider the following Bernstein operators:(59)Bm,nf;x,y=∑k=0m∑j=0nfkm,jnCmkxkCnjyj1−yn−j,where x,y∈0,1×0,1.
LetLm,nf0;x,y=1,Lm,nf1;x,y=x, Lm,nf2;x,y=y, and Lm,nf3;x,y=x2+y2+x−x2/m+y−y2/n. Then, by Corollary 4, we obtain(60)P−limm,nLm,nf−f∞=0,for all f∈CD. Lm,n:CD⟶CD with Lm,nf;x,y=1+sm,nBm,nf;x,y, where(61)sm,n=1,ifmandnare even,0,otherwise.
Letλm,n=1. The double sequence sm,n is neither P-convergent nor statistically convergent, but sm,n is statistically summable Mλm,n to zero. Bm,n1;x,y=1, Bm,nx;x,y=x, Bm,ny;y,x=y, and Bm,nx2+y2;x,y=x2+y2+x−x2/m+y−y2/n, and the double sequence Lm,n satisfies condition (48) for i=0,1,2,3. Hence, we get(62)Mλm,nst−limm,nLm,nf−f∞=0.
We haveLm,nf,0,0=1+sm,nBm,nf;0,0 since Bm,nf;0,0=f0,0, and hence,(63)Lm,nf;x,y−fx,y∞≥Lm,nf;x,y−fx,y≥sm,nf0,0.
It is easy to see thatLm,n does not satisfy the conditions of the classical Korovkin-type theorem since limsm,n and st2−limsm,n do not exist; this proves the claim.


## 3. Conclusion

In this paper, we introduceMλm,n-statistical convergence for double sequences and give the inclusion results for different λm,n’s. These new results can be viewed as a generalization of previously known results. The new concept can be applied to the approximation theory, Fourier analysis, topology, and so on. The Mλm,n-statistically Cauchy sequence can be described, and its properties can be studied. Also, ideal convergent sequence spaces can be given.



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