Deferred Statistical Cluster Points of Real Valued Sequences

In this paper, the concept of deferred statistical cluster points of real valued sequences is defined and studied by using deferred density of the subset of natural numbers. For p(n) and q(n) satisfying certain conditions, we give some results for the set of deferred statistical cluster points ΓDp,q (x). We provide some counter examples regarding ΓDp,q (x). Also we obtain some inclusion results for ΓDp,q (x). At last we consider the case q(n) = λ(n) and p(n) = λ(n − 1) where the sequence λ = {λ(n)} is strictly increasing sequence of positive natural numbers with λ(0) = 0.


Introduction and notations
The concept of statistical convergence was introduced by H. Fast [8] and I.J. Steinhaus [23] independently in 1951. Since then, this subject was applied in different areas of mathematics such as in number theory by P. Erdös-G. Tenenbaum [7] and summability theory by A.R. Freedman-J.J. Sember-M. Raphael [9], etc.
This subject is closely related to the subject of asymptotic (natural) density of the subset of natural numbers [2] and its root goes to A. Zygmund [25].
By using asimptotic density, the concept of statistical cluster points of real valued sequences was first introduced by J.A. Fridy [11]. Some generalizations of this concept have been studied by using regular summability methods in [5,6,13,16,20,24].
In 1932, R.P. Agnew [1] defined the deferred Cesaro mean D p,q of the sequence x = (x k ) by (D p,q x) n := 1 q(n) − p(n) q(n) p(n)+1 (1) Let K be a subset of N, and denote the set {k : p(n) < k ≤ q(n), k ∈ K} by K p,q (n). The deferred density of K is defined by whenever the limit exists. The vertical bars in (2) indicate the cardinality of the set K p,q (n). Because of δ p,q (K) does not exists for all K ⊂ N, it is convenient to use upper deferred asymptotic density of K, defining by It is clear that, exists for every ε > 0.(see [14,15]).
Definition 1.1. The number γ is called deferred statistical cluster point of x = (x k ), for every p(n) ad q(n) satisfying (1), if for every ε > 0 the set does not have deferred density zero i.e., and the set of deferred statistical cluster points of the sequence x = (x k ) is denoted by Γ Dp,q (x), i.e.,  In this section, some topological properties of deferred statistical cluster points of the real valued sequences are giong to be investigated.
Theorem 2.1. If the sequence x = (x n ) is deferred statistical convergence to l, then Γ Dp,q (x) contains only the elements l.
Proof. Assume that the sequence x = (x n ) is deferred statistical convergence to l. Then, for every ε > 0, the limit relation Therefore, l ∈ Γ Dp,q (x). Now let us assume that the set Γ Dp,q (x) contains l which different from l, i.e., l = l . Take into consider ε = 1 2 |l − l |. Since x = (x n ) is deferred statistical convergence to l, then (5) is hold for this ε. It means that deferred asymptotic density of the elements x = (x n ) belonging to the ε−neigborhood of l is 1. Consequently, the deferred asymptotic density of the elements of (x n ) belonging to the ε−neighborhood of l is zero. That is, This is contradiction to assumption on l .
Remark 2.1. The inverse of Theorem 2.1 is not true.
There exists a sequence such that the set of deferred statistical cluster points has unique elements but it is not deferred statistical convergence to this point. Let us consider the sequence x = (x n ) where x n := 1 n , n even, n, n odd.
It is clear that 0 ∈ Γ D0,n (x) but it is not deferred statistical convergence to zero.
Proof. Here we will give the proof for only the monotone increasing sequence. From the definition of supremum for any ε > 0 there exists a n 0 ∈ N such that the following inequality Since the sequence is monotone increasing, then we have for all n > n 0 . It means that for any ε > 0 there exist a n 0 (ε) ∈ N such that the inequality |x n − sup x n | < ε holds for all n > n 0 . From this discussion the following inclusion Since δ p,q (N − {1, 2, ..., n 0 }) = 1, then δ p,q ({n : |x n − sup x n | < ε}) = 0. This gives the desired proof.

Recall that the distance between
Proof. Assume that Γ Dp,q (x) = ∅. Let us consider an arbitrary element y ∈ Γ Dp,q (x). Then, we have for an arbitrary positive ε, So, the set A ε := {x k : |x k − y| < ε} has at least countable elements of x = (x n ) for an arbitrary positive ε. Therefore, is hold. This gives the desired proof.
Proof. Let us assume that Γ Dp,q (x) = ∅ for any p (n) and q (n) . It is enough to show that R\Γ Dp,q (x) is an open set. Let y ∈ R\Γ Dp,q (x) be an arbitrary point. Since y / ∈ Γ Dp,q (x), then there exists an ε > 0 such that If we denote the open interval (y − ε, y + ε) by A, then we have If we choose ε y := 1 2 inf {|x k − y| : x k ∈ A} , then it is clear that ε y < ε and (y − ε y , y + ε y ) ⊂ R\Γ Dp,q (x) . It means that y is an arbitrary interior point of R\Γ Dp,q (x). Therefore R\Γ Dp,q (x) is an open set.
Theorem 2.4. Let x = (x n ) be a real valued sequence and γ ∈ R be an arbitrary fixed point. If d(γ, x) = 0, then γ / ∈ Γ Dp,q (x) for any p (n) and q (n) .
Proof. From the hypthesis we have From this assumption the inequality It means that the open interval (γ − m, γ + m) has no elements of the sequence x = (x n ). So, we have therefore, if we choose an arbitrary ε < m then the relation δ p,q ({p(n) < k ≤ q(n) : |x k − γ| < ε}) = 0 hold. Otherwise it contradicts with (6) since the inclusion Let us consider the sequence x = (x n ) = ( 1 n ) for all n ∈ N. If we take γ = 1 2 , then d( 1 2 , 1 n ) = 0 but 1 2 / ∈ Γ Dp,q (x) = {0} when q(n) = n and p(n) = 0. Proof. If the subset A ⊂ R is a singleton, then the proof is obtained from Theorem 2.4. Let a * ∈ A be an arbitrary element. There is m > 0 such that |a * − x k | > m since d(A, x) > 0. So, the intervals (a * − m, a * + m) has no elements of x = (x n ). Therefore, if we choose ε < m, the set (a * − m, a * + m) contains no element of the sequence. Consequently, we have Theorem 2.6. If the set F \ F is finite and hold. Then, δ p,q (K) = 0 implies δ p,q (K) = 0 for every K ⊆ N.
Proof. Since the set F \ F is finite, then there exists a positive natural number N such that For n ≥ N let j (n) be a strictly increasing sequence such that q (n) := q(j(n)). If δ p,q (K) = 0 then the relation holds. So, we have .
and δ * p,q (K) = lim sup Hence δ p,q (K) = 0 and the proof is obtained.
hold. Then, the followings are true: Theorem 2.7. If E \ E is finite and hold. Then, δ Dp,q (K) = 0 implies δ D p ,q (K) = 0 for every K ⊆ N. For n ≥ N let j (n) be monotone increasing such that p (n) = p(j(n)). If δ Dp,q (K) = 0 then From this we have and δ * Dp,q (K) = lim sup It gives δ D p ,q (K) = 0 and we obtained desired result.
Corollary 2.2. Under the assumption of Theorem 2.7 the following statements are true: Theorem 2.8. Let us assume that hold. Then, δ D p ,q (K) = 0 implies δ Dp,q (K) = 0 for every K ⊆ N.
Proof. If δ D p ,q (K) = 0, then we have and the relation hold. Therefore, Hence δ Dp,q (K) = 0, and the proof is ended.
Let us note that if q(n) = n for all n ∈ N, the condition (7) is ommitted in the Theorem 2.9.
Proof. If the set G − G is finite, then there exists a positive natural number N such that the inclusion {λ (n) : n ≥ N } ⊂ G hold. It means that there exists a monotone increasing sequence (j(n)) n∈N tending infinity such that λ (n) = λ(j(n)).