7_kucukaslan.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 5, No. 2, 2012, 174-186
ISSN 1307-5543 – www.ejpam.com

On Statistical Boundedness of Metric Valued Sequences

Mehmet Küçükaslan∗, Uğur Değer

Department of Mathematics, Faculty of Science and Literature, Mersin University, Mersin, Turkey

Abstract. In this work, statistical boundedness is defined in a metric space and, statistical boundedness
of metric valued sequences and their subsequences are studied. The interplay between the statistical
boundedness and boundedness in a metric spaces are also studied, and it is shown that boundedness
imply statistical boundedness and if the number of elements of the metric space is finite then these two
concepts coincide. Moreover, here is given analogy of Balzano-Weierstrass Theorem.

2010 Mathematics Subject Classifications: 40A35, 30LXX, 40G15, 46A19, 28AXX, 43A85
Key Words and Phrases: Statistical convergence, metric spaces, statistical boundedness, asymptotic
density

1. Introduction and Definitions

The statistical convergence of real or complex valued sequences was first introduced by
Fast [7], but the idea of statistical convergence goes back to Zygmund [16]. In recent years,
statistical convergence has become popular research area for many mathematicians [2, 3, 8,
9, 10, 12, 13, 14, 15] etc.

On the other hand, analysis on metric spaces has rapidly developed in present time [see
11]. This development is usually based on some generalizations of the differentiability.

Some approaches which based on the convergence of the metric valued sequences have
been studied in [6] [see also 1, 4, 5]. But in these studies it is not deductive clear that the
usual convergence is the best possible way to obtain the smooth structure for arbitrary metric
space.

A lot of different convergence methods were defined (Cesaro, Nörlund, Weighted Mean,
Abel etc.) and applied to many branches of mathematics. Almost all convergence methods
depend on the algebraic structure of the space. It is clear that metric space does not have the
algebraic structure in general. So, the generalization of boundedness by using these methods
for metric valued sequences is impossible. However, the notion of statistical convergence is
easy to extend for arbitrary metric spaces.

∗Corresponding author.

Email addresses: mku
ukaslan�mersin.edu.tr(mkkaslan�gmail.
om) (M. Küçükaslan),udeger�mersin.edu.tr(degpar�hotmail.
om) (U. Değer)

http://www.ejpam.com 174 c© 2012 EJPAM All rights reserved.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 175

The concept d-statistical convergence for metric valued sequences was given first time in
[12]. In this work, d- statistical boundedness of metric valued sequences is defined, and the
relations between usual boundedness, convergence, d- statistical convergence and d-statistical
boundedness are investigated.

Let (X , d) be a metric space. For convenience denote by X̃ the set of all sequences of points
from X . That is, X̃ = { x̃ = (xn) : xn ∈ X }.

Let us remember the usual definition of convergence and boundedness in any arbitrary
metric space:

Definition 1 (In usual case). Let x̃ = (xn) ∈ X̃ be a sequence.

(i) x̃ is called convergent to a point a ∈ X if for every ε > 0 there exist an n0 = n0(ε) ∈ N
such that

d(xn, a) < ε (1)

for every n≥ n0.

(ii) x̃ is called bounded if for every x ∈ X there is a M > 0 such that

d(xn, x)< M (2)

for all n ∈ N.

The set of convergent and bounded sequences in X̃ will be denoted by C(X̃ ) and B(X̃ ),
respectively. It is clear that C(X̃ ) ⊂ B(X̃ ). But, in general the inverse is not true: To see this,
take into consider the sequence x̃ = (xn) with

xn =

(
x , n even

y, n odd

for an arbitrary x , y ∈ X . It is clear that x̃ is bounded, but it is not convergent.

Definition 2 (Statistical case). Let x̃ = (xn) ∈ X̃ be a sequence.

(i) x̃ is called d−statistical convergent to a point a ∈ X if, for every ε > 0,

lim
n→∞

1

n

���k : k ≤ n and d(xk, a)≥ ε	
�� = 0 (3)

is satisfied.

(ii) x̃ is called d-statistical bounded if, for an arbitrary x ∈ X there is a M > 0 such that

lim
n→∞

1

n

���k : k ≤ n and d(xk, x)≥ M
	��= 0 (4)

is satisfied.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 176

In (3) and later |B| denotes the number of elements of the set B.
The set of d-statistical convergent sequences and d-statistical bounded sequences in X̃ will

be denoted by C d
st(X̃ ) and Bd

st(X̃ ), respectively.

Definition 3 (Asymptotic Density).

(i) Let K ⊆ N and

δ(K) := lim
n→∞

1

n
|K(n)| ,

where K(n) := {k ∈ K : k ≤ n}. If the limit exists and finite, then the number δ(K) is called

asymptotic density of the set K [13].

(ii) If δ(K) = 1 then the set K ⊆ N is called a statistical dense subset of N [7].

2. Main Theorems and Their Proofs

The relation between bounded sequences and convergent sequences in an arbitrary metric
space is known. How will be it for d− statistical boundedness and d− statistical convergence?

In this section, we will answer this question and give some relations between d− statistical
boundedness and d− statistical convergence and with usual boundedness and convergence.

Theorem 1. Let (X , d) be a metric space and x̃ = (xn) ∈ X̃ . The following statements hold:

(i) If x̃ is bounded, then x̃ is d- statistical bounded.

(ii) If x̃ is d- statistical convergent to a ∈ X , then x̃ is d- statistical bounded.

Proof.

(i) If x̃ is bounded, then for an arbitrary x ∈ X there is M > 0 such that for all n ∈ N

d(xn, x)< M .

That is,
{k : k ≤ n, d(xk, x)≥ M} = ;.

Hence, we have

lim
n→∞

1

n

���k : k ≤ n and d(xk, x)≥ M
	�� = 0.

(ii) For any arbitrary ε > 0 and large M > 0 we have

�
k : k ≤ n and d(xk, a) ≥ M

	 ⊂ �k : k ≤ n and d(xk, a) ≥ ε	 .

This inclusion gives

|�k : k ≤ n and d(xk, a)≥ M
	 | ≤ |�k : k ≤ n and d(xk, a) ≥ ε	 |.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 177

From this inequality, we have

lim
n→∞

1

n

���k : k ≤ n and d(xk, a) ≥ M
	�� = 0

for every ε > 0. This gives the proof.

Remark 1. The inverse of (i) in Theorem 1 is not true in general.

Example 1. Let us take X = R with usual metric d(x , y) := |x − y| and consider the sequence

x̃ = (xn) = (1,1,−1,2,−1,1,−1,1,3, . . .) with

xn :=

(
k, n= k2,

(−1)n, n 6= k2.

It is clear that x̃ is not bounded in usual case. Let’s show that x̃ is d-statistical bounded.
For this aim, choose x = 0 and a sufficiently large M > 0. Then,

1

n
|{k : k ≤ n, |xk| ≥ M}| = 1

n
|{k : k = m2 ≤ n, |m| ≥ M}| ≤ [|

p
n|]− [|M |] + 2

n

This calculation shows x̃ = (xn) is d-statistical bounded.

Remark 2. The inverse of (ii) in Theorem 1 is not true in general.

Example 2. Assume that x , y ∈ X are distinct points and let us define the sequence x̃ = (xn)

with

xn :=

(
x , if n= 2k+ 1, k ∈ N
y, if n= 2k

that is,

x̃ = (
1
bx , y,

3
bx , y,

5
bx , y,

7
bx , . . .).

Let us choose M > 0 satisfying

M > 2 max{d(x , z), d(y, z)}
for an arbitrary fixed z ∈ X . Then we have

{k : k ≤ n, d(xk, z) ≥ M} = ;.
So, we obtain

lim
n→∞

1

n

���k : k ≤ n and d(xk, z) ≥ M
	��= 0.

This shows that x̃ is d–statistical bounded.
Now let us show that x̃ is not d–statistical convergent to x(or y). For this aim, set ε < d(x , y).
Then we have

|{k : k ≤ n, d(xk, x(or y))≥ ε}| =
(
= n

2
, n even;

< n

2
, n odd.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 178

According to this,

lim
n→∞

1

n

���k : k ≤ n and d(xk, x(or y))≥ ε	
��=

1

2
6= 0.

So, x̃ is not d–statistical convergent to x(or y).
Examples 1 and 2 show that the inclusions

B(X̃ )⊂ Bd
st(X̃ ) and C d

st(X̃ )⊂ Bd
st(X̃ )

are sharp.

Corollary 1. If x̃ = (xn) is convergent to a point a ∈ X , then x̃ is d- statistical bounded.

Proof. From [12] we know that convergence implies d- st convergence in an arbitrary
metric space. Therefore, taking into account (ii) in Theorem 1, we get the proof.

Another way, if x̃ is convergent to a ∈ X , x̃ is bounded. By using (i) in Theorem 1, we
have the result.

Corollary 2. Let (X , d) be an arbitrary metric space. Then the following diagram holds:

B(X̃ )
(2)

// Bd
st(X̃ )

C(X̃ )

(1)

OO

(5)

;;
x

x
x

x
x

x
x

x
x

(3)
// C d

st(X̃ )

(4)

OO

where A→ B means that A⊂ B.

Proof. The inclusion (5) is obtained directly from Corollary 1. The inclusions (2) and (4)
are obtained from Theorem 1-(i) and (ii), respectively. The inclusion (3) is obtained from
Proposition 2.1 in [12].

The inverse of all inclusions in the diagram are not true. Take into consider Example 1 in
[12] for the inverse of (3), and in this work Example 1 and Example 2 for the inverse of (2)
and (4). The following theorem gives necessary and sufficient condition on X for which the
inverse of the inclusion (2) in the Corollary 2 is true:

Theorem 2. Let (X , d) be a metric space with X 6= ;. The following two statements are equiva-

lent:

(i) The set of all bounded sequences x̃ = (xk) ∈ X̃ is the same as the set of all d – statisti-

cal bounded sequences x̃ ∈ X̃ .

(ii) The cardinality of X is finite.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 179

Proof. (ii) ⇒ (i) The cardinality of X is finite. That is, there exist a number n0 ∈ N
such that X = {x0, x1, . . . , xn0

}, since the cardinality of X is finite. From the hypothesis, the
diameter of X ,

d(X ) := sup{d(xk, x l) : xk, x l ∈ X ; k, l = 0, n0}
is finite.

If x̃ is bounded, then from Theorem 1-(i), x̃ is also d–statistical bounded.
So, we must only show the inverse is true under the hypothesis. Let’s take d– statistical

bounded sequence x̃ = (xk) ∈ X̃ , i.e., ∃M > 0 and x ∈ X such that

lim
n→∞

1

n
|{k : k ≤ n, d(xk, x)≥ M}| = 0. (5)

Especially, if we take M := 2d(X ), we have, from (5),

{k : k ≤ n, d(xk, x)≥ 2d(X )}= ;.
In other words,

{k : k ≤ n, d(xk, x)< 2d(X )}= n.

Therefore,

lim
n→∞

1

n
|{k : k ≤ n, d(xk, x)< 2d(X )}|= 1.

This shows that x̃ is a bounded sequence.
(i)⇒ (ii) Suppose that |X |=∞. Then, there exist at least one sequence that d– statistical

bounded but not bounded (see Example 1). That’s why, our assumption is not true.

Corollary 3. Let (X , d) be a metric space with finite number elements. Then, the following

diagram holds:

B(X̃ )
(2)

// Bd
st(X̃ )

oo

C(X̃ )

(1)

OO

(5)

;;
x

x
x

x
x

x
x

x
x

(3)
// C d

st(X̃ )

(4)

OO

Corollary 4. Let (X , d) be a metric space with X 6= ;. Then, the following two statements are

equivalent:

(i) B(X̃ ) = C(X̃ ) = Bd
st(X̃ ) = C d

st(X̃ )

(ii) The set X is a singleton.

Proof. The proof is open from above the Theorem 2 and Theorem 2.2 in [12].

Theorem 3. Let X be a set which has at least two elements and endowed ρ discrete metric. Then,

we have

X̃ = B(X̃ ) = B
ρ
st(X̃ ) (6)

and

C(X̃ )⊂ C
ρ
st(X̃ ). (7)



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 180

Proof. We know that any space with discrete metric has finite diameter. Therefore, any
sequence in this space is bounded. That is, there is a M > 0 and arbitrary x ∈ X such that
ρ(xn, x)< M for every x̃ = (xn) ∈ X̃ . So, we have

{k : k ≤ n,ρ(xk, x)≥ M} = ;.

Hence,

lim
n→∞

1

n
|{k : k ≤ n,ρ(xk, x)≥ M}| = 0.

This shows that x̃ ∈ B
ρ
st(X̃ ) for all x̃ ∈ X̃ .

Let us take an arbitrary sequence x̃ ∈ B
ρ
st . If we choose M > 1, then

{k : k ≤ n,ρ(xk, x) ≥ M} = ;. It means that ρ(xk, x) < M for all k ∈ N. Thus, we see that
x̃ ∈ B(X̃ ).

Now let us see that C(X̃ ) ⊂ C
ρ
st(X̃ ). Let x̃ ∈ C(X̃ ). Then for an arbitrary ǫ > 0 there exist

n0 = n0(ǫ) ∈ N such that xn = xn0
for all n≥ n0. Therefore, we have

lim
n→∞

1

n
|{k : k ≤ n,ρ(xk, xn0

) ≥ ǫ}| ≤ lim
n→∞

n0

n
= 0.

This shows that x̃ ∈ C
ρ
st(X̃ ).

If we consider the sequence x̃ = (xn) with

xn :=

(
x if n= k2, k ∈ N
y if n 6= k2

then x̃ is ρ- statistical convergent to x but not convergent. So, the inclusion in (7) is sharp.

Corollary 5. Let X be a set which has at least two elements and endowed bounded metric. Then,

(6) and (7) hold.

Theorem 4. Let (X , d) be a metric space, x̃ = (xn) ∈ X̃ and let x̃ ′ = (xnk
) be a subsequence of

x̃. If x̃ is d-statistical bounded then x̃ ′ is also d-statistical bounded.

Proof. Suppose that x̃ is d – statistical bounded. It is clear that there is a number M > 0
and x ∈ X such that

{nk : nk ≤ n, d(xnk
, x)≥ M} ⊂ {k : k ≤ n, d(xk, x)≥ M}.

Then, since
|{nk : nk ≤ n, d(xnk

, x)≥ M}| ≤ |{k : k ≤ n, d(xk, x)≥ M}|,
we have

0≤ lim
n→∞

1

n

���
¦

nk : nk ≤ n and d(xnk
, x)≥ M
©���≤ 0.

The last inequality shows that x̃ ′ is d – statistical bounded.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 181

Lemma 1. Let (X , d) be a metric space and ex = (xn) ∈ eX . Then, the sequence x̃ is d−st bounded

if and only if the real sequence
�
d(xn, x)
�

is statistical bounded for an arbitrary x ∈ X .

It can be obtained directly from the Definition 1.2-(ii). So, the proof is omitted here.

Lemma 2. Let (X , d) be a metric space and ex = (xn) ∈ eX be a d−statistical bounded sequence.

Then, the sequence ex has at least one bounded subsequence.

Proof. From the Lemma 1 the real sequence
�
d(xn, x)
�

is statistical bounded in real num-
bers. That is, there is a positive number M such that δ(A) = 1,δ(B) = 0 where

A :=
�

k : d(xk, x)< M
	

, B :=
�
k : d(xk, x)≥ M

	
.

Let k1 ∈ N be the minimal element of A and d(xk1
, x)< M . Since δ(A) = 1, it can be choosen

k2 ≥ k1 such that the minimal element of the set
�
k : k > k1, k ∈ A

	

satisfying d(xk2
, x)< M .

In the n-th step we can choose kn ≥ kn−1 which is the minimal element of the set
�

k : k ≥ kn−1, k ∈ A
	

such that d(xkn
, x)< M .

So, we obtain non-decreasing sequence
�
kn

�
n∈N such that ex ′ = (xkn

) is the subsequence
of x̃ satisfying d(xkn

, x)< M for all kn ∈ N. This shows that the subsequence ex ′ is bounded.

The following theorem gives the analogy of Balzano-Weierstrass Theorem:

Theorem 5. Let (X , d) be an arbitrary metric space. Every d−statistical bounded sequence has

at least d−statistical convergent subsequence.

Proof. Let us take d−statistical bounded sequence ex = (xn) ∈ eX and denote the real
sequence (d(xn, x)) for arbitrary x ∈ X by (yn). From Lemma 1, the real sequence (yn) is
statistical bounded with usual metric on R. Take into account the following sets for sufficiently
large M > 0,

A([0, M] , n) :=
�

k : k ≤ n, yk ∈ [0, M)
	

,

B ([0, M] , n) :=
�

k : k ≤ n, yk ∈ [M ,∞)	

such that A∪ B = {1,2,3, . . . , n} and δ(A) = 1, δ(B) = 0.
Also, denote I0 := [0, M] with the length l(I0) = M and divide it into two parts as

I1
0 :=

�
0,

M

2

�
and I2

0 :=

�
M

2
, M

�
.

It is clear that asymptotic density of these sets satisfy

0≤ δ(I1
0 )≤ 1, 0≤ δ(I2

0 ) ≤ 1.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 182

If δ(I1
0 ) = 0 (or δ(I2

0 ) = 0), consider I2
0 (or consider I1

0 ) otherwise consider the interval having
big asymptotic density and denote it by I1. Note that the length of I1 is l(I1) =

M

2
and I1 ⊂ I0.

If we divide the new closed interval into two parts as I1
1 and I1

2 , we can choose a new one
and denote this interval by I2 from the above explanation such that the length of this interval
l(I2) =

M

22 and I2 ⊂ I1. After continuing this procedure we obtain closed nested intervals which
length tends to zero. From the nested Theorem, we get

∞⋂

n=1

In =
�

y∗
	

. (8)

Now our aim to construct a subsequence of (yn) such that it is convergent to y∗. For this
turn to begin of the proof and choose k1 which is the minimal element of I0, choose k2 ≥ k1

which is the minimal element of I1 and so on. If continue this process we obtain kn ≥ kn−1

which is the minimal element of In. Also, we can choose kn+1 ≥ kn that is minimal element
of In+1. Otherwise the number of elements of I1

n (or I2
n) is at most kn. This is contradiction

to assumption on I1
n (or I2

n). So, non-decreasing sequence (kn) gives the subsequence (ykn
) of

the sequence (yn) satisfying
��ykn
− y∗
�� < l(Ik) =

M

2n
. (9)

For every ǫ > 0, there is a kn0
= kn0

(ǫ) ∈ N such that 1
2kn
< ǫ for every kn ≥ kn0

. So, from (9)
we have ��ykn

− y∗
�� < ǫ. (10)

The equation (10) shows (ykn
) is convergent to y∗ in usual case. It means that there is an

element x∗ ∈ X such that the sequence yn := d(xkn
, x) convergent to y∗ = d(x∗, x).

In [12], we know that this implies the sequence d(xkn
, x) is d−statistical convergent to

d(x∗, x). Hence, we have
¦

kn : kn ≤ n, d(xkn
, x∗)≥ ǫ
©
⊂
¦

kn : kn ≤ n,
��d(xkn

, x)− d(x∗, x)
��≥ ǫ
©

.

From this inclusion and the Definition 2-(i) we have

0≤ lim
n→∞

1

n

���
¦

kn : kn ≤ n, d(xkn
, x∗)≥ ǫ
©��� ≤ lim

n→∞
1

n

���
¦

kn : kn ≤ n,
��d(xkn

, x)− d(x∗, x)
��≥ ǫ
©��� = 0.

The last inequality says the sequence x̃ has a d-statistical convergent subsequence.

The following theorem gives the relation between statistical equivalence and statistical
boundedness for sequences.

Theorem 6. Let (X , d) be a metric space, x̃ and ỹ belong to X̃ and assume that x̃ be a d –

statistical bounded sequence. If x̃ ≍ ỹ , then ỹ is also d – statistical bounded.

Proof. We can follow the method in the proof of Lemma 3.2 given in [12]. Let us define a
subset A ofN as "n ∈ A⇔ xn 6= yn". Then, subsetN\A is statistical dense from the Definition 2-
(ii). This implies that

lim
n→∞

1

n
|{m ∈ A : m ≤ n}| = 0. (11)



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 183

Let M > 0 be a sufficiently large number. According to the definition of A we have

{m ∈ A : m≤ n and d(ym, a) ≥ M}
⊆ {m ∈ A : m ≤ n} ∪ {m ∈ A : m ≤ n and d(xm, a)≥ M},

for all n ∈ N and arbitrary point a ∈ X . By using this inclusion and equality (11), we get

lim sup
n→∞

��{m ∈ N : m ≤ n and d(ym, a) ≥ M}
��

n

≤ lim sup
n→∞

|{m ∈ A : m ≤ n}|
n

+ lim sup
n→∞

��{m ∈ N : m ≤ n and d(xm, a)≥ M}
��

n

= lim sup
n→∞

��{m ∈ N : m ≤ n and d(xm, a)≥ M}
��

n
.

Since x̃ is d – statistical bounded, we have

lim sup
n→∞

��{m ∈ N : m ≤ n and d(xm, a)≥ M}
��

n
= 0.

Consequently the inequality

lim sup
n→∞

��{m ∈ N : m ≤ n and d(ym, a)≥ M}
��

n
≤ 0 (12)

holds. By using (12) we obtain

0≤ lim inf
n→∞

��{m ∈ N : m ≤ n and d(ym, a)≥ M}
��

n

≤ lim sup
n→∞

��{m ∈ N : m ≤ n and d(ym, a) ≥ M}
��

n
≤ 0.

Hence the limit relation

lim
n→∞

��{m ∈ N : m≤ n and d(ym, a) ≥ M}
��

n
= 0

holds. Therefore, this gives the desired result.

Suppose (X , d) and (Y, d ′) are two metric spaces. We say that Y is a metric subspace of X

and X is a metric superspace of Y if, and only if Y is a subset of X and d ′ is a restriction of d ,
i.e.

d ′(x , y) := dY (x , y).

Theorem 7. Let (X , d) be a metric space and (Y, d ′) be a metric subspace. Then, the following

statements hold:

(i) If (yn) ∈ eY is d ′−statistical bounded then (yn) is d−statistical bounded.



M. Küçükaslan, U. Değer / Eur. J. Pure Appl. Math, 5 (2012), 174-186 184

(ii) If (xn) ∈ eX is d−statistical bounded, then (xn) d ′−statistical bounded in subspace that

contains all terms of (xn).

Proof.

(i) From the definition of subspace metric and statistical boundednes and for arbitrary
y∗ ∈ Y and sufficiently large positive M > 0 we have

�
k : k ≤ n, d ′(yk, y∗)≥ M

	
=
�

k : k ≤ n, d(yk, y∗)≥ M
	

.

Therefore,

0= lim
n→∞

1

n

���k : k ≤ n, d ′(yk, y∗)≥ M
	��= lim

n→∞
1

n

���k : k ≤ n, d(yk, y∗)≥ M
	�� .

(ii) It can be prove by using the same arguments in (i).

Let X 6= ; be a metric space with equivalent metrics d1 and d2. That is, there are positive
constant m1 and m2 such that

m1d1(x , y)≤ d2(x , y) ≤ m2d1(x , y) (13)

for every x , y ∈ X .

Theorem 8. Let X be a metric space with equivalent d1 and d2 metrics. Then, the following

statements hold:

(i) B
d1
st (eX ) = B

d2
st (eX ),

(ii) C
d1
st (eX ) = C

d2
st (eX ).

Proof.

(i) Assume that x̃ = (xn) ∈ eX is an arbitrary d1- statistical bounded sequence, i.e.
x̃ ∈ B

d1
st (eX ) . Then, there is a positive M > 0 and arbitrary x∗ ∈ X such that

lim
n→∞

1

n

���k : k ≤ n, d1(xk, x∗)≥ M
	�� = 0.

Also, from (13) we have

�
k : k ≤ n, d1(xk, x∗)≥ M

	 ⊂ �k : k ≤ n, d2(xk, x∗)≥ m1M
	

⊂
�

k : k ≤ n, d1(xk, x∗)≥ m1

m2
M

�

and
���k : k ≤ n, d1(xk, x∗)≥ M

	�� ≤
���k : k ≤ n, d2(xk, x∗)≥ m1M

	��



REFERENCES 185

≤
����
�

k : k ≤ n, d1(xk, x∗)≥ m1

m2
M

����� .

By using this inequality, we obtain

lim
n→∞

1

n

���k : k ≤ n, d2(xk, x∗)≥ m1M
	��= 0.

This shows that x̃ = (xn) ∈ eX is d2- statistical bounded sequence, i.e. x̃ ∈ B
d2
st (eX ) . To

obtain inverse, it is enough to consider (13) as follows:

1

m2
d2(x , y)≤ d1(x , y)≤ 1

m1
d2(x , y)

for every x , y ∈ X .

(ii) The proof is clear from Theorem 2.3 in [12].

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