ON THE STATISTICAL CONVERGENCE OF NESTED SEQUENCES OF SETS Текст научной статьи по специальности «Математика»

Probl. Anal. Issues Anal. Vol. 11 (29), No3, 2022, pp. 3-14

DOI: 10.15393/j3.art.2022.11850

3

UDC 515.124

H.Albayrak, F. Babaarslan, O.Olmez, S.Aytar

ON THE STATISTICAL CONVERGENCE OF NESTED SEQUENCES OF SETS

Abstract. In this paper, we show that Wijsman convergence and statistical Wijsman convergence are equivalent to each other if we choose the sequences of sets as monotone. Then, we show that every statistical Wijsman convergent monotone sequence of sets is not only Hausdorff convergent but also statistical Hausdorff convergent to the same set.

Key words: nested sequences of sets, statistical Wijsman convergence, statistical Hausdorff convergence

2020 Mathematical Subject Classification: 40A05, 40A35, 54A20

1. Introduction and Background. There are three basic convergence types for the sequences of sets: Kuratowski convergence, Hausdorff convergence and Wijsman convergence. First, Painleve introduced the concept of convergence of sequences of sets by defining the outer and inner limits of sets. If the inner and outer limits are equal to each other, then this convergence type is known as Kuratowski convergence [7].

Second well-known convergence type for sequences of closed sets was given by Hausdorff as follows: Let (X,px) be a metric space and Cl(X) denote the nonempty closed subsets of X. The Hausdorff distance between two sets A and B of Cl(X) is defined by

H (A, B) = sup ld(x, A) - d(x, B)| ,

xex

where d(-,A): X —> [0, to) is the distance function defined by d(x, A) = inf[px(x, y): y E A} [12].

Equivalently, the Hausdorff distance is given by

H (A,B) = max [h(A, B),h(B, A)} , © Petrozavodsk State University, 2022

where h(A,B) = sup d(a,B) is the Hausdorff excess of the set A with

aeA

respect to the set B [3].

It is clear that a sequence (An) of sets is Hausdorff convergent to the set A if

lim H(An, A) = 0.

In this case, we write An —> A as n ^ <x> [6].

If a sequence (An) of sets is Hausdorff convergent to A, then the sequence [d(-,An)}neN of distance functions is uniform convergent to the distance function d(-,A). Finally, if we replace this type of convergence with the pointwise convergence, we get the Wijsman convergence as follows:

Let An C X for each n E N; a sequence (An) of sets is said to be Wijsman convergent to the set A if

lim d(x, An) = d(x, A) for all x E X.

In this case, we write An A as n ^ ro ( [12], [13]).

A sequence (An) of sets is said to be monotonically increasing if An C An+1 for each n E N and it is said to be monotonically decreasing if An+\ C An for each n E N [7].

The relations among these three types of set convergence have been investigated by several authors. In 1979, Salinetti and Wets [9] showed that every Hausdorff convergent sequence of sets is Kuratowski convergent to the same set. Then, Beer [2] examined that the Wijsman convergence implies the Kuratowski convergence for nonempty closed sets. Also, he gave the conditions for these convergences to be equivalent to each other in [2, Theorem 1]. In 2001, Apreutesei [1] observed that Wijsman convergence and Hausdorff convergence are equivalent to each other if the sequence of compact sets is monotone. In [3], the Hausdorff limit of a sequence of sets was obtained as intersection of closure of the union of terms of this sequence.

In 1951, Fast [4] introduced the concept of statistical convergence. In 2012, the theory of set convergence was generalized to the theory of statistical convergence by Nuray and Rhoades [8]. They investigated the relation between statistical Hausdorff convergence and statistical Wijsman convergence. Furthermore, Talo et al. [10] defined statistical inner and outer limits of sequences of closed sets and they compared the statistical Kuratowski convergence with the statistical Hausdorff convergence.

Recently, Ulusu and Gulle [11] examined statistical Wijsman convergence and statistical Hausdorff convergence of order a for double sequences of sets.

In this paper, we first show that Wijsman convergence and statistical Wijsman convergence are equivalent to each other if we choose the sequences of sets as monotone. Then, we show that every statistical Wijsman convergent monotone sequence of sets is not only Hausdorff convergent, but also statistical Hausdorff convergent to the same set. Finally, we characterize the statistical Hausdorff limit by using the concept of ideal. Note that we used the similar proof tecniques of [1].

2. Main Results. Before giving our results, we recall some definitions about the statistical convergence.

Let K be a subset of the set of positive integers N and l{k ^ n : k E K}| denote the number of elements of K if [1, n]. The natural density of K is defined by

8(K) = lim 1 l{k ^ n: k E K}|

n^x n

if this limit exists. It is clear that any finite subset of N has zero natural density and 5(Kc) = 1 - 5(K), where Kc := N \ K [5].

Let A,An E Cl(X). The sequence (An) is said to be statistical Hausdorff convergent to the set A if for every e > 0,

lim 1 l{k ^ n: max{h(Ak,A),h(A,Ak)} ^ e}| = 0.

n^x n

In this case, we write An A [8].

Let An C X for each n E N. Then the sequence (An) of sets is said to be statistical Wijsman convergent to a set A if

lim 1 l{k ^ n : ld(x,Ak) - d(x,A)l ^ e}| =0

for each x E X. In this case, we write An —A [8].

Theorem 1. Let A,An E Cl(X) (n E N).

(i) Let A

i C A2 C ... C A,n.... Then the sequence (A,n) is Wijsman convergent to the set A if and only if it is statistical Wijsman convergent to the set A.

(ii) Let A1 D A2 D ... D An

D .... Then the sequence (An) is Wijsman convergent to the set A if and only if it is statistical Wijsman convergent to the set A.

Proof. The necessity parts of (i) and (ii) are provided for all sequences of sets (see [8]).

i) Assume that An A. Firstly, we show that An C A for every n G N. Fix n G N and let u G An. From An A, we have 8 (K(u,£)) = 1 for each e > 0, where

K(u, e) := {m G N : ld(u, Am) - d(u, A) I < e] .

For each e > 0, there exists an ms G N, which is ms G K(u, e) and ms ^ n. Since (An) is monotonically increasing, we have An C Ams and u G Ams. Therefore, we get d(u, Ams) = 0 and so,

d(u, A) = Id(u, Ams) — d(u, A) I < e

for each e > 0. We get u G A, since the set A is closed.

Now, let us take any x G X. Hence, we can write d(x, A) ^ d(x, An) and so:

d(x,An) - d(x,A) ^ 0 (1)

for each x G X and each n G N.

We show that for each e > 0 there exists an n0 G N, such that d(x, An) — d(x, A) < e for every n ^ n0. Let e > 0. Since An A, we have 8 (L(x,e)) = 1, where

L(x,e) : = {n G N : Id(x,An) - d(x,A)I < e] . (2)

Let n0(x,e) := minL(x,e). By the monotonicity of the sequence (An), we have

d(x,An) ^ d(x,Ano) (3)

for every n ^ n0. If we combine the expressions (2) and (3), we get

d(x, An) — d(x, A) ^ d(x, Ano) — d(x, A) < e (4)

for every n ^ n0.

By the inequalities (1) and (4), we get

Id(x, An) — d(x, A) I < e

for every n ^ n0.

Since x is arbitrary, we obtain

lim d(x, An) = d(x, A)

for every x E X and An A.

ii) Let us assume that An A. Fix n E N and take u E A. Since An A, we have 8 (K(u, e)) = 1 for each e > 0, where

K(u, e) := {m E N : ld(x, Am) - d(x, A)l < e} .

Then there exists an m£ E N, which is m£ E K(u,e) and m£ ^ n for each e > 0. Since (An) is monotonically decreasing, we have Ams C An and d (u,An) ^ d (u,Ams). Moreover, since d(u,A) = 0, we have

d(u,Ams) = ld(u,Ams) - d(u,A)l < £

for each e > 0. Hence, we get d (u, An) < £ for every e > 0 and, so, u E An since the sets An are closed.

Hence, we can write d(x,An) ^ d(x,A) and, so,

d(x,An) - d(x,A) ^ 0 (5)

for each x E X and each n E N.

Let x E X and e > 0. Since An A, we have 8 (L(x, e)) = 1, where

L(x,£):= {n E N: ld(x,An) - d(x,A)l < £} . (6)

Define n0(x, e) := min L(x, e). Using the monotonicity of the sequence (A„), we have

d(x,Ano) ^ d(x,An) (7)

for every n ^ n0. By the expressions (6) and (7), we get

- e < d(x, Ano) - d(x, A) ^ d(x, An) - d(x, A) (8)

for every n ^ n0.

Using the inequalities (5) and (8), we get

ld(x, An) - d(x, A) l < £

for every n ^ n0.

Since x is arbitrary, we obtain

lim d(x, A„) = d(x, A).

Consequently, we get An —> A. □ Corollary.

(i) Let A\ C A2 C ... C An... (n G N). If there exists a compact set A, such that An s—— A, then

st—H

An C A for every n G N and An —> A.

(ii) Let Ax D A2 D ... D An... (n G N). If there exists a closed set A, such that An s—— A, then

st—H

A C An for every n G N and An —> A.

Proof. i) Assume that An s—— A. By Theorem 1(i), we write An —— A. From [1, Theorem 3.1.(i)], we obtain An C A for every n G N and An —— A. Since An —— A implies An —-— A, the proof is completed. ii) By Theorem 1(ii) and [1, Theorem 3.1.(ii)], the proof is obvious. □

Remark. Assume that the hypotheses of Corollary (i) (or (ii)) are valid. If An s—— A, then we have An —— A.

As can be seen from the following theorem, the hypothesis of Corollary can be weakened using the concept of natural density.

Theorem 2. Let A,An G Cl(X), n G N and K = {nx <n2 <...<nk <...] be a subset of N, such that 5 (K) = 1.

i) Let the subsequence (Ank) keN of (An)neN be monotonically increasing according to the inclusion relation, i. e., Ank C Ank+1 for each k G N.

If An A and A is compact, then

Ank C A for every k G N and An —— A.

(ii) Let the subsequence (Ank) k N of (An) neN be monotonically decreasing according to the inclusion relation. If Ank's are compact and

An S—-— A, then

A C Ank for every k G N and An S—-H A. Proof. i) Fix n*k G K and u G An*k. We have u G Ank for every nk G K

with nk ^ n*k; hence, we have d(u,Ank) = 0. Since An s—— A, we have 5 (L (u, e)) = 1 for every e > 0, where

L (u, e) := {n G N : Id(u, An) — d(u, A) I < e] .

Then, for each e > 0 there exists an nk E K if L (u, e), with nk ^ n*k such that

d(u, A) = ld(u, Ank) — d(u, A) I < e.

Hence, we get d(u, A) < e for every e > 0. So, u E A since the set A is closed. Therefore, we obtain Ank C A for every nk E K. Then we get

h(Ank, A) = sup [d(x, A) : x E Ank} = 0 (9)

for every nk E K.

We have d(x, An ) ^ d(x, An ) for every k\,k2 E N with k2 ^ k\ and for every x E X. Therefore, we have

d(a,Ank ) ^ d(a,Anki) for every a E A

and for every k\,k2 E N with k2 ^ k\ and, so, we get

h(A,Ank2) ^ h(A,Anki).

Define ak = h(A,Ank) for each k E N. Since (ak)keN is a decreasing sequence of positive real numbers, it is convergent. Let us call this "the limit as a ^ 0". Since the function d(.,Ank) is continuous and A is compact, for every k E N there exists an ak E A, such that

ak = sup [d(a, Ank): a E A} = d(a,k, Ank).

By the compactness of A, the sequence (ak)keN has a subsequence (akj) converging to a point a0 E A. By the triangle inequality, we get

0 ^ d(ak , Ank.) ^ P(akj,ao) + d(ao,A„k.).

jj

Since An A, we have

Id(ao,An) — d(ao,A)I < ^

for every n E K f L (a0,e). Since d(a0, A) = 0, we get Id(a0, Ank )| < e/2. By lim akj = a0, there exists a j0 E N, such that

S

p(akj ,a,0) < - for every j ^ ]0. j 2

Then we get

0 ^ d(akj, Ankj) < e

for every nkj E K if L (a0,£) with j ^ j0. Hence, the subsequence (akj) of (ak) is convergent to 0, therefore, we obtain a = 0. Then we have

st—H

lim h (A,Ank) = 0. Consequently, we get An —h A.

k—y^o

ii) Let u E A. Fix nk E K. Since An A, we have 6 (L (u,e)) = 1 for every e > 0 where

L (u,e) := [n E N: ld(u,An) - d(u,A)l < £} .

Then, for each e > 0, there exists an n£k E K f L (u, e) with n£k ^ nk, such that

d(u, Ans) = \d(u, Ans) — d(u, A)| < e.

Also, we have d(u,Ank) ^ d(u,Ans), since nk ^ nk. Hence, we get d(u,Ank) < e for every e > 0 and, so, u E Ank from the closeness of Ank. Therefore, we obtain A C Ank for every nk E K. Hence, we get

h(A,Ank) = sup [d(x,Ank): x E A} = 0 (10)

for every nk E K.

Take = h(Ank ,A) for each k E N. Since h(An^ ,A) ^ h(An^ ,A) for every k\,k2 E N with k2 ^ k\, the sequence (ftk) keN is a decreasing sequence of positive real numbers and therefore it is convergent. Say its limit ¡3 ^ 0. Since the function d(., A) is continuous and Ank are compact, for every k E N there exists an ak E Ank, such that

/3k = sup [d(a, A): a E A,nk} = d(ak, A).

Hence, we have ak E Ani for every k E N due to monotone decrease of (Ank). From the compactness of Ani, the sequence (ak) keN has a subsequence (akj )jeN converging to the point a0 E Ani. Since the subsequence (Ank ) is decreasing and Ank are closed, we get

lim ak. = a0 E Ank, for every j E N.

j—<x j j

Using the triangle inequality, we have

0 ^ d(ak., A) ^ p(ak. ,oq) + d(a0,A)

for each j E N.

Take e > 0. Since An A, we have

Id(a0,A,n) — d(a0,A)I < ^

for every n E K f L (a0,e). Since d(a0,Ank ) = 0 for every j E N we get

Id(a0,A)I < e/2. Also, from lim ak. = a0, there is a j0 E N, such that

j

p(akj ,00) < ^ for every j ^ j0.

Then we get

0 ^ d(ak] ,A) < £

for every nk. E K f L (a0,£) with j ^ j0. Hence, the subsequence (fik.) of (fik) is convergent to 0, and, therefore, we obtain ¡3 = 0. Then we have

st—H

lim h (Ank, A) = 0. Consequently, we get An —h A. □

Theorem 3. Let = {I C N : 8(1) = 0} be an ideal connected with statistical convergence. If An A, then we have

(i) A = p| U Am and

lels m<EN\I

(ii) A ^n U n B(Am,£), where B(Am,£) = {x E X : d(x,Am) ^e}.

e>01els meN\i

Proof. i) Define B := P| (J Am. Firstly, we show that A C B. Let

ieis meN\i st—H

x E A and I E Xs. Since An —> A, we have

I£ := {n E N: h (A,An) ^ e or h (An,A) ^ £} E Xs

for every e> 0. Moreover, since IUle EXS, the set N\(IUle) = (N\I)f(N\Ie) is non-empty. Hence, there exists an m E (N \ I) f (N \ I£) for each e > 0, and we get

h (A,Am) < £

and, so,

d (x, Am) < £.

Since Am is a closed set, there exists an xm E Am, such that

P (x,xm) = d (x,Am) < £.

Also, since xm E U we get x E (J Am. Since I is arbitrary, we

m&i\I m&i\I

get x E B.

Now let us show that B C A. Take x E B = P| (J Am. Then we

iels m€N\/

--st—H

have x E (J Am for each I e!s. Since An —h A, we have

m&i\I

I£ := [n E N: h (A, An) ^ e or h (An, A) ^ e} E Is

for every e > 0. Take e > 0. Then we have x E (J Am. Hence, there

m&i\Is

exists a sequence (xn)nEN C |J Am, such that xn —> x. In this case,

m£N\Is

there exists an n0 (e) E N, such that p (xn, x) < £ for every n ^ n0 (e). We can choose an n\ ^ n0 and an m\ E N \ I£, such that xni E Ami. Then we get

d (x, A) ^ p (x, xni) + d (xni, A) ^

^ p (x,xni) + d (Xni ,Ami) + h (Ami, A) <

< £ + 0 + £ = 2£.

Since £ is arbitrary and A is closed, we get x E A.

ii) Define B := Q U 0 b(Am,£). Let x E A. Choose £ > 0 e>o ieig meN\i

arbitrarily. Since An S-—H A, we have

I£ := [n E N : h (A, An) ^ £ or h (An, A) ^ £} E Xs.

We can write

h (A, Am) < £ =^ d (x, Am) < £ =^ x E b(Am, £),

for every m E N \ I£. Consequently, we obtain x E B and, therefore, A C B.

Now, we show that B C A. Let x E B. Hence, for every £ > 0 there exists an I£ e Is, such that x E B(Am, £) for every m E N \ I£. Also, from

An S-—H A we have

J£ := [n E N: h (A, An) ^ £ or h (An, A) ^ £} E Is.

Then I£ U Js E and there exists an m0 = m0 (e) E N \ (I£ U J£). By x E B(Amo ,e), we have

d (x,Amo ) < £. Since Amo is a closed set, there is a y E Amo, such that

P (%,y) <£. (11)

By h (An, A) ^ e, we have

d (y,A) < e.

Since A is a closed set, there is a z E A, such that

p (y,z) < £. (12)

Using the inequalities (11) and (12), we obtain

P (x, z) ^ P (x, y)+ P (У, z) < 2£. Since £ is arbitrary and A is closed, we get x E A. □

References

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DOI: https://doi.org/10.1090/S0002-9904-1964-11072-7

[13] Wijsman R. A. Convergence of sequences of convex sets, cones and functions. II. Trans. Amer. Math. Soc., 1966, vol. 123, pp. 32-45.

DOI: https://doi.org/10.1090/S0002-9947-1966-0196599-8

Received May 20, 2022. In revised form, August 14, 2022. Accepted August 17, 2022. Published online September 15, 2022.

H. Albayrak

Suuleyman Demirel University,

Department of Statistics, 32260, Isparta, Turkey

E-mail: [email protected]

F. Babaarslan Yozgat Bozok University,

Department of Mathematics, 66100, Yozgat, Turkey E-mail: [email protected]

Ou . Ou lmez

Suuleyman Demirel University,

Department of Mathematics, 32260, Isparta, Turkey E-mail: [email protected]

S. Aytar

Suuleyman Demirel University,

Department of Mathematics, 32260, Isparta, Turkey E-mail: [email protected]