Statistical limit point theorems

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©Hindawi Publishing Corp.

STATISTICAL LIMIT POINT THEOREMS

JEFF ZEAGER

(Received 4 August 1998)

Abstract.It is known that given a regular matrixAand a bounded sequencexthere is a subsequence (respectively, rearrangement, stretching)yofxsuch that the set of limit points ofAyincludes the set of limit points ofx. Using the notion of a statistical limit point, we establish statistical convergence analogues to these results by proving that every complex number sequencexhas a subsequence (respectively, rearrangement, stretching) ysuch that every limit point ofxis a statistical limit point ofy. We then extend our re-sults to the more generalA-statistical convergence, in whichAis an arbitrary nonnegative matrix.

Keywords and phrases. Statistical convergence, statistical limit point, subsequence, re-arrangement, stretching.

2000 Mathematics Subject Classiﬁcation. Primary 40A05.

1. Introduction. In [2, 3] Buck characterized convergence by proving that ifx is a nonconvergent sequence, then no regular matrix can sum every subsequence ofx. This result was extended by Agnew [1] who showed that given a regular matrixAand a bounded sequencex, there is a subsequenceyofxsuch that the set of limit points ofAy includes the set of limit points ofx. Analogues to these results were given by Dawson [6] and Fridy [9] by replacing subsequence with stretching and rearrange-ment, respectively. In [17], statistical convergence analogues andA-statistical conver-gence analogues to Buck’s theorem and its variants are given. Now we generalize the constructions in [17], providing statistical convergence analogues andA-statistical convergence analogues to Agnew’s theorem and its variants.

IfKis a subset of the natural numbersN, letKndenote the set{k≤n:k∈K}and |Kn|denote the cardinality ofKn. Thenatural orasymptotic density ofK (see [13, Chapter 11]) is given byδ(K)=limn(1/n)|Kn|, if the limit exists. A complex number sequencexis said to bestatistically convergent toLif for every positiveε,

δk:xk−L≥ε=0. (1.1)

In this case, we write st-limx=L. This notion was ﬁrst introduced by Fast [7] for real sequences and has since been studied as a regular summability method by several authors (cf. [5, 10, 14]). Using natural density, Fridy [11] deﬁned an analogue to the notion of a limit point of a sequencex. A subsequencey= {x}Kofxis said to be

non-thinifKdoes not have natural density zero, and the numberλis said to be astatistical limit point ofxif there exists a nonthin subsequence ofxthat converges toλ.

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ofKis given by

δA(K)=_nlim_→∞

k∈K

an,k (1.2)

if the limit exists. This notion was used by Kolk in [12] to extend statistical convergence as follows. A complex number sequencexis said to beA-statistically convergent toL

if, for every positiveε,

δAk:xk−L≥ε=0. (1.3)

In this case, we write stA-limx=L. Connor and Kline [4] replaced natural density with

A-density in Fridy’s deﬁnition of a statistical limit point. So a subsequencey= {x}Kof the sequencexis calledA-nonthinifKdoes not haveA-density zero and the number

λis called anA-statistical limit pointofxif there is anA-nonthin subsequence ofx

that converges toλ.

Before we can state our main results, we must give the following two deﬁnitions.

Deﬁnition1.1. A sequencezis called arearrangement of the sequencex pro-vided that there is a bijectionπ:N→Nsuch that for eachk,zk=xπ(k).

Remark1.2. The word “permutation” is reserved for the reordering of a ﬁnite sequence.

Deﬁnition1.3(Dawson [6]). Let{m(p)}∞

p=0be an increasing sequence of integers

such thatm(0)=1. We call the sequencew thestretching ofxinduced by{m(p)} providedwq=xp, wheneverm(p−1)≤q < m(p).

Remark1.4. The sequencewhas also been called adilutionofxby Sledd [15].

2. Statistical limit point theorems. In [1] Agnew proved that, given a regular matrix

Aand a bounded sequencex, there is a subsequenceyofxsuch that the set of limit points ofAy includes the set of limit points ofx. Ifx is bounded but divergent, it has at least two distinct limit points and, so,Ay also has at least those same two limit points. Thereforey is notA-summable and Buck’s theorem [2, 3] (for bounded sequences) follows from Agnew’s theorem.

It is shown in [10, Theorem 1] that a sequencexis statistically convergent if and only ifxis a sequence for which there exists aconvergentsequenceysuch thatxk=yk for almost allk, that is, for everykin a setKwithδ(K)=1. This implies that ifx

is statistically convergent toλthen the set of statistical limit points is the singleton set{λ}. In the bounded cases of [16, Theorems 2.1, respectively 2.2, 2.3], we created a subsequence (respectively, rearrangement, stretching) that had two distinct statistical limit points and was therefore not statistically convergent. In this section, we general-ize those constructions in much the same way that Agnew’s work generalgeneral-ized Buck’s theorem. We begin with the following lemma.

Lemma2.1. Ifxis a complex number sequence with a countably inﬁnite set of (ﬁnite) limit pointsD= {λj}∞j=1, then there is a subsequenceyofxsuch that everyλjinDis

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Proof. For everyλj inD, there is a subsequence{x}K(j) ofxsuch that{x}K(j) converges toλj. We construct yby choosing blocks of terms from the sets{x}K(j) in the following manner. The ﬁrst block is chosen from{x}K(1). The second block is

selected from{x}K(1) and the third block comes from{x}K(2). In general, the index

of the set from which a block is selected follows the pattern 1,1,2,1,2,3,1,2,3,4,.... The blocks are chosen so that the length of each block (after the initial one) is equal to the number of terms ofy which precede it. We begin by selecting the ﬁrst block {y1}(a single term) to be the ﬁrst term of{x}K(1). Then we choose the second block {y2}to be the second term of{x}K(1). The third block consists of two terms,{y3,y4},

and is chosen from{x}K(2)so as to have the index of thexjtaken fory3larger than the indices used fory1andy2, and to havey4be any term in{x}K(2)afterxj. For example, if the ﬁrst two terms of{x}K(1)arex12andx30, we must choose fory3and

y4termsxjandxkin{x}K(2)such thatj≥31 andk > j. Otherwise,ywould not be a subsequence ofxbecause the original order of the chosen terms would not have been preserved. The fourth block of terms,{y5,...,y8}is chosen from{x}K(1)so that the indices of the terms used are larger than those of any previously chosenxj’s. The ﬁfth block,{y9,...,y16}and the sixth block of terms,{y17,...,y32}, are chosen from {x}K(2)and{x}K(3), respectively, with each term’s index being larger than the indices

of all of thexj’s which precede it. Having selected thenth block of terms from{x}K(s), that is, having constructed{y1,...,yq}withq=2n−1, we choose the(n+1)st block of terms{yq+1,...,Y2q}from the set{x}K(i), where

i=       

1, ifn= r

t=1

tfor somerinN,

s+1, otherwise.

(2.1)

Here again, we must have each chosen term’s index larger than the indices of all of the previously selectedxj’s. This construction ofy= {xn(k)}guarantees that{n(k)} is a strictly increasing sequence of indices; soyis, indeed, a subsequence ofx. We now must show that eachλj inDis a statistical limit point of y. Consider a ﬁxed but arbitraryjinN. By selecting the blocks of terms as above, we have ensured that inﬁnitely many blocks are chosen from{x}K(j). By concatenating these blocks in the order in which they appear iny, we get a subsequencewofywhich converges toλj. This makesλja limit point ofy. To see thatλjis astatisticallimit point ofy, we show thatwis a nonthin subsequence ofy. Notice that by the construction ofy, whenever

n≥2 the length of thenth block of terms{yq+1,...,y2q}is the same as the portion ofy which precedes it, namely,{y1,...,yq}. So, for any block of terms chosen from {x}K(j)with ending termyd, we have

1

dk≤d:yk∈ {x}K(j)≥

1

2. (2.2)

Thereforeδ({k:yk∈ {x}K(j)})≠0, and sowis a nonthin subsequence ofy.

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Theorem2.2. Ifxis a complex number sequence andLxis the set of (ﬁnite) limit

points ofx, then there is a subsequenceyofxsuch that everyλinLxis a statistical

limit point ofy.

Proof. IfLx is a finite set, sayLx= {e1,...,en}, letDbe the countably infinite sequence{e1,...,en,e1,...,en,...}. IfLxis infinite, we use the separability of complex plane to find a countably infinite subsetDofLxsuch that the closureDofDisLx. In either case, letybe the subsequence ofxcreated in the proof of Lemma 2.1 usingD. We only need to prove the result for the case whenLx is uncountable. Let λ0 be a fixed but arbitrary element ofLx. BecauseD=Lx,there is a sequence{λn(i)}∞i=1inD

that converges toλ0. By the construction ofy,there are, wheneveri≥1, infinitely many blocks of terms ofyfrom{x}K(n(i)).Also, since each{x}K(n(i))is a subsequence ofxwhich converges to λn(i),there are, for any positive numberε,infinitely many blocks of terms inysuch that|yj−λn(i)|< ε/2 for everyyjin each block. Moreover, since limiλn(i)=λ0,we can find, for any positiveε,ann0inNsuch that whenever

i≥n0,|λn(i)−λ0|< ε/2.Therefore, given a positiveε,we can find infinitely many blocks of terms ofysuch that|yj−λ0|< εfor everyyjin each block. This then allows us to construct a nonthin subsequencew ofywhich converges toλ0.We choose as our first block ofwany block ofysuch that|yj−λ0|<1 for allyjin the block. We then choose the second block ofwto be any block ofywith|yj−λ0|<1/2 for allyjin the block and whose terms have indices larger than those of the first block. (We must concatenate the selected blocks in the order in which they appear iny,otherwise,wis not a subsequence ofy.) Having chosen the firstn−1 blocks ofw,we choose thenth block ofwto be any block ofy,beyond those already chosen, with|yj−λ0|<1/n for allyjin the block. Clearly,wis a subsequence ofywith limqwq=λ0.To see that

w is a nonthin subsequence ofy,recall from the construction ofy in the proof of Lemma 2.1, that the length of any block ofyis the same as the length of the portion ofy which precedes it. LetJ be the index set ofwso thatw= {y}J.If we consider any block chosen in the construction ofw(say, with ending termyd), then

1

dk≤d:yk∈ {y}J≥

1

2. (2.3)

Therefore,δ(J)≠0 andwis a nonthin subsequence ofy.

We now show that a sequencexhas a rearrangementzsuch that every limit point ofx is astatistical limit point ofz. The construction ofzis similar in nature to the construction ofy in the proof of Lemma 2.1, the major diﬀerence being that with a rearrangement, we must use every term ofxexactly once. As was the case in the proof of Theorem 2.2, a lemma for the countable case is used in the proof of the general result.

Lemma2.3. Ifxis a complex number sequence with a countably inﬁnite set of (ﬁnite) limit pointsD= {λj}∞j=1,then there is a rearrangementzofxsuch that everyλjinD

is a statistical limit point ofz.

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with the indexifollowing this pattern: 1,1,2,1,2,3,1,2,3,4,... .Each block is chosen so its length is the same as that of the portion ofz which precedes it. In between theith and(i+1)st chosen blocks, we must use the terms ofx that were skipped over while selecting theith block. We begin by choosing the ﬁrst block{z1}to be any termxn(1)of{x}K(1),wheren(1)≥2.We then let{z2,...,zn(1)}be any permutation of the unused terms{x1,...,xn(1)−1}.The second block of terms{zn(1)+1,...,z2n(1)} is selected from{x}K(1)\{x1,...,xn(1)},wherexn(2)has the largest index of the cho-sen terms. Then let{z2n(1)+1,...,zn(2)}be any permutation of the termsxj,where

n(1)+1≤j≤n(2)−1,that were not selected for the second block. We pick the third block of terms{zn(2)+1,...,z2n(2)}from {x}K(2)\{x1,...,xn(2)}with n(3)being the largest index of the selectedxj’s. Let{z2n(2)+1,...,zn(3)}be any permutation of the unusedxj’s betweenxn(2)+1 andxn(3)−1. Once{z1,...,zn(k)}has been constructed, with thekth block of terms coming from{x}K(s),we select the(k+1)st block of terms {zn(k)+1,...,z2n(k)}from{x}K(i)\{x1,...,xn(k)}, where

i=       

1, ifk= r

t=1

tfor somerinN,

s+1, otherwise.

(2.4)

Letn(k+1)be the largest index of the selected termsxj,and let{z2n(k)+1,...,zn(k+1)} be any permutation of the unused termsxjbetweenxn(k)+1andxn(k+1)−1.

By the construction,{z1,...,zn(k)}is a permutation of{x1,...,xn(k)}wheneverk≥1. Thuszis indeed a rearrangement ofx. We now must show that eachλjinDis a sta-tistical limit point ofz. Consider a ﬁxed but arbitrary j in N. In constructingz as above, we have ensured that inﬁnitely many blocks are chosen from{x}K(j). By con-catenating these blocks in the order in which they appear inz,we get a subsequence

yofzwhich converges toλj.Notice that by the construction ofz, wheneveri≥2,the length of theith block of terms isn(i−1); which is precisely the length of the portion ofzwhich precedes theith block. So for any block of terms chosen from{x}K(j)with ending termzd, we have

1

dk≤d:zk∈ {x}K(j)≥

1

2. (2.5)

Therefore,δ({k:zk∈ {x}K(j)})≠0,soyis a nonthin subsequence ofzconverging toλj. Thusλjis a statistical limit point ofz.

The following theorem is a statistical convergence analogue to Fridy’s rearrangement version of Agnew’s matrix result. (See [9, Theorem 3].)

Theorem2.4. Letxbe a complex number sequence and letLxbe the set of (ﬁnite)

limit points ofx,then there is a rearrangementzof x such that everyλ inLx is a

statistical limit point ofz.

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complex plane to find a countably infinite subsetDofLxsuch that the closureDofD isLx. In either case, letzbe the rearrangement ofxcreated in the proof of Lemma 2.3 usingD. We only need to prove the result for the case whenLxis uncountable. Letλ0 be a fixed but arbitrary element ofLx. Using observations similar to those made in the proof of Theorem 2.2, we can find, for any positive numberε,infinitely many blocks of terms ofzsuch that|zk−λ0|< εfor everyzkin each block. This then allows us to create a subsequenceyofzby concatenating blocks ofz,in the order in which they appear, where|zk−λ0|<1/nfor allzkin thenth block ofy. Clearly,yconverges to

λ0. To show thatλ0is a statistical limit point ofz, we need to show thatyis a nonthin subsequence ofz. We recall from the construction ofzin Lemma 2.3, that the length of any block inzis equal to that of the portion ofzwhich precedes it. LetJbe the index set ofyso thaty= {z}J.If we consider any block chosen in the construction ofy(say, with ending termzd), then

1

dk≤d:zk∈ {z}J≥

1

2. (2.6)

Therefore,δ(J)≠0 andyis a nonthin subsequence ofz. Thusλ0is a statistical limit point ofz.

To end this section, we give a result concerning stretching that is analogous to Theorems 2.2 and 2.4. However, before we can state and prove the result, we need to establish more notation. Letx= {xn}be a complex number sequence, let{xn(k)}be a subsequence ofx, and letwbe the stretching ofxinduced{m(p)}∞

p=0. Let

M= ∞

k=1

mn(k)−1,...,mn(k)−1. (2.7)

Deﬁnition2.5. The subsequencey[xn(k)]= {w}Mofwis called thesubsequence

corresponding to{xn(k)}inw.

Notice thatwq=xn(k)wheneverm(n(k)−1)≤q < m(n(k)).Thus if limkxn(k)=L, then limky[xn(k)]=L.

Example2.6. Consider the sequencex=1,2,3,4,.... The stretchingwofxgiven byw=1,1,1,2,3,3,4,4,4,4,... is induced by the sequencem=1,4,5,7,11,...; that is,m(0)=1,m(1)=4,m(2)=5, and so on. Letxn(k)=1,4,9,... be the subsequence ofxconsisting of the squares, i.e.,xn(k)=k2. Using the notation from Deﬁnition 2.5 above,y[xn(k)]= {1,1,1,4,4,4,4,...} = {w}M, where

M= {1,2,3}∪{7,8,9,10}∪···. (2.8)

Lemma2.7. Ifxis a complex number sequence, then{2p_}∞

p=0induces a stretching wofxin which for any subsequence{xn(k)}ofx,y[xn(k)]is a nonthin subsequence

ofw.

Proof. Let{xn(k)}be any subsequence ofxand let

M= ∞

k=1

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Wheneverk≥1, 1

2n(k)₋₁q≤2n(k)−1 :q∈M≥ 1

2n(k)₋₁q≤2n(k)−1 :wq=xn(k) ≥2n(k)−1−2n(k)−1+1

2n(k)₋₁ ≥

2n(k)−1

2n(k)₋₁> 1 2.

(2.10)

Thusδ(M)≠0 and hencey[xn(k)]= {w}Mis a nonthin subsequence ofw.

Here then is the statistical convergence analogue to Dawson’s stretching version of Agnew’s theorem. (See [6, Theorem 3].)

Theorem2.8. Ifxis a complex number sequence, then{2p_}∞

p=0induces a stretching wofxin which every (ﬁnite) limit point ofxis a statistical limit point.

Proof. Letλ be a (ﬁnite) limit point of x, and let w be the stretching ofx in-duced by{2p_}∞

p=0. Then there is a subsequence{xn(k)}ofx that converges toλ. By Lemma 2.7,y[xn(k)]is a nonthin subsequence ofwwhich converges toλ. Thusλis a statistical limit point ofw.

It should be noted that{2p_}_{is independent of}_x_{, that is, given}_any_sequence_x_{, the} stretching w induced by{2p_}_{has the desired statistical limit points. However, the} constructions of a subsequenceyand rearrangementzwith the appropriate statistical limit points depended on the given sequencex.

3. A-statistical limit point theorems. In [12, Theorem 2.3], it is shown that for a nonnegative regular matrixA, a sequencexisA-statistically convergent toλif and only if there is an inﬁnite index setK withδA(K)=1 such that{x}K converges to

λ. This then implies that if x is A-statistically convergent to λ, then λ is the only

A-statistical limit point ofx. Now, we generalize the constructions given in [17, The-orem 2, 4, 6] to giveA-statistical convergence analogues to Agnew’s theorem [1] and its analogues for rearrangements [9, Theorem 3] and stretchings [6, Theorem 3]. We begin by proving the following lemma.

Lemma3.1. Letx be a complex number sequence with a countably inﬁnite set of (ﬁnite) limit pointsD= {λj}∞j=1. Given a nonnegative regular matrixA, there exists a

subsequenceyofxsuch that everyλjinDis anA-statistical limit point ofy.

Proof. Throughout the proof,Ais a ﬁxed but arbitrary nonnegative regular ma-trix. For everyλjinDthere is a subsequence{x}M(j)ofxsuch that{x}M(j)converges toλj. Using this countable collection of sets, we construct the subsequencey in a manner very similar to the construction found in the proof of Lemma 2.1. Blocks of terms foryare chosen from the sets{x}M(j)with the indexjfollowing this pattern: 1,1,2,1,2,3,1,2,3,4,.... Using the regularity ofA, we select two strictly increasing sequence of indices{n(i)}∞

i=0and{k(i)}i∞=0such thatk(0)=0 and

k(i+1) k=1+k(i)

an(i+1),k>1₂ foralli=0,1,2,... . (3.1)

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Let{y1+k(1),...,yk(2)}be the nextk(2)−k(1)terms of{x}M(1), taken in order. Select {y1+k(2),...,yk(3)}as the ﬁrstk(3)−k(2)terms of{x}M(2), taken in order, whose

in-dices are larger than that of thexp chosen asyk(2). The next three blocks{y1+k(3),

...,yk(4)},{y1+k(4),...,yk(5)}and{y1+k(5),...,yk(6)}are chosen from{x}M(1),{x}M(2) and{x}M(3), respectively, with each term’s index being larger than the indices of all

of thexj’s which precede it.

Having selected themth block of terms{y1+k(m−1),...,yk(m)}from{x}M(s), we then choose the(m+1)st block of terms{y1+k(m),...,yk(m+1)}from{x}M(i), where

i=       

1, ifk= r

t=1

tfor somerinN,

s+1, otherwise.

(3.2)

Here again, we must have the index of each chosen term larger than the indices of all of the previously selectedxj’s so thaty is a subsequence ofx. We must show that eachλjinDis anA-statistical limit point ofy. Consider a ﬁxed but arbitraryj inN. By selecting the blocks of terms as above, we have ensured that inﬁnitely many blocks are chosen from{x}M(j). By concatenating these blocks in the order in which they appear iny, we get a subsequencezofy which converges toλj. Thusλjis a limit point ofy. To see thatλjis anA-statisticallimit point ofy, we show thatzis an

A-nonthin subsequence ofy. LetK= {k:yk∈ {x}M(j)}. Notice that by the selection of{n(i)}and{k(i)}and by the construction ofy, for indicesn(d), whereyk(d)is the last term of a block chosen from{x}M(j), we have

k∈K

an(d),k≥ k(d)

k=1+k(d−1)

an(d),k>1₂. (3.3)

Thus for inﬁnitely many indicesnwe havek∈Kan,k>1/2. ThereforeδA(K)≠0, and sozis anA-nonthin subsequence ofy.

We now state and prove theA-statistical convergence analogue to Agnew’s theo-rem [1].

limit points ofx. Given a nonnegative regular matrixA, there exists a subsequencey

ofxsuch that everyλinLxis anA-statistical limit point ofy.

Proof. IfLx is a finite set, sayLx= {e1,...,en}, letDbe the countably infinite sequence{e1,...,en,e1,...,en,...}. IfLxis infinite, we use the separability of the com-plex plane to find a countably infinite subsetDofLxsuch that the closureDofDis

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A-nonthin subsequence ofy. LetJbe the index set ofzso thatz= {y}J. For indices

n(d), whereyk(d)is the last term of a block chosen forz, we have

k∈J

an(d),k≥ k(d)

k=1+k(d−1)

an(d),k>1₂. (3.4)

Thus for inﬁnitely many indicesn, we havek∈Jan,k>1/2.Therefore

δA(J)=δAk:yk∈ {y}J≠0, (3.5) and sozis anA-nonthin subsequence ofy.

Our next goal is to show that, given a ﬁxed nonnegative regular matrixA, a sequence

xhas a rearrangementzsuch that every limit point ofxis anA-statisticallimit point ofz. Here again, the construction of the rearrangementzis similar in nature to the construction of the subsequenceyabove, the major diﬀerence being that with a re-arrangement, we must use every term ofxexactly once. We ﬁrst prove the following lemma which is used in the proof of the general result.

Lemma3.3. Letx be a complex number sequence with a countably inﬁnite set of (ﬁnite) limit pointsD= {λj}∞j=1. Given a nonnegative regular matrixA, there exists a

rearrangementzofxsuch that everyλjinDis anA-statistical limit point ofz.

Proof. Throughout the proof,Ais a ﬁxed but arbitrary nonnegative regular ma-trix. For everyλjinDthere is a subsequence{x}M(j)ofxsuch that{x}M(j)converges toλj. We constructzby choosing blocks of terms from the sets{x}M(j)with the in-dexjfollowing this pattern: 1,1,2,1,2,3,1,2,3,4,... .In between theith and(i+1)st chosen blocks, we must use the terms ofxthat were skipped over while selecting the

ith block. LetJ(1)=1. ChooseN(1)and thenK(1)such thatK(1) > J(1)and K(1)

k=J(1)

aN(1),k>1₂. (3.6)

Select{zJ(1),...,zK(1)}as any permutation of the ﬁrstK(1)terms of{x}M(1). Letm

be the largest index of the termsxkchosen from{x}M(1)for the block and setJ(2)=

m+1. Notice thatJ(2) > K(2). Next, let{zK(1)+1,...,zJ(2)−1}be any permutation of

the unused termsxj, whereJ(1)≤j≤J(2)−1. SelectN(2)and thenK(2)such that

N(2) > N(1),K(2) > J(2)and

K(2) k=J(2)

aN(2),k>1₂. (3.7)

Choose {zJ(2),...,zK(2)} as any permutation of the ﬁrst K(2)−J(2)+1 terms of {xM(1)}\{x1,...,xJ(2)−1}. Letmbe the largest index of the terms chosen from{x}M(1) thus far and setJ(3)=m+1.Select{zK(2)+1,...,zJ(3)−1}as any permutation of the

unused termsxj, where J(2)≤j≤J(3)−1. ChooseN(3)and thenK(3)such that

N(3) > N(2),K(3) > J(3)and

K(3) k=J(3)

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Let {zJ(3),...,zK(3)} be a permutation of the ﬁrst K(3) − J(3) + 1 terms of {xM(2)}\{x1,...,xJ(3)−1}. Let m be the largest index of the terms xk chosen from {x}M(2) for the block and set J(4)=m+1. Choose {zK(3)+1,...,zJ(4)−1}to be any

permutation of the unused termsxj, whereJ(3)≤j≤J(4)−1.

Having constructed{z1,...,zJ(n)−1}, with the(n−1)st block {zJ(n−1),...,zK(n−1)} coming from {x}M(s), and thus having chosen indices {N(p)}pn=−11, {J(p)}np=1 and {K(p)}n−1

p=1such that

N(1) < N(2) <···< N(n−1),

J(1) < K(1) < J(2) <···< J(n−1) < K(n−1) < J(n), (3.9)

chooseN(n)and thenK(n)such thatN(n) > N(n−1),K(n) > J(n)and K(n)

k=J(n)

aN(n),k>1₂. (3.10)

Thenth block of terms{zJ(n),...,zK(n)}is chosen as any permutation of the ﬁrst

K(n)−J(n)+1 terms of{xM(i)}\{x1,...,xJ(n)−1}, where

i=       

1, ifk= r

t=1

tfor somerinN,

s+1, otherwise.

(3.11)

Letmbe the largest index of the termsxkchosen from{xM(i)}for the block and set

J(n+1)=m+1. Select the block of terms{zK(n)+1,...,zJ(n+1)−1}as any permutation

of the unused termsxj, whereJ(n)≤j≤J(n+1)−1.

By the construction,{z1,...,zJ(n)−1}is a permutation of{x1,...,xJ(n)−1}whenever n≥1. Thuszis indeed a rearrangement ofx. We now must show that eachλjinDis anA-statistical limit point ofz. Consider a ﬁxed but arbitraryjinN. In constructingz

as above, we have ensured that inﬁnitely many blocks are chosen from{x}M(j). By con-catenating these blocks in the order in which they appear inz, we get a subsequence

yofzwhich converges toλj. Thusλjis a limit point ofz. LetK= {k:zk∈ {x}M(j)}. Notice that by the selection of{N(p)}∞

p=1,{J(p)}∞p=1and{K(p)}∞p=1and by the

con-struction ofz, that for indicesN(d),J(d)andK(d), where{zJ(d),...,zK(d)}is a block ofzchosen from{x}M(j), we have

k∈K

aN(d),k≥ K(d)

k=J(d)

aN(d),k>1₂. (3.12)

Thus for inﬁnitely many indicesn, we havek∈Kan,k>1/2. ThereforeδA(K)≠0, and soyis anA-nonthin subsequence ofzandλjis anA-statistical limit point ofz.

We now can state and prove theA-statistical convergence analogue to Fridy’s re-arrangement version of Agnew’s theorem. (See [9, Theorem 3].)

limit points ofx. Given a nonnegative regular matrix A, there exists a rearrangement

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Proof. IfLxis a finite set, sayLx= {e1,...,en}, letDbe the countably infinite se-quence{e1,...,en,e1,...,en,...}. IfLxis infinite, we use the separability of the complex plane to find a countably infinite subsetDofLxsuch that the closureDofDisLx. In either case, letz be the rearrangement of x created in the proof of Lemma 3.3 usingD. We only need to prove the result for the case whenLxis uncountable. Con-sider an arbitrary but fixedλ0inLx. Using observations similar to those made in the proof of Theorem 2.2, we can find, for any positive numberε, infinitely many blocks of terms ofzsuch that|zk−λ0|< εfor everyzkin each block. This then allows us to create a subsequenceyofzby concatenating blocks ofz, in the order in which they appear, where|zk−λ0|<1/nfor allzkin thenth block ofy. Clearly,y converges toλ0. To show thatλ0is anA-statistical limit point ofz, we need to show thatyis anA-nonthin subsequence ofz. LetKbe the index set ofyso thaty= {z}K. If we consider any block chosen in the construction ofy, say{zJ(d),...,zK(d)}, then

k∈K

aN(d),k≥ K(d)

k=J(d)

aN(d),k>1₂. (3.13)

Thus for inﬁnitely many indicesn, we havek∈Kan,k>1/2. Therefore

δA(K)=δAk:zk∈ {z}K≠0, (3.14)

and soyis anA-nonthin subsequence ofz.

To ﬁnish the section, we give anA-statistical convergence analogue to Dawson’s stretching version of Agnew’s theorem. (See [6, Theorem 3].)

limit points ofx. Given a nonnegative regular matrixA, there exists a stretchingwof

xsuch that everyλinLxis anA-statistical limit point ofw.

Proof. Using the regularity ofA, we choose strictly increasing sequences of in-dices{m(p)}∞

p=0and{N(p)}p∞=1such thatm(0)=1 and

m(p)−1

m(p−1)

aN(p),k>1₂, (3.15)

wheneverp≥1. Letw be the stretching induced by{m(p)}∞

p=0. Given a ﬁxed but

arbitrary elementλ0ofLx, there is subsequence{xn(j)}∞j=1ofxthat converges toλ0.

Let

M= ∞

j=1

mn(j)−1,...,mn(j)−1. (3.16)

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k∈M

aN(n(j)),k≥

m(n(j))−1

k=m(n(j)−1)

aN(n(j)),k>1₂. (3.17)

HenceδA(M)≠0 and thus{w}M is anA-nonthin subsequence ofw.

Acknowledgement. This paper is based on the author’s doctoral dissertation, written under the supervision of Prof. J. A. Fridy at Kent State University, 1998.

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Zeager: Division of Science and Mathematics, Lorain County Community College, Elyria, Ohio44035, U SA