©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,
δ
k: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 benon- thinifKdoes not have natural density zero, and the numberλis said to be astatistical limit point ofxif there exists a nonthin subsequence ofxthat converges toλ.
Natural density was generalized by Freedman and Sember [8] by replacingC1with an arbitrary nonnegative regular matrixA. Thus, ifKis a subset ofN, then theA-density
ofKis given by
δA(K)= lim
n→∞
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,
δA
k: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 generalized 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 a statistical limit point ofy.
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.
We can now state and prove the statistical convergence analogue to Agnew’s original matrix result in [1].
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.
Proof. For everyλj inD, there is a subsequence{x}K(j) ofxsuch that{x}K(j)
converges toλj.We constructzin a fashion similar to the construction of the subse- quenceyin the proof of Lemma 2.1. We choose blocks of terms from the sets{x}K(i)
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.
Proof. If Lx is a finite set, say, Lx = {e1,...,en}, let D be the countably infi- nite sequence{e1,...,en,e1,...,en,...}. IfLxis infinite, we use the separability of the
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
m
n(k)−1 ,...,m
n(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
2n(k)−1,...,2n(k)−1
. (2.9)
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 ofx, that is, givenanysequencex, 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>1
2 foralli=0,1,2,... . (3.1) With these indices, we construct the subsequence y as follows. Let {y1,...,yk(1)} be the first k(1) terms of {x}M(1) taken in the order in which they appear in x.
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>1
2. (3.3)
Thus for infinitely many indicesnwe have
k∈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].
Theorem3.2. Letxbe a complex number sequence and letLxbe the set of (finite) 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 Lx. In either case, lety be the subsequence ofx created in the proof of Lemma 3.1 usingD. We need only prove the result for the case whenLx is 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 ofysuch that|yk−λ0|< εfor everyykin each block. This then allows us to create a subsequencezofyby concatenating blocks ofy, in the order in which they appear, where|yk−λ0|<1/nfor allyk in thenth block ofz. Clearly,zconverges toλ0.To show thatλ0is anA-statistical limit point ofy,we need to show thatzis an
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>1
2. (3.4)
Thus for infinitely many indicesn, we have
k∈Jan,k>1/2.Therefore δA(J)=δA
k: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>1
2. (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>1
2. (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)
aN(3),k>1
2. (3.8)
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)}n−1p=1, {J(p)}np=1 and {K(p)}n−1p=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>1
2. (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>1
2. (3.12)
Thus for infinitely many indicesn, we have
k∈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].)
Theorem3.4. Letxbe a complex number sequence and letLxbe the set of (finite) limit points ofx. Given a nonnegative regular matrix A, there exists a rearrangement zofxsuch that everyλinLxis anA-statistical limit point ofz.
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>1
2. (3.13)
Thus for infinitely many indicesn, we have
k∈Kan,k>1/2. Therefore δA(K)=δA
k: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].)
Theorem3.5. Letxbe a complex number sequence and letLxbe the set of (finite) 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>1
2, (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
m
n(j)−1 ,...,m
n(j)
−1
. (3.16)
(See Example 2.6 as an example of this notation.) The subsequence{w}M ofw con- verges toλ0becausewq=xn(j), wheneverm(n(j)−1)≤q < m(n(j)), soλ0is a limit point ofw. To show thatλ0is anA-statisticallimit point ofw, we need to show that {w}M isA-nonthin inw. Notice that wheneverj≥1,
k∈M
aN(n(j)),k≥
m(n(j))−1
k=m(n(j)−1)
aN(n(j)),k>1
2. (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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