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Subsequences and Cluster Points

Module by: Lawrence Baggett. E-mail the author

Summary: An introduction of the convergence of subsequences, the bolzano-weierstrass theorem, cluster sets, suprema, infima, and the catchy criterion. Multiple exercises and proofs are included.

Definition 1:

Let {an}{an} be a sequence of real or complex numbers. A subsequence of {an}{an} is a sequence {bk}{bk} that is determined by the sequence {an}{an} together with a strictly increasing sequence {nk}{nk} of natural numbers. The sequence {bk}{bk} is defined by bk=ank.bk=ank. That is, the kkth term of the sequence {bk}{bk} is the nknkth term of the original sequence {an}.{an}.

Exercise 1

Prove that a subsequence of a subsequence of {an}{an} is itself a subsequence of {an}.{an}. Thus, let {an}{an} be a sequence of numbers, and let {bk}={ank}{bk}={ank} be a subsequence of {an}.{an}. Suppose {cj}={bkj}{cj}={bkj} is a subsequence of the sequence {bk}.{bk}. Prove that {cj}{cj} is a subsequence of {an}.{an}. What is the strictly increasing sequence {mj}{mj} of natural numbers for which cj=amj?cj=amj?

Here is an interesting generalization of the notion of the limit of a sequence.

Definition 2:

Let {an}{an} be a sequence of real or complex numbers. A number xx is called a cluster point of the sequence {an}{an} if there exists a subsequence {bk}{bk} of {an}{an} such that x=limbk.x=limbk. The set of all cluster points of a sequence {an}{an} is called the cluster set of the sequence.

Exercise 2

  1. Give an example of a sequence whose cluster set contains two points. Give an example of a sequence whose cluster set contains exactly nn points. Can you think of a sequence whose cluster set is infinite?
  2. Let {an}{an} be a sequence with cluster set S.S. What is the cluster set for the sequence {-an}?{-an}? What is the cluster set for the sequence {an2}?{an2}?
  3. If {bn}{bn} is a sequence for which b=limbn,b=limbn, and {an}{an} is another sequence, what is the cluster set of the sequence {anbn}?{anbn}?
  4. Give an example of a sequence whose cluster set is empty.
  5. Show that if the sequence {an}{an} is bounded above, then the cluster set SS is bounded above. Show also that if {an}{an} is bounded below, then SS is bounded below.
  6. Give an example of a sequence whose cluster set SS is bounded above but not bounded below.
  7. Give an example of a sequence that is not bounded, and which has exactly one cluster point.

Theorem 1

Suppose {an}{an} is a sequence of real or complex numbers.

  1. (Uniqueness of limits) Suppose liman=L,liman=L, and liman=M.liman=M. Then L=M.L=M. That is, if the limit of a sequence exists, it is unique.
  2. If L=liman,L=liman, and if {bk}{bk} is a subsequence of {an},{an}, then the sequence {bk}{bk} is convergent, and limbk=L.limbk=L. That is, if a sequence has a limit, then every subsequence is convergent and converges to that same limit.

Proof

Suppose liman=Landliman=M.liman=Landliman=M. Let ϵϵ be a positive number, and choose N1N1 so that |an-L|<ϵ/2|an-L|<ϵ/2 if n≥N1,n≥N1, and choose N2N2 so that |an-M|<ϵ/2|an-M|<ϵ/2 if n≥N2.n≥N2. Choose an nn larger than both N1andN2.N1andN2. Then

| L - M | = | L - a n + a n - M | ≤ | L - a n | + | a n - M | < ϵ . | L - M | = | L - a n + a n - M | ≤ | L - a n | + | a n - M | < ϵ .
(1)

Therefore, since |L-M|<ϵ|L-M|<ϵ for every positive ϵ,ϵ, it follows that L-M=0L-M=0 or L=M.L=M. This proves part (1).

Next, suppose liman=Lliman=L and let {bk}{bk} be a subsequence of {an}.{an}. We wish to show that limbk=L.limbk=L. Let ϵ>0ϵ>0 be given, and choose an NN such that |an-L|<ϵ|an-L|<ϵ if n≥N.n≥N. Choose a KK so that nK≥NnK≥N. (How?) Then, if k≥K,k≥K, we have nk≥nK≥N,nk≥nK≥N, whence |bk-L|=|ank-L|<ϵ,|bk-L|=|ank-L|<ϵ, which shows that limbk=L.limbk=L. This proves part (2).

REMARK The preceding theorem has the following interpretation. It says that if a sequence converges to a number L,L, then the cluster set of the sequence contains only one number, and that number is L.L. Indeed, if xx is a cluster point of the sequence, then there must be some subsequence that converges to x.x. But, by part (2), every subsequence converges to L.L. Then, by part (1), x=L.x=L. Part (g) of Exercise 2 shows that the converse of this theorem is not valid. that is, the cluster set may contain only one point, and yet the sequence is not convergent.

We give next what is probably the most useful fundamental result about sequences, the Bolzano-Weierstrass Theorem. It is this theorem that will enable us to derive many of the important properties of continuity, differentiability, and integrability.

Theorem 2: Bolzano-Weierstrass

Every bounded sequence {an}{an} of real or complex numbers has a cluster point. In other words, every bounded sequence has a convergent subsequence.

The Bolzano-Weierstrass Theorem is, perhaps not surprisingly, a very difficult theorem to prove. We begin with a technical, but very helpful, lemma.

Lemma 1

Let {an}{an} be a bounded sequence of real numbers; i.e., assume that there exists an MM such that |an|≤M|an|≤M for all n.n. For each n≥1,n≥1, let SnSn be the set whose elements are {an,an+1,an+2,...}.{an,an+1,an+2,...}. That is, SnSn is just the elements of the tail of the sequence from nn on. Define xn=supSn=supk≥nak.xn=supSn=supk≥nak. Then

  1. The sequence {xn}{xn} is bounded (above and below).
  2. The sequence {xn}{xn} is non-increasing.
  3. The sequence {xn}{xn} converges to a number x.x.
  4. The limit xx of the sequence {xn}{xn} is a cluster point of the sequence {an}.{an}. That is, there exists a subsequence {bk}{bk} of the sequence {an}{an} that converges to x.x.
  5. If yy is any cluster point of the sequence {an},{an}, then y≤x,y≤x, where xx is the cluster point of part (4). That is, xx is the maximum of all cluster points of the sequence {an}.{an}.

Proof

Since xnxn is the supremum of the set Sn,Sn, and since each element of that set is bounded between -M-M and M,M, part (1) is immediate.

Since Sn+1⊆Sn,Sn+1⊆Sn, it is clear that

x n + 1 = sup S n + 1 ≤ sup S n = x n , x n + 1 = sup S n + 1 ≤ sup S n = x n ,
(2)

showing part (2).

The fact that the sequence {xn}{xn} converges to a number xx is then a consequence of (Reference).

We have to show that the limit xx of the sequence {xn}{xn} is a cluster point of {an}.{an}. Notice that {xn}{xn} may not itself be a subsequence of {an},{an}, each xnxn may or may not be one of the numbers ak,ak, so that there really is something to prove. In fact, this is the hard part of this lemma. To finish the proof of part (4), we must define an increasing sequence {nk}{nk} of natural numbers for which the corresponding subsequence {bk}={ank}{bk}={ank} of {an}{an} converges to x.x. We will choose these natural numbers {nk}{nk} so that |x-ank|<1/k.|x-ank|<1/k. Once we have accomplished this, the fact that the corresponding subsequence {ank}{ank} converges to xx will be clear. We choose the nknk's inductively. First, using the fact that x=limxn,x=limxn, choose an nn so that |xn-x|=xn-x<1/1.|xn-x|=xn-x<1/1. Then, because xn=supSn,xn=supSn, we may choose by (Reference) some m≥nm≥n such that xn≥am>xn-1/1.xn≥am>xn-1/1. But then |am-x|<1/1.|am-x|<1/1. (Why?) This mm we call n1.n1. We have that |an1-x|<1/1.|an1-x|<1/1.

Next, again using the fact that x=limxn,x=limxn, choose another nn so that n>n1n>n1 and so that |xn-x|=xn-x<1/2.|xn-x|=xn-x<1/2. Then, since this xn=supSn,xn=supSn, we may choose another m≥nm≥n such that xn≥am>xn-1/2.xn≥am>xn-1/2. This mm we call n2.n2. Note that we have |an2-x|<1/2.|an2-x|<1/2.

Arguing by induction, if we have found an increasing set n1<n2<...<nj,n1<n2<...<nj, for which |ani-x|<1/i|ani-x|<1/i for 1≤i≤j,1≤i≤j, choose an nn larger than njnj such that |xn-x|<1/(j+1).|xn-x|<1/(j+1). Then, since xn=supSn,xn=supSn, choose an m≥nm≥n so that xn≥am>xn-1/(j+1).xn≥am>xn-1/(j+1). Then |am-x|<1/(j+1)|am-x|<1/(j+1), and we let nj+1nj+1 be this m.m. It follows that |anj+1-x|<1/(j+1).|anj+1-x|<1/(j+1).

So, by recursive definition, we have constructed a subsequence of {an}{an} that converges to x,x, and this completes the proof of part (4) of the lemma.

Finally, if yy is any cluster point of {an},{an}, and if y=limank,y=limank, then nk≥k,nk≥k, and so ank≤xk,ank≤xk, implying that xk-ank≥0.xk-ank≥0. Hence, taking limits on k,k, we see that x-y≥0,x-y≥0, and this proves part (5).

Now, using the lemma, we can give the proof of the Bolzano-Weierstrass Theorem.

Proof

If {an}{an} is a sequence of real numbers, this theorem is an immediate consequence of part (4) of the preceding lemma.

If an=bn+cnian=bn+cni is a sequence of complex numbers, and if {an}{an} is bounded, then {bn}{bn} and {cn}{cn} are both bounded sequences of real numbers. See (Reference). So, by the preceding paragraph, there exists a subsequence {bnk}{bnk} of {bn}{bn} that converges to a real number b.b. Now, the subsequence {cnk}{cnk} is itself a bounded sequence of real numbers, so there is a subsequence {cnkj}{cnkj} that converges to a real number c.c. By part (2) of Theorem 1, we also have that the subsequence {bnkj}{bnkj} converges to b.b. So the subsequence {ankj}={bnkj+cnkji}{ankj}={bnkj+cnkji} of {an}{an} converges to the complex number b+ci;b+ci; i.e., {an}{an} has a cluster point. This completes the proof.

There is an important result that is analogous to the Lemma above, and its proof is easily adapted from the proof of that lemma.

Exercise 3

Let {an}{an} be a bounded sequence of real numbers. Define a sequence {yn}{yn} by yn=infk≥nak.yn=infk≥nak. Prove that:

  1. {yn}{yn} is nondecreasing and bounded above.
  2. y=limyny=limyn is a cluster point of {an}.{an}.
  3. If zz is any cluster point of {an},{an}, then y≤z.y≤z. That is, yy is the minimum of all the cluster points of the sequence {an}.{an}. HINT: Let {αn}={-an},{αn}={-an}, and apply the preceding lemma to {αn}.{αn}. This exercise will then follow from that.

The Bolzano-Wierstrass Theorem shows that the cluster set of a bounded sequence {an}{an} is nonempty. It is also a bounded set itself.

The following definition is only for sequences of real numbers. However, like the Bolzano-Weierstrass Theorem, it is of very basic importance and will be used several times in the sequel.

Definition 3:

Let {an}{an} be a sequence of real numbers and let SS denote its cluster set.

If SS is nonempty and bounded above, we define lim supanlim supan to be the supremum supSsupS of S.S.

If SS is nonempty and bounded below, we define lim infanlim infan to be the infimum infSinfS of S.S.

If the sequence {an}{an} of real numbers is not bounded above, we define lim supanlim supan to be ∞,∞, and if {an}{an} is not bounded below, we define lim infanlim infan to be -∞.-∞.

If {an}{an} diverges to ∞,∞, then we define lim supanlim supan and lim infanlim infan both to be ∞.∞. And, if {an}{an} diverges to -∞,-∞, we define lim supanlim supan and lim infanlim infan both to be -∞.-∞.

We call lim supanlim supan the limit superior of the sequence {an}{an}, and lim infanlim infan the limit inferior of {an}.{an}.

Exercise 4

  1. Suppose {an}{an} is a bounded sequence of real numbers. Prove that the sequence {xn}{xn} of the lemma following Theorem 2 converges to lim supan.lim supan. Show also that the sequence {yn}{yn} of Exercise 3 converges to lim infan.lim infan.
  2. Let {an}{an} be a not necessarily bounded sequence of real numbers. Prove that
    lim supan=infnsupk≥nak=limnsupk≥nak.lim supan=infnsupk≥nak=limnsupk≥nak.
    (3)
    and
    lim infan=supninfk≥nak=limninfk≥nak.lim infan=supninfk≥nak=limninfk≥nak.
    (4)
    HINT: Check all cases, and use Lemma 1 and Exercise 3.
  3. Let {an}{an} be a sequence of real numbers. Prove that
    lim supan=-lim inf(-an).lim supan=-lim inf(-an).
    (5)
  4. Give examples to show that all four of the following possibilities can happen.
    1.  lim supanlim supan is finite, and lim infan=-∞.lim infan=-∞.
    2.  lim supan=∞lim supan=∞ and lim infanlim infan is finite.
    3.  lim supan=∞lim supan=∞ and lim infan=-∞.lim infan=-∞.
    4. both lim supanlim supan and lim infanlim infan are finite.

The notions of limsup and liminf are perhaps mysterious, and they are in fact difficult to grasp. The previous exercise describes them as the resultof a kind of two-level process, and there are occasions when this description is a great help. However, the limsup and liminf can also be characterized in other ways that are more reminiscent of the definition of a limit. These other ways are indicated in the next exercise.

Exercise 5

Let {an}{an} be a bounded sequence of real numbers with

lim supan=Llim supan=L and lim infan=l.lim infan=l. Prove that LL and ll satisfy the following properties.

  1. For each ϵ>0,ϵ>0, there exists an NN such that an<L+ϵan<L+ϵ for all n≥N.n≥N. HINT: Use the fact that lim supan=Llim supan=L is the number xx of the lemma following Theorem 2.8, and that xx is the limit of a specific sequence {xn}.{xn}.
  2. For each ϵ>0,ϵ>0, and any natural number k,k, there exists a natural number j≥kj≥k such that aj>L-ϵ.aj>L-ϵ. Same hint as for part (a).
  3. For each ϵ>0,ϵ>0, there exists an NN such that an>l-ϵan>l-ϵ for all n≥N.n≥N.
  4. For each ϵ>0,ϵ>0, and any natural number k,k, there exists a natural number j>kj>k such that aj<l+ϵ.aj<l+ϵ.
  5. Suppose L'L' is a number that satisfies parts (a) and (b). Prove that L'L' is the limsup of {an}.{an}. HINT: Use part (a) to show that L'L' is greater than or equal to every cluster point of {an}.{an}. Then use part (b) to show that L'L' is less than or equal to some cluster point.
  6. If l'l' is any number that satisfies parts (c) and (d), show that l'l' is the liminf of the sequence {an}.{an}.

Exercise 6

  1. Let {an}{an} and {bn}{bn} be two bounded sequences of real numbers, and write L=lim supanL=lim supan and M=lim supbn.M=lim supbn. Prove that lim sup(an+bn)≤lim supan+lim supbn.lim sup(an+bn)≤lim supan+lim supbn. HINT: Using part (a) of the preceding exercise, show that for every ϵ>0ϵ>0 there exists a NN such that an+bn<L+M+ϵan+bn<L+M+ϵ for all n≥N,n≥N, and conclude from this that every cluster point yy of the sequence {an+bn}{an+bn} is less than or equal to L+M.L+M. This will finish the proof, since lim sup(an+bn)lim sup(an+bn) is a cluster point of that sequence.
  2. Again, let {an}{an} and {bn}{bn} be two bounded sequences of real numbers, and write l=lim infanl=lim infan and m=lim infbn.m=lim infbn. Prove that lim inf(an+bn)≥lim infan+lim infbn.lim inf(an+bn)≥lim infan+lim infbn. HINT: Use part (c) of the previous exercise.
  3. Find examples of sequences {an}{an} and {bn}{bn} for which lim supan=lim supbn=1,lim supan=lim supbn=1, but lim sup(an+bn)=0.lim sup(an+bn)=0.

We introduce next another property that a sequence can possess. It looks very like the definition of a convergent sequence, but it differs in a crucial way, and that is that this definition only concerns the elements of the sequence {an}{an} and not the limit L.L.

Definition 4:

A sequence {an}{an} of real or complex numbers is a Cauchy sequence if for every ϵ>0,ϵ>0, there exists a natural number NN such that if n≥Nn≥N and m≥Nm≥N then |an-am|<ϵ.|an-am|<ϵ.

REMARK No doubt, this definition has something to do with limits. Any time there is a positive ϵϵ and an N,N, we must be near some kind of limit notion. The point of the definition of a Cauchy sequence is that there is no explicit mention of what the limit is. It isn't that the terms of the sequence are getting closer and closer to some number L,L, it's that the terms of the sequence are getting closer and closer to each other. This subtle difference is worth some thought.

Exercise 7

Prove that a Cauchy sequence is bounded. (Try to adjust the proof of (Reference) to work for this situation.)

The next theorem, like the Bolzano-Weierstrass Theorem, seems to be quite abstract, but it also turns out to be a very useful tool for proving theorems about continity, differentiability, etc. In the proof, the completeness of the set of real numbers will be crucial. This theorem is not true in ordered fields that are not complete.

Theorem 3: Cauchy Criterion

A sequence {an}{an} of real or complex numbers is convergent if and only if it is a Cauchy sequence.

Proof

If liman=aliman=a then given ϵ>0,ϵ>0, choose NN so that |ak-a|<ϵ/2|ak-a|<ϵ/2 if k≥N.k≥N. From the triangle inequality, and by adding and subtracting a,a, we obtain that |an-am|<ϵ|an-am|<ϵ if n≥Nn≥N and m≥N.m≥N. Hence, if {an}{an} is convergent, then {an}{an} is a Cauchy sequence.

Conversely, if {an}{an} is a cauchy sequence, then {an}{an} is bounded by the previous exercise. Now we use the fact that {an}{an} is a sequence of real or complex numbers. Let xx be a cluster point of {an}.{an}. We know that one exists by the Bolzano-Weierstrass Theorem. Let us show that in fact this number xx not only is a cluster point but that it is in fact the limit of the sequence {an}.{an}. Given ϵ>0,ϵ>0, choose NNso that |an-am|<ϵ/2|an-am|<ϵ/2 whenever both nn and m≥N.m≥N. Let {ank}{ank} be a subsequence of {an}{an} that converges to x.x. Because {nk}{nk} is strictly increasing, we may choose a kk so that nk>Nnk>N and also so that |ank-x|<ϵ/2.|ank-x|<ϵ/2. Then, if n≥N,n≥N, then both nn and this particular nknk are larger than or equal to N.N. Therefore, |an-x|≤|an-ank|+|ank-x|<ϵ.|an-x|≤|an-ank|+|ank-x|<ϵ. this completes the proof that x=liman.x=liman.

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