2.3.1 Definition. A subsequenceof a sequence{an}∞n=1 inRis a sequence {ank}∞k=1, where the indices satisfy 1≤n1 < n2<ã ã ã. The limit in Rof a subsequence is called acluster point of {an}. ♦ For example, in the following sequence the underlined terms define the beginning of a subsequence{ank} withn1= 3,n2= 4,n3= 6, etc.
a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a12, a13, a14, a15, . . . Note that the indicesnk of a subsequence satisfynk≥k.
Examples. (a)The sequence nh1−(−1)b(n−1)/2cio
={0,0,2,2,0,0,2,2. . .}
is a subsequence of
{1−(−1)n}={2,0,2,0, . . .}, which has cluster points 0 and 2.
(b)The sequence{nsin (nπ/2)} has cluster points 0 and±∞.
(c) Let{r1, r2, . . .} be an arbitrary enumeration of the rational numbers (see Appendix A). Then every real number is a cluster point of{rn}. Indeed,
since every interval of the form (x−1/n, x+ 1/n) contains infinitely many terms of the sequence, we may choosen1≥1 such that |x−rn1|<1, n2> n1 such that |x−rn2|<1/2, etc. In this way we may construct a subsequence inductively such that|x−rnk|<1/kfor allk, hencernk→x. ♦ Notation. It is occasionally convenient to use the following alternate method to describe a subsequence: If we setbk =ank and then change the index in {bk}∞k=1 ton, then {bn}may be used to denote the subsequence {ank}. This provides a convenient way to denote a subsequence of a subsequence. In this regard, note that if{cn} is a subsequence of{bn} and{bn} is a subsequence of{an}, then{cn} is a subsequence of{an}.
The following proposition shows that a convergent sequence has a single cluster point.
2.3.2 Proposition. If{an}is a sequence inRandan→a∈R, thenank →a for any subsequence{ank} of{an}.
Proof. We prove the proposition for the casea∈Rand leave the other cases for the reader. Givenε >0, chooseN such that|an−a|< ε for alln≥N. Sincenk≥k,|ank−a|< εfor allk≥N. Therefore,ank →a.
Numerical Sequences 39 2.3.3 Example. We calculate limn(1 + 1/n2)3n2+5 by writing
1 + 1
n2 3n2+5
=
"
1 + 1
n2 n2#3
1 + 1
n2 5
.
The term in the square brackets is a subsequence of (1 + 1/n)n and hence converges toe(see 2.2.4). It follows that (1 + 1/n2)3n2+5→e3. ♦ The following result will have important consequences in later chapters.
2.3.4 Bolzano–Weierstrass Theorem. Every bounded sequence in R has a convergent subsequence.
Proof. The proof is based on the observation that if a union of two sets contains infinitely many terms of a sequence, then at least one of the sets must contain infinitely many of the terms of the sequence.
Let {an} be a bounded sequence, sayc0 ≤an ≤d0 for all n. Bisect the interval I0 := [c0, d0]. By the preceding observation, one of the resulting subintervals, call itI1, contains infinitely many terms of the sequence. Choose one such term, sayan1. Now bisectI1. Again, one of the resulting subintervals, call itI2, contains infinitely many terms of the sequence. Choose one such term an2 withn2 > n1. By repeating this procedure, we produce a subsequence {ank}∞k=1 of{an}and a sequence of intervalsIk= [ck, dk], k= 0,1, . . ., such that
c0≤ck−1≤ck ≤ank≤dk≤dk−1≤d0, anddk+1−ck+1= 12(dk−ck). Since{ck}and{dk}are monotone and boundedck→c anddk →dfor some c, d∈R. Sincedk−ck = 2−k(d0−c0)→0,c=d. By the squeeze principle, ank →c.
I0
I1
an1
I3
I2
an2
an3
c1 d1
c2 d2
c3 d3
c0 d0
...
FIGURE 2.3: Interval halving process.
The Bolzano–Weierstrass theorem may be extended as follows:
2.3.5 Theorem. Every sequence inRhas a subsequence that converges in R.
Proof. If{an}is bounded, then the Bolzano–Weierstrass theorem applies. Sup- pose that{an} is unbounded above. Then for eachk∈Nthere exist infinitely many indicesnsuch thatan> k. We may then construct a subsequence{ank} withank> kfor allksoank→+∞.
40 A Course in Real Analysis
2.3.6 Corollary. A sequence {an} inRhas a limit inRiff it has exactly one cluster point inR.
Proof. The necessity is 2.3.2. For the sufficiency, suppose that{an}has exactly one cluster point a ∈ R. Consider first the case a = +∞. We claim that an →+∞. If not, then there exists M ∈Rsuch that an ≤M for infinitely many n, hence there exists a subsequence {ank} of {an} withank ≤M for allk. By 2.3.5,{ank} has a cluster pointb∈R. Butb≤M < a, so{an} has more than one cluster point, a contradiction. Therefore,an→+∞, as claimed.
The case a=−∞is treated similarly.
Now supposea∈R. Thenan→a. If not, then there existsε >0 such that
|an−a| ≥ εfor infinitely many n, so there is a subsequence {ank} of {an} with|ank−a| ≥ε for allk. By 2.3.4,{ank} has a cluster pointb inR. But then|b−a| ≥ε, so again{an} has more than one cluster point.
2.3.7 Definition. A sequence {an} is said to be Cauchy if for each ε >0 there exists an indexN such that|an−am|< εfor allm, n≥N. We express this condition by writing
limm,n(an−am) = 0. ♦
The definition asserts that the terms of a Cauchy sequence get closer to one another. Thus the following result is not surprising.
2.3.8 Proposition. Every convergent sequence is Cauchy.
Proof. Let an →a. Givenε >0, choose N such that |an−a|< ε/2 for all n≥N. Then forn, m≥N,
|an−am|=|(an−a) + (a−am)| ≤ |an−a|+|am−a|< ε.
It is of fundamental importance that the converse of 2.3.8 is true. To prove this, we need the following lemma.
2.3.9 Lemma. A Cauchy sequence is bounded.
Proof. Let{an} be a Cauchy sequence. ChooseN such that|an−am|<1 for allm, n≥N. Then|an| ≤ |an−aN|+|aN|<1 +|aN|for alln≥N, hence
|an| ≤max{1 +|aN|,|a1|,|a2|, . . . ,|aN−1|} for alln.
2.3.10 Cauchy Criterion. Every Cauchy sequence in Rconverges.
Proof. By 2.3.9 and the Bolzano–Weierstrass theorem, a Cauchy sequence {an} has a convergent subsequence, sayank→a∈R. We claim thatan→a. Letε >0 and chooseNsuch that|an−am|< εfor allm, n≥N. In particular,
|an−ank| < ε for n, k ≥N. Fixing n≥N and lettingk → ∞ in the last inequality yields|an−a| ≤ε, verifying the claim.
Numerical Sequences 41 Exercises
1. Find all cluster points of{an}, wherean= (a)S (−1)n
2n+ 1 4n+ 3
sin2nπ 3
. (b) (−1)n
2n+ 1 n+ 5
cos2nπ 4
. (c)S (−1)bn/3c(1 + 1/n)2+ (−1)bn/4c(2 + 1/n)2+ (−1)bn/5c(3 + 1/n)2. (d) (−1)nrn+r2n, where rk is the remainder on division ofkby 3.
2. Construct a sequence with precisely the cluster points 1, 2, 3, +∞. 3. Letk∈N. Use the fact that limn(1 + 1/n)n=e(2.2.4) to find limnan
foran= (a)
1 + 1 kn
n
. (b)
1 + 1 k+n
n
. (c) 1
k +1 n
n . (d)S
1 + 1
2n+k kn
. (e)
1 + 1
3n3+ 5 7n3−4
.
4. Let{an} and{bn} be bounded sequences. Show that there exist conver- gent subsequences of{an}and{bn} with thesameindices.
5.SProve that a sequence contained in a finite set has a constant subsequence.
6. Let −∞< an < r ≤+∞ withan →r. Show that {an} has a strictly increasing subsequence.
7. Show that every sequence of distinct real numbers has a strictly monotone subsequence.
8.S Letk∈Nand suppose that the seriesP∞
n=1|an+k−an|converges (see Chapter 6). Prove that{an}has a convergent subsequence.
9. Let a0, a1 be arbitrary and define an+1= (an+an−1)/2,n≥1. Show directly that{an}is a Cauchy sequence. (Its limit may be found from Exercise 1.5.14.)
10.S Let 0< p≤q andan >0 for all n. Setbn =aqn/(1 +apn). Show that an→0 iff bn→0. Is the assertion true if 0< q < p?
11. LetI be an open interval and let{an}have the property that each open subinterval J ofI containsan for infinitely many n. Prove that every point ofIis a cluster point of{an}. Give an example of such a sequence.
12. Suppose that the cluster points of{an}form a sequence{bn}. Show that every cluster pointbof {bn}is a cluster point of{an}.Hint.Choose a subsequence {bnk} such that|bnk−b|<1/k.
42 A Course in Real Analysis