subsequences - qmplus.qmul.ac.uk

Subsequences

Intuitively and informally, a subsequence of (xn)1n=1 = x1, x2, x3, x4, . . . is asequence like x1, x3, x5, x7, . . . or x2, x3, x5, x7, x11, . . .. We take just some

of the terms of the original sequence but we take infinitely many of themand we take them in the same order as the original sequence. In theexamples above, we took the xn where n is odd or the ones where n is aprime, but in general there does not have to be such a rule!

Definition 2.31

A subsequence of (xn)1n=1 is a sequence xr1 , xr2 , xr3 , . . . where rj 2 N foreach j 2 N and rj+1 > rj for each j 2 N. We also denote this as (xrj )

1j=1.

MTH5104 Convergence and Continuity October 25, 2021 128 / 264

Examples 2.32

Example (i). x2, x4, x6, . . . is a subsequence of (xn)1n=1. Here, we tookr1 = 2, r2 = 4, r3 = 6, etc. So rn = 2n and the subsequence can bewritten as (x2n)1n=1.

Example (ii). x1, x4, x9, x16, . . . is another subsequence. Here we havechosen rn = n

2, so the subsequence can be expressed as (xj2)1j=1.

Very often a sequence which does not converge has subsequences whichdo converge! For example xn = (�1)n. Here x1, x3, x5, . . . is a subsequencewhich converges to �1 (in fact all the terms are �1) and x2, x4, x6, . . . is asubsequence which converges to +1 (in fact all the terms are +1).


jjI I

i. i

Bolzano-Weierstrass, Version 1

Theorem 2.33

Every bounded sequence contains a convergent subsequence.

Proof. By Theorem 2.26 (“increasing bounded sequences converge”) andits analogue for decreasing sequences, it will be enough to prove that everysequence of real numbers has a monotonic subsequence.Suppose we are given a sequence (xn)1n=1. There are two possibilities:

Case 1: For every N 2 N the set {xn : n � N} has a maximum.

Case 2: For some N 2 N the set {xn : n � N} does not have amaximum.


Bolzano-Weierstrass (continued)

If we are in Case 1,

Case 1: For every N 2 N the set {xn : n � N} has a maximum.


these are diminishing sets

let " bend that "rim-"✗ {"i ""-Let rz be the smallest integer such thatrn> rn and Kr

,= max {xn : u > re}←

Let rs be the smallest integer such that

rzsrz and Kr,= max {xu : n > rz}←

- - -

and so on.

Note that Kr,2 ✗re? Kr

,? - - -- i

Bolzano-Weierstrass (continued)

If we are in Case 2

Case 2: For some N 2 N the set {xn : n � N} does not have amaximum.

This proof is not required knowledge for the final examination.


Let r,be N

.

Since Xp,is not the maximum , we choose r,

to be the smallest number wilt rn > r, and xrixr;since xr

,is not the maximum , we choose r,

to be smallest s.t.rs > re and Kris scrz .

By construction : Er, < Kr, < scr, < - -.

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Accumulation point

Definition 2.34

A real number x 2 R is called an accumulation point of a sequence (xn) iffor every " > 0 there are infinitely many elements of the sequence whichlie "-close to x (i.e. satisfy |xn � x | < "). This means x is an accumulationpoint of (xn) if and only if

8" > 0 8N 2 N 9n > N : |xn � x | < ". (10)


-"

for infinitely many n c- IN"

Limit is an accumulation point

Lemma 2.35

If (xn) converges to x 2 R, then x is an accumulation point.

Proof. We need to prove (10). So we are given " > 0 and N 2 N (pickedby the Demon) and we need to find n > N such that |xn � x | < ".

Since xn ! x as n ! 1, we know by definition of convergence that forthe given " > 0, there exists some N such that 8n > N : |xn � x | < ".(Here we relabelled N to N in the definition of convergence, because N isalready used with a di↵erent meaning.)

Now, we can pick n := max{N, N}+ 1. This choice of n is allowedbecause n > N. From n > N, we conclude that |xn � x | < ", which iswhat we wanted to show.


In general, a sequence can have an accumulation point even if it does notconverge and it can have several accumulation points.

Example 2.36

Let xn = (�1)n. Then we know that (xn) does not converge. But bothx = +1 and x

0 = �1 are accumulation points for (xn).

Check. For x = +1: Given any " > 0 and any N 2 N, let n := 2N so thatn > N and n even. Then |xn � 1| = 0 < ".For x 0 = �1: Given any " > 0 and any N 2 N, let n := 2N + 1 so thatn > N and n odd. Then |xn � (�1)| = 0 < ".

We also know that the subsequence (x2n)1n=1 converges to +1 and thesubsequence (x2n+1)1n=1 converges to �1. So both accumulation pointsare actually limits of subsequences. This is true in general. . .


Convergent subsequences

Lemma 2.37

The real number x 2 R is an accumulation point of (xn)1n=1 if and only if

there is a subsequence (xrn)1n=1 converging to x .

Proof. Suppose there is a subsequence (xrj )1j=1 converging to x . Then for

any " > 0, there is J 2 N such that all the elements of the subsequence(xrj ) with j > J satisfy |xrj � x | < ". In particular, there are infinitelymany. (One can write this more formally, as we have done in the proof ofLemma 2.35.)


Proof (continued). Conversely, if x is an accumulation point, then wecan construct a subsequence converging to x as follows:

From (10), for " = 1 and N = 1, we can find r1 > 1 such that|xr1 � x | < " = 1. Then, using (10) again, this time with " = 1/2 andN = r1, we can find r2 > r1 with |xr2 � x | < " = 1/2. We then iterate this,i.e. if xrj�1 is already constructed, we use (10) with " = 1

j and N = rj�1 to

find some rj > rj�1 with |xrj � x | < 1j .

Clearly, the subsequence (xrj )1j=1 then converges to x .


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Bolzano-Weierstrass, Version 2

We can now restate Theorem 2.33.

Theorem 2.38

Every bounded sequence of real numbers has an accumulation point.

Proof. This follows by combining the first version of the theorem withLemma 2.37.


(sink 1)I.,

Accumulation pts : -1,0 , + I

suppose Cxn7n? , is a bounded sequence .

Then by them 2-33, there is a convergent subsequence

By Lemma 2.37 , there is an accumulation point ,

Cauchy Sequences

How can we prove that a sequence is convergent if we don’t have acandidate for the limit?

Lemma 2.39

Suppose that (xn)1n=1 converges to some x 2 R. Then the sequence

(yn)1n=1 defined by yn = xn+1 � xn converges to zero.

Proof. We first observe that limn!1 xn+1 = limn!1 xn = x . (This is“obvious”, but we should still make sure we can formally prove it! See theCoursework.)

Then, using Theorem 2.19 (ii), we have

limn!1

yn = limn!1

xn+1 � limn!1

xn = x � x = 0.


Genta )n% = (xnÑ=e

"hi:b (xn-u-x.nl

"

Examples 2.40

Example (i). Let xn = n�1n . Then we know that xn ! 1 as n ! 1. So

we obtain yn = xn+1 � xn = nn+1 � n�1

n ! 0 as n ! 1.

Example (ii). Let xn = (�1)n. Then |yn| = |xn+1 � xn| = 2 for all n 2 Nand hence (yn) does not converge to zero. By Lemma 2.39, (xn) can thusnot converge to any real number x 2 R (as we have already seen before).


"

ñ-¥¥"=n¥, ← In→ 0 In→a)

Yn = Kate - Rn = {- l - l if u even1 + 1 if nodd

⇒ lyn 1=2 .

The converse of Lemma 2.39 is false!

Examples 2.41

Example (i). Let xn =pn. Then the sequence (xn) does not converge to

any real number. But the sequence yn = xn+1 � xn =pn + 1�

pn does

converge to zero.


Note true - F) (Fei + Tn)= (ntl ) - n = I

⇒ ont - rn = ¥+rñ £ In → 0

In→a)

Example (ii). Let xn = 1 + 12 + 1

3 + 14 + . . .+ 1

n =Pn

k=11k . This

sequence does not converge (we have actually already seen this in Chapter0, but will see it again more precisely below). Neverthelessyn = xn+1 � xn = 1

n+1 does converge to zero.

This means that for (xn)1n=1 to converge to some x 2 R it is necessarythat the sequence (xn+1 � xn)1n=1 converges to zero, but this condition isnot su�cient.

The next definition gives a necessary and su�cient condition.


Definition of Cauchy sequence

Lemma 2.39 says that if (xn) converges to some x 2 R, then

8" > 0 9N 2 N 8n > N : |xn+1 � xn| < ".

(This is nothing else than the definition of yn = xn+1 � xn ! 0 forn ! 1.) The following definition strengthens this idea, testing not onlythe distance between xn and xn+1 (for n > N) but actually all thedistances between xn to xm for n and m both greater than N.

Definition 2.42 (Cauchy sequence)

A sequence (xn)1n=1 is called a Cauchy sequence if and only if

8" > 0 9N 2 N 8n > N 8m > N : |xm � xn| < ". (11)


Equivalence of Cauchy sequences and convergent

sequences

Theorem 2.43

Let (xn)1n=1 be a sequence of real numbers. Then (xn)1n=1 converges if and

only if it is a Cauchy sequence.

Proof. We first show: Convergent sequences are Cauchy sequences.Assume that there is some x 2 R such that xn ! x as n ! 1. We haveto prove that (xn) is a Cauchy sequence. So, we are given " > 0 (by theDemon). Then by the definition of convergence, letting " = "

2 , we can findsome N such that for all n > N we have |xn � x | < " = "

2 . We pick exactlythis N. Then for all n,m > N, we have

|xm � xn| |xm � x |+ |x � xn| < "+ " = "2 + "

2 = ".


Proof (continued). Now we prove: Cauchy sequences do converge. Wedo this in several steps.

Step 1. Cauchy sequences are bounded: For " = 1 (in the definition of aCauchy sequence), we can find N1 such that for all n,m > N1 we have|xm � xn| < 1. In particular, letting n = N1 + 1 be fixed, we find that forall m > N1 we have |xm| < |xN1+1|+ 1. Thus the sequence is bounded:

|xm| < M = max{|x1|, |x2|, . . . , |xN1 |, |xN1+1|+ 1} for all m 2 N.

Step 2. We can find a convergent subsequence: Indeed, this is just anapplication of the Theorem of Bolzano-Weierstrass (Theorem 2.33). So inparticular, we have a subsequence (xrj )

1j=1 and some number x 2 R, such

that xrj ! x as j ! 1.


-

Assume Gen /% ,is Cauchy

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Proof (continued).

Step 3. The whole sequence converges to x : As xrj ! x , by definition ofconvergence there is some N2 such that 8rj > N2 : |xrj � x | < " = "

2 .

Moreover, by the definition of a Cauchy sequence, there is some N3 suchthat 8m, n > N3 : |xm � xn| < " = "

2 . Thus for anyn > N := max{N2,N3}, we can pick some r` > N and estimate

|xn � x | |xn � xr` |+ |xr` � x | < "+ " ="

2+

"

2= ".

So xn ! x as n ! 1.

This proof is non-examinable.


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Remarks

While this theorem seems hard at first look, Cauchy sequences actuallyhave a lot of advantages:

We have a condition that applies to all sequences (not just monotonicones) and tells us whether the sequence converges or not.

The statement is “if and only if”, so the Cauchy criterion is necessaryand su�cient.

The theorem applies to sequences in Rn and C and Cn, not just to R.(In Rn, by |xm � xn| we mean the distance between xm and xn.)

Finally, there is an easier proof of the Bolzano-Weierstrass Theoremusing Cauchy sequences and it works in Rn, not just in R.


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