1.5 Bounded Subsets

Denition 12. 1. A subset S R is bounded above if R such that x x S.

2. A subset S R is bounded below if R such that x x S.3. A subset S R is said to be bounded if it is both bounded above and bounded

below.

Remark. If S = then S is bounded. Here every real number is an upper bound as wellas lower bound.

Examples 13. 1. S = N. Then S is bounded below since 0 or any negative integer isa lower bound. However S is not bounded above.

2. S = {x R | x 1}. Then S is bounded above but not below.3. S = {x R | x2 < 2} Then S is bounded. What are the bounds?

1.6 Supremum and Inmum

Corollary 14. Any nite set is bounded.

Remark. If is a lower bound of a set S then any real number < is a lower boundfor S.

Notation: We write UB(S) for set of all upper bounds of S and LB(S) for set of alllower bounds of S.

Thus above remark can be written as LB(S) and < then LB(S).Similarly, UB(S) and > then UB(S).

Denition 15. A real number M is called supremum of a set S if1. M UB(S)2. M for all UB(S). i.e. M is least among all upper bounds of S.

M is also called as least upper bound OR l.u.b. of S, written as sup(S).

A real number m is called inmum of a set S if1. m LB(S)2. m for all LB(S). i.e. M is greatest among all lower bounds of S.

m is also called as greatest lower bound OR g.l.b. of S, written as inf(S).

Example 16. Let S = {x R | 0 < x 1}. Then sup(S) = 1 and inf(S) = 0.Remark. If sup(S) S then it is also called as maximum of S.

If inf(S) S then it is also called as minimum of S.

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1.7 Completeness Property of Real Numbers

Result 11 (Completeness Property of R). Every non-empty subset of R having an upperbound has a supremum. This is also known as lub property or supremum property.

The analogous property for inma can be deduced as follows:

Corollary 17. Every non-empty subset of R having a lower bound has an inmum.

Proof. Since S has a lower bound, then the set S = {x | x S} has an upper boundand hence by completeness property has a supremum. Now show that if M = sup(S )then M = inf(S).

Here is an easy (and well-known?) application.

Result 12 (Archimedian Property). If x R then there exists n N such that x < n.Corollary 18. If x R then there exists m N such that m < x.Corollary 19. Let S = { 1

n| n N} then inf S = 0.

Here is another application or alternate version of Completeness Property:

Result 13. For any real number x, there exists a sequence of rationals rn such that

limn

rn = x.

1.8 Monotonic Sequences

Denition 20. 1. A sequence (xn) is calledmonotonically increasing sequence if xn+1 xn for all n 1.

2. A sequence (xn) is called monotonically decreasing sequence if xn+1 xn for alln 1.

3. A sequence (xn) is called monotonic sequence if it is either monotonically increasingor monotonically decreasing sequence.

4. A sequence (xn) is called strictly monotonic sequence if the inequality in the de-nition above is strict.

Examples 21. Examples of monotonically increasing sequence:xn = n xn =

1n

Examples of monotonically decreasing sequence:xn =

1n

Corollary 22. A sequence (xn) is monotonically decreasing if and only if (xn) is mono-tonically increasing.

Remark: The sequence (1)n

nis not monotonic.

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1.9 Monotone Convergence Theorems

Result 14 (Monotone Convergence Theorem). A monotonically increasing sequence isconvergent if and only if it is bounded above.

Proof. We have already proved that if a sequence is convergent then it is bounded.Conversely, suppose (xn) is monotonically increasing and bounded sequence. Then

the set S = {xn | n N} is bounded above. Therefore by completeness property, S hasa supremum, say x.

Claim: x = limn

xn.

Let > 0. Then x is not an upper bound of S. Why? Therefore there exists somexk such that x < xk. But (xn) is increasing, therefore xk xn for all n > k. Thus,

x < xk xn x < x+ for all n > k.

Thus |xn x| < for all n > k. Now is arbitrary, therefore (xn) x.Result 15. 1. A monotonically decreasing sequence is convergent if and only if it is

bounded below.

2. A bounded monotonic sequence is convergent.

Examples 23. 1. limn

1n

= 0.

2. xn+1 = xn +1

n+1. Is this convergent? If yes, what is the limit?

Exercise:

1. If x1 = 1 and xn+1 =13(xn + 1) for all n N, then show that (xn) is convergent

sequence and nd it limit.

2. Let x1 = 1 and xn+1 =2xn for all n N then show that (xn) is convergent

sequence and nd its limit.

1.10 Subsequences

Denition 24 (Subsequences). Let X = (xn) be a sequence of real numbers and n1 0 be given. Then there exists N N such that |xnx| < for all n > N .Note that for any k 1, nk k. Prove it! Therefore for k > N , we have nk k > Nand hence |xnk x| < . So xnk converges to x.Corollary 26 (Divergence Criterion). A sequence X = (xn) diverges if either of thefollowing holds:

1. X is unbounded.

2. X has two convergent subsequences whose limits are distinct.

Examples 27. Below are some examples of divergent sequences. Justify the reason ofdivergence for each of them.

1.(1)n 2. 1 + 1

2+ 1

n

3. x1 = 1, x2 = 1, xn =xn1 + xn2

Result 17 (Monotone Subsequence Theorem). Every sequence has a monotonic subse-quence.

Proof. Easy exercise!

Result 18 (Bolzano-Weierstrass Theorem). A bounded sequence has a convergent subse-quence.

Proof. Let X = (xn) be a bounded sequence. Then by Monotone Subsequence Theorem,X has a monotonic subsequence, say X = (xnk). But since X is bounded, X

is alsobounded. Therefore by Monotone Convergence Theorem, X is convergent.

1.11 Cauchy Sequences

Denition 28 (Cauchy Sequences). A sequence (xn) is said to be Cauchy if for every >0 there exists a natural number N such that for allm,n > N , xn, xm satisfy |xnxm| < .

Examples 29. 1. The sequence (xn) =1

nis Cauchy.

2. The sequence (xn) = sin(n2) is not Cauchy.

Lemma 30. Every convergent sequence is Cauchy.

Proof. Let (xn) be a convergent sequence with limn

xn = x. Let > 0. Then N Nsuch that for all n > N , |xn x| < . Therefore,

|xn xm| |xn x|+ |xm x| < + = 2.But is arbitrary therefore (xn) is Cauchy sequence.

Result 19. Every Cauchy sequence is bounded.

Proof. Let (xn) be a Cauchy sequence. Choose = 1. Then there exists N N suchthat |xn xN | < for all n > N . Therefore by Triangle inequality |xn| |xN |+ 1 for alln > N .

LetM = max{|x1|, |x2|, , |xN1|, |xN |+1}. Then |xn| M for all n N and hence(xn) is bounded.

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Result 20 (Cauchy Convergence Criterion). A sequence is convergent if and only ifCauchy.

Outline. We have already seen that a convergent sequence is Cauchy.Let (xn) be a Cauchy sequence. We will prove that (xn) is convergent. We have also

seen that Cauchy bounded. Now applying Bolzano-Weierstrass, we get a convergentsubsequence X = (xnk) x.Exercise:Show that (xn) converges to x.

Exercise:Using Cauchy Convergence Theorem show that the following sequences are convergent.Also nd their limits.

1. xn = 1 +11!+ 1

1!+ 1

2!+ + 1

n!

2. x1 = 1 and xn+1 =1

2+xnfor all n 2.

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