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18.100A PS9

Author: Eric Emer

March 13, 2013

Reading 6.5-7.7Collaborators: None.

1 Exercise 6.5 #1

1.1 a

S = {sin n/6 : n Z}

max S = sup S = 1

min S = infS = 1

1.2 b

S = {(cos n)/n : n N}

max S = sup S = 1/2 (max when n = 2)

min S = infS = 1 (min when n = 1)

1.3 c

S =

1

n+ cos

n

2: n N

max S = sup S = 5/4 (max when n = 4)

infS = 1

1.4 d

S =

n2

n

: n N

max S = sup S = 1/2 (max when n = 1, 2)

infS = 0

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2 Exercise 6.5 #3

2.1 a

By the Completeness Property for sets, we know that sup A and sup B exist. By definition, for alla A, a sup A. Also, for all b B, b sup B. Because A B, this means that also, for all

a A, a sup B. Transitively, a sup A sup B. So, A B = sup A sup B.

2.2 b

By the Completeness Property for sets, we know that infA and infB exist. By definition, for alla A, a infA. Also, for all b B, b infB. Because A B, this means that also, for alla A, a infB. So by definition of infimum, infA infB. So, A B = infA infB.

2.3 c

We let x cA. We notice that for some a A, x = ca. For all a A, a = sup A, by definitionof least upper bound. By simple algebra, it must also be that ca c. Consider least upper bound

k = sup cA for cA. Then it must be that x k for all x cA. Consider again the case wherex = ca, then ca k, and a k

c. We know that is the least upper bound of A, so it must be that

k/c. Concluding that c k. So c sup A is the least upper bound as well. The supremum ofa set is unique, so it must be that c sup A = sup cA.

2.4 d

We let x cA. We notice that for some a A, x = ca. For all a A, a = infA, by definitionof greatest lower bound. By simple algebra, it must also be that ca c. Consider greatest lowerbound k = infcA for cA. Then it must be that x k for all x cA. Consider again the case wherex = ca, then ca k, and a k

c. We know that is the greatest lower bound of A, so it must be

that k/c. Concluding that c k. So c infA is the greatest lower bound as well. The infimumof a set is unique, so it must be that c infA = infcA.

2.5 e

Allow = infA. Then by definition a for all a A. By sign-change law, it also holds that a for all a A. Now we can apply the definition of an upper bound, and see that is an upper bound for the set A. We define k as any upper bound on A. Then it would holdthat k a for all a A. Also by the sign-change law, k a for all a A. Because is thegreatest lower bound of A, it must also be that k . By the sign-change law, k . Sincek was any upper bound, this means that is the least upper bound. We know from our designthat = infA, and we conclude from the conditions and the definition of supremum as lowest

upper bound that is the is the supremum of A. So sup(A) = infA.

2.6 f

Allow = sup A. Then by definition a for all a A. By sign-change law, it also holds that a for all a A. Now we can apply the definition of a lower bound, and see that isa lower bound for the set A. We define k as any lower bound on A. Then it would hold thatk a for all a A. Also by the sign-change law, k a for all a A. Because is the leastupper bound of A, it must also be that k . By the sign-change law, k . Since k was

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any lower bound, this means that is the greatest lower bound. We know from our design that = sup A, and we conclude from the conditions and the definition of infimum as greatest lowerbound that is the is the infimum of A. So inf (A) = sup A.

2.7 g

We examine the definition of supremum, and see that for all a A, a sup A. Similarly, for allb B, b sup B. Therefore, we can add these terms, to show that a + b sup A + sup Bfor all a A, b B. This happens to fulfill the definition of an upper bound, so thereforesup(A + B) sup A + sup B.

2.8 h

From (g) we know thatsup(A B) sup(A) + sup (B)

and from (e)sup(A) = infA

Combiningsup((A + B)) infA infB

inf (A + B) infA infB

inf (A + B) infA + infB

3 Problem 6-2

3.1 a

We know m = sup S. Consider, bn = m 1

nfor all n N. By sup-2, this is not an upper bound

on S, since it is less than m, the least upper bound. Therefore, we conclude that there exists anan S such that m

1

n< an m. Because limn(m

1

n) = m, for this an, it is clear from the

Squeeze Theorem that as n , so an m.

3.2 b

We know from part (g) of the previous problem that:

sup(A + B) sup A + sup B

We examine the definition of supremum, and see that for all a A, a sup A. Similarly, forall b B, b sup B. Therefore, we can add these terms, to show that a + b sup A + sup B for

all a A, b B. However, by definition of supremum, we see that ifa + b has a least upper bound,that least upper bound is sup (A + B). Therefore a + b sup(A + B) for all a A, b B. Becausethe inequality holds for all b B, we can also say from definition of least upper bound/supremumthat a + sup B sup(A + B). Similarly, the inequality holds for all a A, so we can say fromdefinition of least upper bound that sup A + sup B sup(A + B). Because we have proven theinequality in both directions, sup (A + B) = sup A + sup B.

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4 Problem 7-7

4.1 a

c2k =1

2, and c2k+1 =

1

2+

1

2(k + 1)

lim c2k = lim c2k+1 = C = 12

4.2 b

The Cesaro Sum in terms of the ak term,

cn =

a1 + . . . +

an

n=

n

k=1

ak

n

4.3 c

We consider problem 3-1b, where

bn = (a1 + . . . + an)/n

We can show that an L = bn L.

Proof. We wish to show that if an L then bn L. Recall:

bn =a1 + a2 + . . . + an

n

We know that if an L, then an L 0. Therefore, we examine:

limn

(a1 L) + (a2 L) + . . . + (an L)

n = 0

We would expect that the above equation would go to zero because the numerator would go to zero(we know that an L 0). We can transform the equation:

limn

(a1 L) + (a2 L) + . . . + (an L)

n= lim

n(bn L) = 0

Now that we know that bn L 0 as n , we can conclude by addition that bn L asn .

Proof. In the Cesaro Summation, we now suppose that for sn =

an, we have that sn C. Weconsider,

cn =s1 + . . . + sn

nand so, as above, sn C = cn C.

Proof. For a convergent

an, we have that as n , so

an C. Therefore,

lim cn = lims1 + . . . + sn

n= lim

n

ann

= limnC

n= lim C = C =

an

So the Cesaro Sum is the same as the regular sum.

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