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Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 1


Exercise 1.8

Solution:

(a) Because a<b, we have baP. Similarly, we have dcP. We have(b+d)(a+c)=b+dac by Axiom 6 (a)=(ba)+(dc)P.Hence a+c<b+d.

(b) We give a counterexample. Let F=R. Set a=c=1 and b=d=0, then a<b and c<d. But ac=1>bd=0.


Exercise 1.10

Solution: We remark the following identity1k1k+1=(k+1)kk(k+1)=1k(k+1).Therefore, we havek=1n1k(k+1)=k=1n(1k1k+1)=(112)+(1213)++(1n11n)+(1n1n+1)=1(1212)(1313)(1n1n)1n+1=11n+1=nn+1


Another solution: We prove it by induction on n.

If n=1, the identity is clearly true.

Suppose the identity is true for n, we prove it for n+1. By induction hypothesis, we have(1)k=1n1k(k+1)=nn+1.Then we havek=1n+11k(k+1)=k=1n1k(k+1)+1(n+1)(n+2)use (1)=nn+1+1(n+1)(n+2)=n(n+2)+1(n+1)(n+2)=n2+2n+1(n+1)(n+2)=(n+1)2(n+1)(n+2)=n+1n+2.Hence the identity is true for n+1.

By Mathematical Induction, we are done!


Exercise 1.20

Solution:

(a)Let supA=α and supB=β. Since α<β, we have βα>0. By Theorem 1.24 and setting ϵ=βα, there exists bB such that β(βα)<bα<b.Hence b is an upper bound of A.

(b) We give an example. Let A={1} and B={n/(n+1):nN}. Then supA=supB=1. But no element in B is an upper bound of A since they are all smaller than one.


Exercise 1.21

Solution: Since A is bounded below, there exists M such that Mx for all xA. Hence M is a lower bound of A and MinfA. Let α=infA.

Since A is non-empty, we have an element x in A and hence an element x in A. This implies that A.

Note that Mx for all xA. Multiplying both sides by 1, we obtain Mx for all xA. This implies that My for all yA. Therefore A is bounded above.

We then show that sup(A)=α. We use Theorem 1.24. We have

(a) xα for all xA, and hence xα for all xA. This implies α is an upper bound of A;

(b) for any ϵ>0, there exists xA such that x<αϵ, and thus x>α+ϵ. This implies that α+ϵ is not an upper bound of A.

Therefore, according to Theorem 1.24, we have sup(A)=α=infA.


Exercise 1.22

Solution:


Exercise 1.23

Solution: Let In=(1,1+1/n), then each In is open and non-empty. Moreover,I1I2I3I4.We show that n=1In=.

We argue it by contradiction. Suppose n=1In, then there exist an=1In. Hence aIn for all n, that is (2)1<a<1+1n. Since a>1, we have a1>0. By the Archimedean Principle, we can find mN such that 1m<a1. Therefore 1+1m<a which contradicts with (2) if we take n=m.

Hence n=1In=.


Exercise 1.24

Solution: We use Theorem 1.24. Let α=supA and β=supB. Then for any aA and bB, we have aα and bβ. Adding them together, we conclude that a+bα+β for all aA and bB. Hence α+β is an upper bound of A+B. (This is part (a) of Theorem 1.24 we would like to show.)

By Theorem 1.24, for any ϵ>0, there exist aA and bB such thatαϵ2<aα,βϵ2<bβ.Add them together, we get thatα+βϵ<a+bα+β.Hence we obtain part (b) of Theorem 1.24.

Using Theorem 1.24, we get that α+β=sup(A+B), i.e. supA+supB=sup(A+B).


Exercise 1.25

Solution: We give a counter example. Let A=B={1,2}, then we have supA=supB=1. We also haveAB={1,2,4}.Hence sup(AB)=4, but supAsupB=1sup(AB).


Exercise 1.26

Solution: We shall use Definition 1.29 to show that the set is not dense in R. Let S={mn:m,nZ,1n10}.Then 0S and 110S. We show that there is no element aS satisfing 0<a<110.

We argue it by contradiction. Suppose 0<a<110 and a=mn, where 1n10, then we have m0, mZ, and10m<n.Because m1, we have n>10m10 which is impossible. Hence 0<a<110 cannot happen. By Definition 1.29, the set S is not dense in R.

Linearity

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