Understanding Analysis, Stephen Abbott, Chapter 1.2 Exercises

This is a fairly popular analysis book with many, many questions and no answers. So I will gradually attempt to answer some of them here. I may be wrong sometimes, so feel free to correct me if you can. If you are using this to cheat, then shame on you (this blog entry became the second most viewed after just one week; what's up with that?)!


Claim: 3 is irrational.

Let 3 be rational. Then 3=p/q, where p,qZ, q0, and gcd(p,q)=1. This last assumption can be made without loss of generality. It follows that

3=p2q23q2=p2.

p2 cannot be even because it would imply that p is even and that q2 is even, which would in turn imply that q is even. However, p and q cannot both be even because we assumed that gcd(p,q)=1. Therefore, p2 must be odd, which implies that p is odd and that q2 is odd, which in turn implies that q is odd. Let p=2m+1, and q=2n+1, as per the definition of an odd number, where m,nZ. Then

3(2n+1)2=(2m+1)23(4n2+4n+1)=4m2+4m+112n2+12n+2=4m2+4m6n2+6n+1=2m2+2m2(3n2+3n)+1=2(m2+m).

Nevertheless, this last expression is absurd; an odd number cannot be equal to an even number. Therefore, 3 is irrational. This proves the claim.



Claim: If for every nN, AnAn+1 and An contains infinitely-many elements, then nNAn contains infinitely-many elements.

Let An={n,n+1,n+2,} for every nN. Then the hypothesis holds. Moreover, let xnNAn. However, xAx+1. Therefore, nNAn=. This dismisses the claim.



Claim: If for every nN, AnAn+1 and An is a nonempty set containing a finite number of real numbers, then nNAn is a nonempty set containing a finite number of real numbers.

Let, for every nN, AnAn+1, An be a nonempty set containing a finite number of real numbers, and |A1|=m. Then there exists an NN such that An=An+1 for every nN. Therefore, nNAn=AN, which is a nonempty set containing a finite number of real numbers. This proves the claim.



Claim: A(BC)=(AB)C.

Let A={1,2,3}, B={2,3,4}, and C={3,4,5}. Then A(BC)={2,3}, and (AB)C={2,3,4,5}, which are different. Therefore, the claim is dismissed.



Claim: A(BC)=(AB)C.

Let xA(BC). Then xA or xBC. xBC implies that xB or xC. xA or xB implies that xAB. xAB or xC implies that x(AB)C. Therefore, A(BC)(AB)C. The converse can be proved in the same manner.



Claim: If A,BR and g:RR, then g(AB)g(A)g(B).

Lemma: ABf(A)f(B), A,BR, f:RR.

ABAf(AB)f(A).
ABBf(AB)f(B).
f(AB)f(A)f(B).

Given a function f:DR and a subset BR, let f1(B) be the set of all points from the domain D that get mapped into B; that is, f1(B)={xD:f(x)B}. This set is called the preimage of B.

Let f(x)=x2. If A is the closed interval [0,4] and B is the closed interval [1,1], find f1(A) and f1(B).

f1(A)=[2,2], andf1(B)=[1,1].

Does f1(AB)=f1(A)f1(B) in this case?

AB=[0,1],f1(AB)=[1,1], andf1(A)f1(B)=[1,1].

Therefore, they are equivalent.

The good behavior of preimages just demonstrated is completely general. Show that for an arbitrary function g:RR, it is always true that g1(AB)=g1(A)g1(B) and g1(AB)=g1(A)g1(B) for all sets A,BR.

Let xg1(AB). Then g(x)AB. This implies that g(x)A and g(x)B. This implies that xg1(A) and xg1(B). This implies that xg1(A)g1(B).



To be continued...