1、MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 1Generally, a solution“ is something that would be acceptable if turned in in theform presented here, although the solutions given are often close to minimal in thisrespect. A solution (sketch)“ is too sketchy to be considered a complete solutionif turne
2、d in; varying amounts of detail would need to be fllled in.Problem 1.1: If r 2 Q nf0g and x 2 R n Q, prove that r + x; rx 62 Q.Solution: We prove this by contradiction. Let r 2 Qnf0g, and suppose that r+x 2Q. Then, using the fleld properties of both R and Q,wehavex =(r +x)r 2 Q.Thus x 62 Q implies r
3、 + x 62 Q.Similarly, if rx 2 Q, then x =(rx)=r 2 Q. (Here, in addition to the fleldproperties of R and Q,weuser 6= 0.) Thus x 62 Q implies rx 62 Q.Problem 1.2: Prove that there is no x 2 Q such that x2= 12.Solution: We prove this by contradiction. Suppose there is x 2 Q such thatx2= 12. Write x =mni
4、n lowest terms. Then x2= 12 implies that m2=12n2.Since 3 divides 12n2, it follows that 3 divides m2. Since 3 is prime (and by uniquefactorization in Z), it follows that 3 divides m. Therefore 32divides m2=12n2.Since 32does not divide 12, using again unique factorization in Z and the fact that3 is pr
5、ime, it follows that 3 divides n. We have proved that 3 divides both m andn, contradicting the assumption that the fractionmnis in lowest terms.Alternate solution (Sketch): If x 2 Q satisfles x2= 12, thenx2is in Q and satisflesx22= 3. Now prove that there is no y 2 Q such that y2= 3 by repeating the
6、proof thatp2 62 Q.Problem 1.5: Let A R be nonempty and bounded below. Set A = fa: a 2Ag. Prove that inf(A)=sup(A).Solution: First note that A is nonempty and bounded above. Indeed, A containssome element x, and then x 2 A; moreover, A has a lower bound m,andm isan upper bound for A.We now know that
7、b = sup(A) exists. We show that b = inf(A). That b isa lower bound for A is immediate from the fact that b is an upper bound for A.To show that b is the greatest lower bound, we let cb and prove that c is nota lower bound for A.Nowcc. Then x 2 A and x1, flxed throughout the problem.Comment: We will
8、assume known that the function n 7! bn,fromZ to R,isstrictly increasing, that is, that for m; n 2 Z,wehavebm0. We will also assume that the usual laws of exponents areknown to hold when the exponents are integers. We cant assume anything aboutfractional exponents, except for Theorem 1.21 of the book
9、 and its corollary, becausethe context makes it clear that we are to assume fractional powers have not yetbeen deflned.(a) Let m; n; p; q 2 Z, with n0andq0. Prove that ifmn=pq, then(bm)1=n=(bp)1=q.Solution: By the uniqueness part of Theorem 1.21 of the book, applied to thepositive integer nq, it suc
10、es to show thath(bm)1=ninq=h(bp)1=qinq:Now the deflnition in Theorem 1.21 implies thath(bm)1=nin= bmandh(bp)1=qiq= bp:Therefore, using the laws of integer exponents and the equation mq = np,wegeth(bm)1=ninq=hh(bm)1=niniq=(bm)q= bmq= bnp=(bp)n=hh(bp)1=qiqin=h(bp)1=qinq;as desired.By Part (a), it make
11、s sense to deflne bm=n=(bm)1=nfor m; n 2 Z with n0.This deflnes brfor all r 2 Q.(b) Prove that br+s= brbsfor r; s 2 Q.Solution: Choose m; n; p; q 2 Z, with n0andq0, such that r =mnands =pq. Then r + s =mq+npnq. By the uniqueness part of Theorem 1.21 of the book,applied to the positive integer nq, it
12、 suces to show thathb(mq+np)=(nq)inq=h(bm)1=n(bp)1=qinq:Directly from the deflnitions, we can writehb(mq+np)=(nq)inq=hb(mq+np)i1=(nq)nq= b(mq+np):Using the laws of integer exponents and the deflnitions for rational exponents, wecan rewrite the right hand side ash(bm)1=n(bp)1=qinq=hh(bm)1=niniqhh(bp)
13、1=qiqin=(bm)q(bp)n= b(mq+np):This proves the required equation, and hence the result.(c) For x 2 R, deflneB(x)=fbr: r 2 Q (1;xg:Prove that if r 2 Q, then br= sup(B(r).Solution: The main point is to show that if r; s 2 Q with r0andq0, such that r =mnand s =pq. ThenMATH 413 513 (PHILLIPS) SOLUTIONS TO
14、 HOMEWORK 1 3also r =mqnqand s =npnq, with nq 0, sobr=(bmq)1=(nq)and bs=(bnp)1=(nq):Now mq x.Ifr 2 Q (1;x,then br2 B(k) so that br bkby Part (c). Thus bkis an upper bound for B(x).This shows that the deflnition makes sense, and Part (c) shows it is consistent withour earlier deflnition when r 2 Q.(d
15、) Prove that bx+y= bxbyfor all x; y 2 R.Solution:In order to do this, we are going to need to replace the set B(x)abovebythesetB0(x)=fbr: r 2 Q (1;x)g(that is, we require r1andN = f1; 2; 3;:g.)Proof: Clearly 1 is a lower bound. (Indeed, (b1=n)n= b1=1n,sob1=n 1.) Weshow that 1+x is not a lower bound
16、when x0. If 1+x were a lower bound, then1+x b1=nwould imply (1 + x)n (b1=n)n= b for all n 2 N. By Lemma 1, wewould get 1 + nx b for all n 2 N, which contradicts the Archimedean propertywhen x0.Lemma 3. supfb1=n: n 2 Ng =1.Proof: Part (b) shows that b1=nb1=n= b0= 1, whence b1=n=(b1=n)1. Sinceall numb
17、ers b1=nare strictly positive, it now follows from Lemma 2 that 1 is anupper bound. Suppose x 1, this contradicts Lemma 2. Thus supfb1=n: n 2Ng = 1, as claimed.Lemma 4. bx= sup(B0(x) for x 2 R.Proof: If x 62 Q, then B0(x)=B(x), so there is nothing to prove. If x 2 Q,then at least B0(x) B(x), so bx s
18、up(B0(x). Moreover, Part (b) shows thatbx1=n= bxb1=nfor n 2 N. The numbers bx1=nare all in B0(x), andsupfbxb1=n: n 2 Ng = bxsupfb1=n: n 2 Ngbecause bx 0, so using Lemma 3 in the last step givessup(B0(x) supfbx1=n: n 2 Ng = bxsupfb1=n: n 2 Ng = bx:Now we can prove the formula bx+y= bxby. We start by
19、showing that bx+ybxby, which we do by showing that bxbyis an upper bound for B0(x + y). Thus letr 2 Q satisfy rzare both false.Case 2: Re(w) Re(z). Then wz, but w = z and wzare both false.Case 3.2: Im(w) Im(z). Then wz, but w = z and wzand wzis true, as desired.Now we prove transitivity. Let s 0.Pro
20、blem 1.13: Prove that if x; y 2 C, then jjxjjyjj jx yj.Solution: The desired inequality is equivalent tojxjjyjjx yj and jyjjxjjx yj:We prove the flrst; the second follows by exchanging x and y.Set z = x y. Then x = y + z. The triangle inequality gives jxjjyj + jzj.Substituting the deflnition of z an
21、d subtracting jyj from both sides gives the result.Problem 1.17: Prove that if x; y 2 Rn, thenkx + yk2+ kx yk2=2kxk2+2kyk2:Interpret this result geometrically in terms of parallelograms.Solution: Using the deflnition of the norm in terms of scalar products, we have:kx + yk2+ kx yk2= hx + y;x + yi +
22、hx y;x yi= hx; xi + hx; yi + hy;xi + hy;yi+ hx; xihx; yihy;xi + hy;yi=2hx; xi +2hy;yi =2kxk2+2kyk2:The interpretation is that 0;x;y;x+ y are the vertices of a parallelogram, andthat kx+yk and kxyk are the lengths of its diagonals while kxk and kyk are each6 MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEW
23、ORK 1the lengths of two opposite sides. Therefore the sum of the squares of the lengthsof the diagonals is equal to the sum of the squares of the lengths of the sides.Note: One can do the proof directly in terms of the formula kxk2=Pnk=1jxkj2.The steps are all the same, but it is more complicated to
24、 write. It is also lessgeneral, since the argument above applies to any norm that comes from a scalarproduct.MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 2Generally, a solution“ is something that would be acceptable if turned in in theform presented here, although the solutions given are often clos
25、e to minimal in thisrespect. A solution (sketch)“ is too sketchy to be considered a complete solutionif turned in; varying amounts of detail would need to be fllled in.Problem 2.2: Prove that the set of algebraic numbers is countable.Solution (sketch): For each flxed integer n 0, the set Pnof all po
26、lynomialswith integer coecients and degree at most n is countable, since it has the samecardinality as the set f(a0;:;an): ai2 Ng = Nn+1. The set of all polynomialswith integer coecients isS1n=0Pn, which is a countable union of countable setsand so countable. Each polynomial has only flnitely many r
27、oots (at most n fordegree n), so the set of all possible roots of all polynomials with integer coecientsis a countable union of flnite sets, hence countable.Problem 2.3: Prove that there exist real numbers which are not algebraic.Solution (Sketch): This follows from Problem 2.2, since R is not count
28、able.Problem 2.4: Is RnQ countable?Solution (Sketch): No. Q is countable and R is not countable.Problem 2.5: Construct a bounded subset of R with exactly 3 limit points.Solution (Sketch): For example, use1n: n 2 N“1+1n: n 2 N“2+1n: n 2 N“:Problem 2.6: Let E0denote the set of limit points of E. Prove
29、 that E0is closed.Prove that E0= E0.Is(E0)0always equal to E0?Solution (Sketch): Proving that E0is closed is equivalent to proving that (E0)0E0.Soletx 2 (E0)0and let “0. Choose y 2 E0 (N“(x) nfxg). Choose = min(d(x; y);“ d(x; y) 0. Choose z 2 E (N(y) nfyg). The triangleinequality ensures z 6= x and
30、z 2 N“(x). This shows x is a limit point of E.Here is a difierent way to prove that (E0)0 E0.Letx 2 (E0)0and “0.Choose y 2 E0 (N“=2(x) nfxg). By Theorem 2.20 of Rudin, there are inflnitelymany points in E (N“=2(y) nfyg). In particular there is z 2 E (N“=2(y) nfyg)with z 6= x.Nowz 2 E (N“(x) nfxg).To
31、 prove E0= E0, it suces to prove E0 E0. We flrst claim that if A and Bare any subsets of X, then (A B)0 A0 B0. The fastest way to do this is toassume that x 2 (A B)0but x 62 A0, and to show that x 2 B0. Accordingly, letx 2 (A B)0n A0. Since x 62 A0, there is “0 0 such that N“0(x) A contains noDate:
32、8 October 2001.12 MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 2points except possibly x itself. Now let “0; we show that N“(x) B contains atleast one point difierent from x.Letr = min(“; “0) 0. Because x 2 (A B)0,there is y 2 Nr(x)(AB) with y 6= x. Then y 62 A because r “0. So necessarilyy 2 B, an
33、d thus y is a point difierent from x and in Nr(x) B. This shows thatx 2 B0, and completes the proof that (A B)0 A0 B0.To prove E0 E0, we now observe thatE0=(E E0)0 E0 (E0)0 E0 E0= E0:An alternate proof that E0 E0can be obtained by slightly modifying either ofthe proofs above that (E0)0 E0.For the th
34、ird part, the answer is no. TakeE = f0g1n: n 2 N“:Then E0= f0g and (E0)0= ?. (Of course, you must prove these facts.)Problem 2.8: If E R2is open, is every point of E a limit point of E? What ifE is closed instead of open?Solution (Sketch): Every point of an open set E R2is a limit point of E. Indeed
35、,if x 2 E, then there is “0 such that N“(x) E, and it is easy to show that x isa limit point of N“(x).(Warning: This is not true in a general metric space.)Not every point of a closed set need be a limit point. Take E = f(0; 0)g,whichhas no limit points.Problem 2.9: Let Edenote the set of interior p
36、oints of a set E, that is, theinterior of E.(a) Prove that Eis open.Solution (sketch): If x 2 E, then there is “0 such that N“(x) E. Since N“(x)is open, every point in N“(x) is an interior point of N“(x), hence of the bigger setE.SoN“(x) 2 E.(b) Prove that E is open if and only if E= E.Solution: If
37、E is open, then E = Eby the deflnition of E.IfE = E, then E isopen by Part (a).(c) If G is open and G E, prove that G E.Solution (sketch): If x 2 G E and G is open, then x is an interior point of G.Therefore x is an interior point of the bigger set E.Sox 2 E.(d) Prove that X n E= X n E.Solution (ske
38、tch): First show that X n E X n E.Ifx 62 E, then clearly x 2X n E. Otherwise, consider x 2 E n E. Rearranging the statement that x fails tobe an interior point of E, and noting that x itself is not in X nE, one gets exactlythe statement that x is a limit point of X n E.Now show that X n E X nE.Ifx 2
39、 X nE, then clearly x 62 E.Ifx 62 X nEbut x is a limit point of X nE, then one simply rearranges the deflnition of a limitpointtoshowthatx is not an interior point of E.(e) Prove or disprove:E= E.MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 2 3Solution (sketch): This is false. Example: take E =(0;
40、1)(1; 2). We have E= E,E =0; 2, andE=(0; 2).Another example is Q.(f) Prove or disprove: E= E.Solution (sketch): This is false. Example: take E =(0; 1) f2g. Then E =0; 1 f2g, E=(0; 1), and E=0; 1.The sets Q and f0g are also examples: in both cases, E= ?.Problem 2.11: Which of the following are metric
41、s on R?(a) d1(x; y)=(x y)2.Solution (Sketch): No. The triangle inequality fails with x =0,y =2,andz =4.(b) d2(x; y)=pjx yj.Solution (Sketch): Yes. Some work is needed to check the triangle inequality.(c) d3(x; y)=jx2 y2j.Solution (Sketch): No. d3(1;1) = 0.(d) d4(x; y)=jx 2yj.Solution (Sketch): No. d
42、4(1; 1) 6= 0. Also, d4(1; 6) 6= d4(6; 1).(e) d5(x; y)=jx yj1+jx yj.Solution (Sketch): Yes. Some work is needed to check the triangle inequality. Youneed to know that t 7!t1+tis nondecreasing on 0;1), and that a; b 0 impliesa + b1+a + ba1+a+b1+b:Do the flrst by algebraic manipulation. The second isa
43、+ b1+a + b=a1+a + b+b1+a + ba1+a+b1+b:(This is easier than what most people did the last time I assigned this problem.)MATH 413 513 (PHILLIPS) SOLUTIONS TO HOMEWORK 3Generally, a solution“ is something that would be acceptable if turned in in theform presented here, although the solutions given are
44、often close to minimal in thisrespect. A solution (sketch)“ is too sketchy to be considered a complete solutionif turned in; varying amounts of detail would need to be fllled in.Problem 2.14: Give an example of an open cover of the interval (0; 1) R whichhas no flnite subcover.Solution (sketch): f(1
45、=n; 1): n 2 Ng. (Note that you must show that this works.)Problem 2.16: Regard Q as a metric space with the usual metric. Let E = fx 2Q:2 3. Assume x2 3. (The other case is handled similarly.) Letr = jxjp3 0. Then every z 2 Nr(x) satisflesjzjjxjjx zj jxjr0;which implies that z2 (jxjr)2= 3. This show
46、s that z 62 E, which contradictsthe assumption that x is a limit point of E.The fast way to see that E is not compact is to note that it is a subset of R,but is not closed in R. (See Theorem 2.23.) To prove this directly, show that, forexample, the setsy 2 Q:2+1n0. SetA = fx 2 X : d(x; x0) g:Prove t
47、hat A and B are separated.Solution (Sketch): Both A and B are open sets (proof!), and they are disjoint. Sothis follows from Part (b).(d) Prove that if X is connected and contains at least two points, then X isuncountable.Solution: Let x and y be distinct points of X.LetR = d(x; y) 0. For eachr 2 (0
48、;R), consider the setsAr= fz 2 X : d(z;x) rg:They are separated by Part (c). They are not empty, since x 2 Arand y 2 Br.Since X is connected, there must be a point zr2 X n(ArBr). Then d(x; zr)=r.Note that if r 6= s, then d(x; zr) 6= d(x; zs), so zr6= zs.Thusr 7! zrdeflnes aninjective map from (0;R)t
49、oX. Since (0;R) is not countable, X cant be countableeither.Problem 2.20: Let X be a metric space, and let E X be a connected subset.Is E necessarily connected? Is int(E) necessarily connected?Solution to the flrst question (sketch): The set int(E) need not be connected. Theeasiest example to write down is to take X = R2andE = fx 2 R
