Christopher Heil Introduction to Real Analysis: Solutions Manual Last Updated: April 21, 2019 c 2019 by Christopher Heil (cid:13) Solutions to Exercises and Problems These are my solutions to most of the exercises and problems from my test “An Introduction to Real Analysis.” Additionally, there are detailed verifi- cations of claims stated in the text without proof in the course of various lemmas, theorems, and so forth. Of course, many problems have solutions other than the ones I sketch here In particular, there could very well be easier solutions than the ones I give (and over the years, my students have often shown me better solutions, many of which have been incorporated here). These solutions have not been proofread as carefully as has the text proper, so the probability of errors is correspondingly higher. Please send comments and corrections to “heil@math.gatech.edu”. Non-Distribution Notice. This solutions file is meant for the use of course instructors only. Please do not repost or further distribute this file without the permission of theauthor.Itcanbehelpfultoshareafewselected solutions withstudents afterthey haveworkedthe problemsthemselves,but eventhenitisbesttokeepthedistributionlimited.Inordertolearn,students have to struggle to find their own solutions, there simply is no substitute for this effort, and looking at a solution too early takes this opportunity away from the student. Christopher Heil. Solutions to Exercises and Problems from Chapter 1 1.1.6 “ .”AssumethatE isclosedandthereexistpointsx E suchthat n ⇒ ∈ x x X. If x =x for some n then x=x E, so we may assume that n n n → ∈ ∈ x =x for every n. In this case x is an accumulation point of E. n 6 Suppose that it was the case that x / E. Then EC is an open set that ∈ contains x, so there must exist some r >0 such that B (x) EC. But then r ⊆ every point of E is at least a distance r from x. In particular, d(x ,x) > r n 1 2 Solutions (cid:13)c2019ChristopherHeil for every, which contradicts the fact that x x. Therefore we must have n → x E. ∈ “ .”SupposethatthelimitofeveryconvergentsequenceofpointsfromE ⇐ belongstoE.IfE isnotclosedthenEC isnotopen,sotheremustexistsome pointx EC suchthatnoopenballB (x)centeredatf isentirelycontained r ∈ in EC. Considering r = 1 in particular, this tells us that there must exist n a point x B (x) that is not in EC. But then x E, and we have n 1/n n ∈ ∈ d(x ,x) < 1, and therefore x x. By hypothesis, it follows that x E, n n n → ∈ which is a contradiction since x belongs to EC. Consequently E must be closed. 1.1.7 (a) Let F be the set of all possible limits of elements of E: F = y X :there exist x E such that x y . n n ∈ ∈ → (cid:8) (cid:9) We must show that F =E. Choose any point y F. Then, by definition, there exist points x E n ∈ ∈ such that x y. Since E E, the points x all belong to E. Hence y is a n n → ⊆ limit of elements of E. But E is a closed set, so it must contain all of these limits. Therefore y belongs to E, so we have shown that F E. ⊆ In orderto provethat E is a subset of F, we will firstprovethat FC is an open set. To do this, choose any point y FC. We must show that there is ∈ a ball centered at y that is entirely contained in FC. That is, we must show that there is some r>0 suchthat B (y) containsno limits ofelements of E. r Suppose that for each k N, the ball B (y) contained some point from 1/k ∈ E, say x B (y) E. Then these x are points of E that converge to k 1/k k ∈ ∩ y (why?). Hence y is a limit of points of E, which contradicts the fact that y / F. Hence there must be at least one k such that B (y) contains no 1/k ∈ points of E. We will show that r = 1/k is the radius that we seek. That is, we will show that the ball B (y), where r = 1/k, not only contains no r elements of E but furthermore contains no limits of elements of E. Suppose that B (y) did contain some point z that was a limit of elements r of E, i.e., suppose that there did exist some x E such that x z n n ∈ → ∈ B (y). Then, since d(y,z) < r and since d(z,x ) becomes arbitrarily small, r n by choosing n large enough we will have d(y,x ) d(y,z) + d(z,x ) < r. n n ≤ But then this point x belongs to B (y), which contradicts the fact that n r B (y) contains no points of E. r Thus, B (y) contains no limits of elements of E. Since F is the set of all r limits of elements of E, this means that B (y) contains no points of F. That r is, B (y) FC. r ⊆ In summary, we have shown that each point y FC has some ball B (y) r ∈ that is entirely contained in FC. Therefore FC is an open set. Hence, by definition, F is a closed set. We also know that E F (why?), so F is ⊆ one of the closed sets that contains E. But E is the smallest closed set that contains E, so we conclude that E F. ⊆ Solutions(cid:13)c2019ChristopherHeil 3 (b) “ .” Assume that E is dense in X. Then E =X, so part (a) implies ⇒ that every point x X is a limit of elements of E. ∈ “ .” Suppose that every point x X is a limit of elements of E. Then ⇐ ∈ part (a) implies that E =X, so E is dense in X. 1.1.14 Let V be any open cover of f(K). Each set U = f−1(V ) i i∈J i i { } is open, and U is an open cover of K. Since K is compact, this cover i i∈J { } musthaveafinitesubcover U ,...,U .Butthen V ,...,V isafinite { i1 iN} { i1 iN} subcover of f(K), so f(K) is compact. 1.1.15 ForthisproofweletBX(x)andBY(y)denoteopenballsinX andY, r s respectively. (a) (b). Suppose that f is continuous, and choose any point x X ⇒ ∈ and any ε > 0. Then the ball V = BY f(x) is an open subset of Y, so ε U =f−1(V) mustbe anopensubsetofX.As x U,there existssome δ >0 (cid:0) (cid:1)∈ such that BX(x) U. If y X is any point that satisfies d (x,y)<δ, then δ ⊆ ∈ X we have y BX(x) U, and therefore ∈ δ ⊆ f(y) f(U) V = BY f(x) . ∈ ⊆ ε (cid:0) (cid:1) Consequently d f(x),f(y) <ε. Y (b) (c).Ass(cid:0)ume thatst(cid:1)atement(b) holds,chooseanypointx X,and ⇒ ∈ suppose that x X are such that x x. Fix any ε>0, and let δ >0 be n n ∈ → the number whose existence is given by statement (b). Since x x, there n → mustexistsomeN >0suchthatd (x,x )<δ foralln>N.Statement(b) X n thereforeimpliesthatd f(x),f(x ) <εforalln>N,soweconcludethat Y n f(x ) f(x) in Y. n → (cid:0) (cid:1) (c) (a).Supposethatstatement(c)holds,andletV beanyopensubset ⇒ of Y. Suppose that f−1(V) were not open in X. Then there is some point x f−1(V) such that there is no radius r >0 for which the open ball B (x) r ∈ is a subset of f−1(V). In particular, we have for each n N that the ball ∈ B (x) is not contained in f−1(V), and therefore some point x B (x) 1/n n 1/n ∈ such that x / f−1(V). As a consequence, d(x,x ) < 1/n for every n, but n n ∈ f(x ) / V for any n. n ∈ Now,x f−1(V),sof(x)doesbelongtoV.SinceV isopen,thereissome ∈ open ball centered at f(x) that is entirely contained in V. That is, there is some radius ε>0 such that B (f(x)) V. ε ⊆ On the other hand, we have x x, so by applying statement (c) we n → must have f(x ) f(x). Consequently, there is some N > 0 such that n → d(f(x),f(x ))<ε for all n N. But then n ≥ f(x ) B (f(x)) V, N ε ∈ ⊆ whichcontradictsthefactthatf(x ) / V.Thisisacontradiction,sof−1(V) N ∈ must be open, and therefore f is continuous. 4 Solutions (cid:13)c2019ChristopherHeil 1.1.19 (a) Suppose that x=y, and let r =d(x,y)/2. By the definition of a 6 metric, we have r >0. If z B (x) B (y), then r r ∈ ∩ d(x,y) d(x,z)+d(z,y) < 2r = d(x,y), ≤ whichisacontradiction.ThereforeB (x) B (y)=∅,sowecansimplytake r r ∩ U =B (x) and V =B (y). r r (b) Suppose that x x and x y. Then n n → → 0 d(x,y) d(x,x )+d(x ,y) 0 as n . n n ≤ ≤ → →∞ Consequently d(x,y)=0, so x=y. 1.1.20 Suppose(x ) isaCauchysequenceandthereexistsasubsequence n n∈N (x ) that converges to x. Then given ε > 0, there exists some K > 0 nk k∈N such that d(x ,x) < ε for all n > K. Also, there exists an N such that nk k d(x ,x )<ε for all m, n>N. Suppose that n>N. Then since the n are m n k strictly increasing, there exists some n that is greater than both K and N. k For this n we have k d(x ,x) d(x ,x )+d(x ,x ) < ε+ε = 2ε. n ≤ n nk nk n This is true for all n>N, so x x. n → 1.1.21 (a) We are given that d(x ,x ) < 2−n for every n. Choose any n n+1 ε>0, and let N be large enough that 2−N+1 <ε. If n>m>N, then n−1 ∞ 1 1 1 d(x ,x ) d(x ,x ) = < < ε. m n ≤ k k+1 ≤ 2k 2m−1 2N−1 k=m k=m X X Hence x is Cauchy. n n∈N { } (b) Let x be a Cauchy sequence in X. Then there exists an integer n n∈N { } N such that 1 1 m,n N = d(x ,x )< . 1 m n ≥ ⇒ 2 Set n =N . Then there exists an N such that 1 1 2 1 m,n N = d(x ,x )< . ≥ 2 ⇒ m n 22 Withoutlossofgenerality,wemayassumethatN >N .Letn =N .Then 2 1 2 2 we have n , n N , so 1 2 1 ≥ 1 d(x ,x ) < . n2 n1 2 Continuing in this way, we can construct n 0 such that givenany N we can find an n>N such that d(x ,x) > ε. Iterating this, we can find a subsequence (x ) such that n nk k∈N d(x ,x)>ε for everyk. Butthen no subsequence of(x ) canconverge nk nk k∈N to x, which is a contradiction. 1.1.23 “ .” Suppose that x x as t 0+, and let (t ) be any t k k∈N ⇒ → → sequence of real numbers in (0,c) such that t 0. Choose any ε>0. Then k → since x x, thereexists aδ with 0<δ 0 such that 0 < t < δ for all k k → k N. Hence for every k N we have d(x ,x)<ε, and thereforex x. ≥ ≥ tk tk → “ .” Suppose that x x as t 0+. Then there exists an ε > 0 such t ⇐ 6→ → thatforeachδ =1/k wecanfindarealnumbert (0,c)with00. We are given that ∈ h(x) = inf g(y):y B (x) r ∈ (cid:8) (cid:9) and h(x)= . Since h(x) is a real number, by definition of inf there must 6 −∞ be some z such that z x < r and g(z) h(x)+ε. | − | ≤ Let δ =r z x, and suppose that y x <δ. Then −| − | | − | y z y x + x z < δ+ x z = r, | − | ≤ | − | | − | | − | so z B(y). Therefore ∈ h(y) = inf g(t):t B(y) g(z) h(x)+ε. ∈ ≤ ≤ (cid:8) (cid:9) Since this is true for all y with y x <δ, we conclude that h is usc at x. | − | The result can fail if h(x)= at some point. For example, if −∞ 0, x 0, g(x) = ≤ ( 1/x, x>0, − then 6 Solutions (cid:13)c2019ChristopherHeil 0, x 1, ≤− h(x) = , 11, − − and h is not usc at x=1.  (b) “ .” Suppose that f is continuous at x. Then given ε >0, there is a ⇒ δ >0 such that x y <δ = f(x) ε f(y) f(x)+ε. | − | ⇒ − ≤ ≤ Hence f is both usc and lsc at x. “ .” Suppose f is both usc and lsc at x and fix ε > 0. By definition of ⇐ usc, there is a δ >0 such that 1 x y <δ = f(y) f(x)+ε. 1 | − | ⇒ ≤ Similarly, by definition of lsc, there is a δ >0 such that 2 x y <δ = f(y) f(x) ε. 2 | − | ⇒ ≥ − Taking δ = min δ ,δ , we see that if x y < δ, then f(x) ε f(y) 1 2 { } | − | − ≤ ≤ f(x)+ε, which implies that f(x) f(y) ε. Hence f is continuous at x. | − |≤ (c) Fix ε>0. Since g(x) = inf f (x), α α∈J the definition of infimum implies that there must exist an α J such that ∈ f (x) g(x)+ε. α ≤ Since f is usc at x, there exists a δ >0 such that α x y <δ = f (y) f (x)+ε. α α | − | ⇒ ≤ Hence, if y satisfies x y <δ, then | − | g(y) f (y) f (x)+ε g(x)+2ε. α α ≤ ≤ ≤ Therefore g is usc at x. (d)“ .”Assumethatf isuscateverypointx Rd.Toshowthat f a ⇒ ∈ { ≥ } is closed, suppose that x f a and x x as n . Given ε > 0, n n ∈ { ≥ } → → ∞ we have by definition that there is some δ >0 such that x y <δ = f(y) f(x)+ε. | − | ⇒ ≤ Since x converges to x, there is some N N such that n ∈ Solutions(cid:13)c2019ChristopherHeil 7 x x < δ, n N. n | − | ≥ Therefore a f(x ) f(x)+ε, n N. n ≤ ≤ ≥ Hence a f(x)+ε, and since ε is arbitrary, it follows that a f(x). Thus ≤ ≤ x f a , which implies that f a is closed. ∈{ ≥ } { ≥ } “ .”Nowsupposethat f a is closedforeacha R.Then f 0.Ifweseta=f(x)+εthenwe ∈ have f(x) < a. Therefore x f < a , which is an open set. Consequently, ∈ { } thereexistssomeδ >0suchthattheopenballB (x)iscontainedin f n. Since K is compact, n n ∈ the sequence x must have a convergent subsequence, i.e., there exist n n∈N { } integers n and a point x K such that x x . But f is usc and k 0 ∈ nk → 0 f(x )>n , so nk k f(x) limsup f(x) limsupf(x ) = . ≥ ≥ nk ∞ x→x0,x∈K k→∞ This contradicts the fact that f was assumed to be finite at every point. 1.2.4 (a) Given x, y X, ∈ x = (x y)+y x y + y , k k k − k ≤ k − k k k so x y x y . By reversing the roles of x and y we obtain the k k−k k ≤ k − k inequality y x x y , so we conclude that x y x y . k k−k k ≤ k − k k k−k k ≤k − k (b) Assume xn x and fix ε > 0. Then there is(cid:12)an N > (cid:12)0 such that → (cid:12) (cid:12) x x <ε for all n>N. Therefore, if m, n>N then n k − k x x x x + x x < 2ε, m n m n k − k ≤ k − k k − k so x is Cauchy. n n∈N { } (c) Suppose that x is Cauchy. Then there exists an N > 0 such n n∈N { } that x x <1 for all m, n N. Therefore, for n N we have m n k − k ≥ ≥ x = x x +x x x + x 1+ x . n n N N n N N N k k k − k ≤ k − k k k ≤ k k Hence, for an arbitrary n we have x max x ,..., x , x +1 . n 1 N−1 N k k ≤ k k k k k k (cid:8) (cid:9) 8 Solutions (cid:13)c2019ChristopherHeil (d) By the Reverse Triangle Inequality, if x x, then n → x x x x 0. n n k k−k k ≤ k − k → (cid:12) (cid:12) (cid:12) (cid:12) (e) If xn x and y(cid:12)n y, then(cid:12) → → (x+y) (x +y ) = (x x )+(y y ) x x + y y 0. n n n n n n k − k k − − k ≤ k − k k − k → (f) Suppose x x and c c. Then C =sup c < , so n n n → → | | ∞ cx c x cx c x + c x c x n n n n n n k − k ≤ k − k k − k = c c x + c x x n n n | − |k k | |k − k c c x +C x x 0. n n ≤ | − |k k k − k → 1.2.11 If y, z are any two points in B (x) and 0 t 1, then r ≤ ≤ x (1 t)y+tz = (1 t)(x y)+t(x z) − − − − − (cid:13) (cid:0) (cid:1)(cid:13) ((cid:13)1 t) (x y) +t x z(cid:13) (cid:13) (cid:13) ≤ (cid:13) − k − k k −(cid:13)k < (1 t)r+tr = r. − Therefore (1 t)y+tz B (x), so B (x) is convex. r r − ∈ 1.2.12 “ .” Suppose that Y is a closed subspace of X, and let x be n n∈N ⇐ { } a Cauchy sequence in Y. Then x is a Cauchy sequence in X, so there n n∈N { } exists some vector x X such that x x. Since Y is closed, this vector x n ∈ → must belong to Y. Therefore every Cauchy sequence in Y converges to an element of Y, so Y is complete. “ .” Suppose that Y is a Banach space with respect to the norm of X. ⇒ Suppose that x Y and x x X. Then x is a Cauchy sequence n n n n∈N ∈ → ∈ { } in Y and hence must converge to some vector y Y. Thus, x is a n n∈N ∈ { } sequence in X such that x x and x y. By uniqueness of limits, we n n → → conclude that x=y Y. Exercise 1.1.6 therefore implies that Y is closed. ∈ 1.2.13 The partial sums s = N x of the series converge to x in N n=1 n norm, so by the continuity of the norm and the Triangle Inequality (which P by induction applies to any finite sum), we see that ∞ x = x = lim s (continuity of the norm) n N k k N→∞k k (cid:13)n=1 (cid:13) (cid:13)X (cid:13) (cid:13) (cid:13) N (cid:13) (cid:13) = lim x (definition of s ) n N N→∞ (cid:13)n=1 (cid:13) (cid:13)X (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)

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