Rudin s Principles of Mthemticl Anlysis: Solutions to Selected Exercises Sm Blinstein UCLA Deprtment of Mthemtics Mrch 29, 2008
Contents Chpter : The Rel nd Complex Number Systems 2 Chpter 2: Bsic Topology 0 Chpter 3: Numericl Sequences nd Series 6 Chpter 4: Continuity 23 Chpter 5: Differentition 32 Chpter 6: The Riemnn-Stieltjes Integrl 39 Chpter 7: Sequences nd Series of Functions 47
The following re solutions to selected exercises from Wlter Rudin s Principles of Mthemticl Anlysis, Third Edition, which I compiled during the Winter of 2008 while grdute student in Mthemtics t UCLA. Equtions re numbered within ech Chpter nd their lbels correspond to the question number nd eqution number, so (2.3) refers to the third eqution in question #2 of the current Chpter. Chpter : The Rel nd Complex Number Systems. If r Q, r 0, nd x is irrtionl, prove tht r + x nd rx re irrtionl. Solution: Suppose tht r + x Q. Since Q is field, r Q nd we hve tht r + r + x = x Q, contrdiction. Therefore r + x R\Q, i.e. r + x is irrtionl. Similrly, suppose tht rx Q. Since Q is field, r Q nd we hve tht r rx = x Q, contrdiction. Therefore rx R\Q, i.e. rx is irrtionl. 2. Prove tht there is no rtionl number whose squre is 2. Solution: Suppose there exist m, n Z such tht neither re divisible by 3 (i.e. we ssume their rtio is in simplified form) nd m n = 2 = m 2 = 2n 2 = 2 2 3n 2. Therefore, 3 m 2 = 3 m becuse 3 is prime. Therefore, 3 2 m 2 = 3 2 2n 2 = 3 n 2 = 3 n, gin becuse 3 is prime. This contrdicts tht m nd n were chosen such tht their rtio is in simplified form becuse both re divisible by 3. We cn investigte this further, discovering tht Q lcks the gretest/lest upper bound property, s is done nlogously in Exmple. on pge 2 of the text. Let A = {p Q p > 0, p 2 < 2} nd B = {p Q p > 0, p 2 > 2}. Define Then, we hve tht q = p p2 2 p + 2 = p(p + 2) p2 + 2 2(p + ) = p + 2 p + 2. (2.) q 2 2 = 44(p + )2 2(p + 2) 2 (p + 2) 2 = 32(p2 2) (p + 2) 2. (2.2) Now, if p A then p 2 2 < 0 so by Eqution (2.) q > p nd by Eqution (2.2) q 2 < 2 so q A. Therefore, A contins no lrgest number. Similrly, if p B then p 2 2 > 0 so by Eqution (2.) 0 < q < p nd by Eqution (2.2) q 2 > 2 so q B. Therefore, B contins no smllest number. The elements of B re precisely the upper bounds of A, therefore, since B hs no lest element, A hs no lest upper bound. Similrly, the elements of A re the lower bounds of B nd since A hs no gretest element, B hs no gretest upper bound. 3. Prove the following using the xioms of multipliction in field. () If x 0 nd xy = xz then y = z. (b) If x 0 nd xy = x then y =. (c) If x 0 nd xy = then y = /x. (d) If x 0 the /(/x) = x. 2
Solution: To show prt () we simply use ssocitivity of multipliction nd the existence of multiplictive inverses. ( y = y = )y x x = x (xy) = x ( ) x (xz) = x z = z = z. The sttement of (b) nd (c) follow directly from () if we let z = nd z = /x, respectively. Prt (d) follows from (c) if we replce x with /x nd y with x. 4. Let E be nonempty subset of n ordered set; suppose α is lower bound of E nd β is n upper bound of E. Prove tht α β. Solution: Since E is nonempty, let x E. Then, by definition of upper nd lower bounds, we hve tht α x β. 5. Let A be nonempty set of rel numbers which is bounded below. Let A be the set of ll numbers x, where x A. Prove tht inf A = sup( A). Solution: Let β = sup( A), which exists becuse A is nonempty subset of the rel numbers which is bounded below, hence A is nonempty subset of the rel numbers which is bounded bove. Thus, x A, x β = x β so β is lower bound for A. Now, suppose there exists γ such tht γ > β nd x A, x γ. This implies tht γ < β nd x γ, x A. But then, γ is smller upper bound (thn β) for A, contrdicting tht β = sup( A). Therefore no such γ exists nd so sup( A) = β = inf(a). The reversing of inequlities throughout follows from Proposition.8 on pge 8 of the text. 6. Fix b >. () If m, n, p, q re integers, n > 0, q > 0, nd r = m/n = p/q, prove tht (b m ) /n = (b p ) /q. Hence it mkes sense to define b r = (b m ) /n. (b) Prove tht b r+s = b r b s if r nd s re rtionl. (c) If x is rel, define B(x) to be the set of ll numbers b t, where t is rtionl nd t x. Prove tht b r = sup B(r) where r is rtionl. Hence, it mkes sense to define for every rel x. b x = sup B(x) (d) Prove tht b x+y = b x b y for every rel x nd y. Solution: () Frst observe tht (b n ) n 2 = b n n 2 for n, n 2 Z by direct expnsion of either term. Let y n = b m so tht y = (b m ) /n. Then, y nq = b mq = b np = (y q ) n = (b p ) n, becuse m/n = p/q = mq = np. By Theorem.2 on pge 0, since n th roots re unique, we hve y q = b p. Then, (b m ) /n = y = (b p ) /q. 3
(b) Let r, s Q nd write r = m/n nd s = p/q for m, n, p, q Z. Then, r+s = m/n+p/q = (mq + np)/nq. Note, for n, n 2 Z we hve b n b n 2 = b n +n 2 by direct expnsion. Then, by (), b r+s = (b mq+np ) /nq = (b mq b np ) /nq = (b mq ) /nq (b np ) /nq, where the lst equlity follows from the Corollry to Theorem.2 on pge of the text. Agin using the result of () we cn simplify this lst expression nd obtin (b mq ) /nq (b np ) /nq = (b m ) /n (b p ) /q = b r b s, showing (b). (c) Let t, r Q nd write t = p/q, r = m/n, where p, q, m, n Z. Then, t < r = p/q < m/n = pn < mq. Thus, since b >, we hve tht b pn < b mq = b p < (b mq ) /n = (b p ) /q < ((b mq ) /n ) /q. Using the result of prt (), we cn simplify the lst expression to obtin b t = (b p ) /q < (b m ) /n = b r, showing tht nd proving prt (c). b r > {b t t < r} = b r = sup {b t t r} = B(r) (d) We first observe tht b r b t = b r+t by prt (b) nd so 7. Omitted. b x b y = sup b r b t = r x,t y sup b r+t r x,t y sup b r+t = b x+y. r+t x+y Here, sup r x,t y b r+t sup r+t x+y b r+t becuse we re tking the sup over more restricted set on the right since r x, t y = r + t x + y but r + t x + y doesn t imply r x nd t y. But, we in fct hve equlity becuse if sup r x,t y b r+t < sup r+t x+y b r+t then there exist r, t Q such tht r + t x + y nd sup b ρ+τ < b r +t sup b r+t. (6.) ρ x,τ y r+t x+y If r + t = x + y we get contrdiction to Eqution (6.) by choosing ρ = x, τ = y (this follows from Exercise bove). Otherwise, r + t < x + y nd then we cn choose ρ < x, τ < y such tht r + t < ρ + τ < x + y, becuse Q is dense in R. But then, b ρ+τ > b r +t, lso contrdicting Eqution (6.). Therefore we hve equlity nd prt (d) is shown. 8. Prove tht no order cn be defined in the complex field tht turns it into n ordered field. Hint: is squre. Solution: Suppose there is n ordered defined on the complex field tht turns it into n ordered field. Then, we get contrdiction to Proposition.8(d) on pge 8 becuse i 0 yet i 2 = < 0, where < 0 is forced by prt () of the sme Proposition. 9. Suppose z = + bi, w = c + di. Define z < w if < c, nd lso if = c but b < d. Prove tht this turns the set of ll complex numbers into n ordered set. (This type of order reltion is clled dictionry order, or lexicogrphic order, for obvious resons.) Does this ordered set hve the lest-upper-bound property? Solution: Let z = + bi, w = c + di C. Then, since R is n ordered set, either < b, = b, or > b. If < b, then z < w. If > b then z > w. If = b we check the reltionship between b nd d nd similrly will conclude tht either z < w if b < d, z > w if b > d, or z = w if b = d. To check trnsitivity, let u = e + fi C nd suppose tht z < w nd w < u. Thus < c or = c nd b < d. Also, c < e or c = e nd d < f. If < c nd c < e then < e becuse R is n 4
ordered set nd in this cse z < u. If < c nd c = e with d < f then < c = e so z < u. If = c nd b < d with c < e then = c < e nd z < u. If = c nd b < d with c = e nd d < f then = e but b < d < f so b < f becuse R is n ordered set nd therefore z < u. Thus, C is n ordered set under this order. This ordered set does hve the lest-upper-bound property. To see this, let E C nd define α = sup { +bi E} nd β = sup {b +bi E}. Then, it is cler α+βi C becuse R hs the lest-upper bound property nd furthermore +bi E, +bi α+βi, (becuse α nd b β) so α + βi is n upper bound for E. Now, suppose there exists γ + δi C such tht γ + δi < α + βi nd γ + δi is n upper bound for E. If γ < α, then we hve γ < α such tht + bi E, contrdicting the definition of α. An identicl contrdiction with the definition of β occurs if γ = α nd δ < β. Therefore, α + βi = sup(e) C nd this ordered set does indeed hve the lest-upper-bound property. 0. Suppose z = + bi, w = u + vi, nd ( ) w + u /2 ( ) w u /2 =, b =. 2 2 Prove tht z 2 = w if v 0 nd tht ( z) 2 = w if v 0. Conclude tht every complex number (with one exception!) hs two complex squre roots. Solution: First, suppose tht v 0. Then ( (u z 2 = ( + bi)( + bi) = ( 2 b 2 2 + v 2 ) /2 ) ( + u (u 2 + v 2 ) /2 ) u ) + 2bi = + 2bi 2 2 ( (u 2 + v 2 ) u 2 ) /2 = u + 2 i 4 = u + vi = w where we needed to use tht v 0 becuse (v 2 ) /2 = v. If v 0 then ( (u ( z) 2 = ( bi)( bi) = ( 2 b 2 2 + v 2 ) u 2 ) /2 ) 2bi = u 2bi = u 2 i 4 = u v i = w. Then it is cler ech complex number (except 0 of course!) hs two complex roots becuse we cn tke either both positive or both negtive squre roots defining nd b bove since if we choose nd b, the minus signs vnish in w = ( 2 b 2 )±2bi. It is cler we cn t hve more thn 2 complex roots becuse if we mixed tking the positive squre root defining nd the negtive squre root defining b, or vice vers, the minus signs would not cncel in the ±2bi term.. If z is complex number, prove tht there exists n r 0 nd complex number w with w = such tht z = rw. Are w nd r lwys uniquely determined by z? Solution: Simply let w = z/ z nd r = z. w is not lwys uniquely determined since if z = 0 we cn choose ny w C such tht w = nd r = 0. r is lwys uniquely 5
determined since z = rw = r w = r = r becuse r 0 by hypothesis. If z 0 then w is uniquely determined becuse it must lie on the rdius which z lies on, where it intersects the unit circle. To see this, let z = + bi nd w = c + di. Then, + bi = rc + rdi nd therefore c = /r nd d = b/r. Hence, since r is uniquely determined, so is w provided z 0. 2. If z,..., z n re complex, prove tht z + + z n z + + z n. Solution: We proceed by induction on n. For n = there is nothing to show nd the cse n = 2 is the tringle inequlity which is given by Theorem.33 on pge 4 of the text. Suppose the result holds for n. Then z + + z n = z + (z 2 + + z n ) z + z 2 + + z n z + z 2 + + z n where the lst inequlity follows by the inductive hypothesis pplied to the term z 2 + + z n. 3. If x, y re complex, prove tht x y x y. Solution: Let x, y C nd let z = y x. Then, by the tringle inequlity we hve tht x + z x + z = x + z x z, hence substituting in the expression for z we obtin tht Now, set z = x y nd proceeding nlogously we find tht y x y x. (3.) y + z y + z = y + z y z, hence substituting in the expression for z we obtin tht x y x y = y x, thus combined with Eqution (3.) bove we see tht x y x y, s desired. Alternte Solution: Let x = + bi nd y = c + di with, b, c, d R. By the tringle inequlity (Theorem.33(e) on pge 4 of the text) we hve tht x + y x + y x + y 2 x 2 + y 2 + 2 x y. Substituting our expressions for x nd y nd noting tht by Definition.32 on pge 4 of the text we hve tht x = ( 2 + b 2 ) /2 we find 2 + c 2 + 2c + b 2 + d 2 + 2bd = ( + c) 2 + (b + d) 2 = ( + c) + (b + d)i 2 = x + y 2 x 2 + y 2 + 2 x y = 2 + b 2 + c 2 + d 2 + 2( 2 + b 2 ) /2 (c 2 + d 2 ) /2 6
which immeditely implies tht c + bd ( 2 + b 2 ) /2 (c 2 + d 2 ) /2. (3.2) Therefore, x y 2 = ( 2 + b 2 ) /2 (c 2 + d 2 ) /2 2 = [ ( 2 + b 2 ) /2 (c 2 + d 2 ) /2] 2 = 2 + b 2 + c 2 + d 2 2 ( ( 2 + b 2 ) /2 (c 2 + d 2 ) /2) 2 + b 2 + c 2 + d 2 2(c + bd) = 2 + c 2 2c + b 2 + d 2 2bd = ( c) 2 + (b d) 2 = ( c) + (b d)i 2 = ( + bi) (c + di) 2 = x y 2 hence we hve x y x y. Since the tringle inequlity ws the fundmentl trick in both solutions, we might suspect tht there is geometric interprettion. There is nd it is esy to visulize if we tret x nd y s vectors in R 2. Then, this merely sttes tht the mgnitude of the vector given by their difference is greter thn or equl to the difference in their mgnitudes. This mkes sense if we consider x nd y to point in opposite directions (i.e. y = αx where α > 0). Then the mgnitude of the vector given by their difference is the sum of their mgnitudes which is greter thn or equl to the difference of their mgnitudes becuse mgnitudes re positive or zero. When x nd y point in the sme direction, we get equlity, otherwise we hve inequlity. When they point in opposite directions we get mximl inequlity. 4. If z is complex number such tht z =, tht is, such tht zz =, compute Solution: Write z = + bi with, b R. Then + z 2 + z 2. + z 2 + z 2 = ( + ) + bi 2 + ( ) + bi 2 = ( + ) 2 + b 2 + ( ) 2 + b 2 = + 2 + 2 + b 2 + + 2 2 + b 2 = 2 + 2 2 + 2b 2 = 2( 2 + b 2 ) + 2 = 2 z 2 + 2 = 4. 5. Under wht conditions does equlity hold in the Schwrz inequlity? Solution: If,..., n nd b,..., b n re complex numbers then the Schwrz inequlity sttes tht n 2 n n i b i i 2 b j 2. 7
We clim tht equlity holds if there exists λ C such tht b i = λ i i =,..., n. Supposing this condition holds we hve n 2 n 2 ( n 2 n )( n ) i b i = i i λ = λ 2 i 2 = λ 2 i 2 i 2 ( n )( n ) = i 2 λ 2 i 2 ( n )( n ) = i 2 λ i 2 n n = i 2 b i 2. 6. Suppose k 3, x, y R k, x y = d > 0, nd r > 0. Prove: () If 2r > d, there re infinitely mny z R k such tht (b) If 2r = d, there is exctly one such z. (c) If 2r < d, there re no such z. z x = z y = r. How must these sttements be modified if k is 2 or? Solution: () If z R k such tht z x = z y = r then z B(x, r) B(y, r), where B(x, r) = {z R k z x = r} is the boundry of the bll of rdius r centered t x. If 2r > d = r > d/2 then the two boundries intersect in surfce (of dimension k 2) nd thus there re infinitely mny points in this intersection which stisfy the given eqution becuse k 3. (b) If r = d/2 then the two boundries intersect in single point, the midpoint of the segment joining x nd y. (c) If r < d/2 then the boundries never intersect nd so no z R k cn be on both boundries simultneously. If k = then there re no z R stisfying both equtions simultneously unless r = d/2. This is becuse there re only two points tht re distnce r from x for given x R. Without loss of generlity we cn ssume x < y. If r = d/2, we see tht z = x + r is the only point tht is distnce r from both x nd y. If r > d/2 or r < d/2 then the boundries (set s of two points) never intersect. This is becuse the points of distnce r from x re x + r nd x r. But, y (x + r) = y x r = d r < d d/2 = d/2 < r so x + r is not of distnce r from y. Similrly y (x r) = y x + r = d + r > r so x r is lso not of distnce r from y. Anlogous results hold when r < d/2 with inequlities reversed. If k = 2 then when r > d/2 insted of infinitely mny points in the intersection of the boundries, there re just two since in this cse the boundries of the bll s round x nd y re circles nd two different circles cn mximlly intersect in only 2 points. The cses r = d/2 nd r < d/2 give the sme results s they did for k 3. 8
7. Prove tht x + y 2 + x y 2 = 2 x 2 + 2 y 2 if x, y R k. Interpret this geometriclly, s sttement bout prllelogrms. Solution: If we write x = (x,..., x k ) nd y = (y,..., y k ) then expnding the left hnd side gives x + y 2 + x y 2 = = = = = 2 k (x i + y i ) 2 + k (x i y i ) 2 k (x i + y i ) 2 + (x i y i ) 2 k (x 2 i + 2x i y i + yi 2 + x 2 i 2x i y i + yi 2 ) k (2x 2 i + 2yi 2 ) k x 2 i + 2 = 2 x 2 + 2 y 2. If we consider the vectors x nd y s representing two sides of prllelogrm then x y nd x + y represent the digonls. The eqution bove is then seen to be nothing more then the generlized Pythgoren Theorem. Considering the cse k = 2 the eqution is merely the sum of the ordinry Pythgoren Theorem for the two digonls. Since for both digonls the sides of the tringle for which they re the hypotenuse re x nd y we hve tht k y 2 i x + y 2 = x 2 + y 2 x y 2 = x 2 + y 2 hence our originl eqution is just the sum of these two. 8. If k 2 nd x R k, prove tht there exists y R k such tht y 0 but x y = 0. Is this lso true if k =? Solution: We cn see right wy tht the sttement must fil for k = becuse in this cse the dot product is nothing more thn multipliction of rel numbers nd since the rels re field xy = 0 = x = 0 or y = 0, i.e. fields hve no zero divisors. This mens there is no notion of perpendiculrity in -dimensionl spce s we d expect since perpendiculrity implies liner independence nd hence extr dimension. For k 2 let x = (x,..., x k ) nd without loss of generlity suppose x 0 becuse otherwise ny non-zero y R k will suffice. We consider two cses bsed on the prity of k. If k is even then we let y = (x 2, x, x 4, x 3,..., x j+, x j,..., x k, x k ) where x j+ is in the j th slot. We then hve x y = x x 2 x 2 x + + x j x j+ x j+ x j + + x k x k x k x k = 0 nd y 0 becuse x 0 by hypothesis. Now, suppose tht k is odd. Since x 0 by hypothesis we hve tht x j 0 for some j k. Supposing tht x j is not the only non-zero entry, 9
we repet the sme rgument nd choose y = (x 2, x,..., x j+, 0, x j,..., x k, x k ) where the 0 is in the j th slot nd y 0 becuse we hve supposed tht x j is not the only non-zero entry. Then we hve x y = x x 2 x 2 x + + x j x j+ + 0 x j+ x j + + x k x k x k x k = 0. If x j is the only non-zero entry then we simply choose y = (,...,, 0,,..., ) where the 0 is in the j th slot. Chpter 2: Bsic Topology. Prove tht the empty set is subset of every set. Solution: Let A be set nd let be the empty set. To show A we need to show tht p, p A. Since there re no p, this sttement is vcuously true nd therefore A. 2. A complex number z is sid to be lgebric if there re integers 0,..., n not ll zero such tht 0 z n + z n + + n z + n = 0. Prove tht the set of ll lgebric numbers is countble. Solution: Let { n } P n = i t i i Z, i = 0,..., n i=0 so P n is the set of ll degree n or smller polynomils with integer coefficients. It is cler tht P n Z n+ given by n i t i ( 0,..., n ) i=0 where mens bijective s sets, so tht in prticulr, P n is countble for ech n. Now, let R n = { C p() = 0 for some p P n } so tht R n is the set of ll complex roots of ll degree n or smller polynomils with integer coefficients. Since P n is countble we hve tht P n N so tht in prticulr to ech polynomil in P n we cn ssocite unique nturl number. Since every degree n or smller polynomil hs t most n complex roots we see tht R n { m,..., n m } m=0 where m N is the index for the elements of the countble set P n nd the set { m,..., n m } is set of n elements for ech m representing the mximum number of roots of ny degree n or smller polynomil. Since the set bove on the right is countble union of countble sets it is lso countble by Theorem 2.2 on pge 20 of the text nd so we hve tht R n is countble. But since the set of ll complex roots of ll polynomils with integer coefficients is simply n=0 R n, it is lso countble by the sme Theorem. 0
3. Prove tht there exist rel numbers which re not lgebric. Solution: We see tht the set of ll rel lgebric numbers is t most countble becuse it is subset of the set of ll complex lgebric numbers which is countble by Exercise 3 bove. But, since R is uncountble (see Theorem 2.4 on pge 30 of the text) there must exist rel numbers which re not lgebric. 4. Is the set of ll irrtionl rel numbers countble? Solution: We hve by the Corollry to Theorem 2.3 on pges 29-30 of the text tht Q is countble. Since R is uncountble (see Theorem 2.4 on pge 30 of the text) we must hve tht the irrtionls re uncountble becuse the union of two countble sets is countble by Theorem 2.2 on pge 29 of the text. 5. Construct bounded set of rel numbers with exctly three limit points. Solution: Consider the set S = {/n n N} { + /n n N} {2 + /n n N}. Then S is bounded becuse d(p, 0) < 3 p S. It is lso cler tht 0,, nd 2 re limit points of S. No other points re limit points of S since round ny other rel number we cn choose smll enough neighborhood such tht it excludes ll the points of S, except possibly itself, since we cn lwys choose rdius smller then n n+ for ny fixed n. 6. Let E be the set of ll limit points of set E. Prove tht E is closed. Prove tht E nd E hve the sme limit points. (Recll tht E = E E ). Do E nd E lwys hve the sme limit points? Solution: Let p be limit point of E. Thus, for ech n N there exists p n E such tht p n p nd d(p, p n ) < 2n. Since ech p n E we hve tht for ech p n there exists point s n E such tht s n p n nd d(p n, s n ) < min{ 2n, d(p n, p)}. Since d(p n, s n ) < d(p n, p) we hve tht s n p. By the tringle inequlity we then hve tht d(p, s n ) d(p, p n ) + d(p n, s n ) < 2n + 2n = n. Thus, every neighborhood of p contins point in E not equl to p itself so p is limit point of E nd hence p E. Therefore, E is closed. Now, let p be limit point of E. Then ny neighborhood of p contins point q E such tht q p. Since E E we hve tht q E so tht in prticulr p is lso limit point of E. Now, suppose tht p is limit point of E. Then, for ech n N there exists point p n E such tht p n p nd d(p, p n ) < 2n. If p n / E then p n E becuse p n E nd we replce p n with n element q n E such tht q n p n nd d(p n, q n ) < min{ 2n, d(p n, p)}. Then q n p becuse d(p n, q n ) < d(p n, p) nd d(p, q n ) d(p, p n ) + d(p n, q n ) < 2n + 2n = n. Therefore, p is lso limit point of E nd so E nd E hve the sme limit points. We cn see tht E nd E do not lwys hve the sme limit points by the simple exmple E = (0, ). Here, the limit points of E re 0 nd so tht E = {0, }. But, E itself hs no limit points by the Corollry to Theorem 2.20 on pges 32-33 of the text becuse it is finite point set. We cn see this directly becuse if s 0, then ny neighborhood of s of rdius r < min{d(s, 0), d(s, )} contins no elements of E nd ny neighborhood of 0 or of rdius less thn contins no other points of E except 0 nd, respectively, hence E hs no limit points. 7. Let A, A 2,... be subsets of metric spce.
() If B n = n A i, prove tht B n = n A i, for n =, 2,... (b) If B = A i, prove tht B A i. Show, by n exmple, tht the inclusion in (b) cn be proper. Solution: () We use the nottion B for the set of limit points of B. For (), suppose tht p B n. Then, either p B n or p B n (or both). If p B n = n A i = p A i for some i hence p A i = p n A i. So, suppose tht p B n so tht p is limit point of B n. Now, suppose tht p / A i i =,..., n, i.e. tht p is not limit point for ny of the A i s. This mens tht for ech i ε i > 0 such tht N(p, ε i ) A i {p}. Here N(p, ε i ) is the bll round p of rdius ε i. Tht is, there is some non-zero neighborhood of p which intersects A i in t most p itself. Letting 0 < ε < min i {ε i } we see tht N(p, ε) A i {p} i =,..., n = N(p, ε) n A i = B n {p} contrdicting tht p is limit point of B n. Therefore, we must hve tht p A i for t lest one i hence p n A i nd we obtin the inclusion B n n A i. Now, let p n A i. Then we hve tht p A i for some i. If p A i B n then p B n. If p A i then every neighborhood of p contins point q A i such tht q p. Since A i B n, we hve tht q B n lso nd hence p B n showing the other inclusion n A i B n nd completing prt (). (b) The proof is identicl to the second inclusion just shown in (). As n exmple, if we let A i = {/i} nd B = A i we see tht 0 B yet 0 / A i for ny i becuse for ech i we cn choose 0 < ε < /i nd then N(0, ε) A i =. Hence the inclusion in prt (b) is proper. 8. Is every point of every open set E R 2 limit point of E? Answer the sme question for closed sets in R 2. Solution: Let p E nd let δ > 0 be such tht B(p, δ) E so tht in prticulr there exists q B(p, δ) such tht q p nd of course q E becuse B(p, δ) E. For exmple, if p = (p, p 2 ) we cn choose q = (p + δ/2, p 2 ). Now, let ε > 0 be given. If ε = δ then by construction B(p, ε) contins point of E distinct from p. If ε < δ then B(p, ε) B(p, δ) E nd so p q = (p + ε/2, p 2 ) B(p, ε) E so tht B(p, ε) contins point of E distinct from p. If ε > δ then B(p, δ) B(p, ε) nd so choosing q = (p +δ/2, p 2 ) B(p, δ) B(p, ε) shows tht B(p, ε) contins point in E distinct from p. Thus every neighborhood of p contins point of E distinct from p hence p is limit point of E nd so every point of E is limit point of E. The sme is not true of closed sets. Consider A = {(n, 0) n Z} R 2. Then A is closed becuse A contins no limit points t ll. This is becuse ny neighborhood of ny point in A with rdius less thn contins no other points of A. Yet, since A is not empty, it contins points which re not limit points. 9. Let E denote the set of ll interior points of set E. [See Definition 2.8(e) on pge 32 of the text; E is clled the interior of E.] () Prove tht E is lwys open. (b) Prove tht E is open if nd only if E = E. (c) If G E nd G is open, prove tht G E. (d) Prove tht the complement of E is the closure of the complement of E. (e) Do E nd E lwys hve the sme interiors? 2
(f) Do E nd E lwys hve the sme closures? Solution: () Let p E. Then by definition there exists δ > 0 such tht N(p, δ) E. By Theorem 2.9 on pge 32 of the text we know tht every neighborhood is open nd thus if q N(p, δ) there exists ε > 0 such tht N(q, ε) N(p, δ) E. Therefore q E nd so N(p, δ) E nd hence E is open. (b) If E = E then E is open by prt (). Now, suppose tht E is open. If p E then there exists δ > 0 such tht N(p, δ) E nd therefore p E by definition. Thus we hve the inclusion E E. By definition, interior points of E re elements of E becuse ny neighborhood of point contins tht point. Therefore E E nd we obtin E = E. (c) Now, let p G E. If G is open, there exists δ > 0 such tht N(p, δ) G E. Thus, p E by definition nd we obtin G E, showing prt (c). (d) The result follows from p (E ) c ε > 0, N(p, ε) E c p E c or ε > 0, q N(p, ε) E c such tht q p p E c or p (E c ) p E c. (e) Consider Q, which hs no interior points becuse ny neighborhood of rtionl must contin irrtionls becuse irrtionls re dense in R, hence ny such neighborhood cnnot be contined entirely in Q. Thus Q =. But, since Q = R (becuse the limit points of Q re precisely irrtionls) we see tht (Q) = R = R (becuse R is open in itself) hence Q (Q). (f) The sme exmple of Q shows tht (f) is lso flse. Q = R = = Q. 0. Let X be n infinite set. For p, q X, define { if p q d(p, q) = 0 if p = q. Prove tht this is metric. Which subsets of the resulting metric spce re open? Which re closed? Which re compct? Solution: To show it is metric, only the tringle inequlity is not obvious from the definition. So, let p, q, r X. Then if p = q we hve tht d(p, q) = 0 d(p, r) + d(r, q) becuse d(p, r), d(r, q) 0. If p q we hve tht d(p, q) = d(p, r) + d(r, q) becuse if r = q we hve tht r p nd then d(p, r) = nd d(r, q) = 0 so the inequlity holds. If r = p then we similrly get tht r q so d(p, r) = 0 nd d(r, q) = so the inequlity still holds. If we finlly hve tht r p, q then both d(p, r) = = d(p, q) nd the inequlity holds. Thus, this is metric. First, note tht N(p, r) = X if r for ny p X nd N(p, r) = {p} if r < for ny p X. Now, let A X nd let p A. Then N(p, 0.5) = {p} A thus every point in A hs neighborhood entirely contined in A, thus A is open nd so every subset of X is open. Since set is closed if nd only if its complement is open by Theorem 2.23 on pge 34 of the text, we see tht every subset of X is lso closed. It is cler from the definition tht ny finite subset of X will be compct. But, if A X is infinite we see tht it cnnot be compct since by Theorem 2.37 on pge 38 of the text n infinite subset of compct set 3
hs limit point in the compct set, yet from this metric we see tht there re no limit points for ny sets. This is becuse ny neighborhood of ny point of rdius less then contins no other points. Hence, the only compct subsets of X re finite ones.. For x R nd y R, define d (x, y) = (x y) 2 d 2 (x, y) = x y d 3 (x, y) = x 2 y 2 d 4 (x, y) = x 2y d 5 (x, y) = x y + x y. Determine, for ech of these, whether it is metric or not. Solution: For ech of these we must determine whether d i (x, y) > 0 for x y, d i (x, x) = 0 x R, d i (x, y) = d i (y, x), nd d i (x, y) d(x, z) + d(z, y). d (x, y) is not metric becuse it does not stisfy the tringle inequlity since d (0, 2) = 4, d (0, ) =, nd d (, 2) = thus 4 = d (0, 2) > d (0, ) + d (, 2) = + = 2. d 2 (x, y) is metric from the following: d 2 (x, y) > 0 if x y nd d 2 (x, x) = 0 for ll x R. Also, d 2 (x, y) = x y = y x = d 2 (y, x). Finlly, since x y x z + z y we hve tht x y x z + z y x z + y z. The lst inequlity follows since + b + b + 2 b = ( + b) 2 = + b + b. d 3 (x, y) is not metric since d(, ) = 0. d 4 (x, y) is not metric since d 4 (, ) = 0. To show tht d 5 (x, y) is metric we prove the more generl result tht whenever d(x, y) is metric then d d(x, y) (x, y) = + d(x, y) is lso be metric. Since d(x, y) = x y is the ordinry metric on R, it will follow tht d 5 (x, y) is metric. Becuse d(x, y) is metric we hve tht d (x, y) > 0 if x y, d (x, x) = 0 for ll x R, nd d (x, y) = d (y, x). Now, let p = d(x, y), q = d(x, z), r = d(z, y). Since d(x, y) is metric we hve tht p, q, r 0 nd p q + r = p q + r + 2qr + pqr Thus, d (x, y) is metric. = p + pq + pr + pqr (q + pq + qr + pqr) + (r + pr + qr + pqr) = p( + q)( + r) q( + r)( + p) + r( + q)( + p) = p + p q + q + r + r = d (x, y) d (x, z) + d (z, y). 2. Let K R consist of 0 nd the numbers /n, for n =, 2, 3,... Prove tht K is compct directly from the definition (without using the Heine-Borel theorem). Solution: Let {U α } be n open cover of K. Let 0 U α0. Since U α0 is open there exists δ > 0 such tht B(0, δ) U α0 hence we cn choose ny n such tht /n < δ nd then 4
/n B(0, δ) U α0. Thus, we lso hve tht /m B(0, δ) U α0 for ll m > n. Then, let /i U αi for i = n, n 2,...,. Since there re only finitely mny i s we hve tht {U αj } n j= U α 0 is finite subcover. 3. Construct compct set of rel numbers whose limit points form countble set. Solution: Consider the set A m = {m + /n n =, 2,...}. Then A m is compct for ll m N by Exercise 2 bove (it is merely trnsltion of the set in tht exercise) nd the set of limit points of A m is {m}. Therefore, if we let A = A is countble union of countble sets so it is countble by Theorem 2.2 on pge 29 of the text. The set of limit points of A is then precisely N so is countble lso. 4. Give n exmple of n open cover of the segment (0, ) which hs no finite subcover. Solution: Consider the open cover {(/n, )} n=2. This covers (0, ), becuse if r (0, ), let n 2 be such tht /n < r. Then r (/n, ). Now, consider ny finite collection in this open cover {(/i, )} i I where I <. Let n 2 be such tht n > mx i I {i}. Then we hve tht /n (0, ) yet /n < /i i I nd thus /n / i I (/i, ) hence {(/i, )} i I is not n open cover of (0, ). Therefore, no finite subcover of this cover exists. 5. Show tht Theorem 2.36 nd its Corollry become flse (in R, for exmple) if the word compct is replced by closed or by bounded. Solution: Theorem 2.36, on pge 38 of the text, sttes tht If {K α } is collection of compct subsets of metric spce X such tht the intersection of every finite subcollection of {K α } is nonempty, then K α is nonempty. Its Corollry sttes tht if {K n } is sequence of nonempty compct sets such tht K n K n+ (n =, 2,...), then K n is nonempty. Consider the sets K n = (0, /n). Then ech K n is bounded nd for ny finite subcollection {K i } i I where I < let n = min i I {i}. Then i I K i = (0, /n). Yet we hve tht K n = since if 0 < r R we cn choose n N such tht /n < r. But then r / (0, /n) = r / K n. Now, let K n = {m N m n}. Then ech K n is closed since it hs no limit points (becuse it is discrete), so it vcuously contins them. Then, if {K i } i I is finite subcollection where I <, let m N such tht m > mx i I {i}. Then, m K i i I nd so m i I K i. But, we hve tht K n = becuse if m N, let i N be such tht i > m. Then m / K i nd so m / K n. 6. Omitted. 7. Omitted. 8. Omitted. 9. Omitted. 20. Omitted. A i 5
2. Omitted. 22. A metric spce is clled seprble if it contins countble dense subset. Show tht R k is seprble. Hint: Consider the set of points which hve only rtionl coordinte. Solution: Let Q k = {(x,..., x k ) R k x i Q i =,..., k}. If (x,..., x k ) R k let ε > 0 be given. Then, since Q is dense in R, choose p,..., p n Q such tht x i p i < k ε. Then, writing p = (p,..., p n ) Q k nd x = (x,..., x n ) R k we see tht ( k ) /2 ( k ) d(p, x) = x i p i 2 /2 < k ε2 = ε. Therefore, Q k is dense in R k nd Q k is countble becuse it cn be relized s the disjoint union of Q, k times. 23. A collection {V α } of open subsets of X is sid to be bse if the following is true: For every x X nd every open set G X such tht x G, we hve x V α G for some α. In other words, every open set in X is the union of subcollection of {V α }. Prove tht every seprble metric spce hs countble bse. Hint: Tke ll neighborhoods with rtionl rdius nd center in some countble dense subset of X. Solution: Since X is seprble, by definition it hs countble dense subset S X. Let S = {p, p 2,...} nd enumerte Q = {q, q 2,...}. Let V ij = N(p i, q j ). Then, {V ij } i,j= is countble collection of open sets in X by Theorem 2.2 on pge 29 of the text. Let x X nd let G X be open such tht x G. Since G is open, there exists δ > 0 such tht N(x, δ) G. Let p i S such tht d(p i, x) = ε < δ/2 nd then choose q j Q such tht ε < q j < δ/2, both of which re possible becuse S is dense in X nd Q is dense in R. Then, consider V ij = N(p i, q j ). Since d(x, p i ) = ε < q j we see tht x V ij. Now, let V ij. Then, d(, x) d(, p i ) + d(p i, x) < q j + δ/2 < δ. Therefore N(x, δ) G nd so we hve tht x V ij N(x, δ) G showing tht {V ij } i,j= is countble bse. Chpter 3: Numericl Sequences nd Series. Prove tht convergence of {s n } implies convergence of { s n }. Is the converse true? Solution: Suppose tht s n s. Let ε > 0 be given nd let N be such tht s n s < ε n N. Then, by Exercise #3 in Chpter bove, we hve tht sn s sn s < ε n N Thus s n s. The converse is flse becuse if we let s n = ( ) n then s n converges, yet s n does not. 2. Omitted. 3. If s = 2 nd s n+ = 2 + s n prove tht {s n } converges, nd tht s n < 2 for ll n. 6
Solution: s = 2 < 2. Suppose inductively tht s n < 2. Then, s n+ = 2 + s n < 2 + 2 < 2 becuse 2 + 2 < 2 + 2 = 4 = 2 + 2 < 4 = 2. Thus we see tht s n < 2 for ll n. We lso hve tht s 2 = Suppose inductively tht s n > s n. Then s n+ = 2 + 2 > 2 = s. 2 + s n > 2 + s n = s n nd therefore we hve tht s n > s n for ll n by induction. Thus the sequence is incresing nd since > s n > s n > > s > 0 we hve n incresing sequence of nonnegtive terms which is bounded bove. Hence by Theorem 3.24 on pge 60 of the text {s n } converges. 4. Find the upper nd lower limits of the sequence {s n } defined by s = 0, s 2m = s 2m, s 2m+ = 2 2 + s 2m. Solution: If we write out the first few terms we obtin s = 0 s 2 = 0 s 3 = 2 s 4 = 4 s 5 = 2 + 4 s 6 = 4 + 8 thus we re led to write s 2m+ = s 7 = 2 + 4 + 8 s 8 = 4 + 8 + 6 s 2m = m ( ) k 2 m m ( ) k 2 m 2. k= k=2 (4.) To prove this we proceed by induction on m. For m = we hve tht s 3 = /2 = k= (/2)k. For m = 2 we hve tht s 5 = /2 + /4 = 2 k= (/2)k nd s 4 = /4 = 2 k=2 (/2)k. Now, 7
suppose inductively tht Eqution (4.) holds for m 2. Then, s 2(m+)+ = 2 + s 2(m+) = 2 + s 2m+ 2 s 2(m+) = s 2m+ 2 = 2 m k= ( ) k = 2 = 2 + m m k= ( 2 k= ( 2 ) k+ m+ = ) k+ = m+ k=2 ( 2 k= ) k ( ) k 2 nd so Eqution (4.) holds. We thus see tht there re only two subsequentil limits lim s 2m = m lim s 2m+ = m ( ) k = (4.2) 2 2 k=2 ( ) k = (4.3) 2 where the two limiting vlues re given by Theorem 3.26 on pge 6 of the text (geometric series). These re the only possible limits becuse if we hve subsequence {s nk } then either {n k } hs infinitely mny evens, infinitely mny odds, or infinitely mny of both. If there re infinitely mny of both even nd odd numbers, then {s nk } wouldn t be Cuchy sequence becuse for ll N we could find n k, n k2 N such tht n k ws even nd n k2 ws odd nd then we would hve tht s nk s nk2 /2. If there re only finitely mny odds in {n k } then we would hve tht lim k s nk = lim m s 2m. Specificlly, letting N > mx k {n k odd } we see tht s nk s nj = n k k=n j (/2) k for ll n k n j N precisely becuse both n j nd n k re even. Since k=2 (/2)k converges, it is Cuchy sequence by Theorem 3. on pge 53 of the text nd so n k k=n j (/2) k cn be mde rbitrrily smll for lrge enough n j nd n k. If there re only finitely mny evens in {n k } then we would hve tht lim k s nk = lim m s 2m+. Specificlly, letting N > mx k {n k even } we see tht s nk s nj = n k k=n j (/2) k for ll n k n j N precisely becuse both n j nd n k re odd. Since k= (/2)k converges, it is Cuchy sequence by Theorem 3. on pge 53 of the text nd so n k k=n j (/2) k cn be mde rbitrrily smll for lrge enough n j nd n k. Since Equtions (4.2) nd (4.3) re the only two subsequentil limits we find tht the upper nd lower limits re k= lim inf s m = m 2 lim sup s m =. m 5. For ny two rel sequences { n }, {b n }, prove tht lim sup( n + b n ) lim sup( n ) + lim sup(b n ), n n n provided the sum on the right is not of the form. Solution: If either lim sup on the right side bove is infinite, we re done, so we cn ssume tht both re finite. Assume lso tht the left side is finite (hence our discussion here is incomplete). We mke some generl remrks bout this sttement, but we will prove it by 8
ctully using n equivlent definition of lim sup which is nonetheless esier to work with thn Rudin s. First, let = lim sup n ( n ), b = lim sup n (b n ), nd c = lim sup n ( n + b n ). Then, let nk + b nk γ be convergent subsequence of the sequence { n + b n }. If the component subsequences { nk } nd {b nk } both converge, sy to α nd β, respectively, then by Theorem 3.3 on pge 49 of the text, we hve tht γ = lim k ( n k + b nk ) = lim k n k + lim k b n k = α + β + b. where the lst inequlity follows by the definition of nd b s lim sup s. Thus, we need only consider subsequences { nk +b nk } where one, or both, of the the component subsequences do not converge. But, we cn ctully do better thn this becuse we cn show tht if one of the component subsequences converges, then both must. To see this, without loss of generlity, let nk + b nk γ nd suppose tht b nk β. Then, by the tringle inequlity, nk (γ β) nk (γ b nk ) + (γ b nk ) (γ β) = ( nk + b nk ) γ + (γ b nk ) (γ β). Since b nk β we see tht γ b nk γ β becuse γ is constnt nd so both terms on the right bove cn be mde rbitrrily smll for lrge enough k. This shows tht nk γ β nd therefore if one of the component subsequences of the convergent subsequence { nk + b nk converges, both must. Hence, we re reduced to considering only those convergent subsequences { nk + b nk } in which both component subsequences do not converge. Tht is, sequences such s nk = ( ) k nd b nk = ( ) k+ so tht neither converges individully, but their sum does. Now, we would be done if we could show tht γ = lim k ( nk + b nk ) = lim sup k ( nk + b nk ) lim sup k ( nk ) + lim sup k (b nk ) + b which is the sttement of the problem, but restricted to the subsequence { nk + b nk }. Thus, we could sy repet the bove rgument nd for every convergent sub-subsequence nkj + b nkj such tht neither component sub-subsequence converges, repet the rgument gin. Eventully, it will end becuse eventully some sub- -subsequence will hve only convergent component sequences, since we re ssuming tht, b, nd c re finite. But, this is hrd to mke rigorous nd so we will now prove the result by resorting to different, but equivlent, definition of the lim sup. Consider sequence { n } nd let A k = sup{ m m k} nd define A = lim k A k (llowing it to be ± ). Now, choose n such tht n A <, which is possible by the definition of A s the sup, so n. Now, choose nk inductively such tht nk A k < /k nd such tht nk > nk. Tht is, by definition of A k, we hve tht there exists K such tht m A k < /k for ll m K. So, simply require tht n k > mx{k, n k }. Then, this defines subsequence of { n } which by construction converges to A, thus we hve tht lim sup n ( n ) A. Now, by Theorem 3.7() on pge 56 of the text we see tht lim sup n ( n ) is ctully subsequentil limit of the sequence { n } hence there exists some subsequence nk lim sup n ( n ). Since by construction we hve tht nk A nk tking the limit s k of both sides gives lim sup n ( n ) A thus A = lim sup n ( n ) nd our A is n equivlent formultion of the lim sup to Rudin s. The proof of the bove inequlity is now nerly trivil since if nk + b n+k is ny subsequence of { n + b n } whose limit exists then we of course hve tht nk + b nk sup m k { nm } + sup m k {b nm } = A nk + B nk. Tking the limit of both sides gives lim k ( nk +b nk ) A+B, where B is defined nlogously to A bove. Since { nk +b nk } ws ny subsequence with limit we see tht lim sup n ( n +b n ) A + B = lim sup n ( n ) + lim sup n (b n ). 9
6. Omitted. 7. Prove tht the convergence of n implies the convergence of if n 0. n n, Solution: By the Cuchy-Schwrz inequlity (Theorem.35 on pge 5 of the text), we hve tht n ( n )( n j b j j 2 b j ). 2 j= j= j= Thus, for ech n =, 2,... we hve n j= n ( n n n n)( j= j= ) ( )( n 2 n j= j= ) n 2 < where the second inequlity follows becuse ll terms re positive nd the lst inequlity follows becuse both sequences on the right re convergent. Thus the prtil sums, s n = n n n, form bounded sequence. Since ll the terms re nonnegtive, by Theorem 3.24 on pge 60 of the text the series n n converges. 8. If n converges, nd if {b n } is monotonic nd bounded, prove tht nb n converges. Solution: Since n converges, its prtil sums form bounded sequence by definition. By Theorem 3.4 on pge 55 of the text, we see tht {b n } converges becuse it is monotonic nd bounded, so let b n b. Since {b n } is monotonic, it is either incresing or decresing. If it is incresing, define c n = b b n nd in this cse the sequence {c n } is decresing nd we hve tht c c 2 nd lso c n 0 becuse b n b. On the other hnd, if {b n } is decresing then let c n = b n b nd in this cse {c n } is still decresing sequence so c c 2 nd we still hve tht c n 0. Thus, in either cse, by Theorem 3.42 on pge 70 of the text, we hve tht the sequence nc n converges. First, suppose tht {b n } ws incresing nd so c n = b b n. Then, the prtil sums C m = m n c n = m m ( n b n b n ) = b n re convergent sequence, sy C m c. But then m n b n b n c, hence nb n converges. In the other cse we would obtin m n b n c + b m n b n where here c = nc n nd c n = b n b nd so nb n still converges. 20 n
9. Omitted. 0. Omitted.. Suppose tht n > 0 nd s n = + + n, nd n diverges. () Prove tht n= n + n diverges. (b) Prove tht N+ s N+ + + N+k s N+k nd deduce tht n s n diverges. (c) Prove tht n s 2 n nd deduce tht n converges. s 2 n (d) Wht cn be sid bout s n s n s N s N+k n + n n nd n + n 2 n? Solution: () Suppose tht n + n converges. Observe tht n 0 + n n + 0 n 0. n Since we re supposing tht n + n converges we hve by Theorem 3.23 on pge 60 0f the text tht n + n 0 = n 0 by the bove. Hence, there exists N such tht n < for ll n N. Now, let ε > 0 be given. Then there exists N 2 such tht m + m + + n + n < ε 2 n, m N 2. Letting N = mx{n, N 2 } we hve tht ε 2 > m + + n > m + m + n + + + n + = m 2 + + n 2 for ll n, m N thus ε > m + + n n, m N = n converges, contrdiction. Thus n + n diverges. (b) Becuse n > 0 for ll n we hve tht s n+ > s n for ll n. Thus, N+ s N+ + + N+k s N+k > N+ + + N+k s N+k = s N+k s N s N+k = s N s N+k. 2
Suppose now tht n s n converges nd let ε > 0 be given. Then, there exists n N such tht for ll n, m N we hve tht ε > m s m + + n s n. In prticulr, for m = N + nd n = N + k we obtin tht ε > N+ + + N+k > s N. s N+ s N+k s N+k Now, since n diverges nd n > 0 so the s n re incresing, we see tht s N+k s k. Thus we cn choose k lrge enough such tht s N+k > 2s N = s N s N+k < 2 becuse N is fixed. But then, we would obtin ε > s N s N+k > 2 = 2, contrdicting tht ε cn be chosen rbitrry, i.e. just choose ε = 2 t the beginning. Thus, n s n diverges. (c) As in prt (b), since n > 0 for ll n we hve tht s n > s n for ll n. Thus Then since s n > 0 for ll n, thus s n s n = s n s n s n s n k n=2 n s 2 n < k n=2 so the prtil sums of n s 2 n we hve tht n s 2 n > s n s n s 2 n = n s 2 n n 2. ( ) = < =, (.) s n s n s s k s k n s 2 n < 2 form bounded sequence nd since n > 0 for ll n, converges by Theorem 3.24 on pge 60 of the text. We could lso tke the limit of both sides of Eqution (.) s k nd note tht s k becuse n diverges nd n > 0 for ll n so s n. (d) n +n n my either diverge or converge. If we let n = n then n + n n = /n 2 = 2 n =. On the other hnd, if we let n = n log(n) p where p > nd n 2 then n + n n = n log(n) p + n ( n log(n) p ) = n log(n) p ( + log(n) p ) = n log(n) p + n < n log(n) p 0 s k 22
thus On the other hnd, n +n n converges by Theorem 3.29 on pge 62 of the text for this choice of n. n + n 2 n = / n + n 2 n 2 where the inequlity follows becuse n > 0 for ll n. Thus, n +n 2 n converges. Chpter 4: Continuity. Suppose f is rel function defined on R which stisfies lim[f(x + h) f(x h)] = 0 h 0 for every x R. Does this imply f is continuous? Solution: No, this does not imply f is continuous becuse this sttement merely sys tht lim t x + f(t) = lim t x f(t), tht is, the right nd left hnded limits of f re equl, but it sys nothing bout whether these ctully equl f(x) itself. Tht is, by Theorem 4.6 on pge 86, f is continuous t x lim t x f(t) = f(x) nd by the comment in Definition 4.25 on pge 94 we see tht lim t x f(t) exists lim t x + f(t) = lim t x f(t) = lim t x f(t). So, if we define { if x Z f(x) = 0 if x / Z then t ny x Z we hve tht lim h 0 f(x + h) = lim t x + f(t) = 0 = lim t x f(t) = lim h 0 f(x h) but of course lim t x + f(t) = lim t x f(t) = 0 = f(x). Tht is, lim t x f(t) = 0 = f(x) so f is not continuous t ny x Z. 2. If f is continuous mpping of metric spce X into metric spce Y, prove tht f(e) f(e). Show by n exmple tht f(e) cn be proper subset of f(e). Solution: If E is empty then the inclusion holds trivilly, so suppose tht it is not empty. Let x E so f(x) f(e) f(e). Thus, x f (f(e)) nd since x ws rbitrry we see tht E f (f(e)). But, f(e) is closed in Y (Theorem 2.27() on pge 35 of the text) nd since f is continuous we hve tht f (f(e)) is then closed in X by the Corollry to Theorem 4.8 on pge 87 of the text. But, this implies tht E f (f(e)) becuse E is the smllest closed set contining E so it is subset of ny closed set contining E, i.e. this follows from Theorem 2.27(c) on pge 35 of the text. But, this is precisely the sttement tht f(e) f(e)). To see tht this inclusion cn be proper, let E = Z nd define f : Z R by f(n) = /n. Then, f is continuous (in fct, it is uniformly continuous if we tke δ < becuse then d(n, m) < = n = m when n, m Z nd thus d(f(n), f(m)) = 0 < ε ε > 0). See the comment t the end of the proof of Theorem 4.20, on pge 92 of the text. But, since Z is closed, we hve tht f(z) = f(z) = {/n n Z} f(z) = {/n n Z} = {/n n Z} {0}, hence the inclusion is proper. Alternte Solution: If f(e) is empty then the conclusion holds trivilly, so suppose tht it is not nd let y f(e). Thus, there exists point p E such tht f(p) = y. If p E then we hve tht y = f(p) f(e) f(e). So, suppose tht p E, so p is limit point of E. Since f is continuous t p, for ll ε > 0 there exists δ > 0 such tht d(x, p) < δ = d(f(x), f(p)) < ε. But, since p is limit point of E we see tht there lwys exists x N(p, δ) for some x E nd for ny δ > 0. Thus, for such n x, we hve tht d(f(x), f(p)) < ε which implies tht 23
f(p) is limit point of f(e) since we cn lwys find points in f(e) rbitrrily close to it. Thus y = f(p) f(e) nd so f(e) f(e). 3. Let f be continuous rel function on metric spce X. Let Z(f) (the zero set of f) be the set of ll p X t which f(p) = 0. Prove tht Z(f) is closed. Solution: Let p n p in Z(f) so tht p n Z(f) for ll n =, 2,... Of course p X nd since f is continuous on X we see tht lim q p f(q) = f(p) by Theorem 4.6 on pge 86 of the text. Then, by Theorem 4.2 on pge 84 of the text, we see tht 0 = lim n f(p n ) = f(p) (where p n p n becuse otherwise we re done) hence p Z(f) nd so Z(f) is closed. Another pproch is to show tht Z(f) c = X\Z(f) is open. To this end, let x E = Z(f) c so tht f(x) 0. Without loss of generlity, suppose tht f(x) > 0. Since f is continuous on X we see tht for ε = f(x)/2 > 0 there exists δ > 0 such tht d(f(x), f(y)) = f(x) f(y) < ε = f(x)/2 whenever d(x, y) < δ. But, this implies tht f(y) 0 becuse otherwise f(x) f(y) = f(x) > ε, contrdiction. This mens tht N(x, δ) E so tht in prticulr, E is open. Thus, Z(f) = E c is closed. If we hd tht f(x) < 0 we would simply let ε = f(x)/2 > 0. 4. Let f nd g be continuous mppings of metric spce X into metric spce Y, nd let E be dense subset of X. Prove tht f(e) is dense in f(x). If g(p) = f(p) for ll p E, prove tht g(p) = f(p) for ll p X. (In other words, continuous mpping is determined by its vlues on dense subset of its domin.) Solution: First, let y f(x), let y = f(x) nd let ε > 0. Then, becuse f is continuous, there exists δ > 0 such tht d(f(x), f(p)) < ε whenever d(x, p) < δ. But, since E is dense in X there exists p E such tht d(x, p) < δ, which gives tht d(y, f(p)) = d(f(x), f(p)) < ε hence f(e) is dense in f(x) becuse we cn find n element of f(e) rbitrrily close to ny element of f(x). Now, let ε > 0 be given nd suppose tht g(p) = f(p) for ll p E. Let q X. Becuse f is continuous t q, there exists δ > 0 such tht d(f(p), f(q)) < ε/2 whenever d(p, q) < δ. Since g is lso continuous t q let δ 2 > 0 be such tht d(g(p), g(q)) < ε/2 whenever d(p, q) < δ 2 nd let δ = min{δ, δ 2 }. Since E is dense in X we cn find p E such tht d(p, q) < δ. Then, d(g(q), f(q)) d(g(q), g(p)) + d(g(p), f(p)) + d(f(p), f(q)) = d(g(q), g(p)) + d(f(p), f(q)) < ε 2 + ε 2 = ε, where p E so g(p) = f(p). Thus, since ε ws rbitrry we must hve tht g(q) = f(q) hence they gree on ll of X. Another wy to see this is tht if q X\E then we cn find sequence p n q such tht p n E for ll n nd p n q for ll n becuse E is dense in X, i.e. every element of X is point or limit point of E, so q must be limit point of E. But then becuse g nd f re both continuous we cn interchnge limits (by Theorem 4.6 on pge 86 nd Theorem 4.2 on pge 84 of the text) so we obtin, g(q) = g( lim n p n) = lim n g(p n) = lim n f(p n) = f( lim n p n) = f(q) becuse g(p n ) = f(p n ) n since they gree on E. 24
5. If f is rel continuous function defined on closed set E R, prove tht there exist continuous rel functions g on R such tht g(x) = f(x) for ll x E. (Such functions g re clled continuous extensions of f from E to R.) Show tht the result becomes flse if the world closed is omitted. Extend the result to vector-vlued functions. Hint: Let the grph of g be stright line on ech of the segments which constitute the complement of E (compre Exercise 29, Chpter 2). The result remins true if R is replced by ny metric spce, but the proof is not so simple. Solution: By Exercise 20 in Chpter we see tht every open set of R is n t most countble union of disjoint open sets. Hence, since E is closed, we cn write E c = n ( i, b i ) where i < b i < i+ < b i+ nd we llow n to be. Then, on ech ( i, b i ) define g(x) = f( i ) + (x i ) f(b i) f( i ) b i i (5.) for x ( i, b i ). Tht is, the grph of g is nothing more thn the stright line connecting the points f( i ) nd f(b i ), where i, b i E becuse i, b i / E c. We let g(x) = f(x) for x E. Then it is cler tht g is continuous on the interior of E nd since it is liner function on E c it is continuous there lso. Hence, we need only check tht it is continuous t the boundry of E, nmely the points { i, b i } n. But, from Eqution (5.) we see tht lim t i g(t) = f( i ) = g( i ) becuse i E, hence g is continuous t i by Theorem 4.6 on pge 86 of the text. Similrly lim t bi g(t) = f(b i ) = g(b i ) becuse b i E hence g is continuous extension of f. The fct tht E is closed is essentil becuse if we consider f(x) = /x on E = (0, ), for exmple, then f is continuous on E yet there is no continuous extension g of f to R such tht g = f on E yet g(0) R since s x 0 we hve tht f(x) hence we would lso hve tht g(x) s x 0. The vector-vlued cse is simply nturl extension of the single vrible cse since if f is continuos vector vlued function on R we cn write f = (f,..., f n ) nd so f i is continuous for ech i =,..., n becuse f is continuous. Hence, by the bove, we cn extend ech f i to continuous function g i on R nd letting g = (g,..., g n ) we see tht g is then continuous extension of f to R n. 6. Omitted. 7. Omitted. 8. Let f be rel uniformly continuous function on the bounded set E in R. Prove tht f is bounded on E. Show tht the conclusion is flse if boundedness of E is omitted from the hypothesis. Solution: Since f is uniformly continuous, there exists single δ > 0 such tht whenever x y < δ we hve tht f(x) f(y) < for ll x, y E. Since E is bounded there exists M > 0 such tht M < x < M for ll x E. Now, note tht if N 2M/δ is n integer, then N neighborhoods of size δ put side by side cover t lest ll but finite number of points in ( M, M) (missing the points tht lie between the boundries of two 25