Chapter 3 Selected Exercises (Rudin)

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1 W. Rudi, Priciples of Mathematical Aalysis, rd Editio. Prove that covergece of {s } implies covergece of { s }. Is the coverse true? (i) Suppose {s } coverges to some s, the for ε > 0, there exists a iteger N such that s s < ε for all N. It follows that s s s s < ε for N, ad the { s } coverges to s. (ii) No. For example, let s = ( ) for =,,... ( ). Calculate lim +. We have [ ( ( ) ] lim + ) = lim ( + ) = lim ++ ( ) = lim ++ = lim + + =. If s =, ad s + = + s ( =,,,...) prove that {s } coverges, ad that s < for =,,,... First, it is easy to prove that s < for =,,..., ad that {s } is bouded above. To show it, use iductio o. At first, s = < is immediate. Now suppose s < for some positive iteger, the s + = + s < + <

2 Heces < forall. Next, weshowthat{s }ismootoicallyicreasig, ad so the covergece of {s } follows by Theorem.4, completig this exercise. Use iductio agai, at first, clearly s < s. Now suppose s s + for some positive iteger, the s + = + s + s + = s + Hece {s } is mootoically icreasig. 4. Fid the upper ad lower limits of the sequece {s } defied by s = 0; s m = s m ; s m+ = +s m By some observatio, we show that, for m =,,..., s m = m ad s m+ = m Note that the secod equality is a direct cosequece of the first oe, so we oly have to prove the first. Use iductio o m, for m =, we have s = s = 0 =. Now suppose s m = m for some iteger m, the s (m+) = s m+ = s m+ = ( ) +s m = ( ) m = m+ Hece s m = m for all m. The above argumet gives the followig results lim sups = ad limif s = 5. For ay two real sequeces {a }, {b }, prove that lim sup (a +b ) limsup a +limsupb

3 provided the sum o the right is ot of the form. W.l.o.g., we may cosider the followig three cases, where the first two follow by the assumptio that (ad + ) is exceptioal. Case : If limsup a = + or limsup b = +, the the r.h.s. of the iequality must be either of the form + or +r, for some real umber r. Hece the result is trivial. Case : If limsup a = ad {b } has a upper boud M. The for each real umber x, there exists N such that N implies a < x, ad so a +b a +M < x+m Sicexisarbitrary,wehavelim (a +b ) =,i.e.,limsup (a +b ) =. Hece the result follows. Case : If limsup a = a ad limsup b = b, where a,b R. Suppose limsup (a +b ) > a+b, the we ca choose a subsequece { k } of icreasig positive itegers such that lim k (a k +b k ) > a + b. Choose a umber x 0 such that lim k (a k +b k ) > x 0 > a + b, the there exists K such that k K implies a k +b k > x 0. O the other had, sice x 0 + a b > a ad x 0 + b a > b, by Theorem.7(b), there exist K ad K such that k K implies a k < x 0 + a b, ad k K implies b k < x 0 + b a. Put K = max{k,k }, the k K implies a k +b k < ( x0 + a b ) + ( x0 + b a ) = x 0 Put M = max{k,k }, the k M implies both a k +b k > x 0 ad a k +b k < x 0. This gives a cotradictio. 6. Ivestigate the behavior (covergece or divergece) of a if (a) a = + ; + ; (b) a = (c) a = ( ) ; (d) a = z, for complex values of z. (a) Sice m a = = ( ) ( + )+ + ( m+ m ) = m+

4 we have a = lim m ( m+ ) =, i.e., a diverges. (b) Sice + = ( ++ ) = The series a coverges by the compariso test with the p-series where p =. (c) Sice lim ( a = lim ) = = 0 where the third equality follows by Theorem.0(c). By the root rest, a coverges. (d) We skip this questio. 7. Prove that the covergece of a implies the covergece of if a 0. a If a 0 for =,,..., the by arithmetic-geometry iequality, we have a (a + ) So if a coverges, the addig the covergece of, ad by Theorem.47, that (a + ) = ( a + ) ( a + ) also coverges. Hece a coverges by the com- the series pariso test., 8. If a coverges, ad if b is mootoic ad bouded, prove that a b coverges.

5 Sice {b } is mootoic ad bouded, we may oly cosider the followig two cases. Case : If b ց b, put c = b b. The c ց 0, which follows by Theorem.4 that a c coverges. Also, by Theorem.47, ba coverges. Fially sice we ca write a b = a (c +b) = (a c +ba ) By Theorem.47 agai, a b coverges. Case : If b ր b, put d = b b. The the argumet is similar to that i Case. 0. Suppose that the coefficiet of the power series a z are itegers, ifiitely may of which are distict from zero. Prove that the radius of covergece is at most. Sice the sequece {a } cotais ifiitely may ozero umbers, we may let { k } be a subsequece of positive itegers such that a k 0 for each k. Moreover, sice a is a iteger for each, we have a k for each k, ad thus k ak. Hece limsup a or R = lim sup a. Suppose a > 0, s = a + +a, ad a diverges. (a) Prove that a +a diverges. (b) Prove that ad deduce that a s diverges. (c) Prove that a N+ s N+ + + a N+k s N+k s N s N+k a s s s ad deduce that a coverges. s (d) What ca be said about a +a ad a + a?

6 (a) To show this, we oly have to cosider the followig two cases. a Case : If {a } is ot bouded above, the lim +a 0. Hece a +a diverges. Case : If {a } is bouded above, choose a fixed umber M > 0 such that a M for all. The a +a a +M for all. Sice a +M apparetly diverges, the divergece of a +a follows by the compariso test. (b) Sice a > 0 for each, the sequece {s } is mootoically icreasig. Thus a N+ s N+ + a N+ s N+ + + a N+k s N+k a N+ +a N+ + +a N+k s N+k = s N+k s N s N+k = s N s N+k Now, to show that a s diverges, let t = a s + a s + + a s. Sice {s } is mootoically icreasig, ad sice a diverges, {s } is ot bouded above. So give a positive iteger N, there is a aother positive iteger M with M > N, such that M implies s > s N, or s N s <. It follows that t M t N = a N+ s N+ + a N+ s N+ + + a M s M s N s M > = ad the the sequece {t } is ot Cauchy. So {t } diverges, i.e., a s diverges. (c) First, the iequality follows by writig a s = s s s s s s s = s s Now, to show that a s coverges, observe that a k s k= k = a a + a k s k= k a + ( k= s k s k ) = a s Sice s ր, the r.h.s. of the above iequality coverges to a as (this is a telescopig series ( e. wikipedia. org/ wiki/ Telescopig_ series)). So we coclude that a s coverges. (d) (i) The series a +a may be either coverget or diverget, we shall give

7 two examples for them. Example : Let a = for =,,... The a diverges, ad a also diverges. Example : Let if is a perfect square; a = if otherwise a +a Note that a still diverges. However, if =,4,9,6,5,...; if otherwise which follows that a +a coverges. (ii) The series a + a apparetly coverges, sice. Suppose a > 0 ad a coverges. Put r = m= a m (a) Prove that a m + + a r r m r r m if m <, ad deduce that a r diverges. (b) Prove that a < ( r r + ) r ad deduce that a r coverges. (a) First, the iequality follows by writig, for m <, a m r m + a m+ r m+ + + a r a m +a m+ + +a r m a + a a a =. > a m +a m+ + +a r m = r m r r m = r r m +a =

8 Now, to show that a r diverges, let s = a r + a r + + a r. Sice r ց 0 (if we let t = m= a m ad a = t, the t ր t ad t + r = t, so r = t t ց 0), give a positive iteger M, there exists a aother positive iteger N with N > M +, such that N implies r < r M+ r, or r M+ <. It follows that s N s M = a M+ r M+ + a M+ r M+ + + a N r N > r N r M+ > ad the the sequece {s } is ot Cauchy. So {s } diverges, i.e., a r diverges. (b) First, the iequality follows by writig a r = r r + r = < r + r + r ( r r + ) r + r r ( r r + ) = ( r r + ) Now, to show that a r coverges, observe that k= a k rk < ( rk ) r k+ = ( r r + ) k= Sice r ց 0, the r.h.s. of the above iequality coverges to r as. So we coclude that a r coverges.. Prove that the Cauchy product of two absolutely coverget series coverges absolutely. Let a ad b betwoabsolutelycovergetseries, ad c bethecauchy product of the two give series. Note that c = a k b k a k b k k=0 Sice both a ad b coverge absolutely, if we let d = k=0 a k b k, the by Theorem.50, d coverges. Sice c d, by the compariso test, c coverges. k=0

9 4. If {s } is a complex sequece, defie its arithmetic meas σ by σ = s 0 +s + +s + ( = 0,,,...) (a) If lims = s, prove that limσ = s. (b) Costruct a sequece {s } which does ot coverges, although limσ = 0. (c) Ca it happe that s > 0 for all ad that limsups =, although limσ = 0? (d) Put a = s s, for. Show that s σ = + ka k Assume that lim(a ) = 0 ad that {σ } coverges. Prove that {s } coverges. [This gives a coverse of (a), but uder the additioal assumptio that a 0.] (e) Derive the last coclusio from a weaker hypothesis: Assume M <, a M for all, ad limσ = σ. Prove that lims = σ, by completig the followig outlie: If m <, the For these i, k= s σ = m+ m (σ σ m )+ m s s i ( i)m i+ i=m+ ( m )M m+ (s s i ) Fix ε > 0 ad associate with each the iteger m that satisfies m ε +ε < m+ The (m+)/( m) /ε ad s s i < Mε. Hece Sice ε was arbitrary, lims = σ. lim sup s σ Mε (a) Give ε > 0, there exists a positive iteger N such that N implies s s < ε. Also, there exists a positive iteger N such that N implies s 0 +s + +s N Ns + < ε

10 So max{n,n } implies σ s = s 0 +s + +s s + s 0 +s + +s N Ns + + (s N s)+(s N+ s)+ +(s s) + < ε+ + ( s N+ s + s N+ s + + s s ) N + < ε+ ε + < ε Hece lim σ = s. (b) For example, let s = ( ) for = 0,,,... (c) For example, let s = if is a perfect cube; if otherwise The clearly limsup s =. However, give, if we let k be the largest perfect cube satisfyig k, the σ +(++ +k) k + as k. Hece lim σ = 0. (d) Observe that for m <, = + k +k k + < k +k +4 k 0 s s m = (s s )+(s s )+ +(s m+ s m ) = a +a + +a m+ = k=m+ a k

11 Thus s σ = s s 0 +s + +s + = + [(s s 0 )+(s s )+ +(s s )] ( ) = a k + a k + + a k + = + k= ka k k= k= Now, assume that lim (a ) = 0 ad that {σ } coverges, we wat to show {s } also coverges. By the previous result, s = σ + + ka k Also, if we let s 0 = 0 ad s = a for =,,..., ad let σ = s 0 +s + +s +. The lim s = 0, ad the apply part (a), lim σ = 0. So lim s = lim σ +σ ( ) = lim σ + lim σ = lim σ ad we coclude that {s } coverges. (e) To complete the details of the outlie, we prove the followig three facts. (i) We show that Sice m k= s σ = m+ m (σ σ m )+ m i=m+ (s s i ) = s m i=m+ s i k= i=m+ = s (+)σ (m+)σ m m (s s i ) = s ( m)σ (m+)σ m +(m+)σ m = s m+ m (σ σ m ) σ

12 the result follows. (ii) We show that, for these i, We ca directly write s s i ( i)m i+ ( m )M m+ s s i = (s s )+(s s )+ +(s i+ s i ) = a +a + +a i+ a + a + + a i+ M + M + + M i+ ( i)m i+ ad the result follows by takig i = m+. (iii) To show that m+ m ε, observe that m ε +ε implies ad m+ ( ε)+(+ε) +ε = + +ε m ε +ε = +ε +ε = ε(+) +ε +ε Thus the result follows. Next, to show that s s i < Mε, observe that ε < m+ implies +ε ad m < ε +ε = +ε +ε = ε(+) +ε +ε Thus by (ii), the result follows. m+ = (m+)+ > ε + + = +ε +ε 6. Fix a positive umber α. Choose x > α, ad defie x,x,x 4,..., by the recursio formula x + = ) (x + αx

13 (a) Prove that {x } decreases mootoically ad that limx = α. (b) Put ε = x α, ad show that so that, settig β = α, ε + = ε x < ε α ε + < β ( ) ε β ( =,,,...) (c) This is a good algorithm for computig square roots, sice the recursio formula is simple ad the covergece is extremely rapid. For example, if α = ad x =, show that ε /β < 0 ad that therefore ε 5 < 4 0 6, ε 6 < 4 0 (a) First, we show that {x } is decreasig, ad that x > α for =,,... Startig x > α, ad suppose x > α holds for some. The sice α, we have α x < α < x, ad x < x + = ) (x + αx < (x +x ) = x ( ) Otheotherhad,bythearithmetic-geometriciequality,x + > x αx = α, where the strict iequality symbol > follows by the fact α x x (sice α x < x ). Next, we show that lim x = α. By the above result, {x } is mootoically decreasig with lower boud α. This meas that lim x = L for some real L, ad the L = ( L+ α L) or L = α. (b) Expad the followig two values ε + = x + α = ) (x + αx α = x + α α x ad ε = (x α) x x = x +α x α = x x + α α x

14 we obtai ε + = ε x < ε α. Settig β = α, the recursively ε + < β ε < β β ε4 < β β β 4 ε8. < β β β 4 β ε ( ) ε = β β (c) Note that β = ad ε =, compute ε β = = Sice + > + =, ad ε β ( + ) ( = )( + ) = < 0 we coclude that ε β < 0. Therefore by part (b), ε 5 < β ad similarly ε 6 < Fix α >. Take x > α, ad defie ( ) 4 ε < 0 6 < β x + = α+x +x = x + α x +x (a) Prove that x > x > x 5 >. (b) Prove that x < x 4 < x 6 <. (c) Prove that limx = α. (d) Compare the rapidity of covergece of this process with the oe described

15 i Exercise 6. (a) Observe that x + = α+x + +x + = α+ α+x +x + α+x +x = α+(α+)x α++x ( ) α x = x + α++x Note that x > α. Now suppose x > α for some odd umber, the clearly x + < x. O the other had, sice α >, we have α+ > α. It follows that ( ) α x x + = x + α++x ( ) α x > x + α+x = x + ( α+x )( α x ) α+x = α So we coclude that x > x > x 5 >. (b) First, sice α x < 0 ad α >, we have x = x + α x < x + α x = α +x α+x Now suppose x < α for some eve umber, the clearly x + > x [by the first equality i part (a)]. O the other had, use a similar argumet to that i part (a), x + < α. So we coclude that x < x 4 < x 6 <. (c) By part (a) ad (b), the sequeces {x k } ad {x k } (k =,,...) are mootoically decreasig ad icreasig, respectively. Also, α serves as a loweradupperboudof{x k }ad{x k },respectively. Soweletlim k x k =

16 L ad lim k x k = L. Note that we may assume ( ) α L L = L + or L = α α++l ad similarly L = α. It is easy to coclude that lim x = α. (d) We imitate a aalysis i Exercise 6(b). Put ε = x α ad β = α +x, the observe that < +x (sice x = α+x +x > 0) ad +x +x for all [by part (a) ad (b)]. It follows that ad therefore ε + = x + α = α+x α +x = x α α(x α) +x = α +x ε βε ε βε β ε β ε So roughly the speed of covergece of the process is ot more rapid tha the oe i Exercise Replace the recursio formula of Exercise 6 by x + = p p x + α p x p+ where p is a fixed positive iteger, ad describe the behavior of the resultig sequece {x }. Observe that if the limit of {x } exists, sayig L, the L = p p L+ α p L p+ or L = α p Now, we take x > α p, ad suppose x > α p for some positive iteger, the x p < α ad x + < p p x + α p α x = x

17 O the other had, by the arithmetic-geometric iequality, x + > p x p αx p+ = α p where the strict iequality symbol > follows by the fact x αx p+ (sice x p < α or αx p < ). So we coclude that {x } decreases mootoically with lower boud αp, ad hece {x } coverges to L = αp, i.e., lim x = α p. Remark. Exercise 6 is a special case that p =. 9. Associate to each sequece a = {a }, i which a is 0 or, the real umber x(a) = Prove that the set of all x(a) is precisely the Cator set described i Sec..44. We will use the otatios E,E,E,... ad P that ca be foud i Sec..44 (p.4). Let x(a) = = a, where a is 0 or. Cosider the first term of x(a), i.e., a, the either a = 0 or a =. It is clear that the first term of x(a) is located at the lower edpoit of oe of the two itervals i E, i.e., a P. Now, suppose that the partial sum m = a of x(a) is located at the lower edpoit of oe of the m itervals i E m, sayig I m. The m = a P, ad the partial sum m+ = a must be located at the lower edpoit of oe of the two itervals i I m E m+, i.e., m+ = a P. Cotiue this process, we see that either x(a) P or x(a) is a limit poit of P, ad hece x(a) P (sice P is closed). To show the coverse, use the fact that (as also show i the text) every poit of P ca be writte as a limit of a sequece {x m } of lower edpoits of some iterval I m E m (actually we ca write x m = m = a, where a is 0 or ), for m =,, Suppose {p } is a Cauchy sequece i a metric space X, ad some subsequece {p i } coverges to a poit p X. Prove that the full sequece {p } coverges to p. Give ε > 0. Sice {p i } coverges to p, there is a positive iteger N such that i N implies d(p i,p) < ε. Sice {p } is Cauchy, there is a positive iteger N such that d(p m,p ) < ε for all m N ad N. Choose N N such that N = i for some i N, the for all N, = α d(p,p) d(p,p N )+d(p N,p) < ε+ε = ε

18 Hece {p } coverges to p.. ProvethefollowigaalogueofTheorem.0(b): If{E }isasequeceofclosed oempty ad bouded sets i a complete metric space X, if E E +, ad if lim diame = 0 the E cosists of exactly oe poit. Deote E = = E. Suppose E cosists of at least two poits, the diame > 0. But sice E E for all, we have diame diame for all, which cotradicts the assumptio that lim diame = 0. Next, we show that E. Sice E for all, we ca choose a poit x E for each, ad form the sequece {x }. We claim that {x } is Cauchy. Proof of the claim. Give ε > 0, the sice lim diame = 0, there exists a positive iteger N such that N implies diame < ε. For ay m N ad N, by assumptio, x m E m E N ad x E E N,. So d(x m,x ) diame N < ε. Now, sice X is complete ad {x } is Cauchy, we have lim x = x for some x X. Toshowthatsuchx E, supposeot, thex / E M orx EM c forsome positive iteger M. Sice E M is closed, EM c is ope, ad there exists a δ > 0 such that N δ (x) EM c, or N δ(x) E M =. Also, ote that x E E M for all M. It follows that d(x,x ) δ for all M, cotradictig the assumptio lim x = x.. Suppose X is a oempty complete metric space, ad {G } is a sequece of dese ope subsets of X. Prove Baire s theorem, amely, that G is ot empty. (I fact, it is dese i X.) Hit: Fid a shrikig sequece of eighborhoods E such that Ē G, ad apply Exercise. Sice X, we begi with some poit x 0 X ad give r > 0. Sice G is dese ad ope, the set N r (x 0 ) G is oempty ad ope. Choose x N (x 0 ) G ad there exists a eighborhood E of x such that Ē N r (x 0 ) G ad diame r Havig chose oempty eighborhoods E,E,...,E, such that Ēi G i, E i+ E i, ad diame i i r. Sice G is dese ad ope, the set E G is oempty ad ope. Choose x E G ad there exists a eighborhood E of x such that Ē E G ad diame r

19 Cotiue this process, we may costruct the desired shrikig sequece {Ē} such that Ē G for all, ad lim diame = 0. The by Exercise, =Ē cosists of oe poit, sayig x. Observe that =Ē = G, we have x = G ad hece = G. I fact, = G is dese i X, because x Ē N r (x 0 ), where x 0 X ad r > 0 are arbitrarily give.. Suppose {p } ad {q } are Cauchy sequeces i a metric space X. Show that the sequece {d(p,q )} coverges. Hit: For ay m,, it follows that is small if m ad are large. d(p,q ) d(p,q m )+d(p m,q m )+d(p m,q ) d(p,q ) d(p m,q m ) Give ε > 0, the sice {p } ad {q } are Cauchy, there exist positive itegers N ad N such that m, N implies d(p m,p ) < ε, ad m, N implies d(q m,q ) < ε. Put N = max{n,n }, the by the hit, for m, N, d(p,q ) d(p,q m )+d(p m,q m )+d(p m,q ) < d(p m,q m )+ε which follows that d(p,q ) d(p m,q m ) < ε. Thus {d(p,q )} is Cauchy i R, ad hece {d(p,q )} coverges by Theorem.(c). 4. Let X be a metric space. (a) Call two Cauchy sequeces {p }, {q } i X equivalet if lim d(p,q ) = 0 Prove that this is a equivalece relatio. (b) Let X be the set of all equivalece classes so obtaied. If P X, Q X, {p } P, {q } Q, defie (P,Q) = lim d(p,q ) by Exercise, this limit exists. Show that the umber (P, Q) is uchaged if {p } ad {q } are replaced by equivalet sequeces, ad hece that is a distace fuctio i X. (c) Prove that the resultig metric space X is complete.

20 (d) For each p X, there is a Cauchy sequece all of whose terms are p; let P p be the elemet of X which cotais this sequece. Prove that (P p,p q ) = d(p,q) for all p,q X. I other words, the mappig ϕ defied by ϕ(p) = P p is a isometry (i.e., a distace-preservig mappig) of X ito X. (e) Prove that ϕ(x) is dese i X, ad that ϕ(x) = X if X is complete. By (d), we may idetify X ad ϕ(x) ad thus regard X as embedded i the complete metric space X. We call X the completio of X. (a) Let {p }, {q }, ad {r } be Cauchy sequeces i X. (Reflexivity) Sice d(p,p ) = 0 for all, we have lim d(p,p ) = 0. So {p } is equivalet to itself. (Symmetry) If {p } is equivalet to {q }, the sice d(p,q ) = d(q,p ) for all, we have lim d(q,p ) = lim d(p,q ) = 0 i.e., {q } is equivalet to {p }. (Trasitivity) If {p } is equivalet to {q }, ad if {q } is equivalet to {r }. The we have lim d(p,q ) = 0 ad lim d(q,r ) = 0. Also, by the triagular iequality, d(p,r ) d(p,q )+d(q,r ) for all. It follows by the compariso test that lim d(p,r ) lim [d(p,q )+d(q,r )] = lim d(p,q )+ lim d(q,r ) = 0+0 = 0 or lim d(p,r ) = 0. So {p } is equivalet to {r }. (b) Let {p },{p } P ad {q },{q } Q, where P,Q X. The by defiitio, Also, by the triagular iwquality, lim d(p,p ) = lim d(q,q ) = 0 d(p,q ) d(p,p )+d(p,q )+d(q,q )

21 which follows that lim d(p,q ) [ lim d(p,p )+d(p,q )+d(q,q ) ] = lim d(p,p )+ lim d(p,q )+ lim d(q,q ) = lim d(p,q ) ad by symmetry lim d(p,q ) lim d(p,q ). So lim d(p,q ) = lim d(p,q ). Now, we show that is a distace fuctio i X. ()Bydefiitio, itisclearthat (P,Q) 0forallP,Q X. Also, (P,Q) = 0 if ad oly if lim d(p,q ) = 0 for all {p } P ad {q } Q, if ad oly if {p } is equivalet to {q } for all {p } P ad {q } Q, if ad oly if P = Q. () The equality (P,Q) = (Q,P) holds simply by defiitio. () Give P,Q,R X, choose {p } P, {q } Q, ad {r } R. The by defiitio ad by the triagular iequality (w.r.t. d), (P,Q) = lim d(p,q ) lim [d(p,r )+d(r,q )] = lim d(p,r )+ lim d(r,q ) = (P,R)+ (R,Q) i.e., the triagular iequality (w.r.t. ) holds. By checkig () (), we coclude that is a distace fuctio i X. (c)let{p }beacauchysequeceix, adchoose{p k } P for =,,... For each, there exists a positive iteger K such that k,l K implies d(p k,p l ) <. Defie p = p K for =,,..., we show that {p } is Cauchy i X. Give ε > 0, the sice {P } is Cauchy i X, there exists a positive iteger N such that m, N implies (P m,p ) < ε, or lim k d(p m k,p k ) < ε. Let N be a positive iteger such that N < ε, the m, max{n,n } implies d(p m,p ) = d(p m K m,p K ) d(p m K m,p m k )+d(pm k,p k )+d(p k,p K ) < m +ε+ < ε where the iteger k ca be chose sufficietly large such that k max{k m,k } ad d(p m k,p k ) < ε. Hece {p } is Cauchy i X. Let P X be such that

22 {p } P, it suffices to show that P P, which completes the proof. Give ε > 0, put max{n,n }, k max{n,n,k }, ad m max{k,k k } sufficietly large such that d(p m,p k m) < ε. We have d(p k,p k) = d(p k,pk K k ) d(p k,p m)+d(p m,p k m)+d(p k m,p k K k ) < +ε+ k < ε which follows that lim k d(p k,p k) ε, or (P,P) ε. Hece P P. (d) The result is trivial [by part (b)]. (e) (i) Let P X ad {p } P, ad give ε > 0. Sice {p } is Cauchy, there exists a positive iteger N such that m, N implies d(p m,p ) < ε. It follows that (P,ϕ(p N )) = (P,P pn ) = lim d(p,p N ) ε < ε i.e., the eighborhood of P with radius ε cotais a elemet ϕ(p N ) ϕ(x). So we coclude that ϕ(x) is dese i X. (ii) If X is complete, let P X ad {p } P. Sice {p } is Cauchy, p p for some p X, i.e., we have So P = P p, ad hece X = ϕ(x). (P,P p ) = lim d(p,p) = 0 5. Let X be the metric space whose poits are the ratioal umbers, with the metric d(x,y) = x y. What is the completio of this space? (Compare Exercise 4.) By Theorem.(c), every ratioal Cauchy sequece coverges to a real umber. O the other had, for every x R, there exists a ratioal Cauchy sequece {p } such that lim p = x. So the completio of X should be ϕ(r ).