Math 220A Fall 2007 Homework #2. Will Garner A

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1 Math 0A Fall 007 Homewor # Will Garer

2 Pg 3 #: Show that {cis : a o-egative iteger} is dese i T = {z œ : z = }. For which values of q is {cis(q): a o-egative iteger} dese i T? To show that {cis : a o-egative iteger} is dese i T, we eed oly show that the closure of {cis : a o-egative iteger} = T. Observe that cis cosists of ifiitely may poits o the uit circle. (Repetitio could oly occur if = j + pl, for some iteger l ad p Z.) We claim that the closure of that set will cosist of the uit circle itself, which is precisely T. To prove this claim, though, we shall prove a stroger statemet. Namely, we shall prove: The set {cis(pq)} is dese i T if ad oly if q is irratioal. (*) Notice that i the case where q = /p, we have precisely the problem above. Defie frac(x) = x x, where x deotes the floor of x. Showig (*), is equivalet to showig the followig: The sequece frac(q), frac(q), is dese i [0, ) if ad oly if q is irratioal. (**) The equivalece follows from the fact that cis(pq) is determied by the value pq. So, we essetially wat to show that the agle is dese o the iterval[0, p). However, we eed to cosider pq (mod p), sice the uit circle wraps aroud. Dividig through by p, we are left with showig that q (mod ) is dese i [0, ), which is precisely what (**) loos to be establish with the same if ad oly if statemet. Before we prove (**), we shall prove some lemmas. Lemma : If q is irratioal ad m, the frac(q) frac(mq). Proof: Suppose ot. That is, suppose frac(q) = frac(mq). The we have that q q = θ mθ mq mq, i.e. ( m)q = q mq, i.e. θ =, which is a m cotradictio to q beig irratioal, which completes the proof. Note: We used the fact that m to esure that the deomiator is o-zero.

3 Lemma : Suppose q œ [0, ) is irratioal. The for all e > 0, there exists h, œ Z + such that q h < e. Proof: Subdivide [0, ) ito half-ope itervals of equal legth less tha e. By Lemma, two of the fractioal parts lie i the same subiterval. That is, there exists, m œ Z + (with m) such that frac(q) frac(mq) < e. (This follows from the Pigeohole Priciple.) Thus, we have that ( m)q ( q mq ) < e. Lettig = m ad h = q mq, we complete the proof. Now we have all of the ecessary machiery to prove (**). Oe directio is trivial. If q œ Q, the {frac(q): œ Z + } is fiite ad thus caot be dese i T. (A fiite subset of is closed ad therefore caot be dese i a ifiite set.) Suppose q is irratioal. Without loss of geerality, we may suppose q œ [0, ). Ideed, we may replace q with frac(q) after oticig that frac(q) = frac(frac(q)). Let e > 0 be give ad a œ [0, ). We eed to fid a iteger m such that frac(mq) a < e, sice this will show that {frac(q): œ Z + } is dese i [0, ). Usig the otatio from Lemma, suppose that q > h. (The argumet is almost idetical i the case q < h.) hoose the largest N œ Z + such that frac(q) < /N. osider the sequece: 0, frac(q), frac(q),, frac(nq). I light of frac(mq) = mfrac(q) if ad oly if frac(q) < /m, we see that the above sequece is icreasig ad equally-spaced with commo differece frac(q). (The equal spacig is readily apparet ad the icreasig property follows from the fact that the sequece is ever greater tha.) Thus, we have subdivided the uit iterval ito N + subitervals with the followig partitio: [0, frac(q)), [frac(q), frac(q)), [frac(nq), ). Sice a lies i oe of the subitervals above, it suffices to show the followig claim.

4 laim: [0, frac(q)), [frac(q), frac(q)), [frac(nq), ) divides [0, ) ito subitervals of legth less tha e. Will Garer Proof: By the proof of Lemma, we have that the first N subitervals have legth frac(q) < e. Thus, we eed oly cosider the last subiterval. By the def. of N, we have that frac( θ ) > ad thus we have that N + N Nfrac( θ ) > =. This implies that N + N + 0 < frac( N θ ) < < frac( ) N θ < + ε, provig the claim ad completig the proof. Thus, we have show that the sequece frac(q), frac(q), is dese i [0, ) if ad oly if q is irratioal. From our discussio earlier, though, this is equivalet to showig that the set {cis(pq)} is dese i T if ad oly if q is irratioal. By the proof of the slightly more geeral theorem, we see that {cis(q)} will be dese i T provided q 0 or q mp/, where m, œ Z. This esures that the q will ever be equal to p, thereby esurig that cis(q) maps to ifiitely may poits o the uit circle. Ad by the proof of (**), we have that the set will be dese i T. Pg 7 #4: Prove the followig geeralizatio of Lemma.6. If {D j : j œ J} is a collectio of coected subsets of X ad if for each j ad i J we have D j D «, the D =» {D j : j œ J} is coected. If we suppose that J is coutable (which may ot be the case), the we ca apply the followig iductive argumet. A geeral proof for the case whe J may ot be coutable appears afterwards. For this first (icomplete) proof, suppose that J is coutable. We shall proceed via iductio o J, the umber of D j. Let S = {{D j : j œ J} is a collectio of coected subsets of X ad if for each j ad i J we have D j D «, the D =» {D j : j œ J} is coected}. Notice that œ S, trivially. We also have that œ S, sice if D D «, the there exists a poit x, œ D D. By Lemma.6, we have that D = D» D is coected. 3

5 Suppose that œ S for some. We wat to show that + œ S as well. osider a collectio {D j : j œ J} of coected subsets of X, where J = +. osider D =» {D j : j œ J}. Let D = D» D»» D. By the iductive hypothesis, we have that D is coected. We ca rewrite D as D = D» D +. Sice D j D «for all j,, we have that D D + «. Thus, there exists a poit x,,,+ œ D D +. Agai, by Lemma.6, we have that D» D + is coected, which completes the iductive hypothesis ad hece the proof. Essetially, we ca draw a series of lie segmets through each x j, passig through the D j, thus coectig them. Figure : A illustratio of how the D j may be coected via lie segmets Now, suppose we do ot have ay restrictio o J. That is, J is oly a idex set. Let A be a subset of the metric space (D, d) that is both ope ad closed ad suppose that A «. Our goal is to show that A = D, sice that implies that D is coected. A D j is ope i (D j, d) for each j ad also A D j is closed i (D j, d). (These statemets follow from Exercises.8 ad.9, respectively.) Sice D j is coected, the either A D j = «or A D j = D j. Sice A «, we ow that there exists at least oe such that A D «, i.e. A D = D. By hypothesis, there exists x j œ D j D for all j, œ J. This implies that x j œ D = A D ad x j œ D j, i.e. x j œ A D D j, which implies that x j œ A D j. Hece, we have that A D j «. Thus, it must be that A D j = D j. We ca cotiue this process for each idex j. Thus, A D j = D j, i.e. D j Œ A for each idex j. This gives that D = A, which shows that D is coected. 4

6 #5: Show that if F Õ X is closed ad coected, the for every pair of poits a, b i F ad each e > 0, there are poits z 0, z,, z i F with z 0 = a, z = b ad d(z, z ) < e for. Is the hypothesis that F be closed eeded? If F is a set which satisfies this property the F is ot ecessarily coected, eve if F is closed. Give a example to illustrate this. Fix a. Let e > 0. osider A { z F: { z } j j= 0 with z0 a,... z z s.t. d( zi, zi+ ) ε} = = = <. Our goal is to show that A = F, so we eed to show A is both ope ad closed i F ad A «. Observe that a œ A, so we have that A «. To show A is ope i F, cosider B(z, e). We wat to show that B(z, e) F Œ A. This follows from the fact that if we ca fid a path from a to z, the we ca fid a path to a poit, say x, i B(z, e) by addig oe more path from z to x. Sice d(z, x) < e, we have that the coditios of iclusio i A are met. Thus, A is ope i F. To show that A is closed i F, we shall show that A Œ F is ope. Give x œ A, we eed to show that B(x, e) Œ A. osider y œ B(x, e). We claim that y œ A. Suppose ot. The y œ A. Sice d(x, y) < e, by the same argumet for why A is ope, we have that there is a path from a to x via y, hece x œ A. But this is a cotradictio, hece y œ A, which implies that B(x, e) Œ A, i.e. A is ope i F. Thus A is also closed i F. So, we coclude that A = F ad hece coected. Note: We did ot eed the hypothesis that F is closed i this proof. + + j As a couterexample, cosider F = {} 0 : j,, + j. For ay e >0, we ca fid dyadic ratioals such that d(z i, z i+ ) < e, but F is ot coected. Pg 0 #5: Show that every coverget sequece i (X, d) is a auchy sequece. Suppose {x } (ö x) is a coverget sequece i (X, d). The for every e > 0, there exists a iteger N such that d(x, x ) < e/ wheever N. Recall, a sequece {x } is a auchy sequece if for every e > 0, there is a iteger N such that d(x, x m ) < e, for all, m N. osider d(x, x m ). By the fourth property of metric spaces (the triagle iequality), we have d(x, x m ) d(x, x) + d(x, x m ) < e/ + e/ = e, for all, m N, as desired. 5

7 #8: Suppose {x } is a auchy sequece ad { x } is a subsequece that is coverget. Show that {x } must be coverget. Suppose {x } is a auchy sequece ad { x } is a subsequece that is coverget, say to x. By the defiitio of covergece, for all e/ > 0, there exists a iteger K such that for all K, d( x, x ) < e/. By the defiitio of a auchy sequece, there exists a iteger N such that for all, N, d( x, x ) < e/. Let N = max{k, N} For all N, we have d( x, x) d( x, x ) + d( x, x) < e/ + e/ = e, where the first e/ comes from the fact that we have a auchy sequece ad the secod e/ comes from the fact that we have a coverget sequece. Thus, we have that for all e > 0, there exists a iteger N such that d(x, x) < e wheever N. 6

8 Additioal Problems to Loo At ad Kow Pg 3 #: Which of the followig subsets of are ope ad which are closed: (a) {z: z < }; (b) the real axis; (c) {z: z = for some iteger }; (d) {z œ : z is real ad 0 z < }; (e) {z œ : z is real ad 0 z }? (a) {z: z < } is ope. - - (b) the real axis is closed; x œ (-, ). (It is ot ope sice ay ball is ot completely cotaied i the iterval.) It is closed because a limit of real umbers coverges to a real umber. - (c) {z: z = for some iteger } is closed, sice it cosists of a fiite set of poits. If z = rcisq, the z = r cis(q). Sice z =, we have that z = cis(q), which cosists oly of (fiitely may, actually ) poits o the uit circle. - - (d) {z œ : z is real ad 0 z < } is either ope or closed, sice the left edpoit is icluded but the right ed poit is ot. 0 (e) {z œ : z is real ad 0 z } is closed, sice both edpoits are icluded. 0 7

9 #3: If (X, d) is ay metric space, show that every ope ball is, i fact, a ope set. Also, show that every closed ball is a closed set. Let x ad r be fixed. osider B(x, r) = {y œ X: d(x, y) < r}. Suppose that y œ B(x, r). We wat to show that there exists e > 0 such that B(y, e) Õ B(x, r). Sice y œ B(x, r), there exists a positive umber e such that d(x, y) = r e. For all poits s such that d(y, s) < e, we have that d(x, s) d(x, y) + d(y, s) < r e + e = r e < r, so we have that s œ B(x, r). Thus, B(x, r) is ope. A pictorial descriptio appears to the right. x r y e Let x ad r agai be fixed. osider B( xr, ) = {y œ X: d(x, y) r}. To show that B( xr, ) is a closed set, we eed to show that B( xr, ) = {y œ X: d(x, y) > r} is ope. That is, we wat to show that there exists e > 0 such that B(y, e) Õ B( xr, ). Suppose that y œ B( xr, ). Thus, there exists e > 0 such that d(x, y) = r + e. By the triagle iequality, we have that d(x, y) d(x, s) + d(s, y), i.e. d(x, s) d(x, y) d(y, s). For all poits s such that d(y, s) < e, we have that d(x, s) d(x, y) d(y, s) = r + e e = r + e > r, so we have that s œ B( xr, ). Thus, B( xr, ) is closed. x r e y Agai, a pictorial descriptio appears to the right. B( xr, ). Propositio. Let (X, d) be a metric space. The: (a) The sets X ad «are closed; (b) If F,, F are closed sets i X, the so is F ; = (c) If {F j : j œ J} is ay collectio of closed sets i X, J ay idexig set, the F = {F j : j œ J} is also closed. #5: Prove Propositio.. We begi with the followig, useful lemma. 8

10 Lemma 3: Let {F } be a (fiite or ifiite) collectio of sets F. The (i) ( F) = ( F) ad (ii) ( F) = ( F) Proof: If x ( F ) Will Garer, the x F, hece x F for ay, hece x œ (F ) for every, so that x ( F ). Thus, we have that ( F) ( F). oversely, if x ( F ), the x œ (F ) for every, hece x F for ay, hece x F, so that x ( F ). Thus, ( F) ( F). It follows that ( F) = ( F), which completes the proof of (i). To prove (ii), we tae the complemet of both sides of (i) ad replace F with (F ). (a) Recall, a set is closed if its complemet is ope. Sice X = «ad «= X, each of which is ope, we have that both X ad «are closed. (b) By Lemma 3(i), ( F) = ( F) ad (F ) are ope, sice F are closed. Thus, Propositio.9(b) implies that F = ( F) is ope as well. Thus, = = we have that F is closed, as desired. = (c) By Lemma 3(ii), ( F) = ( F) ad (F ) are ope, sice F are closed. Thus, Propositio.9(c) implies that ( F) = ( F) is ope as well. Thus, we have that F is closed, as desired. 9

11 I.6.7 I.6.8 d(, z z ) = d(, z ) = z z ( + z )( + z ) + z #7: Show that (, d) where d is give by (I.6.7) ad (I.6.8) is a metric space. To show that (, d) is a metric space, we eed to show that d satisfies: (i) d(z, z ) 0 ad d(z, z ) = 0 if ad oly if z = z. d(z, ) 0 ad d(z, ) = 0 if ad oly if z =. Notice that d(z, z ) 0 from its costructio. d(z, z ) = 0 iff z z = 0, which happes if ad oly if z = z. Similarly, d(z, ) 0 ad if we cosider the limit as z ö, we see that d(z, ) = 0 if ad oly if z =, sice the deomiator heads towards ifiity. (ii) d(z, z ) = d(z, z) ad d(z, ) = d(, z). z z ( z z) z z d(, z z ) = = = = d( z,) z. ( + z )( + z ) ( + z )( + z ) ( + z )( + z ) Ad we have that d(z, ) = d(, z) by its costructio. (iii) d(z, z ) d(z, z ) + d(z, z ), d(z, z ) d(z, ) + d(, z ), d(z, ) d(z, z ) + d(z, ). These follow immediately from the fact that this metric represets the Euclidea distace betwee the stereographic images o the Riema sphere ad the triagle iequality holds for Euclidea distace. 0

12 Pg 7 #3: Which of the followig subsets X of are coected: if X is ot coected, what are its compoets: (a) X = {z: z }» {z: z < }, (b) X = [0, )» { : } X = (A» B), where A = [0, ) ad B = {z = rcisq: r = q, 0 q }? +, (c) (a) z defies a uit dis cetered at the origi while z < defied a uit dis (without its perimeter) cetered at (, 0). The subset is coected via the poit (, 0). Thus, we have that X = {z: z }» {z: z < } is coected. - 3 (b) X = [0, )» + : is ot coected. lim + = [0,). If the two pieces were to be coected, they would meet at the poit x = (c) X = (A» B) is ot goig to be coected. Notice that X = {shaded regio}. If we oly cosidered B, the this would be coected. The subtractio of half of the real axis esures that we will have a set of spirals, oe of which are coected to each other..

13 Pg 0 #4: Let z, z be poits i ad let d be the metric o. Show that z z ö 0 if ad oly if d(z, z) ö 0. Also show that if z ö, the {z } is auchy i. (Must {z } coverge i?) Looig at I.6.7, it is clear that if z z ö 0, the d(z, z) ö 0. Similarly, for I.6.8, if z =, the if z ö 0, the d(z, ) ö 0. I I.6.7, suppose that z z 0. We wat to show that d( z, z) 0 as well. There are two cases to cosider: (i) z z (which ecessarily implies that z ö ) ad (ii) z z L, where 0 < L < (which implies that z is fiite). For (i), z z z z z ( z z ) d( z, z) = =. ( + z )( + z ) ( + z )( + z ) z + ( + z ) As ö, we have that z ( z z ) ( z ) ( ) > 0, so we have that + ( + z ) ( + z ) d( z, z) 0. For (ii), sice z z 0, it suffices to verify that the deomiator ( + z )( + z ) <. However, sice z is fixed ad z is fiite, we are doe. I I.6.8, otice that if d(z, ) ö 0, the it must be that z ö, i.e. z ö 0. Now, suppose z ö. We wat to show that for, m sufficietly large, the we have that d(z, z m ) < e for ay e >0. Let e > 0 be fixed. By the triagle iequality, d( z, zm) d( z, ) + d(, zm) = + + z + z Sice z ö, there exists N such that sice z m ö, there exists N such that + z + z m m ε < for all > N. Similarly, ε < for all m > N. Let N = max{n, N }. The d(z, z m ) e/ + e/ = e for all m, > N.

14 #6: Give three examples of o-complete metric spaces. We wat to fid a metric space (X, d) such that there exists a auchy sequece that does ot have a limit i X.. Let X = Q, with the stadard metric of absolute value. osider the (auchy) x sequece defied by x =, x+ = +. The limit, x, of the sequece would x eed to satisfy x =. No ratioal umber solves this equatio.. Let X = R\{0}, agai with the stadard metric. osider the (auchy) sequece {/, =,, }. The limit of this sequece is 0, which is ot i X. 3. Let X be the ope iterval (0, ), agai with the stadard metric. I a similar fashio as i (), cosider the (auchy) sequece {/, =, 3, }. The limit of this sequece is 0, which is ot i the iterval. #7: Put a metric d o R such that x x ö 0 if ad oly if d(x, x) ö 0, but that {x } is a auchy sequece i (R, d) whe x ö. (Hit: Tae ispiratio from.) Let d( x, x) = x ( + x )( + x ) x from Problem #4 (above).. The desired properties hold based o the results 3