Math 104: Introduction to Analysis

Transcription

1 Mth 104: Introduction to Anlysis Contents 1 Lecture The nturl numbers Equivlence reltions The integers The rtionl numbers Lecture The rel numbers by xioms The rel numbers by Dedekind cuts Properties of R Lecture Metric spces Topologicl definitions Some topologicl fundmentls Lecture Sequences nd convergence Sequences in R Extended rel numbers Lecture Compctness Lecture Compctness in R k Lecture Subsequences Cuchy sequences, complete metric spces Aside: Construction of the rel numbers by completion Lecture Tking powers in the rel numbers Toolbox sequences Lecture Series Adding, regrouping series Toolbox series

2 10 Lecture Root nd rtio tests Summtion by prts, lternting series Absolute convergence, multiplying nd rerrnging series Lecture Limits of functions Continuity Lecture Properties of continuity Lecture Uniform continuity The derivtive Lecture Men vlue theorem Lecture L Hospitl s Rule Power series Tylor series Lecture The Riemnn-Stieltjes integrl Some Riemnn-integrble functions Lecture Properties of the integrl Lecture The fundmentl theorem of clculus Lecture Things tht ren t true Uniform convergence Lecture Bsic criteri for uniform convergence Uniform convergence nd continuity Lecture Uniform convergence nd differentition An everywhere continuous but nowhere differentible function Differentition nd integrtion of power series Lecture The Stone-Weierstrss theorem

3 Though it my seem (or not!) s though I put cre into these notes, they re ctully very sloppily written. So I gurntee you tht there will be errors, typos, flt out lies, nd other vrieties of mistkes. You my lert me if you find one, but the best thing to do is probbly to red it with skepticl nd lso not-so-literl eye. You should be especilly skepticl bout nything I sy tht seems like it hs to do with mthemticl logic, since I literlly know nothing bout tht nd m just mking stuff up. Much of the content of these notes is tken from Wlter Rudin s Principles of Mthemticl Anlysis nd to lesser extent Kenneth Ross Elementry Anlysis, though of course the errors re ll mine. 1 Lecture 1 Remrk 1.1 (Proof by contrdiction). This is something you should become cquinted with, if you hve not lredy. 1.1 The nturl numbers We will tke the following s our xiomtic definition of the nturl numbers. There re constructions of the nturl numbers from more bsic principles but this is our strting point. As Bertrnd Russell (mybe) sid: it s turtles ll the wy down. Definition 1.2 (The Peno xioms for the nturl numbers). The set of nturl numbers, denoted N, is defined xiomticlly by: 1. There is distinguished element which we denote 1 P N. It is the first element of the nturl numbers. 2. There is function S : N Ñ N clled the successor function. This should be thought of s the function Spxq x ` 1. In other words, every element hs (unique) well-defined successor. 3. The first element 1 is not successor of ny element, i.e. 1 is not in the imge of S. 4. S is injective, i.e. n element cn succeed t most one element. 5. Let P be property. If 1 hs P, nd lso if x hs P implies Spxq hs P, then every x P N hs P. This is the induction principle. Remrk 1.3. The inductive xiom is importnt. Otherwise, we my be llowed things like two copies of the nturl numbers: N Y N. In some wys it specifies tht there is unique first element. It does more thn tht though: it tells us tht rguments by induction re llowed. If your next question is: wht other kinds of rguments re llowed or not llowed, my nswer is I don t know, nd tht you should consult logicin. I do not worry much bout issues like this. This is good strting point, but we re not only interested in N s set but lso in the ddition nd multipliction opertions on it, which give it some dditionl structure. We should define these in terms of the xioms bove. Definition 1.4 (Addition, multipliction). One cn define ddition on N recursively s follows. ` 1 : Spq ` Spbq : Sp ` bq One cn lso define multipliction on N rescursively s follows. 1 : Spbq : ` b One cn then prove tht ddition nd multipliction s defined bove re commuttive, ssocitive, nd distributive, i.e. tht the following re true 3

4 ` b b ` p ` bq ` c ` pb ` cq b b pbqc pbcq pb ` cq b ` c Since I m tking the nturl numbers s we know them intuitively s strting point, I won t ctully prove these things. If you re interested you cn try it. Remrk 1.5. Sometimes people define the nturl numbers to include zero. I m mking n rbitrry decision here not to invite zero to the prty. (It wsn t very positive number to be round... h h h...) 1.2 Equivlence reltions I will introduce notion which is very useful in mking constructions. Definition 1.6. Given set S, n equivlence reltion on the set S is subset E Ă S ˆ S. If p, bq P E, we often write b. This subset E stisfies the following properties: (reflexivity) for ll s P S, we hve ps, sq P E (i.e. s s), (symmetry) for ll ps, tq P E, we hve pt, sq P E (i.e. if s t then t s), (trnsitivity) for ll ps, tq P E nd pt, uq P E, we hve ps, uq P E (i.e. if s t nd t u then s u). This notion is supposed to generlize equlity, s equlity is very rigid nd literl, but there re situtions in which we might wnt to consider two non-equl things to be the sme. Exmple 1.7. Exmples of equivlence reltions: (1) Equlity is n equivlence reltion. e(2) Similrity nd congruence in Eucliden geometry. (3) Similrity of mtrices in liner lgebr. (4) Congruence modulo n. Definition 1.8 (Equivlence clsses). We will introduce the following nottion first. For s P S, denote rss to be the subset of S contining ll elements equivlent to s, i.e. rss tt P S s tu. An equivlence clss is subset of the form rss for some s P S. Remrk 1.9 (More intrinsic chrcteriztion of equivlence clsses). Another chrcteriztion of equivlence clsses is the following: subset T Ă S is n equivlence clss such tht (1) every t, t 1 P T hs t t 1 ; nd (2) if s R T then for ll t P T we hve t j s. Remrk 1.10 (Equivlence clsses prtition S). Let S be set with n equivlence reltion. Then the equivlence clsses prtition S. Check this s n exercise. Definition 1.11 (Quotients). Let be n equivlence reltion on S. equivlence clsses. Then the quotient S{ is the set of Exmple Exmples of quotients (1) Under equlity, for ny set S, S{ S. (2) Left s exercise. (3) Let M n be the set of ll n ˆ n mtrices. Then M n { is the set of equivlence clsses with representtives ll possible Jordn norml forms, up to reordering. (4) Z{ tr0s, r1s,..., rn 1su 4

5 1.3 The integers We will define the integers from the nturl numbers s follows. The ide is to let pir p, bq represent the quntity b. However, sometimes different pirs will be equl under this ssignment so we wnt to cpture this using n equivlence reltion. Definition The set of integers Z is defined s the quotient pn ˆ Nq{ under the equivlence reltion p, bq pc, dq if ` d c ` b. The set of integers hs n opertion, clled ddition, which is defined by rp, bqs ` rpc, dqs rp ` c, b ` dqs. This opertion is well-defined in the quotient (check s exercise). The set of integers lso hs dditive identity (i.e. zero), which is rp1, 1qs. The set of integers lso hs n opertion clled multipliction defined by p, bq pc, dq pc ` bd, d ` bcq. It lso hs multiplictive identity given by rp2, 1qs. Finlly, there is n opertion clled dditive inversion, where p, bq pb, q. Remrk 1.14 (Motivting the equivlence reltion). We wnt to think of the pir p, bq of nturl numbers s representing the integer b. However, doing this mens tht every integer, e.g. 0, hs mny representtives, e.g. p1, 1q, p2, 2q, p3, 3q,.... Thus we must mke ll these representtives equl using n equivlence reltion. The equivlence reltion tht cptures when two pirs p, bq pc, dq is tht b c d. However, since subtrction is not well-defined in the nturl numbers, we must rewrite this s ` d c ` b. Remrk 1.15 (Checking opertions re well defined). We defined the ddition opertion by rp, bqs ` rpc, dqs rp ` c, b ` dqs. Wht we ve done is, for ech equivlence clss, we chose prticulr representtives nd our opertion returns bck n element of NˆN. Then we tke the equivlence clss of tht element. To show this is well-defined, one must show tht if we chose different representtives, we wht we get by pplying the opertion rule gives us n equivlent element. The set of integers forms wht is clled commuttive ring. Definition A commuttive ring is set R with ddition nd multipliction opertions, denoted ` nd ; dditive nd multiplictive identities denoted 0 nd 1; nd dditive inverses, denoted. The following properties re stisfied: 1. (ssocitivity of ddition) p ` bq ` c ` pb ` cq 2. (commuttivity of ddition) ` b b ` 3. (dditive identity) 0 ` 4. (dditive inverses) ` p q 0 5. (ssocitivity of multipliction) pbqc pbcq 6. (commuttivity of multipliction) b b 7. (multiplictive identity) 1 8. (distributivity) pb ` cq b ` c Remrk This is not very importnt definition for this course; it ppers frequently in lgebr. It will not pper beyond this lecture. You cn think of this definition to just be word which is shorthnd for sying: set which hs ddition nd multipliction nd their respective identity elements which stisfies commuttivity, ssocitivity nd distributtivity properties. Or, set tht hs ll the opertions tht the integers do. Proposition The integers form ring. 5

6 Proof. We will prove one of the properties bove: dditive inverses. The rest re left s exercises. Wht we wnt to show is tht: p`, q ` p, `q p1, 1q Using the bove formul for ddition, we hve: p` `, ` ` q p1, 1q This is true, since using our definition of, we check tht indeed: ` ` ` 1 ` ` 1. Exmple 1.19 (Exmples of other commuttive rings). The (multivrite) polynomils with integer or rtionl or rel or complex coefficients, under the usul ddition nd multipliction of polynomils, form ring, denoted krxs for one vrible, krx, ys for two vribles x nd y, et ceter, where k Z, Q, R, C. 1.4 The rtionl numbers The construction of the rtionl numbers is left s homework exercise. The rtionl numbers form field, which is to sy ll numbers except 0 hve multiplctive inverse. Definition A field k is commuttive ring with multiplictive inverses for ll nonzero elements, denoted 1, i.e. such tht 1 1. We will mke the following definition s well. We could hve mde similr definitions before, but I m lzy. Definition A (totl) order on set S is subset of L Ă S ˆS stisfying the following properties. If p, bq P L then we will write for shorthnd ď b. (ntisymmetry) if ď b nd b ď then b (trnsitivity) if ď b db b ď c then ď c (totlity) t lest one of the following must be true: ď b or b ď. An ordered field k is field with totl order such tht: () if ď b then ` c ď b ` c for ny c P k (b) if ě 0 nd b ě 0 then b ě 0 Remrk There re other notions of orders: for exmple, there re orders on sets where two elements might be incomprble. You cn check Wikipedi if you re interested. We re interested in totl orders, but since there re no other orders we will consider, I might be lzy bout writing totl. Exmple The rtionls nd rels, Q, R, re both ordered fields. The complex numbers, C is field but it is not n ordered field. 2 Lecture The rel numbers by xioms There re three pproches we will tke to the rel numbers in this clss. The first is xiomtic. Definition 2.1 (Axiomtic definition of the rel numbers). The rel numbers re n ordered field with the lest upper bound (LUB) property. The LUB property is s follows: every subset which hs n upper bound hs supremum (lest upper bound). (We will define these terms now). 6

7 Definition 2.2. Let X be n ordered set, nd S Ă X. The mximum of S is n element m P S such tht for ll s P S, s ď m (exercise: prove mximums must be unique). An upper bound on S is n element x P X such tht for ll s P S, s ď x. Remrk 2.3. Notice tht mximums must be in the set S, wheres upper bounds do not hve to be. Further, notice tht upper bounds re usully not unique. Exmple 2.4. Let X R nd S r0, 1s. Then, 1 is mximum of S nd ny number greter thn 1 is n upper bound. Remrk 2.5. All finite sets hve mximums. However, not ll sets in generl hve mximum. For trivil exmple, consider p0, 8q Ă R. This fils to hve mximum, nd fils to hve n upper bound. I cn live with this. Wht s kind of bd is tht not ll bounded sets hve mximums. For exmple, the intervl p 1, 0q is bounded bove by 0 but does not hve mximum. Let m P p 1, 0q be possible mximum. But, 1 2m is lso in the set nd is even bigger (since m is negtive). So there cn t be mximum. In this wy sking for mximums is somehow the wrong thing to do for sets which might be infinite. This will motivte the next definition. Definition 2.6. A lest upper bound on subset S of n ordered set X is n element x P X which is n upper bound on S, nd such tht there is no smller upper bound. More mthemticlly, it is n upper bound such tht if x 1 is nother upper bound, the necessrily, x ď x 1. The lest upper bound is lso clled the supremum nd is denoted suppsq. Likewise, the gretest lower bound is clled the infimum nd is denoted infpsq. Remrk 2.7. In our previous exmple, the supremum is 0. However, it is still not true tht every subset of wellordered set hs supremum. For exmple, consider p0,? 2q s subset of Q (not R!) does not hve supremum. Showing this is ctully somewht involved nd involves some elementry number theoretic rguments. I will just ppel to intuition here. In wy, though, this problem is less of defect of our definition of lest upper bound thn it is defect of the rtionl numbers: the subset p0,? 2q of the rtionl numbers doesn t hve nything tht even resembles generlized mximum if we sty in Q. This motivtes the next definition. Definition 2.8. A well ordered set hs the lest upper bound property if for every bounded bove subset S, S hs supremum (lest upper bound). Remrk 2.9. The lest upper bound property is equivlent to the gretest lower bound property for n ordered field, since for ny set S we cn lwys tke its negtion S, nd the supremum of S is the infimum of S nd vice vers. Remrk Thus the rel numbers re defined s n ordered field which hs the LUB property. Vguely, this mens ll its holes re filled in. However, it s still not cler such n object exists, since we hven t constructed it. In the next section we will do tht. In somewht lter section, we will show tht there is generl construction one cn mke to fill in the holes on certin kinds of spces clled metric spces, nd tht the rel numbers re wht we get when we do this to Q. 2.2 The rel numbers by Dedekind cuts In this section we will give construction of the rel numbers by Dedekind cuts. The wy to think bout Dedekind cuts is tht to rel number α, we ssocite the two subsets p 8, αq nd pα, 8q, s subsets of Q (i.e. intersect with Q). Though this is how we think of these cuts, in mking rguments we cnnot ctully refer to things like p 8, αq for α irrtionl, since this would mke our construction circulr! Definition A Dedekind cut of the rtionl numbers Q is prtition of Q into subsets A nd B tht stisfies the following properties: (1) (definition of prtition) A Y B Q nd A X B H (2) (A is closed downwrd) if P A, then p 8, q Ă A (3) (B is closed upwrd) if b P B then pb.8q Ă B (4) (infinity is not number) neither A nor B re empty nor ll of Q (5) ( number is cut out) A does not hve mximum, nd B does not hve minimum 7

8 We cn give n ordering on the Dedekind cuts s follows. We sy pa, Bq ď pa 1, B 1 q if A Ă A 1 nd B 1 Ă B. There is lso wy to mke Dedekind cuts into field, but this is left s homework exercise. Remrk Note tht the the dt of B is extrneous. Given A tht stisfies the bove properties, it is utomtic tht B Q A does. If this B hs minimum, then remove it from B. (Exercise: show tht this new B cnnot hve minimum. Hint: show tht between ny two rtionl numbers, there is rtionl number.) I mde the bove definition to illustrte why this is cut, but from now on we will only be using the dt of A. Remrk Property (5) is mde becuse otherwise, the cuts p 8, qq nd p 8, qs for q P Q would look like they should be the sme. Try to prove the field xioms without this property nd see why it s necessry. For exmple, we cn t hve two cndidtes for zero in field! Remrk The Dedekind cuts need to be mde into n ordered field. Unfortuntely this cn get kind of hiry for multipliction, nd I won t do it here. Interested reders should refer to the constructions in Rudin. Proposition The ordered set of Dedekind cuts hs the LUB property. In prticulr, let ta i u ipi be collection of Dedekind cuts which is bounded bove. Then, its supremum is Ť ipi A i. Proof. We need to check tht (1) the climed supremum is lso Dedekind cut (in prticulr, it is not ll of Q) nd tht (2) it relly is the supremum. For (1), it is esy to see tht Ť i A i is still closed downwrds. The fct tht the Dedekind cuts re bounded bove mens there is some cut A such tht A i Ă A for ll i P I. This mens tht Ť i A i Ă A so A Q. Finlly, suppose tht Ť i A i hd mximum, sy m. Then m P A p for some fixed p P I, but since it is mximum of the union it would hve to be mximum for A p s well, contrdicting tht A p is Dedekind cut. For (2), Suppose there is some Y such tht A i Ă Y for i P I (A i ă Y for ll i, so Y is n upper bound) but Y Ă Ť ipi A i (Y is lower thn the climed supremum). Let x P Ť ipi A i; then x P A p for some p P I. But this contrdicts the first sttement, so this cnnot be. One cn sum this rgument up in sentence like, the union of collection of subsets is the miniml subset contining ll the subsets in the collection. Remrk 2.16 (Why we like xioms). This construction is nice in ressuring us tht the rel numbers exist (yy!). I think the logic-y thing to sy would be tht Dedekind cuts re model for the rel numbers. The nice thing bout xioms is tht we cn build theories without thinking too hrd bout specific constructions, s long s our xioms re well-designed. The specific model doesn t relly mtter s long s we sty within the bounds of the xioms. One might complin tht if we wnt to cll something the rel numbers, there should be question bout uniqueness, which I know nothing bout. So this question is lso going under the rug. In prctice this is not importnt. 2.3 Properties of R Proposition 2.17 (Archimedin property). Let, b P R. If ą 0 nd b ą 0, then there is positive integer n such tht n ą b. Proof. Suppose otherwise, i.e. tht there is some, b P R such tht n ď b for ll n P N. In prticulr, this mens tht b is n upper bound for S : tn n P Zu. Thus, sup S is number. In homework exercise, you will show tht suppsq ` suppt q supps ` T q. Let S be s bove nd T tu. Then notice tht S ` T S, nd suppt q, so we should hve suppsq ` suppsq which contrdicts tht ą 0. Thus, the set must be unbounded nd there is some n such tht n ą b. Proposition 2.18 (Q is dense in R). For ny, b P R, there is x P Q such tht x P p, bq. Proof. Write x p q, for p, q P Z. Then we wnt to find p, q such tht q ă p ă bq. We cn use the Archimedin property to see tht there is n integer q such tht qpb q bq q ą 1. We clim tht if β α ą 1, then there is n integer in pα, βq. This would complete the proof, letting β bq nd α q, nd p the integer. 8

9 Lemm Let α, β P R. If β α ą 1, then there is n integer p P pα, βq. Proof. If α ă 0 nd β ą 0, we cn prove this result for the smller intervl p0, βq. If α, β ă 0, we cn prove this result for the intervl p β, αq. So, without loss of generlity, we cn ssume tht α, β ě 0: The strtegy is s follows. We wnt to tke the set tn P Z α ă nu nd find its minimum vlue, sy p. By construction we hve α ă p nd since it is the minimum, we kno wtht p 1 ă α nd so p ă α ` 1 ă β s well, completing the proposition. The issue is tht this set my not hve minimum nd it my lso be empty. First, let s ddress the existence of minimum. We cn force set to hve minimum by mking it finite, i.e. by dding some conditions bounding the set: tn P Z α ă n nd n P r0, Nsu for some integer N. This is true so long s α ă N, nd we cn use the Archimedin property to find n integer N such tht N 1 ą mxp α q. 3 Lecture Metric spces Definition 3.1. Let k be totlly ordered field. A k-metric spce is set X long with distnce function d : X ˆ X Ñ k stisfying the following properties: (non-negtivity) dpx, yq ě 0 for ll x, y P X, (identity of indiscernbles) dpx, yq 0 if nd only if x y, (symmetry) dpx, yq dpy, xq for ll x, y P X, (tringle inequlity) dpx, yq ` dpy, zq ě dpx, zq for ll x, y, z P X. Remrk 3.2. Most texts fix k R. In generl, we will ssume k R, but lter when we give consruction of R we will require k Q to void circulr rgument. Remrk 3.3. I should be more creful bout this definition thn I m. I m not sure whether ny completeness properties of R re used in results in the literture on metric spces. I m lso not sure how much in the literture pplies to loosening the restriction on k, i.e. if we consider ordered belin groups or ordered rings insted. I hope tht these notes t lest re internlly consistent. Exmple 3.4 (The Eucliden metric). R n is metric spce under the Eucliden metric. Let x px 1,..., x n q nd y py 1,..., y n q, nd define: dpx, yq px 1 y 1 q 2 ` ` px n y n q 2 One cn think of this is generlized Pythgoren theorem wy of finding distnce. For n 1 it is not hrd to see this is metric. For lrger n, the proofs I know use Cuchy-Schwrtz. Note tht this metric doesn t quite mke sense when k Q for Q n when n 1, since squre roots might not exist in Q. Exmple 3.5 (The discrete metric). Let X be ny set. Then the discrete metric is defined by: One cn esily verify this is metric. dpx, yq # 1 if x y 0 if x y 9

10 3.2 Topologicl definitions Definition 3.6. Let px, dq be metric spce. The open bll of rdius ɛ t x is defined by: B ɛ pxq : tp P X dpx, pq ă ɛu nd the closed bll is defined by: B ɛ pxq : tp P X dpx, pq ď ɛu. We will sometimes use the word neighborhood of x to refer to ny open bll of ny rdius t x. Exmple 3.7. Let X R with the Eucliden metric. Then B ɛ pxq px ɛ, x ` ɛq. Exmple 3.8. Let X hve the discrete metric. Then B ɛ pxq txu if ɛ ă 1 nd B ɛ pxq X otherwise. Definition 3.9. Let px, dq be metric spce, nd E Ă X. A point p P E is clled n interior point or E if there is neighborhood U of x such tht U Ă E. A subset E is clled open if every point is interior. Exmple 3.10 (Open blls re open). Every open bll is open. To show tht every p P B ɛ pxq is n interior point, tke ɛ dpp, xq ą 0 note tht by the tringle inequlity, B ppq Ă B ɛ pxq. Definition Let px, dq be metric spce nd E Ă X. A point p P X is limit point of E if every neighborhood U of p hs point q p such tht q P E. A subset E which contins ll its limit points is clled closed. The closure of E, denote E, is the union of E with ll of its limit points. Exmple 3.12 (Closed blls re closed). Every closed bll is closed. Consider B ɛ pxq nd let p be possible limit point. If dpp, xq ď r then we re hppy. Otherwise, let ă dpp, xq r. Then B ppq nd B ɛ pxq re disjoint by the tringle inequlity, nd thus p cnnot be limit point. Proposition 3.13 (The closure of ny subset is closed). Let px, dq be metric spce nd E Ă X. The closure E is closed. Proof. This might seem obvious, but there is something to prove. One hs to show tht one does not introduce more limit points by including the limit points of E. Suppose p is limit point of E; then every neighborhood N of p intersects E. If N contins point of E we re done. If N contins limit point of E, sy q, then since N is open, q is n interior point, so there is neighborhood of q contined in N which contins point of E, concluding the proof. Remrk The closed bll isn t relly used tht much s topologicl notion, s fr s I know. I include it here for fun. Definition Let px, dq be metric spce. A subset E Ă X is dense if E X. Exmple As we ve shown before, Q is dense in R. To see why, we clim tht every rel number is limit point of rtionl numbers: let P R; every bll B ɛ pq p ɛ, ` ɛq contins rtionl number. 3.3 Some topologicl fundmentls Proposition 3.17 (Complements). A set is open if nd only if its complement is closed. Proof. This proof is mostly just juggling the logic in the definitions. Suppose tht E is open. Let p be limit point of E c. Then every neighborhood of p intersects E c nontrivilly, i.e. every neighorhood is not contined in E, so p R E since it cnnot be n interior point, so p P E c, so E c is closed. Suppose tht E is closed. Let p P E c, so p is not limit point of E, so their is neighborhood of p disjoint from E, so p is n interior point, so E c is open. Proposition 3.18 (Unions nd intersections). Any (possibly infinite) union of open sets is open. Any (possible infinite) intersection of closed sets is closed. Any finite union or intersection of open or closed sets is open or closed, respectively. 10

11 Proof. By the previous proposition, we only need to prove the sttements for open sets. I will leve the sttement tht n infinite union of open sets is open s n exercise. For finite intersection of open sets, tke open sets G 1,..., G n, nd x P Ş n i 1 G i. Since x P G i for ll i is interior, we hve B ɛi pxq Ă G i for ll i nd for some ɛ i. Then, B minpɛiqpxq Ă Ş n i 1 G i nd the result follows. Proposition 3.19 (Closure). The closure E is the intersection of ll closed sets tht contin E. Equivlently, it is the smllest closed set contining E. Proof. Let F Ş EĂV closed V be the intersection of ll closed sets contining E. Clerly, F Ă E since E is closed nd contins E. We wnt to show tht every closed set contining E lso contins E. Let V be closed nd E Ă V. Tking the closure of both sides, we find tht E Ă V. 4 Lecture Sequences nd convergence Remrk 4.1. Wrning: the sequence notion of limit is not quite the sme s thn the notion of limit you lerned in clculus the context is different. Definition 4.2. Let px, dq be metric spce. A sequence tp n u in X converges to p P X if for every ɛ ą 0, there is some N such tht dpp n, pq ă ɛ when n ą N. Equivlently, tp n u něn Ă B ɛ pxq X tp n u. In this cse, we write: lim nñ8 p n p If sequence does not converge to ny point, it diverges. Proposition 4.3. Limits of sequences re unique. Proof. Let p, q be two limits of sequence tp n u. We will show tht dpp, qq ă ɛ for ny ɛ ą 0. Choose ny ɛ ą 0; then there is n N such tht dpp n, pq ă ɛ 2 for n ą N nd n M such tht dpq n, pq ă ɛ 2 for n ą M. Then, for n ą mxpn, Mq, we hve tht dpp, qq ď dpp, p n q ` dpp n, qq ă ɛ s desired. Proposition 4.4 (Connections to topology). If E Ă X nd p is limit point of E, then there is sequence in E converging to p. Proof. For every n, choose some s n P B 1{n ppq X E, which we know is nonempty since p is limit point of E. We clim this sequence converges to p. For ɛ ą 0, choose N such tht 1 N ă ɛ. Then, one sees tht for n ą N, we hve dps n, pq ă 1 N ă ɛ. Definition 4.5. Let px, dq be metric spce. A subset E Ă X is bounded if there is number L such tht tdpp, qq p, q P Eu is bounded bove (in the order-theoretic sense) by L. Proposition 4.6 (Convergent sequences re bounded). Convergent sequences re bounded (considered s sets). Proof. Let tp n u Ñ p be the convergent sequence. We will brek up the sequence into two groups for which we cn find bound on dpp n, p m q: one for m, n ď N nd one for m, n ě N. Fix ny ɛ (for exmple, ɛ 1) nd tke n N such tht n ě N implies dpp n, pq ă ɛ. By the tringle inequlity, dpp n, p m q ă 2ɛ for n, m ě N. This is the first group. For the second, we know tht for n, m ď N, dpp n, p m q ě m : mxptdpp n, p m u n, m ď Nu. Then, using the tringle inequlity, we know tht for ny m, n, estblishing bound. dpp n, p m q ď dpp n, p N q ` dpp N, p m q ď 2 mxpm, 2ɛq 11

12 4.2 Sequences in R Proposition 4.7. Let ts n u nd tt n u be sequences in R, with limits s nd t respectively. Let c P R be constnt. Then: () lim nñ8 pcs n q cs nd lim nñ8 pc ` s n q c ` s. (b) lim nñ8 ps n ` t n q s ` t (c) lim nñ8 ps n t n q st (d) lim nñ8 ps n {t n q s{t if for ll n, s n 0 nd s 0 Proof. The proof of () is esy nd left s n exercise. For (b), fix n ɛ ą 0. We wnt to find n N such tht if n ą N then ps ` tq ps n ` t n q ă ɛ. By the tringle inequlity, we hve tht ps ` tq ps n ` t n q ď s s n ` t t n. By the limits we lredy know, we cn find n N s such tht n ą N s implies s s n ă 1 2 ɛ nd likewise n N t. Then, let N mxpn s, N t q nd we hve the result. For (c), rewrite the expression: st s n t n pt t n qps ` s n q ` t n s s n t pt t n qps ` s n q ` pt n tqs ps n sqt The strtegy is s follows: in ech of the three dditive terms, we hve one fctor tht should go to zero. We lso hve bounds on the other fctors, which gurntees tht their products go to zero. More precisely let L be bound on the sequence ts ` s n u. Fix ɛ ą 0; we wnt to find n N such tht if n ą N then st s n t n ă ɛ. Applying the tringle inequlity to our rewriting, we find tht There is n N such tht for n ą N, we hve s s n ă st s n t n ă ɛ s desired. For (d), we only need to show tht st s n t n ď L t t n ` s t t n ` t s s n lim nñ8 ɛ L ` s ` t nd t t n ă ɛ L ` s ` t. Then, one finds tht 1 s n 1 s. Fix ɛ ą 0. We wnt to find N such tht for n ą N, we hve: 1 s n 1 s s n s s n s ă ɛ The ide is s follows: the denomintor s n s tends towrd s 2 which is constnt we cn ccount for. We wnt to overestimte this quntity with respect to s 2 since it is in the denomintor. So, we cn find N 1 such tht n ą N 1 implies tht 2 s n ą s (check this), nd so s n s ą 1 2 s2. Thus our expression becomes: s n s s n s ă 2 1 s 2 s n s Now, we cn find N 2 such tht s n s ă s2 ɛ 2 nd the result follows. 4.3 Extended rel numbers Definition 4.8. We write lim s n 8 nñ8 to men tht for ech M there is some N such tht n ą N implies tht s n ą M. Likewise, we write lim s n 8 nñ8 to men tht for ech M there is some N such tht n ą N implies tht s n ă N. 12

13 Remrk 4.9. The extended rels re not metric spce, nor field, though it does hve n totl order. However, the following rithmeticl sentences still mke sense for ny x P R: x x ` 8 8, x 8 8, 8 x 8 0 if x ą 0 then x 8 8 nd x p 8q 8 if x ă 0 then x p 8q 8 nd x 8 8 The sttements of the bove proposition re still true in these cses. 5 Lecture Compctness Remrk 5.1 (Induced metric). Let px, dq be metric spce, nd let Y Ă X. Then, py, d Y q is metric spce, where d Y is the distnce function when restricted to Y. Proposition 5.2. Let px, dq be metric spce, nd E Ă X. Let U be open in X. Then U X E is open in E. Proof. This is left s n exercise. Just check the definition, nd note tht open blls in Y re open blls in X intersected with Y. Remrk 5.3 (Openness nd closedness s properties of subsets). Asking whether set is opened or closed only mkes sense once we identify it s n subset of metric spce, nd the nswer depends on its embedding s subset. For exmple, r0, 1s is n open subset of the metric spce r0, 1s, but it is not n open subset of the metric spce R. The notion we will now introduce, compctness, is intrinsic: it does not depend on ny embedding. Definition 5.4. Let px, dq be metric spce. Then, X is compct if every open cover hs finite subcover. An open cover of spce X is (possibly infinite) collection of open subsets of X, sy tu i u such tht they cover X, i.e. Ť U i X. A finite subcover is finite subcollection which is still cover. Exmple 5.5 (Open covers of R). Here re mny mny exmples of open covers of R. Here re some. (1) U n p n, nq for n P Z, (2) U n p 8, nq for n P N, (3) U n p 8, 8q for n P Z, (4) U n p ` 1, 1q for P R, (5) U n p ` ɛ, ɛq for, ɛ P R. I clim tht none of these hve finite subcovers. I ll prove this for (1) nd leve the rest s n exercise. Proof. Suppose there ws finite subcover; then it would be given by finite subset of the index set of integers, cll it I Ă Z. Check tht Ť ipip n, nq p mxpiq, mxpiq q, which is not R. Exmple 5.6 (Not open covers). Let X p0, 1s Ă R. Then p0, 1 nq is not n open cover, since 1 P X is not in ny of those open sets. Corollry 5.7. R is not compct. Proof. To be compct, every open cover must hve finite subcover. Since we ve given n open cover which does not hve finite subcover, we re done. Remrk 5.8 (R hs n open cover with finite subcover). Note tht exmple (3) is n open cover which hs finite subcover. This does not contrdict tht R is not compct, since only one open cover cn brek compctness. Exmple 5.9 (Compct intervls in R). Some more exmples which we cn do explicitly. (1) p 1, 1q is not compct. (2) r0, 1s is compct. (3) p0, 8q is not compct. (4) r0, 8q is not compct. (5) Z is not compct. (6) Q X r0, 1s is not compct. 13

14 Proof. (1) Use the open subcover given by U n p 1 ` 1 n, 1 1 nq for n P N. (2) We will prove generl theorem lter. (3) Use the open subcover p0, nq for n P N. (4) Use the open subcover r0, nq for n P N. Note tht while these re not open in R, they re open in r0, 8q. (5) Use the open subcover tnu for n P Z. (6) This is little trickier. Choose n irrtionl number α P r0, 1s. Let s n be rtionl number in pα 1 n, αq. Then, tke the open subcover Q X pr0, s n q Y pα, 1sq for n P N. Proposition Let px, dq be metric spce nd E Ă X. If E is compct then E is closed nd bounded subset of X. Proof. We cn do the following for ny metric spce. For every x P E (nd x p), tke neighborhood U x which is disjoint from some neighborhood of p. For exmple, one could tke U x B dpx,pq pxq X E. If p E, this is n 2 open cover, since it contins every point in E. By compctness, tke finite subcover, sy indexed by the finite points x 1,..., x n P E. Then, notice tht Ť n i 1 U x n is disjoint from B dpx minp i,pq ppq, so p is not limit point. Thus 2 q E contins ll of its limit points. For boundedness, let x P E be ny point, nd let consider the open cover U n B n pxq X E. By compctness this hs n open subcover, nd let N be the mximum n in this subcover. Then E Ă B N pxq nd is bounded. Remrk It is little remrkble tht n intrinsic property of spce tells us property bout every embedding of tht spce. Proposition Let X be compct metric spce. A closed subset of X is compct. Proof. Let E Ă X be closed. Tke finite subcover U i of E. First, we clim (nd will prove lter) tht n open set of E cn be written s the intersection of E with n open set of X. Given this clim, for ech U i tke V i Ă X open such tht U i V i X E. Now tke the open cover tv i u Y tx Eu of X. Since X is compct, this hs finite subcover if X E is in the finite subcover, remove it to obtin finite subcover of E. Lemm Let px, dq be metric spce nd let E Ă X be subset considered s metric spce. Let U be n open set of E. Then there is n open set V of X such tht U E X V. Proof. Recll tht every open set cn be written s the union of open blls. Tke the sme open blls but in X insted. Proposition Any finite union of compct subsets is compct. Proof. Left s n exercise. 6 Lecture Compctness in R k We will mke the following definition, which will only be used in this section. Definition 6.1. A k-cell is subset of R k of the form i.e. box with given vertices. r 1, b 1 s ˆ r 2, b 2 s ˆ ˆ r k, b k s Lemm 6.2 (Cntor intersection lemm for R k ). Let E 0 Ą E 1 Ą E 2 Ą be descending chin of nested nonempty k-cells of R k. Then, Ş i E i is nonempty. 14

15 Proof. First, we cn prove this for R 1. Consider sequence where E i r i, b i s. To be nested mens 1 ď 2 ď nd b 1 ě b 2 ě. Let supp 1, 2,...q nd b infpb 1, b 2, q. We clim tht ď b. Then ny x P r, bs would suffice. To prove the clim, suppose tht ą b; let x P pb, q. Then there is some n for with x ă n nd lso some m for with x ą b m. Then, b mxpm,nq ă x ă mxpm,nq, which cnnot hppen. To prove this for R k, one mkes similr rgument for ech component of the k-tuple, nd is left s n exercise. Theorem 6.3 (Heine-Borel theorem). A subset of R n is compct if nd only if it is closed nd bounded. Proof. We ve lredy proven one direction. Now let X Ă R n be closed nd bounded. Since X is bounded, it is contined in closed box, T 0 : r N, Ns n for lrge N. Since closed subset of compct set is compct, it suffices to show tht this box is compct. Fix n open cover of the box T 0. If one cuts this box in hlf in ll possible directions, one obtins 2 n smller boxes with hlf the side length of the originl. If every one of the 2 n hlf-boxes hd finite subcover, then T 0 would hve finite subcover (by tking the finite union of the finite subcovers). So, suppose otherwise, nd choose box T 1 which does not hve finite subcover. We cn repet this process to obtin descending chin of nested boxes: T 0 Ą T 1 Ą T 2 Ą. By the previous lemm, this is nonempty nd contins point p. This point p must be in one of the open sets of the cover, sy U, nd thus there is some bll B ɛ ppq Ă U. But, this bll contins ll T i for i ą M for some lrge M, contrdicting tht ech of the T i does not hve finite subcover. This proves the theorem. Exmple 6.4 (Not true for ll metric spces). The Heine-Borel theorem is not true for ll metric spces. For exmple, if X is n infinite metric spce with the discrete topology, then subset is compct if nd only if it is finite. However, every subset is closed nd bounded. Remrk 6.5. This theorem might convince you tht compctness is redundnt notion. In the following sections we will see tht this is very untrue, if we wnt to work in the generl context of metric spces or topologicl spces. 7 Lecture Subsequences I will not give definition of subsequences, nd convergence of subsequence. It is exctly wht you think it is. However, we will mke few definitions nywy. Definition 7.1. A sequence s in R is monotoniclly incresing if s i ď s i`1 for ll i. It is monotoniclly decresing Proposition 7.2. All bounded monotone sequences converge. Proof. Suppose s n is monotoniclly incresing. We clim s n converges to s : suppts n uq. For ɛ ą 0, there is some s N P ps ɛ, sq by definition of supremum. Since the sequence is mootonic, this is true for ll s n nd n ą N. Proposition 7.3. Every sequence hs monotone subsequence. Proof. Let s n be sequence. We sy the kth term, s k, is dominnt if s k ą s i for ll i ą k, i.e. bigger thn ll its subsequent terms. There re two cses to consider. In the first cse, there re infinitely mny dominnt terms; then, the subsequence consisting of dominnt terms is monotoniclly decresing sequence. In the second cse, there re only finitely mny. Then, in choosing subsequence, we follow the rule: never choose dominnt term. In doing so, we cn lwys choose some term in the sequence following ny choice which is bigger, obtining monotoniclly incresing sequence. Theorem 7.4 (Bolzno-Weierstrss for R). Every bounded sequence hs convergent subsequence. Proposition 7.5 (Bolzno-Weierstrss, generliztion). Let X be compct metric spce. Every sequence hs convergent subsequence. 15

16 Proof. Let tx n u be the sequence, nd let E tx 1, x 2,...u i.e. the set of points in the sequence. If E is finite then there must be point tht reoccurs infinitely mny times, nd tht is our convergent subsequence. If E is infinite, then we clim it hs limit point, sy p. Given this clim, for ech bll B 1 ppq we cn choose successive elements n in smller blls converging to p. More precisely, there is some N n such tht s i is in the bll B 1 ppq for i ą N n. n Choose successively i n such tht i n ą i n 1 nd i n ą N n to obtin the subsequence. Lemm 7.6. If E is n infinite subset of compct metric spce X, then E hs limit point in X. Proof. Suppose to the contrry; if no point in X is limit point, every point in x P X hs neighborhood which contins t most one point of E (nmely, the point x itself if x P E). This gives n open cover of X, but it cnnot hve finite subcover since ech open contins t most one point of E. Remrk 7.7. Note tht this proposition is not true for just closed nd bounded subsets. For exmple, tke X Z under the discrete topology, so tht it is closed nd bounded but not compct. The sequence s n n hs no convergent subsequence. So this is nother piece of evidence in fvor of compctness s good notion. Definition 7.8. The limit supremum of sequence s n is defined s The limit infimum is defined similrly lim sup nñ8 lim inf nñ8 lim nñ8 supps n, s n`1, s n`2,...q lim nñ8 infps n, s n`1, s n`2,...q Proposition 7.9 (Chrcteriztion of limit supremum). Let s n be bounded sequence. Let E be the set of subsequentil limits. Then lim sup s n exists nd is equl to suppeq, nd lim inf s n exists nd is equl to infpeq. Proof. I will prove the sttement for lim sup. Since s n is bounded bove, suppeq exists. ehhhhhhhhh Proposition For ny sequence s n we hve tht lim inf s n ď lim sup s n. If they re equl, then the limit exists nd lim nñ8 s n lim inf s n lim sup s n. Conversely, if the limit exists, then lim nñ8 s n lim inf s n lim sup s n. Proof. The first sttement is left s n exercise. For the second, let s lim sup s n lim inf s n. Fix ɛ ą 0. There is n N such tht for n ą N, we hve tht supps n, s n`1,...q P ps ɛ, s ` ɛq. Thus, s n ă s ` ɛ for ll n ą N. A similr rgument with the infimum shows tht s ɛ ă s n for ll n ą N. Thus, s n s ă ɛ for n ą N. For the third sttement, we hve tht for every ɛ ą 0, there is some N such tht for n ą N we hve s n P ps ɛ, s ` ɛq. Thus, lim supps n q ă s ` ɛ nd lim inf ą s ɛ for every ɛ, nd the result follows. Corollry 7.11 (Appliction to sequences: squeezing ). Let s n, t n be sequences in R tht converge to s, t, such tht s n ď t n for n ą N for some N. Then, lim s n ď lim t n. Proof. One cn esily check tht lim sup s n ď lim inf t n. The rest follows by the previous proposition. 7.2 Cuchy sequences, complete metric spces Exmple Let X R t0u be metric spce. The sequence t 1 nu does not converge, becuse 0 R X. But it looks like convergent sequence. We wnt to mke definition tht cptures this phenomenon. Definition A sequence is Cuchy if for every ɛ ą 0 there is n integer N such tht dpp n, p m q ă ɛ for m, n ą N. A metric spce for which every Cuchy sequence is convergent is clled complete. Remrk As we sw bove, not every Cuchy sequence converges, nd in our exmple ws becuse there ws hole. Proposition Cuchy sequences re bounded Proof. Left s n exercise. 16

17 Proposition Every convergent sequence is Cuchy. Proof. Suppose s n converges to s. Then, for every ɛ ą 0 there is n N such tht dps n, sq ă ɛ 2 m, n ą N, dps n, s m q ď dps n, sq ` dps m, sq ă ɛ. for n ą N. Then, for Proposition 7.17 (Cntor intersection lemm). Let E 0 Ą E 1 Ą E 2 Ą be descending chin of nonempty compct subsets of metric spce X. Then, Ş i E i is nonempty. Further, define the dimeter of set E by dimpeq : supptdpx, yq x, y P Eu. If lim nñ8 dimpe i q 0, then Ş E i consists of exctly one point. Proof. For the first sttement, if the intersection is empty, then te 0 E i u ipn is n open cover of E 0 (open becuse E i is compct in compct E 0 nd thus closed). It hs finite subcover, since E 0 is compct. This mens tht eventully the sequence stbilizes, i.e. E i E i`1. This contrdicts tht the E i re nonempty, so the intersection must be nonempty. For the second sttement, this must be true since if the intersection contined two points, sy x, y, then dimpe i q ě dpx, yq for ll i. Proposition Every compct spce is complete. Proof. Let s i be Cuchy sequence. Define E i ts i, s i`1,...u, in the nottion of the previous proposition. Since the sequence is Cuchy, lim nñ8 dimpe n q 0. Thus there is single point in the intersection, sy s. I clim this is the limit (the hrd prt is over, since the filure for Cuchy sequence to converge is somehow due to the point it should converge to not being there ). To see this, fix ɛ ą 0; if dimpe N q ă ɛ then this implies tht dps n, sq ă ɛ for ll n ą N. Corollry R k is complete. Proof. Consider Cuchy sequence; it is bounded (check s exercise) so we my enclose it in bounded nd closed box. This is compct by Heine-Borel, nd thus every bounded Cuchy sequence hs limit. Definition Let px, dq be metric spce. We will define the completion of X, denoted X. As set, the elements of X re Cuchy sequences (indexed by N) in X, modulo the n equivlence reltion which we will define lter. For now, define the distnce function between two Cuchy sequences by: dpx, yq lim n dpx n, y n q Check s n exercise tht this is well-defined since the sequences re Cuchy. Then, two sequences re equivlent x y if dpx, yq 0. This distnce function is lso well-defined on equivlence clsses nd stisfies the xioms of distnce function. 7.3 Aside: Construction of the rel numbers by completion In this section we will bcktrck nd pretend we don t know wht R is. We will tret Q s Q-metric spce. This will be the only time in these notes tht we think of distnces s tking vlues in ny field other thn R. We will use Cuchy sequences in Q to construct R. Definition The set of rel numbers is defined to be the set of Cuchy sequences in Q (using the bsolute vlue metric) modulo the equivlence reltion where x y if lim nñ8 x n y n 0. Addition nd multipliction re defined on Cuchy sequences by px ` yq n x n ` y n nd likewise. The dditive identity is the constnt zero sequence, nd the multiplictive identity is the constnt one sequence. The order is defined by x ď y if x y of if there is some N such tht x n ă y n for ll n ą N. Finlly, the rtionl numbers re considered s subset of R by tking constnt sequences. 17

18 Proof. Check tht ll opertions nd orders re well-defined nd the field xioms s n exercise. One is ssigned s homework. The upper bound property is little more work. Let S be subset of rel numbers, nd let u 0 be n upper bound of S. Choose l 0 such tht there is some s P pl 0, u 0 q X S. One should quickly check tht this cn be done: tke some s P S nd find the N such tht m, n ą N implies s n s m ă 1, nd we cn tke l 0 s n 2. Now, repet the following process to construct sequence: consider m n ln 1`u0 2. If this is n upper bound for S, then set u n m n nd l n l n 1. Otherwise, set u n u n 1 nd l n m n. This defines two sequences, u n nd l n. These re both Cuchy sequences; I will leve this s n exercise. First, notice tht every u n is n upper bound on S by construction. Thus the sequence u thought of s n element of R is n upper bound on S by definition. I clim tht u is the supremum. To see this, suppose there is smller upper bound, u 1. This mens tht lim nñ8 u n u 1 n for some positive number. Notice tht the the quntity u n l n is cut in hlf ech time we increse n by one. Also notice tht by construction, there is lwys element in S greter thn l n. Thus, when u n l n ă, we find tht this contrdicts tht u 1 n is n upper bound. 8 Lecture Tking powers in the rel numbers We hven t ctully define wht it mens to tke powers in the positive rel numbers to numbers other thn integers. We will first define frctionl powers, which we could hve done erlier s soon s we defined Dedekind cuts. Definition 8.1. Let α P R be hve α ą 1 nd p P N. We define α 1{p to be the unique positive rel number x such tht x p α. Constructively, this is the Dedekind cut: tx P Q x ď 1 or x p ă αu. Now, let p P R be positive, nd α ą 1. We define α p to be the Dedekind cut: tx P Q x ď 1 or x q ă α for 0 ă q ă pu. For the cse 0 ă α ă 1, one uses tht α 1 ą 1. One does hve to prove tht this cut does hve the given property, nd tht it is unique in this sense. The first is left s n exercise, nd the second follows from the order xioms. We won t prove the following. Proposition 8.2. For, b ą 0 nd p rel numbers, using the bove definitions, we hve tht pbq p p b p. 8.2 Toolbox sequences Recll: Theorem 8.3 (Binomil theorem). Let n P N. Then, p1 ` xq n 1 ` nx ` Proposition 8.4. The following re useful sequences to know. 1 () If p ą 0, then lim nñ8 n 0. p (b) If p ą 0, then lim? n nñ8 p 1. (c) lim? n nñ8 n 1. n (d) If p ą 0, then lim α nñ8 p1`pq 0. n (e) If x ă 1, then lim nñ8 x n 0. ˆn x 2 ` ` nx n 1 ` x n 2 Proof. For (), we wnt to verify tht we cn mke 1 n p ă ɛ for lrge n. Doing some rerrnging, we wnt n ą p 1 ɛ q1{p. Here we use two fcts: tht 0 ă x ă y implies 0 ă 1 y ă 1 x nd tht the function fpxq xα is monotoniclly incresing for α ą 0 (i.e. x ď y implies x α ď y α ). 18