MATH 215 Mathematical Analysis Lecture Notes
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1 MATH 25 Mathematical Aalysis Lecture Notes Aıl Taş September 2005 M.A. studet at the Departmet of Ecoomics, Bilket Uiversity, Bilket, 06800, Akara, Turkey. This course was give by Prof. Mefharet Kocatepe i the 2005 Bilket Summer School.
2 Cotets The Real Ad Complex Number Systems 3. The Real Number System Exteded Real Numbers The Complex Field Sets Ad Fuctios 2 2. Geeral Coutable Ad Ucoutable Sets Basic Topology 8 3. Metric Spaces Subspaces Compact Sets The Cator Set Coected Sets Sequeces Ad Series Sequeces Subsequeces Cauchy Sequeces Upper Ad Lower Limits Series Operatios With Series Cotiuity Geeral Cotiuity Ad Compactess Cotiuity Ad Coectedess Uiform Cotiuity Sequeces Ad Series Of Fuctios Geeral Uiform Covergece Uiform Covergece Ad Cotiuity Uiform Covergece Ad Itegratio Uiform Covergece Ad Differetiatio Figures 86 2
3 The Real Ad Complex Number Systems. The Real Number System R : The set of real umbers Q : The set of ratioal umbers (Numbers i the form m itegers, 0) where m, are Z : The set of itegers, {0,,, 2, 2, 3, 3,...} N : The set of atural umbers, {, 2, 3, 4,...} Proofs: ) Direct Proofs 2) Idirect Proofs a) Proof by Cotrapositio b) Proof by Cotradictio. Example (Direct Proof). If f is differetiable at x = c the f is cotiuous at x = c. Proof. Hypothesis: f is differetiable at x = c, i.e. f(x) f(c) lim x c x c = f (c) exists ad it is a umber. Claim: f is cotiuous at c, i.e. lim x c f(x) = f(c). ( ) f(x) f(c) lim f(x) = lim (x c) + f(c) x c x c x c f(x) f(c) = lim lim(x c) + lim f(c) x c }{{ x c x c }}{{} x c }{{} f (c) 0 f(c) = f(c).2 Example (Idirect Proof: Proof by Cotrapositio). Let be a iteger. If } 2 is{{ eve} the } is{{ eve}. p q If p the q: p q. This is equivalet to: If ot q the ot p, i.e. q p. If is odd the 2 is odd. 3
4 Proof. Assume is odd the = 2k + for some iteger k. The So 2 is odd. 2 = 4k 2 + 4k + = 2(2k } 2 {{ + 2k } ) + a iteger.3 Example (Idirect Proof: Proof by Cotradictio). Show that 2 is ot a ratioal umber. Proof. If c} 2 {{ = 2} the c} is ot{{ ratioal }. We assume p ad q, proceed ad p q get a cotradictio. So assume c 2 = 2 ad c is ratioal. The c = m where m, Z, 0 ad m, have o commo factors. We have c 2 = m2 the 2 m 2 = 2 2. So m 2 is eve ad m is eve. The m = 2k for some iteger k. We get 4k 2 = 2 2 or 2k 2 = 2. The 2 is eve ad is eve. The = 2l for some iteger l. So m ad have a commo factor. A cotradictio. Some Symbols Ad Notatio p q : if p the q (p implies q) p q : if p the q ad if q the p (p iff q) : such that : therefore, so : for all, for every : for some, there exists.4 Example. Cosider the followig two statemets: () x R y R, y < x (2) y R x R, y < x ot true () says, give ay real umber x we ca fid a real umber y (depedig o x) such that y is smaller tha x. (2) says, there is a real umber y that is smaller tha every real umber x. This is false. 4
5 .5 Example. Let A = {p : p Q, p > 0, p 2 < 2} ad B = {p : p Q, p > 0, p 2 > 2}. Claim: A has o largest elemet ad B has o smallest elemet. Proof. Give ay p Q, p > 0, let q = p p2 2. The q Q ad p+2 q > 0. Let p A, i.e. p 2 < 2. The show q A ad p < q. p+2 = 2p+2 p 2 2 < 0 so q = p p2 2 p + 2 > p ( ) 2 2p + 2 q 2 2 = 2 = 4p p 2p 2 8 8p p + 2 p p = 2(p2 2) (p + 2) 2 < 0 So q 2 < 2. The we have q A ad A has o largest elemet. Properties of R ) R has two operatios + ad with respect to which it is a field. (i) x, y R, x + y R (ii) x, y R, x + y = y + x (commutativity of +) (iii) x, y, z R, x + (y + z) = (x + y) + z (associativity of +) (iv) a elemet (0 elemet) such that x R, x + 0 = x (v) x R a elemet x R such that x + ( x) = 0 (vi) x, y R, (vii) x, y R, (viii) x, y, z R, x y R x y = y x x (y z) = (x y) z (ix) a elemet 0 i R such that x R, x = x (x) x 0 i R there is a elemet x (xi) x, y, z R, x (y + z) = x y + x z i R such that x x =.6 Remark. Note that Q with + ad is also a field. 2) R is a ordered field, i.e. there is a relatio < with the followig properties (i) x, y R oe ad oly oe of the followig is true: x < y, x = y, y < x (Trichotomy Law) 5
6 (ii) x, y, z R, x < y ad y < z x < z (Trasitive Law) (iii) x, y, z R, x < y x + z < y + z (iv) 0 < x ad 0 < y 0 < x y.7 Remark. Note that Q is also a ordered field. 3) R is complete..8 Defiitio. Let E R. We say E is bouded above if there is a elemet b R such that x E we have x b. b is called a upper boud for E..9 Example. E = {p : p Q, p > 0, p 2 < 2}. The b = 3, b = 2, b = 5, 2 b = 00, b = 2 are all upper bouds for E..0 Example. E = N = {, 2, 3, 4,...} is ot bouded above.. Remark. Bouded below ad lower boud are defied aalogously..2 Defiitio. Let E R be bouded above. A umber b R is called a least upper boud (lub) or supremum (sup) of E if (i) b is a upper boud for E ad (ii) if b ay upper boud for E, the b b.3 Remark. sup E eed ot be a member of E. If sup E is i E the it is called the maximum elemet. Completeess Property (or Least Upper Boud Property) of R: Every o-empty set E R that is bouded above has a least upper boud i R..4 Remark. Q does ot have LUB property. Proof. Let E = {p : p Q, p > 0, p 2 < 2} is a o-empty subset of Q that is bouded above but it has o least upper boud i Q. Assume b Q is a least upper boud of E. Sice p = E ad p b we have b. We have two possibilities: 6
7 (i) b E (ii) b / E If b E, the q E such that b < q. The b caot be a upper boud. So b / E. Sice b Q ad 0 < b, we have that b 2 < 2 is ot true. By trichotomy law, we have either b 2 = 2 or 2 < b 2. If b Q, b 2 = 2 caot be true. So 2 < b 2. Let F = {p : p Q, p > 0, 2 < p 2 }. The b F ad there is a elemet q i F such that q < b. Show q is a upper boud for E. Let p E be a arbitrary elemet. The p > 0 ad p 2 < 2. Also q F so q > 0 ad 2 < q 2. p 2 < 2 ad 2 < q 2 p 2 < q 2 p < q So q is bigger tha every p E, i.e. q is a upper boud for E. The b q. Also q < b. Cotradictio..5 Remark. Aalogously we have greatest lower boud (glb) or ifimum (if) ad the Greatest Lower Boud Property..6 Theorem. (a) (Archimedia Property) For every x, y R, x > 0, there is N (depedig o x, y) such that x > y. (b) (Q is dese i R) For every x, y R with x < y, p Q such that x < p < y. Proof. (a) Assume it is ot true. So there are x, y R such that x > 0 for which there is o N such that x > y. So for all N we have x y. Let E = {x : N} = {x, 2x, 3x,...}. The y is a upper boud for E ad E. So E has a least upper boud, say α R. Sice x > 0, α x < α. The α x is ot a upper boud for E. So there is a elemet of E, say mx (where m N) such that α x < mx. The α < (m + )x. This elemet of E is bigger tha α = sup E. Cotradictio. 7
8 (b) Let x, y R be such that x < y. The y x > 0. By part (a), N such that (y x) >, i.e. y > + x. I (a), replace y by x ad x by > 0. So m N such that m > x. Let A = {m : m Z, x < m}. The A sice m A. Sice A is o-empty set of itegers which is bouded below, A has a smallest elemet m 0. The x < m 0 ad m 0 Z. m 0 / A. So we have x m 0. So m 0 x < m 0. The x < m 0, m 0 + x ad + x < y. So x < m 0 < y. If we divide this by > 0, we get x < m 0 < y. m 0 is a ratioal umber. Uiqueess of Least Upper Boud: Assume E R is o-empty ad bouded above. The E has oly oe least upper boud. Proof. Assume b, b are two least upper bouds for E. (i) b = sup E ad b is a upper boud for E b b (ii) b = sup E ad b is a upper boud for E b b The b = b..7 Fact. Let E R be o-empty ad bouded above ad let α R. The α = sup E (i) x E, x α ad (ii) Give ay ε > 0, y E such that α ε < y.2 Exteded Real Numbers I the set of exteded real umbers we cosider the set R {, + }. Preserve the order of R ad x R, set < +. This way every oempty subset E of R {, + } has a least upper boud ad greatest lower boud i R {, + }. For example, if E = N, the sup E = +. Also for x R, we defie the followig x + = +, x + ( ) =, If x > 0, we defie x (+ ) = + ad x ( ) = If x < 0, we defie x (+ ) = ad x ( ) = + 0 ( ) is udefied. 8 x + = x = 0
9 .3 The Complex Field Let C deote the set of all ordered pairs (a, b) where a, b R. We say (a, b) = (c, d) a = c ad b = d Let x = (a, b), y = (c, d) C. We defie x + y = (a + c, b + d) ad x y = (ac bd, ad + bc) Uder these operatios C becomes a field with (0, 0), (, 0) beig the zero elemet ad multiplicative uit. Defie φ : R C by φ(a) = (a, 0). The φ(a + c) = φ(a + c, 0) = (a, 0) + (c, 0) = φ(a) + φ(c) φ(ac) = (ac, 0) = (a, 0)(c, 0) = φ(a) φ(c) φ is - (oe-to-oe), i.e. if a a the φ(a) φ(a ). This way we ca idetify R with the subset {(a, 0) : a R} by meas of φ. Let i = (0, ). The i 2 = (0, ) (0, ) = (, 0) = φ( ). Also if (a, b) C, the φ(a) + iφ(b) = (a, 0) + (0, )(b, 0) = (a, 0) + (0, b) = (a, b) If we idetify φ(a) with a the we idetify (a, b) with a + ib. So C is the set of all imagiary umbers i the form a + ib where a, b R ad i 2 =. If z = x + iy C, x, y R we defie z = x iy (Cojugate of z) x = Re z (Real Part of z) y = Im z (Imagiary Part of z) If z, w C, we have z + w = z + w, zw = z w, Re z = z + z 2, Im z = z z 2i Sice zz = x 2 +y 2 0, we defie the modulus of z as z = zz = x 2 + y 2..8 Propositio. Let z, w C. The (a) z 0 z > 0 ad z = 0 z = 0 (b) z = z (c) zw = z w 9
10 (d) Re z z (i.e. x x 2 + y 2 ) (e) z + w z + w (Triagle Iequality) Proof of (e). z + w 2 = (z + w)(z + w) = (z + w)(z + w) = zz + zw }{{ + wz } +ww = z Re(zw) + w 2 }{{} 2Re(zw) zw = z w z z w + w 2 = ( z + w ) 2 So z + w 2 ( z + w ) 2. If we take positive square root of both sides, we get z + w z + w..9 Theorem (Cauchy-Schwarz Iequality). Let a j, b j C where j =, 2,...,. The 2 ( ) ( ) a j b j a j 2 b j 2 j= j= j= Proof. Let us defie A = a j 2, B = j= b j 2, C = j= a j b j j= We have B 0. If B = 0, the all b j = 0. The iequality becomes 0 0. So assume B > 0. The 0 Ba j Cb j 2 = j= = (Ba j Cb j ) ( ) Ba j C b j j= ( B 2 a j 2 BCa j b j CBb j a j + C 2 b j 2) j= = B 2 A } BCC {{} CBC }{{} + C 2 B B C 2 C 2 B = B ( AB C 2) So 0 B (AB C 2 ) ad sice B > 0, we have AB C 2 0. The C 2 AB. 0
11 Replacig b j by b j ad usig z = z, z = z we get 2 ( ) ( ) a j b j a j 2 b j 2 j= j= j= If a j 0, b j 0 are real umbers the ( ) 2 ( a j b j ) ( ) a 2 j b 2 j j= j= j=
12 2 Sets Ad Fuctios 2. Geeral Let X ad Y be two o-empty sets. A fuctio f : X Y is a rule which assigs to each x X a uique elemet y = f(x) i Y. 2. Example. Give x R, cosider its decimal expasio i which there is o ifiite chai of 9 s. x = is represeted as istead of Let y = f(x) be the fortyith digit after the decimal poit. We have f : R {0,, 2,..., 9}. 2.2 Defiitio. Let f : X Y, A X. We defie the image of A uder f as the set f(a) = {y Y : x A such that y = f(x)}. Let B Y. We defie the iverse image of B uder f as the set f (B) = {x X : f(x) B}. 2.3 Example. f : R R, f(x) = x 2. The A = (, 2] the f(a) = (, 4] A = (, 3) the f(a) = [0, 9) B = [, 4] the f (B) = [ 2, ] [, 2] B = [, 4] the f (B) = [ 2, 2] B = [ 2, ] the f (B) = 2.4 Defiitio. Let f : X Y be a fuctio. The we defie X: Domai of f ad f(x): Rage of f If f(x) = Y, the f is called a surjectio or oto. If x x 2 the f(x ) f(x 2 ) (equivaletly f(x ) = f(x 2 ) x = x 2 ) the f is called oe-to-oe (- ) or a ijectio. If f is both a ijectio ad a surjectio, the f is called a bijectio or a - correspodece. Let f : X Y, A X, B Y. We have (a) f(f (B)) B. f(f (B)) = B for all B f is oto (b) A f(f (A)). A = f(f (A)) for all A f is - (c) f(a A 2 ) f(a ) f(a 2 ). f(a A 2 ) = f(a ) f(a 2 ) for all A ad A 2 f is - 2
13 (d) f(a A 2 ) = f(a ) f(a 2 ) Proof of (b). A f (f(a)) Let x A ad let y = f(x). The y f(a), i.e. f(x) f(a). So x f (f(a)) ad A f (f(a)). A = f (f(a)) for all A X f is - ( ): Assume f is -. Show for all A X, A = f (f(a)), i.e. show A f (f(a)) ad f }{{} (f(a)) A. So show f (f(a)) A. Let x always true f (f(a)). The f(x) f(a). The x A such that f(x) = f(x ). Sice f is -, x = x. So x A ad f (f(a)) A. ( ): Assume A = f (f(a)) for all A X. Show f is -. Assume f is ot -. The there are x, x 2 X such that x x 2 ad f(x ) = f(x 2 ). Let y = f(x ) = f(x 2 ). Let A = {x }. The f(a) = {y} ad f (f(a)) = {x, x 2,...}. The A f (f(a)). Cotradictio. Let {A i : i I} be a arbitrary class of subsets of a set X idexed by a set I of subscripts. We defie A i = {x : x A i for at least oe i I} i I A i = {x : x A i for every i I} i I If I =, the we defie i A i = ad i A i = X. (If we require of a elemet that it belogs to each set i the class ad if there are o sets i the class, the every elemet x X satisfies this requiremet.) If A X we defie A C = {x : x / A} complemet of A. ( ( A C i, (De Morga s Laws) i I A i)c = i I i I A i)c = i I Let f : X Y, let {A i : i I}, {B j : j J} be classes of subsets of X ad Y. ( ) f A i = ( ) f(a i ), f A i f(a i ) i I i I i I ( ) f B j = f (B j ), j J j J For all B Y we have 3 i I A C i f ( j J B j ) = j J f (B j )
14 (a) f (B C ) = (f (B)) C (b) f(a) C f(a C ) for all A X f is oto (c) f(a C ) f(a) C for all A X f is - Proof of (a). Let x f (B C ). The f(x) B C. Show x (f (B)) C. Assume it is ot true, i.e. x f (B) f(x) B. So f(x) B} C {{ B}. Cotradictio. Thus, x (f (B)) C. So, f (B C ) (f (B)) C. Let x (f (B)) C. Show x f (B C ), i.e. f(x) B C. Assume it is ot true, i.e. f(x) B so x f (B). The, x (f (B)) C f (B). }{{} Cotradictio. So, (f (B)) C f (B C ). If f : X Y is - oto, the y Y uique x X such that y = f(x). Sed Y X. This way we get f : Y X (the iverse fuctio of f) f (f(x)) = x x X, f(f (y)) = y y Y. 2.2 Coutable Ad Ucoutable Sets Let C be ay collectio of sets. For A, B C we write A B (ad say A ad B are umerically equivalet) if there is a - correspodece f : A B. has the followig properties (i) A A (ii) A B B A (iii) A C A C 2.5 Example. A = N = {, 2, 3,...}, B = 2N = {2, 4, 6,...}. The A B by f : A B, f() = Defiitio. For =, 2, 3,... let J = {, 2, 3,..., }. Let X be ay set. We say X is fiite if N such that X J. X is ifiite if it is ot fiite. X is coutable (or deumerable) if X N. X is ucoutable if X is ot fiite ad ot coutable. 4
15 X is at most coutable if X is fiite or coutable. 2.7 Example. N, 2N are coutable. 2.8 Example. Z = {..., 3, 2,, 0,, 2, 3,...} is coutable. Defie f : N Z as follows N :, 2, 3, 4, 5,... Z : 0,, -, 2, -2,... So we have f() = { 2 if is eve 2 if is odd 2.9 Example. Q + = {q : q Q, q > 0} is coutable. Give q Q +, we have m, Z m > 0, > 0 such that q = m. List the elemets of Q+ i m\ this order omittig the oes which are already listed before. The we get the followig sequece Defie f : N Q + as follows, 2, 2, 3, 3, 4, 2 3, 3 2, 4 f() =, f(2) = 2, f(3) = 2, f(4) = 3 5
16 Suppose X is coutable, so we have -, oto fuctio f : N X. f() = x. The we ca write the elemets of X as a sequece Let X = {x, x 2, x 3,...} 2.0 Example. Q is coutable. (Similar to the proof of Z is coutable.) 2. Propositio. Let I be a coutable idex set ad assume for every i I, A i is a coutable set. The i I A i is coutable. (Coutable uio of coutable sets is coutable.) 2.2 Example. X = [0, ) is ot coutable. Every x X has a biary (i.e. base 2) expasio x = 0.x x 2 x 3 x 4 where each of x, x 2, x 3,... is 0 or. Assume X = [0, ) is coutable. The write its elemets as a sequece. X = {y, y 2, y 3,...}. The write each of y, y 2, y 3,... i the biary expasio. Defie y = 0.y y 2y 3 y 2 = 0.y 2 y 2 2y 2 3 y 3 = 0.y 3 y 3 2y 3 3. z = { 0 if y = if y = 0 z 2 = { 0 if y 2 2 = if y 2 2 = 0 z = { 0 if y = if y = 0 Let z = 0.z z 2 z 3 [0, ). But z y, z y 2,... Cotradictio. 2.3 Example. [0, ], (0, ), R, (a, b) are all ucoutable. They are all umerically equivalet. 2.4 Example. f : ( π 2, π 2 ) R ad f(x) = ta x gives ( π 2, π 2 ) R. 2.5 Theorem (Cator-Schröder-Berstei). Let X, Y be two o-empty sets. Assume there are - fuctios. f : X Y ad g : Y X. The X Y. 6
17 2.6 Example. Let X = {(x, y) : 0 < x <, 0 < y < }. The X (0, ). Defie g : (0, ) X by g(x) = ( x, 2). Defie f : X (0, ) as follows: Give (x, y) X, write X ad Y i their decimal expasio with o ifiite chai of 9 s. Let x = 0.x x 2 x 3 ad y = 0.y y 2 y 3 f(x, y) = 0.x y x 2 y 2 x 3 y 3 The f, g are - so by the above theorem X (0, ). 7
18 3 Basic Topology 3. Metric Spaces I R, R k = {(x,..., x k ) : x,..., x k R} we have the otio of distace. I R, d(x, y) = x y I R k, x = (x,..., x k ), y = (y,..., y k ). We have d(x, y) = (x y ) (x k y k ) 2 3. Defiitio. Let X be a set. Suppose we have a fuctio with the followig properties d : X }{{ X } R {(x,y):x,y X} (i) x, y X, d(x, y) 0 d(x, y) = 0 x = y (ii) x, y X, d(x, y) = d(y, x) (iii) x, y, z X, d(x, y) d(x, z) + d(z, y) (Triagle Iequality) d is called a metric ad the pair (X, d) is called a metric space. Proof of (iii). X = R k, d(x, y) = (x y ) (x k y k ) 2 the d satisfies (i) ad (ii). For triagle iequality, let x, y, z R k be give. The we have d(x, y) 2 =(x y ) (x k y k ) 2 =((x z ) + (z y )) ((x k z k ) + (z k y k )) 2 =(x z ) (x k z k ) 2 + (z y ) (z k y k ) 2 + 2[(x z )(z y ) + + (x k z k )(z k y k )] (d(x, z)) 2 + (d(z, y)) (x z )(z y ) + + (x k z k )(z k y k ) }{{} ((x z ) 2 + +(x k z k ) 2 ) /2 ((z y ) 2 + +(z k y k ) 2 ) /2 (from Cauchy-Schwarz iequality) (d(x, z)) 2 + (d(z, y)) 2 + 2d(x, z)d(z, y) (d(x, z) + d(z, y)) 2 8
19 3.2 Example. ) X ay set. Give ay x, y X let { if x y d(x, y) = 0 if x = y (i) ad (ii) are trivially true. Check (iii). Proof of (iii). If d(x, y) = 0, the it is true as RHS 0. If d(x, y) =, the we caot have RHS = 0. If RHS = 0 we have, d(x, z) = 0 ad d(z, y) = 0. That is x = z ad z = y. So x = y. The d(x, y) = 0. Cotradictio. This metric is called the discrete metric. 2) Let X = R k = {x = (x,..., x k ) : x,, x k R} d (x, y) = x y + x 2 y x k y k (i) ad (ii) are trivially true. Check (iii). Proof of (iii). Give x, y, z R k d (x, y) = x y + + x k y k = (x z ) + (z y ) + + (x k z k ) + (z k y k ) x z + z y + + x k z k + z k y k = d (x, z) + d (z, y) d is called the l metric o R k. 3) X = R k d (x, y) = max{ x y,..., x k y k } (i) ad (ii) are trivially true. Check (iii). Proof of (iii). Give x, y, z R k d (x, y) = max{ x y,..., x k y k } = x i y i ( i k) = (x i z i ) + (z i y i ) x i z i + z }{{} i y i }{{} max max d (x, z) + d (z, y) 9
20 d is called the l metric o R k. 4) Let S be ay fixed o-empty set. A fuctio f : S R is called bouded if f(s) is a bouded subset of R, i.e. there are two umbers A, B such that s S, A f(s) B. For example, f : R R, f(s) = arcta s is bouded, f(s) = s 2 + is ubouded. Let X = B(s) = all bouded fuctios f : S R. Let f, g B(S), the f g is also bouded. We defie d(f, g) = sup{ f(s) g(s) : s S} Geometrically, it is the supremum of the vertical distaces betwee the two graphs (See Figure ). (i) ad (ii) are trivially true. Check (iii). We eed the followig: Let A, B be two o-empty subsets of R that are bouded above. The, sup(a + B) sup A + sup B. A + B = {x + y : x A, y B}. Let a = sup A ad b = sup B. Let z A + B be arbitrary. The x A, y B such that z = x + y. The we have x + y a + b. Sice z = x + y, z a + b. So a+b is a upper boud for the set A+B. Sice supremum is the smallest upper boud, sup(a+b) a+b. I fact we have sup(a+b) = sup A+sup B. Show sup(a + B) sup A + sup B. Give ε > 0, x A ad y B such that a ε < x ad b ε < y. The a + b 2ε < x + y. Sice x + y A + B, we have x + y sup(a + B). The a + b 2ε < sup(a + B). Here a, b, sup(a + B) are fixed umbers, i.e. they do ot deped o ε > 0. We have a+b < sup(a+b)+2ε true for every ε > 0. The we have a+b sup(a+b). If it is ot true, the a + b > sup(a + B). Let δ = a+b sup(a+b) the δ > 0. 4 So we have Cotradictio. a + b < sup(a + B) + 2δ a + b 2 < sup(a + B) + sup(a + B) < 2 a + b sup(a + B) 2 Proof of (iii). Give f, g, h B(S), let C = { f(s) g(s) : s S} A = { f(s) h(s) : s S} B = { h(s) g(s) : s S} 20
21 The sup C = d(f, g), sup A = d(f, h) ad supb = d(h, g). Let x C. The s S such that x = f(s) g(s). The x = f(s) g(s) = (f(s) h(s)) + (h(s) g(s)) f(s) h(s) + h(s) g(s) }{{}}{{} call y call z = y + z A + B So x C, there is a elemet u A + B such that x u. The, sup C sup(a + B) sup A + sup B. That is, d(f, g) d(f, h) + d(h, g). 3.3 Defiitio. Let (X, d) be a metric space, p X ad r > 0. The The ope ball cetered at p of radius r is defied as the set B r (p) = {x X : d(x, p) < r} The closed ball cetered at p of radius r is defied as the set B r [p] = {x X : d(x, p) r} 3.4 Example. ) X = R k, d 2 (x, y) = (x y ) (x k y k ) 2. (See Figure 2). 2) X, d: discrete metric. The B r (p) = { {p} if r X if r > B r [p] = { {p} if r < X if r 3) X = R k with l metric d. Let X = R 2, p = (p, p 2 ), x = (x, x 2 ). The x p + x 2 p 2 < r. If p = p 2 = 0 we have x + x 2 < r. (See Figure 3). 4) X = R k with l metric d. Let X = R 2, p = (0, 0), x = (x, x 2 ). The max{ x, x 2 } < r. (See Figure 4). 5) X = B(S) with sup metric. Let S = [a, b]. The B r (f) is the set of all fuctios g whose graph is i the shaded area. (See Figure 5). 2
22 3.5 Defiitio. Let (X, d) be a metric space ad E a subset of X. ) A poit p E is called a iterior poit of E if r > 0 such that B r (p) E. (See Figure 6). 2) The set of all iterior poits of E is called the iterior of E. It is deoted by it E or E. We have it E E. (See Figure 7). 3) E is said to be ope if it E = E, i.e. if every poit of E is a iterior poit of E, i.e. p E r > 0 such that B r (p) E. (See Figure 8). 3.6 Propositio. Every ope ball B r (p) is a ope set. Proof. Let q B r (p). Show that s > 0 such that B s (q) B r (p). Sice q B r (p) we have d(q, p) < r. So r d(q, p) > 0. Show B }{{} s (q) B r (p). Let let this be s x B s (q), i.e. d(x, q) < s. The So x B r (p). d(x, p) d(x, q) + d(q, p) < s + d(q, p) < r d(q, p) + d(q, p) < r 4) Let p X. A subset N X is called a eighborhood of p if p it N. (See Figure 9). B r (p) is a eighborhood of p or B r (p) is a eighborhood of all of its poits. 5) A poit p X is called a limit poit (or accumulatio poit or cluster poit) of E if every eighborhood N of p cotais a poit q E with q p. (See Figure 0). 6) A poit p E is called a isolated poit of E if p is ot a limit poit of E, i.e. if there is a eighborhood N of p such that N E = {p}. 3.7 Example. Let X = R, E = {,,,,...}, d(x, y) = x y. Isolated poits of E are,,,,... E has oly oe limit poit that is 0. Give ay ope ball B r (0) = ( r, r), fid N such that <. The x = r B r (0) E ad x 0. 22
23 7) E is the set of all limit poits of E. We defie E = E E. E is called the closure of E. We have p E for every eighborhood N of p we have N E. 8) E is closed if every limit poit of E is a elemet of E, i.e. E E, i.e. E = E. (See Figure ). 9) E is perfect if E is closed ad has o isolated poits. (See Figure 2). 0) E is bouded if m > 0 such that x, y E d(x, y) m. (See Figure 3). ) E is dese i X if E = X, i.e. p X ad for all eighborhood N of p we have N E. 3.8 Example. Let X = R, E = Q. We have Q = R. Give p R ad give a eighborhood B r (p) = (p r, p + r), fid a ratioal umber x such that p r < x < p + r. So x B r (p) Q. 3.9 Theorem. E is ope E C is closed. Proof. ( ): Let E be ope. Show that every limit poit p of E C is a elemet of E C. Assume it is ot true. The E C has a limit poit p 0 such that p 0 / E C. The p 0 E. Sice E is ope r > 0 such that B r (p 0 ) E. Also sice p 0 is a limit poit of E C, every eighborhood N of p 0 cotais a poit q E C such that q p 0. Sice B r (p 0 ) is also a eighborhood of p 0 we have that B r (p 0 ) cotais a poit q E C such that q p 0. The q B r (p 0 ) E but also q E C. Cotradictio. ( ): Let E C be closed. Show that every poit p i E is a iterior poit. Let p E be arbitrary. Sice p / E C, p is ot a limit poit of E C. The p has a eighborhood N such that N does ot cotai ay poit q E C with q p. The N E C = so N E. Sice N is a eighborhood of p, r > o such that B r (p) N. So B r (p) E. 3.0 Theorem. p is a limit poit of E every eighborhood N of p cotais ifiitely may poits of E. Proof. 23
24 ( ): Trivial. ( ): Let p be a limit poit of E ad let N be a arbitrary eighborhood of p. The r > 0 such that B r (p) N. q B r (p) E such that q p. d(q, p) > 0. Let r = d(q, p) < r. q 2 B r (p) E such that q 2 p. d(q 2, p) < r = d(q, p). The q 2 q. d(q 2, p) > 0. Let r 2 = d(q 2, p) < r. Cotiue this way ad get a sequece of poits q, q 2, q 3,..., q,... i E such that q i q j for i + j ad also each q i p ad q i B r (p). 3. Corollary. If E is a fiite set the E has o limit poits. 3.2 Theorem. Let E X. The (a) E is a closed set. (b) E = E E is a closed set. (c) E is the smallest closed set that cotais E, i.e. if F is ay closed set such that E F the E F. Proof. (a) Show (E) C is a ope set. Let p (E) C. The r > 0 such that B r (p) E =. This meas B r (p) E C. Show that actually B r (p) (E) C. If ot true, q B r (p) such that q / (E) C, i.e. q E. The for every eighborhood N of q we have N E. This is also true for N = B r (p), i.e. B r (p) E. But B r (p) E =. Cotradictio. So B r (p) (E) C. (b)( ): E is closed by (a) so E is closed. ( ): Let E be closed. The E is cotais all of its limit poits, i.e. E E, E = E E E. Sice E E is always true, we have E = E. (c) Let E be give ad F be a closed set such that E F. Show E F. Let p E = E E. If p E the p F. If p E show p F. Give ay eighborhood N of p, N cotais ifiitely may poits of E so N cotais ifiitely may poits of F, i.e. p F F. Basic properties: Let (X, d) be a metric space. 24
25 ) The uio of ay collectio of ope sets is ope. 2) The itersectio of a fiite umber of ope sets is ope. 3) The itersectio of ay collectio of closed sets is closed. 4) The uio of a fiite umber of closed sets is closed. 5) E is ope E C is closed. 6) E is closed E = E. 7) E is the smallest closed set that cotais E. 8) E is ope E = it E. 9) it E is the largest ope set that is cotaied i E. 3.3 Example. Itersectio of ifiitely may ope sets may ot be ope. Let X = R, d(x, y) = x y. For =, 2, 3,... let E = (, ) +. All E s are ope but = E = [0, ] is ot ope. 3.4 Propositio. Let E R be bouded above. Let y = sup E. The y E. Proof. Let N be a arbitrary eighborhood of y. The r > 0 such that B r (y) N. y r caot be a upper boud for E. The x E such that y r < x. Also, x y < y +r. So y r < x < y +r, i.e. x B r (y) x N. So x E N, i.e. E N. 3.2 Subspaces 3.5 Defiitio. Let (X, d) be a metric space ad Y be a subset of X. The Y is a metric space i its ow right with the same distace fuctio. I this case we say (Y, d) is a subspace of (X, d). 3.6 Example. X = R 2, Y = {(x, 0) : x R}. Let E = {(x, 0) : < x < 2}. The E Y so E X. As a subset of Y, E is a ope set. As a subset of X, E is ot a ope set. Let E Y X. We say E is ope (closed) relative to Y if E is ope (closed) as a subset of the metric space (Y, d). E is ope relative to Y p E r > 0 such that B Y r (p) }{{} B X r (p) Y E is closed relative to Y Y \ E is ope relative to Y. 25 E.
26 3.7 Theorem. Let E Y X. The (a) E is ope relative to Y there is a ope set F X such that E = F Y. (b) E is closed relative to Y there is a closed set F X such that E = F Y. Proof. (a)( ): Let E be ope relative to Y. The p E r p > 0 such that Br X p (p) Y E. Let F = p E BX r p (p). The F is a ope set i X. Show F Y = E. ( ) F Y = Br X p (p) Y = ( ) Br X p (p) Y E p E p E }{{} E for all p Coversely, let p 0 E. The p 0 Y sice E Y. The p 0 B X r p0 F. So p 0 Y F. ( ): Assume E = F Y where F X is ope i X. Let p E. The p Y ad p F. Sice F is ope i X, r > 0 such that B X r (p) F. The B X r (p) Y F Y = E. (b)( ): Let E Y be closed relative to Y. The Y \ E is ope relative to Y. So there exists a ope set A X such that Y \ E = A Y. The E = Y \ (Y \ E) = Y \ (A Y ) = (A Y ) C Y = (A C Y C ) Y = (A C Y ) (Y C Y ) = A }{{} C Y. A C is closed i X ad we may call it F. ( ): Assume E = F Y where F X is a closed set. The Y \ E = Y \ (F Y ) = Y (F Y ) C = Y (F C Y C ) = Y }{{} F C is ope i X ope relative to Y. 3.3 Compact Sets 3.8 Defiitio. Let (X, d) be a metric space ad E be a o-empty subset of X. A ope cover of X is a collectio of {G i : i I} of ope subsets G i of X such that E i I G i. 26
27 3.9 Example. X = R, d(x, y) = x y, E = (0, ). For every x E let G x = (, x). x E G x = (, ) E. So {G x : x E} is a ope cover of E Example. X = R 2 with d 2 ad E = B (0). For N let G = B (0). + = G = E. So {G : N} is a ope cover of E. 3.2 Defiitio. A subset K of a metric space (X, d) is said to be compact if every ope cover of K cotais a fiite subcover, i.e. give ay ope cover {G i : i I} of K, we have that i,..., i I such that K G i G i Example. X = R, d(x, y) = x y. E = (0, ) is ot compact. Take I = E = (0, ). For x I, G x = (, x). x I G x = (, ) E. So, {G x : x I} is a ope cover for E. This ope cover does ot have ay fiite subcover. Assume it is ot true, so assume x,..., x E such that G x G x E. Let x k = max{x,..., x }. The (0, ) (, x k ) but x k E = (0, ), i.e. x k <. Let x = x k+. The x E but x / (, x 2 k ) Theorem. Let K Y X. The K is compact relative to Y K is compact relative to Y. Proof. ( ): Assume K is compact relative to Y. Let {G i : i I} be ay collectio of sets ope relative to X such that i I G i K. The {Y G i : i I} is a collectio of sets ope relative to Y ad K = K Y ( i I G i) Y = i I (Y G i). So {Y G i : i I} is a ope (relative to Y ) cover of K. Sice K is compact relative to Y, i,..., i I such that K (Y G i ) (Y G i ). The K G i G i. ( ): Let K be compact relative to X. Let {G i : i I} be a arbitrary collectio of sets ope relative to Y such that K i I G i. The we have G i = Y E i for some ope subset E i of X. The K i I (Y E i ) i I E i. So {E i : i I} is a ope cover of K i X. Sice K is compact relative to X, i,..., i such that K E i E i. The K = K Y (E i Y ) (E i Y ) = G i G i Theorem. If K X is compact the K is closed. 27
28 Proof. We will show K C is ope. Let p K C. Show r > 0 such that B r (p) K C q K (sice q p). d(q, p) > 0. Let r q = d(q, p) 2 V q = B rq (p) W q = B rq (q) The V q W q =. Fid V q ad W q q K K q K W q. The collectio {W q : q K} is a ope cover of K. Sice K is compact, q,..., q K such that K W q W q. Let V = V q V q. The V is a ope set ad p V. If r qk = mi{r q,..., r q } the V = B rqk (p). Show V K C. If it is ot true, the z V but z / K C, i.e. z K. The z W qi for some i {,..., }. z V V qi (the same i). So z W qi V qi =. Cotradictio Theorem. Closed subsets of compact sets are compact. Proof. Let F K X where K is compact ad F is closed. (relative to X ad relative to K are the same sice K is closed.) Let {G i : i I} be a set ope i X such that F i I G i. The {G i : i I} {F C } is a ope cover of K. Sice K is compact, i,..., i I such that K G i G i F C. The F = F K (G i F ) (G i F ) (F C F ) G }{{} i G i Corollary. Let F, K X where K is compact ad F is closed. The F K is compact Theorem. Let {K i : i I} be a collectio of compact subsets of X such that the itersectio of every fiite subcollectio of {K i : i I} is o-empty. i I K i. Proof. Assume the cotrary. i I K i =. Fix oe of these sets, K i0. The ( ) ( ) C, K i0 i i K 0 i =. The K i0 K i i0 i i.e. Ki0 i i0 KC i. So {Ki C : i i 0 } is a ope cover of K i0. Sice K i0 is compact, i,..., i such that K i0 Ki C Ki C }{{. But K i0 K i K i =. This cotradicts } (K i K i ) C the hypothesis Corollary. Let {K : N} be a sequece of o-empty compact sets such that K K 2 K 3 The = K. 28
29 3.29 Example. X = R, d(x, y) = x y, E = [, + ) = {x : x R, x }, =, 2, 3,... The E s are closed, E ad E E 2 E 3 We have = E = Theorem (Nested Itervals). Let {I : =, 2,...} be a sequece of o-empty, closed, bouded itervals i R such that I I 2 I 3 The = I. Proof. Let I = [a, b ] ad a b. That is, a a 2 a 3 ad b 3 b 2 b. We also have that k a b k. Give, k N (i) If = k the a b = b k. (ii) If > k the I I k. The a I I k a k a b k. (iii) If < k the I k I. The b k I k I a b k b. Let E = {a, a 2, a 3,...}. The E ad E is bouded above. Let x = sup E. The a x for all. Claim: x b. Assume it is ot true. The such that b 0 < X. The b 0 caot be a upper boud for E. So there is a elemet a k0 E such that a k0 > b 0. Cotradictio. So we have a x b, i.e. x I. So x = I. 3.3 Remark. If also lim (b a ) = 0 the = I cosists of oly oe poit Defiitio. Let a b,..., a k b k be real umbers. The the set of all poits p = (x,..., x k ) R k such that a x b,..., a k x k b k is called a k-cell i R k. (See Figure 4) Theorem. Let k be fixed. Let {I } be a sequece of o-empty k-cells such that I I 2 I 3 The = I Theorem. Let K be a compact subset of a metric space (X, d). The K is bouded. Proof. Fix a poit p 0 K. X = = B (p 0 ) ad K = B (p 0 ). So,... r N such that K B (p 0 ) B r (p 0 ). Let 0 = max{,..., r }. Let p, q K, the p B i (p 0 ) ad q B j (p 0 ) where i, j are oe of,..., r. We have d(p, q) d(p, p 0 )+d(p 0, q) i + j
30 3.35 Theorem (Heie-Borel). A subset K of R k is compact K is closed ad bouded. Proof. ( ): True i all metric spaces. ( ): Let K R k be closed ad bouded. Sice K is bouded, there is a k-cell I such that K I. I is compact, so K beig a closed subset of the compact set I is compact Theorem. Every k-cell is a compact set i R k. Proof. Let E R k be a k-cell. The there are real umbers a, b, a 2, b 2,..., a k, b k such that a b, a 2 b 2,..., a k b k ad E = {p = (x,..., x k ) R k : a x b,..., a k x k b k } If a = b, a 2 = b 2,..., a k = b k the E cosists of oe poit which is compact. So assume there is at least oe j, j k, such that a j < b j. Let δ = (b a ) 2 + (b 2 a 2 ) (b a ) 2 The δ > 0. Assume E is ot compact. So there is a ope cover {G α : α A} such that o fiite subcollectio of G α s covers E. Let c i = a i + b i 2 We divide each side of E ito two parts ad this way we divide E ito 2 k subcells. Call them Q, Q 2,..., Q 2 k. The at least oe of these Q j s caot be covered by fiitely may sets, G α s. Call this Q j E. For all p, q E we have d 2 (p, q) δ ad for all p, q E, d 2 (p, q) δ 2. Next divide E ito 2 k subcells by halvig each side ad cotiue this way. This way we obtai a sequece {E } of k-cells such that (a) E E E 2 E 3 (b) E caot be covered by ay fiite subcollectio of {G α : α A} (c) For all p, q E, d 2 (p, q) δ 2 30
31 By (a), = E. Let p = E. The p E. Sice {G α : α A} is a ope cover of E, there is a α 0 A such that p G α0. Sice G α0 is ope, there is r > 0 such that B r (p ) G α0. Fid a atural umber 0 such that δ < r 2 0 δ, i.e. < r. Now we show that E G α0. p E 0. Let p E 0 be a arbitrary poit. By (c), d 2 (p, p ) δ. Also, δ < r. So d (p, p ) < r. So p B r (p ) G α0. Thus, p E 0 p G α0. So E 0 G α0. This meas E 0 ca be covered by fiitely may sets from {G α : α A} (ideed just by oe set). This cotradicts (b) Theorem. Let (X, d) be a metric space ad K X. The K is compact every ifiite subset of K has a limit poit i K. Proof. ( ): Let K be compact. Assume claim is ot true. The there is a ifiite subset A K such that A has o limit poit i K. So, give ay poit p K, p is ot a limit poit of A. So there is r p > 0 such that B rp (p) cotais o poit of A differet from p. The collectio of {B rp (p) : p K} is a ope cover of K. Sice K is compact, there are p, p 2,..., p K such that K B rp (p ) B rp2 (p 2 ) B rp (p ). Sice A K ad A is a ifiite set, oe of the ope balls o the right had side must cotai ifiitely may poits of A. Cotradictio. ( ): Omitted Theorem (Bolzao-Weierstrass). Every ifiite, bouded subset E of R k has a limit poit i R k. Proof. Sice E is bouded, there is a k-cell I such that E I. The sice I is compact, the ifiite subset E of I has a limit poit p I R k. 3.4 The Cator Set Let us defie E 0 = [0, ] [ E = 0, ] [ ] 2 3 3, [ E 2 = 0, ] [ 2 9 9, ] 3. [ 2 3, 7 ] [ ] 8 9 9, 3
32 Cotiue this way by removig the ope middle thirds of the remaiig itervals. This way we get a sequece E E 2 E 3 E such that E is the uio of 2 disjoit closed itervals of legth. We defie 3 C = which is called the Cator Set. Properties of C () C is compact (2) C (3) it C = (4) C is perfect (5) C is ucoutable = Proof of (3). Assume it C. The there is a o-empty ope iterval (α, β) C ad C does ot cotai itervals of the form ( ) 3k+, 3k+2 3 m 3. Sice m they are removed i the process of costructio. Assume C cotais (α, β) where α < β. Let a > 0 be a costat which will be determied later. Choose a m N such that < β α 3m, i.e. < β α. Let k be the smallest iteger 3 m a such that α < 3k+, i.e. α3m < k. The k α3m. Show 3k+2 < β. 3 m m Show Now we have So let a Cotradictio. k α3m 3 α3 m 3 E k α3m < 3m β 2 3 < 3m β 2 3m α 3 3 < 3m (β α) 3 3 m (β α) = a 3 3, i.e. a 4. Choose a = 4. This way we have that 3 ( 3k + 3, 3k + 2 m 3 m > 3m a 3 m 3 ) (α, β) ( 3k + 3, 3k + 2 m 3 m ) C 32
33 3.5 Coected Sets 3.39 Defiitio. Let (X, d) be a metric space ad A, B X. We say A ad B are separated if A B = ad A B = Example. X = R, A = Q, B = R\Q. We have A B = R (R\Q). The Q, R \ Q i R are ot separated. 3.4 Defiitio. A subset E X is said to be discoected if there are two o-empty separated sets A, B such that E = A B. A subset E X is said to be coected if it is ot discoected, i.e. there are o o-empty separated sets A, B such that E = A B Theorem. A o-empty subset E R is coected E is a iterval. Proof. A iterval is defied as follows: Wheever x < z ad x, z E for all y with x < y < z we have y E. ( ): Let E R be coected. Assume E is ot a iterval. So there are two poits x, z E with x < z, there is y with x < y < z ad y / E. Let A = E (, y), B = E (y, + ). x A, z B. So A, B. Show A B =, A B =. A (, y) A (, y]. So A B (, y) (y, + ) =. Similarly we have A B =. ( ): Omitted. A B = (E (, y)) (E (y, + )) = E ((, y) (y, + )) = E sice y / E }{{} R\{y} 33
34 4 Sequeces Ad Series 4. Sequeces Let (X, d) be a metric space. A sequece i X is a fuctio f : N X. If p = f(), we deote this sequece by (p ) or {p }. ( ) 4. Example. X = R 2, p = where =, 2, 3,... The, ( ) p = (0, ) ( p 2 = 2, ) 2 ( 2 p 3 = 3, ) Defiitio. We say the sequece {p } coverges to p X if for every ε > 0 there is a atural umber 0 (depedig o ε > 0 i geeral) such that for all N with 0 we have d(p, p) < ε, i.e. p B ε (p). We write p p or lim p = p. p p every eighborhood of p cotais all but fiitely may terms p. If {p } does ot coverge to ay p X, we say {p } is diverget. ( 4.3 Example. X = R 2, p = ), ( ). p = (, 0). Show p p. Let ε > 0 be give. Let 0 be ay atural umber such that 2 < ε 0. Let be ay atural umber such that 0. ( 2 ( ( ) d 2 (p, p) = ( )) + ) 2 0 ( ) 2 ( ( ) = = = 2 2 < ε 0 ) 2 34
35 4.4 Remark. {x } = { } coverges to x = 0 i (R, ), but it is diverget i ((0, 2), ). 4.5 Defiitio. We say that the sequece {p } is bouded if the set E = {p, p 2, p 3,...} is a bouded subset of X, i.e. there is a costat M > 0 such that for all p i, p j E we have d(p i, p j ) M. 4.6 Theorem. (a) Let {p } be a sequece such that p p ad p p. The p = p. (b) If {p } is coverget the {p } is bouded. (c) Let E be subset of X. The p E there is a sequece {p } cotaied i E such that p p. Proof. (a) Let p p ad p p. Assume p p. The d(p, p ) > 0. Let ε 0 = d(p,p ) 3 the ε 0 > 0. We have p p so there is N such that for all we have d(p, p) < ε 0. We also have p p so there is 2 N such that for all 2 we have d(p, p ) < ε 0. Let 0 = max{, 2 }. The 0 d(p 0, p) < ε 0 ad 0 2 d(p 0, p ) < ε 0. The 3ε 0 = d(p, p ) d(p, p 0 ) + d(p 0, p ) < ε 0 + ε 0 = 2ε 0. So 3ε 0 < 2ε 0. Sice ε 0 > 0, this caot be true. (b) Let p p. For ε = > 0, there is 0 N such that for all 0 we have d(p, p) <. If i, j 0, the d(p i, p j ) d(p i, p) + d(p, p j ) < + = 2. Let K = max{, d(p, p),..., d(p 0, p)}. The for all N we have d(p, p) K. For ay i, j N, d(p i, p j ) d(p i, p)+d(p, p j ) 2K. (c)( ): {p } i E such that p p. Show p E. Give r > 0 we have 0 N such that for all 0, d(p, p) < r. So p 0 B r (p) E. So B r (p) E. So p E. ( ): Let p E. The for every r > 0, B r (p) E. For r =, fid p B (p) E For r = 2, fid p 2 B (p) E 2. For r =, fid p B (p) E 35
36 The {p } is a sequece i E. Give ε > 0 let 0 be such that < ε 0. The for all 0 we have d(p, p) < 0 < ε. So p p. 4.7 Theorem. Let {s } ad {t } be sequeces i R such that s s ad t t where s, t R. The (a) s + t s + t (b) For all costats c R, cs c s (c) s t s t (d) If s 0 the s s Proof of (a). We have that d(s + t, s + t) = s + t (s + t) = (s s) + (t t) s s + t t Give ε > 0, let ε = ε 2 > 0. We have s s so there is N such that for all we have s s < ε. We also have t t so there is 2 N such that for all 2 we have t t < ε. Let 0 = max{, 2 }. The 0 s s < ε ad 0 2 t t < ε. So for all 0 we have d(s + t, s + t) s s + t t < ε + ε = 2ε = ε Proof of (d). Let ε 0 = s the ε 2 0 > 0. So there is N such that for all we have s s < ε 0. Let, s = s s +s s s + s < s + s 2. So for all, s < s 2. I particular for all, s 0 so s is defied. Ad also s < 2. To show lim s s =, let ε > 0 be s give. Let ε = ε s 2 > 0. We have s 2 s so there is 2 N such that for all 2 we have s s < ε. Let 0 = max{, 2 } ad 0. The s s = s s s s = s s s s < ε s s < ε s 2 s = ε s s 2 = ε
37 4.8 Theorem. (a) Let {p } be a sequece i R k where p = (x,..., x k ) ad p = (x,..., x k ) R k. The p p x x, x 2 x 2,..., x k x k. (b) Let {p }, {q } be two sequeces i R k ad {α } be a sequece i R. Assume p p, q q i R k ad α α i R. The p + q p + q ad α p αp. Proof of (a). We eed the followig: If q = (y,..., y k ) R k the for all i =, 2,..., k y i y 2 + y yk 2 y + y y k y i 2 = y 2 i y 2 + y y 2 k y 2 + y y 2 k ( y + y y k ) 2 Assume p p. Give ε > 0, we have 0 N such that for all 0, d 2 (p, p) < ε. Let 0. The x x (x x ) 2 + (x 2 x 2 ) (x k x k) 2 = d 2 (p, p) < ε x 2 x 2 < ε. Coversely, assume x x, x 2 x 2,..., x k x k. To show p p, let ε > 0 be give. Let ε = ε k > 0. x x so we have such that for all, x x < ε x 2 x 2 so we have 2 such that for all 2, x 2 x 2 < ε. x k x k so we have k such that for all k, x k x k < ε Let 0 = max{, 2,..., k }. For all 0 d 2 (p, p) = (x x ) 2 + (x 2 x 2 ) (x k x k) 2 x x + x 2 x x k x k < ε + ε + + ε = kε = ε 37
38 4.2 Subsequeces 4.9 Defiitio. Let {p } be a sequece i X. Let { k } be a sequece of atural umbers such that < 2 < 3 < The the sequece {p k } is called a subsequece of {p }. 4.0 Example. {p, p 3, p 7, p 0, p 23,...} is a subsequece of {p }. = 2, 2 = 3, 3 = 7, 4 = 0, 5 = 23 ad so o. 4. Propositio. p p every subsequece of {p } coverges to p. Proof. ( ): Sice {p } is a subsequece of itself, p p. ( ): Let p p. Let {p k } be a arbitrary subsequece of {p }. To show p k p, let ε > 0 be give. Sice p p, we have 0 such that for all 0, d(p, p) < ε. If k 0 the k k 0 so d(p k, p) < ε. 4.2 Remark. Limits of subsequeces are limit poits of the sequece. 4.3 Example. X = R 2 ad p = ( +( ) +, ( 2 + is eve p =, ) (2, 0) ( is odd p =, ) (0, 0) So the sequece {p } has limit poits (2, 0) ad (0, 0). ) 4.4 Example. I X = R, x = + ( ) + is eve x = is odd x = 0 We do ot accept + as a limit sice + is ot a member of R. So 0 is the oly limit poit of the sequece {x } but {x } is diverget sice it is ot bouded. 38
39 4.5 Theorem. (a) Let (X, d) be a compact metric space ad {p } be ay sequece i X. The {p } has a subsequece that coverges to a poit p X. (b) Every bouded sequece i R k has a coverget subsequece. Proof. (a) Case : {p } has oly fiitely may distict terms. The at least oe term, say p 0 is repeated ifiitely may times, i.e. the sequece {p } has a subsequece all of whose terms are p. The the limit of this subsequece is p = p 0 X. Case 2: {p } has ifiitely may distict terms. The the set E = {p, p 2, p 3,...} is a ifiite subset of the compact set X. So it has a limit poit p X. The p is the limit of a subsequece of {p }. (b) Sice {p } is bouded, there is a k-cell I such that {p } I. I is compact, so by (a), {p } has a subsequece that coverges to a poit p I. 4.3 Cauchy Sequeces 4.6 Defiitio. Let (X, d) be a metric space. A sequece {p } i X is called a Cauchy sequece if for every ε > 0 we have 0 N such that for all, m 0, d(p, p m ) < ε. 4.7 Example. I X = R x = cos t t dt 2 The {x } is a Cauchy sequece i R. Give, m if = m the x = x m 39
40 so d(x, x m ) = x x m = 0 < ε. If m, assume m <. d(x x m ) = x x m = cos t m t dt cos t 2 t dt 2 = cos t m t dt 2 cos t m t 2 dt We kow cos t t 2 m t dt = 2 t = m + m m 0 < ε = cos t t 2 t 2 Give ε > 0, choose 0 N such that ε < 0. The for all, m N with 0 m we have d(x, x m ) < ε. 4.8 Example. X = R, x =. If = m + the d(x, x m ) = x m+ x m = m + m = m + m = ( m + m + + m m) m + + m = < m + + m m Give ε > 0, choose 0 N such that < ε 2 0. The for all m 0 we have d(x m+, x m ) < ε. So the distace betwee successive terms gets smaller as the idex gets larger but this sequece {x } is ot a Cauchy sequece. For example, for ε =, cosider d(x m, x 3m+ ) = m 3m + = 3m + m = m + 2m + m ( m + ) 2 m m + m 4.9 Theorem. Let (X, d) be a metric space. 40
41 (a) Every coverget sequece i X is Cauchy. (b) Every Cauchy sequece is bouded. Proof. (a) Assume p p. Show {p } is Cauchy. Give ε > 0 let ε = ε 2 > 0. Sice p p, there is 0 N such that d(p, p) < ε for all 0. Let, m 0. The d(p, p m ) d(p, p) + d(p, p m ) < ε + ε = 2ε = ε (b) Let {p } be a Cauchy sequece i X. For ε = there is 0 N such that for all, m 0 we have d(p, p m ) <. Let K = max{, d(p, p 0 ),..., d(p 0, p 0 )} ad M = 2k. The we show that for all, m N, d(p, p m ) M. Case :, m 0. The Case 2:, m < 0. The d(p, p m ) < K < M d(p, p m ) d(p, p 0 ) + d(p 0, p m ) K + K = M Case 3: m < 0 ad 0. The d(p, p m ) d(p, p 0 ) + d(p }{{} 0, p m ) < + K K + K = M }{{} < K Coverse of (a) is ot true i geeral Example. X = (0, ) = {x R : 0 < x < } with d(x, x ) = x x. x = 2. {x } is a Cauchy sequece i X but {x } has o limit i X. 4.2 Defiitio. A metric space (X, d) is said to be complete if every Cauchy sequece i (X, d) is coverget to some poit p X Theorem. (a) Every compact metric space (X, d) is complete. (b) (R k, d 2 ) is complete. ((R k, d ), (R k, d ) are also complete.) 4
42 Proof. (a) Let (X, d) be a compact metric space ad let {p } be a Cauchy sequece i X. The {p } has a subsequece {p k } which coverges to a poit p X. Show p p. Let ε > 0 be give. Let ε = ε > 0. {p 2 } is Cauchy, so there is N N such that for all, m N we have d(p, p m ) < ε. p k p, so there is N 2 such that for all k N 2 we have d(p k, p) < ε. Let N = max{n, N 2 }. The for all N d(p, p) d(p, p N ) + d(p }{{} N, p) < 2ε = ε }{{} <ε <ε (b) Let {p } be a Cauchy sequece i R k. The {p } is bouded so there is a k-cell I such that {p } I. (I, d 2 ) is compact. The by (a), {p } has a limit p I R k Remark. S, B(S) all bouded fuctios f : S R. (B(S), d) is complete. d(f, g) = sup{ f(s) g(s) : s S} 4.24 Theorem. Let (X, d) be a complete metric space ad Y be a subset of X. The subspace (Y, d) is complete Y is a closed subset of X. Proof. ( ): Assume (Y, d) is complete ad show Y is closed, i.e. Y Y. Let p Y. The there is a sequece {p } i Y such that p p. The {p } is coverget i X. So {p } is Cauchy i X. Sice all p Y, {p } is Cauchy i Y. Sice Y is complete, there is a elemet q Y such that p q. The p = q Y. So p Y, i.e. Y Y. ( ): Assume Y is closed. Show (Y, d) is complete. Let {p } be a Cauchy sequece i Y. The {p } is a Cauchy sequece i X. Sice (X, d) is complete, there is p X such that p p. The p is the limit of the sequece {p } i Y. The p Y. Sice Y is closed, Y = Y. So p Y. So (Y, d) is complete Example. X = R 2, Y = {(x, y) : x 0, y 0}. The (Y, d 2 ) is complete. 42
43 Mootoe Sequece Property I R Let {s } be a sequece i R. We say {s } is icreasig if s s 2 s 3 s s + {s } is decreasig if s s 2 s 3 s s + {s } is mootoe if {s } is either icreasig or decreasig Theorem (Mootoe Sequece Property). Let {s } be a mootoe sequece i R. The {s } is coverget {s } is bouded. Proof. ( ): True for all sequeces. ( ): We do the proof for decreasig sequeces. Let s = if{s, s 2, s 3,...}. Show lim s = s. Let ε > 0 be give. The s + ε caot be a lower boud for the set {s, s 2, s 3,...}. The there is s 0 such that s 0 < s + ε. Let 0. s ε < s s s 0 < s + ε. For all 0 s ε < s < s + ε ε < s s < ε s s < ε d(s, s) < ε 4.27 Example. Let A > 0 be fixed. Start with ay x > 0 ad defie x = ( x + A ) = 2, 3, 4,... 2 x The lim x = A. We will show x 2 x 3 x 4 For 2 x 2 A = ) (x 2 + A2 + 2A A 4 x 2 = ) (x 2 + A2 2A 4 x 2 = ( x A ) x 43
44 So x 2 A for all 2. Sice all x > 0, x A for all 2. For 2 x x + = x ) (x + Ax 2 = ) (x Ax 2 = x 2 A 0 2 x So x x + for all 2. So {x } =2 is decreasig ad bouded. So lim x = x exists. The we solve for x. We have that ) x + = (x 2 + Ax ( ) x = 2 x + A x The 2x = x + A x x 2 = A x = A Sice all x > 0, limit x caot be egative. So x = A. 4.4 Upper Ad Lower Limits Let {x } be a sequece i R. We write lim x = + (or x + ) if for every M > 0 we ca fid a atural umber 0 (depedig o M i geeral) such that for all 0 we have M x. We write lim x = (or x ) if for every M < 0 we ca fid a atural umber 0 (depedig o M i geeral) such that for all 0 we have x M. I either case, we say {x } is diverget Example. x = + ad lim x = +. {x } is diverget Defiitio. Let {x } be a sequece i R. Let E be the set of all subsequetial limits of {x }. The E R {, + }. 44
45 4.30 Example. x = + ( ) +. Subsequetial limits are + ad 0. So E = {0, + }. 4.3 Defiitio. Let x = sup E ad x = if E (Cosidered i the set of exteded real umbers.) x is called the upper limit (or limit superior) of {x } ad it is deoted by x = lim sup x = lim x x is called the lower limit (or limit iferior) of {x } ad it is deoted by x = lim if x = lim x 4.32 Example. The fuctio Π : N N N defied by Π(r, s) = 2 r (2s ), is - ad oto. Let s be fixed. N s = {2 r (2s ) : r =, 2, 3,...} = r\s {2s, 2(2s ), 4(2s ),...} The for s s, N s N s =. Also s= N s = N. Defie a sequece {x } i R as follows: Give N, there is a uique s such that N s. Defie x = s. What are the subsequetial + limits of {x }? What are lim sup x ad lim if x? If N = {, 2, 4, 8,...} the x = + If N 2 = {3, 6, 2, 24,...} the x = If N s = { } the x =..... s + s So all, 2, 3,... are subsequetial limits. The we have {, 2, 3,...} E. The sup E = +, i.e. lim sup x = +. 45
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