1.4 Cauchy Sequence in R

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1.4 Cauchy Sequence in R 1.4 Cauchy Sequence in R De¯nition. (1.4.1) A sequence xn 2 R is said to converge to a limit x if ² 8² > 0; 9N s:t: n > N ) jxn ¡ xj < ²: A sequence xn 2 R is called Cauchy sequence if ² 8²; 9N s:t: n > N & m > N ) jxn ¡ xmj < ²: Proposition. (1.4.2) Every convergent sequence is a Cauchy sequence. Proof. Assume xk ! x. Let ² > 0 be given. ² ² 9N s:t: n > N ) jxn ¡ xj < 2 . ² n; m ¸ N ) ² ² jxn ¡ xmj · jx ¡ xnj + jx ¡ xmj < 2 + 2 = ²: 1 Theorem. (1.4.3; Bolzano-Weierstrass Property) Every bounded sequence in R has a subsequence that converges to some point in R. Proof. Suppose xn is a bounded sequence in R. 9M such that ¡M · xn · M; n = 1; 2; ¢ ¢ ¢ . Select xn0 = x1. ² Bisect I0 := [¡M; M] into [¡M; 0] and [0; M]. ² At least one of these (either [¡M; 0] or [0; M]) must contain xn for in¯nitely many indices n. ² Call it I1 and select n1 > n0 with xn1 2 I0. ² Continue in this way to get a subsequence xnk such that ² I0 ⊃ I1 ⊃ I2 ⊃ I3 ¢ ¢ ¢ ¡k ² Ik = [ak ; bk ] with jIk j = 2 M. ² Choose n0 < n1 < n2 < ¢ ¢ ¢ with xnk 2 Ik . 9 ² Since ak · ak+1 · M (monotone and bounded), ak ! x. ¡k ² Since xnk 2 Ik and jIk j = 2 M, we have ¡k¡1 jxnk ¡xj < jxnk ¡ak j+jak ¡ xj · 2 M+jak ¡ xj ! 0 as k ! 1: 2 Corollary. (1.4.5; Compactness) Every sequence in the closed interval [a; b] has a subsequence in R that converges to some point in R. Proof. Assume a · xn · b for n = 1; 2; ¢ ¢ ¢ . By Theorem 1.4.3, 9 9 a subsequence xnk and a · x · b such that xnk ! x. Lemma. (1.4.6; Boundedness of Cauchy sequence) If xn is a Cauchy sequence, xn is bounded. Proof. 9N s.t. n ¸ N ) jxn ¡ xj < 1. Then supn jxnj · 1 + maxfjx1j; ¢ ¢ ¢ ; jxN jg (Why?) Theorem. (1.4.3; Completeness) Every Cauchy sequence in R converges to an element in [a; b]. Proof. Cauchy seq. ) bounded seq. ) convergent subseq. 3 1.5. Cluster Points of the sequence xn De¯nition. (1.5.1; cluster points) A point x is called a cluster point of the sequence xn if ² 8² > 0; 9in¯nitely many values of n with jxn ¡ xj < ² In other words, a point x is a cluster point of the sequence xn i® 8² > 0 & 8N; 9n > N s:t: jxn ¡ xj < ² Example ² Both 1 and ¡1 are cluster points of the sequence 1; ¡1; 1; ¡1; ¢ ¢ ¢ : 1 ² The sequence xn = n has the only cluster point 0. ² The sequence xn = n does not have any cluster point. 4 Proposition. 1. x is a cluster point of the sequence xn i® 9 a subsequence xnk s:t: xnk ! x: 2. xn ! x i® every subsequence of xn converges to x 3. xn ! x i® the sequence fxng is bounded and x is its only cluster points. Proof. 1. ()) Assume x is a cluster point. Then, we can choose 1 n1 < n2 < n3 ¢ ¢ ¢ s:t: jxnk ¡ xj < k . (Why?) This gives a subsequence xnk ! x. 2. Trivial 3. (()If not, 9² and 9 a subseq xnk so that jxnk ¡ xj > ². Since xnk is bounded, 9 a convergent subseq. The limit of that subseq would be a cluster pt of the seq xn di®erent from x, but there are no such pt. Contradiction. 5 De¯nition. (1.5.3; limit superior & limit inferior of seq xn ) De¯ne the limit superior limxn in the following way: ² If xn is bounded above, then lim supn!1 xn = limxn = the largest cluster point limxn = ¡1 if the set cluster point is empty ² If xn is NOT bounded above, then limxn = 1 Similarly, we can de¯ne the limit inferior limxn. Examples ² For the seq 1; 0; ¡1; 1; 0; ¡1; ¢ ¢ ¢ , limxn = 1 and limxn = ¡1. ² If xn = n, then limxn = 1 = limxn n 1+n ² Let xn = (¡1) n . Then limxn = 1 and limxn = ¡1. 6 De¯nition. (1.6.2; Vector space) A real vector space V is a set of elements called vectors, with given operations of vector addition + : V £ V ! V and scalar multiplication ¢ : R £ V ! V such that the followings hold for all v; u; w 2 V and all ¸; ¹ 2 R: 1. v + w = w + v; (v + u) + w = v + (u + w); ¸(v + w) = ¸v + ¸w; ¸(¹v) = (¸¹)v; (¸ + ¹)v = ¸v + ¹v; 1v = v. 2. 90 2 V s.t. v + 0 = v. 9 ¡ v 2 V s.t. v ¡ v = 0. ² A subset of V is called a subspace if it is itself a vector space with the same operations. ² W is a vector subspace of V i® ¸v + ¹u 2 W whenever u; v 2 W and ¸; ¹ 2 R. ² The straight line W = f(x1; x2): x1 = 2x2g is a subspace of R2. 7 Euclidean space Rn & De¯nitions & Properties The Euclidean n-space Rn with the operations (x1; ¢ ¢ ¢ ; xn) + (y1; ¢ ¢ ¢ ; yn) = (x1 + y1; ¢ ¢ ¢ ; xn + yn)& ¸(x1; ¢ ¢ ¢ ; xn) = (¸x1; ¢ ¢ ¢ ; ¸xn) is a vector space of dimension n. ² The standard basis of Rn; e1 = (1; 0; ¢ ¢ ¢ ; 0); ¢ ¢ ¢ ; en = (0; ¢ ¢ ¢ ; 0; 1). n ² Unique representation: x = (x1; ¢ ¢ ¢ ; xn) 2 R can be expressed uniquely as x = x1e1 + ¢ ¢ ¢ + xnen: P ² Inner product of x and y: hx; yi = n x y p i=1 i i ² Norm of x: kxk = hx; yi. ² Distance between x and y: dist(x; y) = kx ¡ yk ² Triangle inequality: kx + yk · kxk + kyk. ² Cauchy-Schwartz inequality: hx; yi · kxk kyk ² Pythagorean theorem: If hx; yi = 0, then kx + yk2 = kxk2 + kyk2. 8 De¯nition. (1.7.1; Metric Space (M; d) equipped with d =distance) A metric space (M; d) is a set M and a function d : M £ M ! R such that 1. d(x; y) ¸ 0 for all x; y 2 M. 2. d(x; y) = 0 i® x = y. 3. d(x; y) = d(y; x) for all x; y 2 M. 4. d(x; y) · d(x; z) + d(z; y) for all x; y 2 M. Example [Fingerprint Recognition] Let M be a data set of ¯ngerprints in Seoul city police department. ² Motivation: Design an e±cient access system to ¯nd a target. ² We need to de¯ne a dissimilarity function stating the distance between the data. The distance d(x; y) between two data x and y must satisfy the above four rules. ² Similarity queries. For a given target x¤ 2 M and ² > 0, arrest all having ¯nger print y 2 M such that d(y; x¤) < ². 9 De¯nition. (1.7.3. Normed Space (V; k ¢ k)) A normed space (V; k ¢ k) is a vector space V and a function k ¢ k : V! R called a norm such that 1. kvk ¸ 0; 8v 2 V 2. kvk = 0 i® v = 0. 3. k¸vk = j¸jkvk; 8v 2 V and every scaler ¸. 4. kv + wk · kvk + kwk; 8v; w 2 V Examples ² V = R and kxk = jxj for all x 2 R. q 2 2 2 2 ² V = R and kvk = v1 + v2 for all v = (v1; v2) 2 R . ² Let V = C([0; 1])=all continuous functions on the interval [a; b]. De¯ne kf k = supfjf (x)j : x 2 [0; 1]g (called supremum norm): 10 Proposition. If (V; k ¢ k) is a normed vector space and d(v; w) = kv ¡ wk , then d is a metric in V. Proof. EASY. Examples ² For V = C([0; 1]), the metric is d(f ; g) = kf ¡ gk = supfjf (x) ¡ g(x)j : x 2 [0; 1]g: The sup distance between functions is the largest vertical distance between their graphs. 11 De¯nition. A vector space V with a function h¢; ¢i : V £ V ! R is called an inner product space if 1. h v; vi ¸ 0 for all v 2 V. 2. h v; vi = 0 i® v = 0. 3. h¸v; wi = ¸hv; wi; 8v 2 V and every scaler ¸. 4. hv + w; hi = h v; h i + h w; h i. 5. h v; w i = h w; v i Examples 2 1. V = R and h v; w i = v1w1 + v2w2. Two vectors v and w are orthogonal if h v; w i = 0. R 2. V = C[0; 1] and h f ; g i = 1 f (x)g(x)dx p 0 3. kvk = h v; v i is a norm on V. 12 Theorem. (Cauchy-Schwarz inequality ) If h ¢; ¢ i is an inner product in a real vector space V, then jh f ; g ij · kf kkgk Proof: ² g Suppose g 6= 0. Let h = kgk . It su±ces to prove that jh f ; h ij · kf k. (Why? jh f ; g ij · kf kkgk i® jh f ; h ij · kf k.) ² Denote ® = h f ; h i. Then 0 · kf ¡ ®hk2 = hf ¡ ®h; f ¡ ®hi = kf k2 ¡ ® hh; f i ¡ ® hf ; hi + j®j2 = kf k2 ¡ j®j2 Hence, j®j = jh f ; h ij · kf k. This completes the proof. 13 Chapter 2: Topology of M = Rn n Throughout this chapter, assumepP M = R ( the Euclidean space ) n 2 with the metric d(x; y) = i=1 jxi ¡ yi j = kx ¡ yk De¯nition. (D(x; ²), open, neighborhood) ² D(x; ²) := fy 2 M : d(y; x) < ²g is called ²-ball (or ²-disk) about x. ² A ½ M is open if 8 x 2 M; 9² > 0 s.t. D(x; ²) ½ A. ² A neighborhood of x is an open set A containing x.
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