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1.4 Cauchy Sequence inRDe¯nition. (1.4.1)A sequence xn2Ris said to converge to a limit x if

² 8² >0;9N s:t:n>N) jxn¡xj< ²:

A sequence x

n2Ris 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

x k!x . Let² >0 be given. 9 N s:t:n>N) jxn¡xj< 2 n;m¸ N jxn¡xmj · jx¡xnj+jx¡xmj 2 2 1

Theorem. (1.4.3; Bolzano-Weierstrass Property)

Everyboundedsequence inRhas asubsequencethat

converges to some point inR.

Proof.Supposexnis a bounded sequence inR.

9Msuch that

¡M·xn·M;n= 1;2;¢¢¢. Selectxn0=x1

BisectI0:= [¡M;M] into

[¡M;0] and [0;M]

At least one of these

(either [¡M;0] or [0;M]) must containxnfor in¯nitely many indicesn.

Call itI1and selectn1>n0withxn12I0.

Continue in this way to get a subsequencexnksuch that I

0¾I1¾I2¾I3¢¢¢

I k= [ak;bk] withjIkj= 2¡kM.

Choosen0

Sinceak·ak+1·M(monotone and bounded),ak!9x.

Sincexnk2IkandjIkj= 2¡kM, we have

jxnk¡xj·2¡k¡1M+

jak¡xj !0 ask! 1: 2

Corollary. (1.4.5; Compactness)

Everysequence in the closed interval[a;b]has asubsequence inRthat converges to some point inR. Proof.Assumea·xn·bforn= 1;2;¢¢¢. By Theorem 1.4.3,9 a subsequencexnkanda· 9 x

·bsuch thatxnk!

x

Lemma. (1.4.6; Boundedness of Cauchy sequence)

If x nis a

Cauchy

sequence, x nis bounded

Proof.9Ns.t.n¸N) jxn¡xj<1. Then

sup njxnj ·1 + maxfjx1j;¢¢¢;jxNjg(Why?)

Theorem. (1.4.3; Completeness)

Every

Cauchy

sequence inRconverges to an element in [a;b]

Proof.

Cauchy seq.

bounded seq. convergent subseq. 3

1.5. Cluster Points of the sequencexnDe¯nition. (1.5.1; cluster points)A point x is called acluster pointof the sequence xnif

² 8² >0;9in¯nitely many values ofnwithjxn¡xj< ² In other words, a point x is a cluster point of the sequence x ni®

8² >0 &8N;9n>N s:t:jxn¡xj< ²

Example

Both 1 and¡1 are cluster points of the sequence

1;¡1;1;¡1;¢¢¢:

The sequencexn=1

n has the only cluster point 0.

The sequencexn=ndoes not have any cluster point.

4

Proposition.

1.x is a cluster point of the sequence x

ni®

9a subsequence xnks:t:xnk!x:

2. x n!x i® every subsequence of xnconverges to x 3. x n!x i® the sequencefxngis bounded and x is its only cluster points.

Proof.

1. ()) Assumexis a cluster point. Then, we can choose n

1 k . (Why?) This gives a subsequencexnk!x. 2.

Trivial

3. If not,9²and9a subseqxnkso thatjxnk¡xj> ². Since x nkis bounded,9a convergent subseq. The limit of that subseq would be a cluster pt of the seqxndi®erent fromx, but there are no such pt.

Contradiction.

5

De¯nition. (1.5.3; limit superior & limit inferior of seqxn)De¯ne the limit superiorlimxnin the following way:

²If xnis bounded above, then

limsup n!1xn=limxn=the largest cluster point limxn =¡1if the set cluster point is empty If x nis NOT bounded above, then limxn=1

Similarly, we can de¯ne the limit inferior

lim x n

Examples

For the seq 1;0;¡1;1;0;¡1;¢¢¢,

limxn= 1 and lim x n=¡1.

Ifxn=n, then

limxn=1= lim x n

Letxn= (¡1)n1+n

n . Then limxn= 1 and lim x n=¡1. 6

De¯nition. (1.6.2;Vector space)

A real vector spaceVis a set of elements calledvectors, with given operations of vector addition + :V £ V ! Vand scalar multiplication

¢:R£ V ! V

such that the followings hold for all v;u;w2 Vand all¸;¹2R: 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.

902 Vs.t. v+ 0 =v.9 ¡v2 Vs.t. v¡v= 0.

A subset ofVis called a subspace if it is itself a vector space with the same operations.

Wis a vector subspace ofVi®

v+ u2 Wwhenever u;v2 Wand

¸;¹2R

The straight lineW=f(x1;x2) :x1= 2x2gis a subspace of R 2. 7 Euclidean spaceRn& De¯nitions & PropertiesThe Euclidean n-spaceRnwith the operations (x1;¢¢¢;xn)+(y1;¢¢¢;yn) = (x1+y1;¢¢¢;xn+yn) &¸(x1;¢¢¢;xn) = (¸x1;¢¢¢;¸xn)is a vector space of dimensionn.

The standard basis ofRn;

e

1= (1;0;¢¢¢;0);¢¢¢;en= (0;¢¢¢;0;1).

Unique representation:

x= (x1;¢¢¢;xn)2Rncan be

Inner product ofxandy:

hx;yi=Pn i=1xiyi

Norm ofx:

kxk=p hx;yi.

Distance betweenxandy:

dist(x;y) =kx¡yk

Triangle inequality:

kx+yk · kxk+kyk.

Cauchy-Schwartz inequality:

hx;yi · kxk kyk

Pythagorean theorem:

Ifhx;yi= 0, then

kx+yk2=kxk2+kyk2. 8

De¯nition. (1.7.1; Metric Space (M;d)equipped withd=distance)A metric space(M;d)is a set M and a function d:M£M!R

such that 1. d(x;y)¸0for all x;y2M. 2. d(x;y) = 0i® x=y. 3. d(x;y) =d(y;x)for all x;y2M. 4. d(x;y)·d(x;z) +d(z;y)for all x;y2M. Example [Fingerprint Recognition]LetMbe a data set of

¯ngerprints in Seoul city police department.

Motivation: Design an e±cient access system to ¯nd a target. We need to de¯ne adissimilarityfunction stating the distance between the data.

The distanced(x;y) between two

dataxandymust satisfy the above four rules.

Similarity queries.

9

De¯nition. (1.7.3. Normed Space(V;k ¢ k))

Anormed space(V;k ¢ k)is a vector spaceVand a functionk ¢ k:V !Rcalled anormsuch that 1. kvk ¸0;8v2 V 2. kvk= 0i®v= 0. 3. k¸vk=j¸jkvk;8v2 Vand every scaler¸. 4. kv+wk · kvk+kwk;8v;w2 V

Examples

V=Randkxk=jxjfor allx2R.

V=R2andkvk=q

v

21+v22for allv= (v1;v2)2R2.

LetV=C([0;1])=all continuous functions on the interval [a;b]. De¯ne kfk= supfjf(x)j:x2[0;1]g(called supremum norm): 10

Proposition.

If(V;k ¢ k)is a normed vector space andd(v;w) =kv¡wk , then d is a metric inV

Proof.EASY.

Examples

ForV=C([0;1]), the metric is

d(f;g) =kf¡gk= supfjf(x)¡g(x)j:x2[0;1]g: The sup distance between functions is the largest vertical distance between their graphs. 11

De¯nition.

A vector spaceVwith a functionh¢;¢i:V£V!Ris called an inner product spaceif1.hv;vi ¸0for allv2 V. 2. hv;vi= 0i®v= 0. 3. h¸v;wi=¸hv;wi;8v2 Vand every scaler¸. 4. hv+w;hi=hv;hi+hw;hi. 5. hv;wi=hw;vi

Examples

1.

V=R2andhv;wi=v1w1+v2w2.

Two vectorsvandware orthogonal ifhv;wi= 0.

2.

V=C[0;1] andhf;gi=R1

0f(x)g(x)dx

3. kvk=p hv;viis a norm onV. 12

Theorem. (Cauchy-Schwarz inequality )

Ifh¢;¢iis an inner product in a real vector spaceV, then jhf;gij · kfkkgk

Proof:

Supposeg6= 0. Let

h=g kgk . It su±ces to prove that jhf;hij · kfk. (Why?jhf;gij · kfkkgki®jhf;hij · kfk.)

Denote

®=hf;hi

. Then

0· kf¡®

h k

2=hf¡®

h ;f¡ h i =kfk2¡ h h ;fi ¡ hf; h i+j j 2 =kfk2¡ j j 2

Hence,j

j=j hf; h i j · kfk. This completes the proof. 13 Chapter 2: Topology ofM=RnThroughout this chapter, assumeM=Rn( the Euclidean space ) with the metricd(x;y) =pP n i=1jxi¡yij2=kx¡yk

De¯nition. (D(x;²), open, neighborhood)

D(x;²) :=fy2M:d(y;x)< ²gis

called²-ball (or²-disk) aboutx A½M is open if8x2M;9² >0s.t. D(x;²)½A. A neighborhood ofx is an open set A containingx.

Open sets: (a;b),D(x;²),f(x;y)2R2: 0 The union of anarbitrary collectionof open subsets ofMis open. (Why?) The intersection ofa ¯nite numberof open subsets ofMis open. (Note that\1n=1(¡1=n;1=n) =f0gis closed. ) 1

2.2 Interior of a setA:int(A)De¯nition. (2.2.1; Interior point & interior ofA)Let(M;d)is a metric space and A½M.xis called aninterior

point ofAif9D(x;²)s.t.D(x;²)½A.

Denote

int(A) :=the collection of all interior points of A:

Examples.Proofs are very easy.

IfA= [0;1], thenint(A) = (0;1).

intf(x;y)2R2: 01g=f(x;y)2R2: 0 1g:

IfAis open, thenint(A) =A.

Let (M;d) be a metric space andx02M.

intfy2M:d(y;x0)

1g=fy2M:d(y;x0)

1g 2 De¯nition. (2.3-4: Closed sets & Accumulation Points ) ²A set B in a metric space M is said to be closed if MnB is open. ²x2M isaccumulation point (or cluster point ) of a set

A½M

if8² >0;D(x;²)containsy2A withy6=x.

Prove the followings:

Closed sets: [a;b],fy2R2:d(y;x0)

1g. The union of ana ¯nite numberof closed subsets ofMis closed. (Note that[1n=1[1=n;2¡1=n] = (0;2) is open. ) The intersection ofan arbitrary familyof closed subsets of

Mis closed.

Why?

Every ¯nite set inRnis closed.

A setA½Mis closed i® the accumulation points ofA belongs toA.

A=f1;1

2 ;1quotesdbs_dbs5.pdfusesText_9