1.4 Cauchy Sequence in R
A sequence xn ? R is said to converge to a limit x if Every convergent sequence is a Cauchy sequence. Proof. Assume xk ? x. Let ? > 0 be given.
Suggested Solution to Homework 1
If (xn) is Cauchy and has a convergent subsequence say
Aronszajns theorem
A function K : X ×X ? R is called a positive definite kernel on X iff it is This shows that for any x the sequence (fn(x))n?0 is Cauchy in R and has ...
Chapter 1. Metric spaces - Proofs covered in class
In a metric space every convergent sequence is a Cauchy sequence. Proof. Suppose that {xn} is a sequence which converges to x and let ? > 0 be given.
Week 3 Solutions Page 1 Exercise (2.4.1). Prove that ) is Cauchy
Assume that (xn)n?N is a bounded sequence in R and that there exists x ? R such that any convergent subsequence (xni )i?N converges to x. Then limn?? xn =
Fixed point theorem and Cauchy-Lipschitz for linear ODEs
7 févr. 2018 of points in X. We say that x is a Cauchy sequence when ... Our goal here is to prove the Cauchy-Lipschitz theorem in the linear case.
ON I-CAUCHY SEQUENCES
In this section we prove a decomposition theorem for ? -convergent sequences. Theorem 1. Let (X p) be a linear metric space
Homework 4 1. Let X and Y be normed spaces T ? B(X
https://www.math.wustl.edu/~wick/teaching/math6338/math6338_hw4.pdf
The University of Sydney Pure Mathematics 3901 Metric Spaces
Let X = (X d) be a metric space. Let (xn) and (yn) be two sequences in X such that (yn) is a Cauchy sequence and d(xn
Basic Topology of R
Now assume that the limit of every Cauchy sequence (or convergent sequence) contained in F is also an element of F. We show F is closed. Let x be any limit
Lecture 2 - Vector Spaces Norms and Cauchy Sequences
Therefore we have the ability to determine if a sequenceis a Cauchy sequence Proposition 3 1If (X;k k)is a normed vector space then a sequence of pointsfXig1 i=1 is a Cauchy sequence i given any >0 there is anN2Nso thati; j > Nimplies kXi Xjk< : Proof Simple exercise in verifying the de nitions
How do I prove a sequence is Cauchy - Mathematics Stack Exchange
Theorem Cauchy sequences converge Homework problems 2 4 1: Show directly from the de nition that n21 n2 0
Is the sequence 1 n Cauchy?
Claim: The sequence { 1 n } is Cauchy. Proof: Let ? > 0 be given and let N > 2 ?. Then for any n, m > N, one has 0 < 1 n, 1 m < ? 2. Therefore, ? > 1 n + 1 m = | 1 n | + | 1 m | ? | 1 n ? 1 m |. Thus, the sequence is Cauchy as was to be shown. Everything you wrote is correct, but I think your point would be better illustrated by = ? 1.
Is x n Cauchy in R?
Examples: 1. (X;d) = Q, as a subspace of R with the usual metric. Take x 0= 2 and defne x n+1= xn 2 +1 xn The sequence continues 3=2, 17=12, 577=408;:::and indeed x n!xwhere x=x 2 +1 x , i.e., x2= 2. But this isn’t in Q. Thus (x n) is Cauchy in R, since it converges to p 2 when we think of it as a sequence in R.
How can we guarantee that (x n) will be Cauchy?
n 1) for n1. This gives a sequence (x n); if it is Cauchy and (X;d) is complete, then x= lim n!1x nexists and xshould solve x= ?(x). How can we guarantee that (x n) will be Cauchy? Note that d(x n;x n+1) = d(?(x n 1);?(x n)), so to get (x n) Cauchy we want ?to shrink distances.
How do you prove that a metric space is Cauchy?
Prove directly that it’s Cauchy, by showing how the nin the de nition depends upon . De nition: A metric space (X;d) is complete if every Cauchy sequence in Xconverges in X (i.e., to a limit that’s in X).
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 if² 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 1Theorem. (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=x1BisectI0:= [¡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 I0¾I1¾I2¾I3¢¢¢
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 Sinceak·ak+1·M(monotone and bounded),ak!9x.
Sincexnk2IkandjIkj= 2¡kM, we have
jxnk¡xjCorollary. (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!
xLemma. (1.4.6; Boundedness of Cauchy sequence)
If x nis aCauchy
sequence, x nis boundedProof.9Ns.t.n¸N) jxn¡xj<1. Then
sup njxnj ·1 + maxfjx1j;¢¢¢;jxNjg(Why?)Theorem. (1.4.3; Completeness)
EveryCauchy
sequence inRconverges to an element in [a;b]Proof.
Cauchy seq.
bounded seq. convergent subseq. 31.5. 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 sequence1;¡1;1;¡1;¢¢¢:
The sequencexn=1
n has the only cluster point 0.The sequencexn=ndoes not have any cluster point.
4Proposition.
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 n1 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
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.
5De¯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=1Similarly, we can de¯ne the limit inferior
lim x nExamples
For the seq 1;0;¡1;1;0;¡1;¢¢¢,
limxn= 1 and lim x n=¡1.Ifxn=n, then
limxn=1= lim x nLetxn= (¡1)n1+n
n . Then limxn= 1 and lim x n=¡1. 6De¯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;
e1= (1;0;¢¢¢;0);¢¢¢;en= (0;¢¢¢;0;1).
Unique representation:
x= (x1;¢¢¢;xn)2Rncan beInner product ofxandy:
hx;yi=Pn i=1xiyiNorm ofx:
kxk=p hx;yi.Distance betweenxandy:
dist(x;y) =kx¡ykTriangle inequality:
kx+yk · kxk+kyk.Cauchy-Schwartz inequality:
hx;yi · kxk kykPythagorean theorem:
Ifhx;yi= 0, then
kx+yk2=kxk2+kyk2. 8De¯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.
9De¯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 VExamples
V=Randkxk=jxjfor allx2R.
V=R2andkvk=q
v21+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): 10Proposition.
If(V;k ¢ k)is a normed vector space andd(v;w) =kv¡wk , then d is a metric inVProof.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. 11De¯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;viExamples
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. 12Theorem. (Cauchy-Schwarz inequality )
Ifh¢;¢iis an inner product in a real vector spaceV, then jhf;gij · kfkkgkProof:
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
. Then0· kf¡®
h k2=hf¡®
h ;f¡ h i =kfk2¡ h h ;fi ¡ hf; h i+j j 2 =kfk2¡ j j 2Hence,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¡ykDe¯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
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: 0IfAis 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 setA½M
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 ofMis closed.
Why?Every ¯nite set inRnis closed.
A setA½Mis closed i® the accumulation points ofA belongs toA.A=f1;1
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