Lecture2:Metric Spaces:Dense Set, Separable Space,Convergence,Cauchy Sequence,Completeness
October11,2009
1Dense Set
Let M be a subset of a metric space X.Then a point x0of X(which may or may not be a point of M)is called an accumulation point of M(or limit point of M)if every neighborhood of x0contains at least one point y∈M distinct from x0.The set consisting of the points of M and the accumulation points of M is called the closure of M and is denoted by
ˉM
It is the smallest closed set containing M.
De?nition1(Dense set)A subset M of a metric space X is said to be dense in X if
ˉM=X.
De?nition2(Dense set)Given a metric space X,A?X,B?X.A is said to be dense in B if for any x∈B and anyδ>0,there exists
δ(x)∩A/∈?.
Theorem1Given a metric space X,A?X,B?X.A is dense in B if and only if
B?ˉA.
Theorem2Given a metric space X,A?X,B?X.A is dense in B if and only if for any x∈B,there exists{x n}?A such that
lim
x n=x.
n→∞
Theorem3Given a metric space X,A?X,B?X,C?X.If A is dense in B and B is dense in C.Then A is dense in C.
Example
The set of all rational numbers is dense in R
2Separable space
De?nition3(separable space)A metric space X is said to be separable if it has a countable subset which is dense in X.
Example
1.Real line R.The real line R is separable.
Proof.The set of all rational numbers is countable and dense in R.
https://www.doczj.com/doc/8f5476740.html,plex plane C.The complex plane C is separable.
Proof.A countable dense subset of C is the set of all complex numbers whose real and imaginary parts are both rational.
3Convergence
De?nition4A sequence(x n)in a metric space X=(X,d)is said to converge or to be convergent if there is an x∈X such that
lim
n→∞
d(x n,x)=0.
x is called the limit of(x n)and we write
lim
n→∞
x n=x
or,simply,
x n→x.
We say that(x n)converges to x or has the limit x.If(x n)is not convergent, it is said to be divergent.
We see that if x n→x,anε>0being given,there is an N=N(ε)such that all x n with n>N lies in theε-neighborhood B(x;ε)of x.
To avoid trivial misunderstandings,we note that the limit of a convergent sequence must be a point the space X.For instance,let X be the open interval (0,1)on R with the usual metric de?ned by d(x,y)=|x?y|.The the sequence
(1 2,1
3
,1
4
,···)is not convergent since0,the point to which the sequence”wants
to converge”,is not in X.
Let us?rst show that two familiar properties of a convergent sequence (uniqueness of the limit and boundedness)carry over from calculus to our present much more general setting.
We call a nonempty subset M?X a bounded set if its diameter
δ(M)=sup
x,y∈M
d(x,y)
is?nite.And we call a sequence(x n)in X is a bounded sequence if the corresponding point set is a bounded subset of X.
Obviously,if M is bounded,then M?B(x0;r),where x0∈X is any point and r is a(su?ciently large)real number.
Lemma1(Boundedness,limit)Let X=(X,d)be a metric space.Then:
(a)A convergent sequence in X is bounded and its limit is unique.
(b)If x n→x and y n→y in X,then d(x n,y n)→d(x,y).
Proof.(a)Suppose that x n→x.Then,takingε=1,we can?nd an N such that d(x n,x)<1for all n>N.Hence by the triangle inequality(M4),for all n we have
d(x n,x)<1+a
where
a=max{d(x1,x),d(x2,x),···,d(x N,x)}.
This shows that(x n)is bounded.Assuming that x n→x and x n→z,we obtain from(M4)
0≤d(x,z)≤d(x,x n)+d(x n,z)→0+0
and the uniqueness x=z of the limit follows from(M2).
(b)By(1),we have
d(x n,y n)≤d(x n,x)+d(x,y)+d(y,y n).
Hence we obtain
d(x n,y n)?d(x,y)≤d(x n,x)+d(y,y n)→0
and a similar inequality by interchanging x n and x as well as y n and y and multiplying by?1.Together,
|d(x n,y n)?d(x,y)|≤d(x n,x)+d(y,y n)→0
as n→∞.
4Cauchy Sequence and Completeness
De?nition5(Cauchy sequence,completeness)A sequence(x n)in a met-ric space X=(X,d)is said to be Cauchy(or fundamental)if for everyε>0 there is an N=N(ε)such that
d(x m,x n)<εfor every m,n>N.
The space X is said to be complete if every Cauchy sequence in X converges.
Note that in calculus,a sequence(x n)of real or complex numbers converges on the real R or in the complex plane C,respectively,if and only if it satis?es the Cauchy convergence criterion,that is,if and only if for every givenε>0 there is an N=N(ε)such that
|x m?x n|<εfor every m,n>N.
Unfortunately,in more general spaces the situation may be more complicated, and there may be Cauchy sequences which do not converge.
Theorem4The real line and the complex plane are complete metric spaces.
For the time being let us mention a few simple incomplete spaces which we can readily obtain.Omission of a point a from the real line yields the incomplete space R?{a},which is incomplete.An open interval(a,b)with the metric induced from R is another incomplete metric space and so on.
It is clear from the de?nition that in an arbitrary metric space,condition(1) may no longer be su?cient for convergence since the space may be incomplete.
Theorem5(Convergent sequence)Every convergent sequence in a metric space is a Cauchy sequence.
Proof.If x n→x,then for everyε>0there is an N=N(ε)such that
d(x n,x)<ε
2
for all n>N.
Hence by the triangle inequality we obtain for m,n>N
d(x m,x n)≤d(x m,x)+d(x,x n)<ε
2
+
ε
2
=ε.
This shows that(x n)is Cauchy.
Theorem6(Closure,closed set)Let M be a nonempty subset of a metric space(X,d)andˉM its closure as de?ned in the previous section.Then:
(a)x∈ˉM if and only if there is a sequence(x n)in M such that x n→x.
(b)M is closed if and only if the situation x n∈M,x n→x implies that
x∈M.
Proof.(a)Let x∈ˉM.If x∈M,a sequence of that type is(x,x,,···).If x/∈M,it is a point of accumulation ofˉM.Hence for each n=1,2,···the ball B(x;1/n)contains an x n∈M and x n→x because1/n→0as n→∞.
Conversely,if(x n)is in M and x n→x,then x∈M or every neighborhood of x contains points x n=x,so that x is a point of accumulation of M.Hence x∈ˉM.
(b)M is closed if and only if x∈ˉM=M,so that(b)follows readily from (a).
Theorem7(Complete subspace)A subspace M of a complete metric space X is itself complete if and only if the set M is closed in X.
Proof.Let M be complete.For every x∈ˉM there is a sequence(x n)in M which converges to x.Since(x n)is Cauchy and M is complete,(x n)converges in M,the limit being unique.Hence x∈M.This proves that M is closed because x∈ˉM was arbitrary.
Conversely,let M be closed and(x n)Cauchy in M.Then x n→x∈M, which implies x∈ˉM and x∈M sinceˉM=M by assumption.Hence the
arbitrary Cauchy sequence(x n)converges in M,which proves completeness of M.
In various applications a set X is given,and X is made into a metric space. This we do by choosing a metric d on X.To prove completeness.we take an arbitrary Cauchy sequence(x n)in X and show that it converges in X.For dif-ferent spaces,such proofs may vary in complexity,but they have approximately the same general patterns:
(i)Construct an element x(to be used as a limit).
(ii)Prove that x is in the space considered.
(iii)Prove convergence x n→x(in the sense of the metric).
Examples
Example1Completeness of R n and C n.Euclidean space R n and unitary space C n are complete.
Proof.We?rst consider R n.We remember that the metric on R n is de?ned
by
d(x,y)=
n
i=1
(ξj?ηj)2
1/2
where x=(ξj)and y=(ηj).We consider any Cauchy sequence(x m)in R n,
writing x m=(ξ(m)
1,···,ξ(m)
n
).Since(x m)is Cauchy,for everyε>0there is
an N such that
d(x m,x r)=
n
i=1
(ξ(m)
j
?ξ(r)
j
)2
1/2
<ε,(m,r>N).(1)
Squaring,we have for m,r>N and j=1,···,n
(ξ(m)
j ?ξ(r)
j
)2<ε2and|ξ(m)
j
?ξ(r)
j
|<ε.
This shows that for each?xed j,(1≤j≤n),the sequence(ξ(1)
j ,ξ(2)
j
,···)
is a Cauchy sequence of real numbers.It converges,say,ξ(m)
j →ξj as m→∞.
Using these n limits,we de?ne x=(ξ1,···,ξn).Clearly,x∈R n.From(1), with r→∞,
d(x m,x)≤ε(m>N).
This shows that x is the limit of(x m)and proves completeness of R n because (x m)was an arbitrary Cauchy https://www.doczj.com/doc/8f5476740.html,pleteness of C n follows by the same method of proof.
Example2Completeness of l∞.The space l∞is complete.
Proof.Let(x m)be any Cauchy sequence in the space l∞,where x m=(ξ(m)
1,···,ξ(m)
n
).
Since the metric on l∞is given by
d(x,y)=sup
j
|ξj?ηj|
where x=(ξj)and y=(ηj).And(x m)is Cauchy,for anyε>0there is an N such that for all m,n>N,
d(x m,x n)=sup
j |ξ(m)
j
?ξ(n)
j
|<ε,(m,n>N).
A fortiori,for every?xed j,
|ξ(m) j ?ξ(n)
j
|≤ε,(m,n>N).(2)
Hence,for every?xed j,the sequence(ξ(1)
j ,ξ(2)
j
,···)is a Cauchy sequence of
numbers.It converges,say,ξ(m)
j →ξj as m→∞.Using these in?nitely many
limitsξ1,ξ2,···,we de?ne x=(ξ1,ξ2,···)and show that x∈l∞and x m→x. From(2)with n→∞,we have
|ξ(m)
j
?ξj|≤ε.(3)
Since x m=(ξ(m)
j )∈l∞,there is a real number k m such that|ξ(m)
j
|≤k m
for all j.Hence by the triangle inequality
|ξj|≤|ξj?ξ(m)
j |+|ξ(m)
j
|≤ε+k m,(m>N).
This inequality holds for every j,and the right-hand side does not involve j. Hence(ξj)is a bounded sequence of numbers.This implies that x=(ξj)∈l∞. Also from(3)we obtain
d(x m,x)=sup
j |ξ(m)
j
?ξj|≤ε,(m,r>N).(4)
This shows that x m→x.Since(x m)was an arbitrary Cauchy sequence,l∞is complete.