To Study the Sequences and Their Convergence in Metric Spaces

Sequences in metric spaces

Objective

To study the sequences and their convergence in metric spaces.

Modules

Module I- The limit of a sequence

Module II- Cluster point of a sequence

Module III- Bounded sequences

Introduction

We have seen in elementary course of real analysis that the study of convergence of real sequences is carried out primarily to analyze the convergence of several types of infinite series of real numbers which, in turn, occur as solutions of differential equations occurring in many practical applications. As a further application, the notion of continuity of a function is characterized in terms of sequences. In metric spaces, the notion of convergence of a sequence plays an enhanced role. To study it in greater detail, we make it explicit in the following definition.

Module I- The limit of a sequence

Definition

Let (X,d) be a metric space, and (xn) be a sequence in X. Then (xn) is said to converge to a point if for each real number , there exist an integer m such that d(xn,x)<for all .

The point x in the above definition is called a limit of the sequence (xn), and we write .

Examples

Let (X,d) be a discrete metric space, and (xn) be a sequence in X which converges to x in X. Let . Since (xn) converges to x, there exist an integer m such that d(xn,x)for all . This means that xn=x for, that the sequence is of the form

(x1,x2,…,xm-1,x,x,x,…) which has the property that it becomes constant after a certain number of terms. Such a sequence is called an eventually constant sequence. Thus the convergent sequences in a discrete metric space are eventually constant sequences.

In R2, the sequenceis a convergent sequence which converges to (0,0): Letbe a real number. If we choose an integer m to be greater than, then

for all .

Let fn:[0,1] R be the function fn(x)=xn. Then the sequence (fn(x)) converges in R for each x in [0,1].

Theorem

A sequence (xn) in a metric space (X,d) converges to a point x in X if and only if for each neighbourhood N of x, there exist an integer m such that for all .

Proof.

If N is a neighbourhood of x, then there exists a real numbersuch that . If (xn) converges to x, then there exist an integer m such that for all , and the necessity part follows.

Conversely, since for each real number ,is a neighbourhood of x, by our assumption, there exist an integer m such that for all . Thus (xn) converges to x.

Theorem

A sequence in a metric space cannot converge to more than one point.

Proof.

Let (X,d) be a metric space and (xn) be a sequence in X which converges to two distinct points x and y in X. Taking, and considering the fact that (xn) converges to x, there exists an integer m1 such that d(xn,x)< for all . Again since (xn) converges to y, there exists an integer m2 such that d(xn,y)< for all . Then by the triangle inequality

for all .

Thus d(x,y)<2. Since is arbitrary, d(x,y)=0 which means that x=y, a contradiction.

Lemma

Let (X,d) be a metric space. Then the following inequality holds

for all x1,x2,y1 and y2in X.

Proof.

By the triangle inequality,

Therefore, . Now by interchanging x1 with x2 and y1 with y2, the desired inequality follows.

Theorem

Let (X,d) be a metric space, and (xn) and (yn) be convergent sequences in X converging to x and y in X, respectively. Then the sequenceconverges in the real line R.

Proof.

By the above lemma, we have

and the required convergence immediately follows.

Theorem

Let (X,d) be a metric space. A sequences (xn) in X converges to x in X if and only if the sequence (d(xn,x)) converges to 0 in R.

Proof.

Letbe a real number. The sequences (xn) in X converges to x in X if and only if there exist an integer m such that d(xn,x)<for all .

i.e., for all.

i.e., (d(xn,x)) converges in R to 0R.

A double sequence in a set X is a mapping. If s(n,m)=xn,m then the double sequence s may be denoted in terms of its values as . A double sequence or simply (xn,m) in a metric space (X,d) is said to converge to a point x in X if for each real number, there exist a positive integer p such that

d(xn,m,x)<for all n,mp.

Now it may be noted that if (xn) is a convergent sequence in a metric space (X,d) converging to x in X, then the double sequence (d(xn,xm)) in R converges to zero since.

Theorem

Let (X,d) be a metric space. A sequences (xn) in X converges to a point x in X if and only if every subsequence (xnk) of (xn) converges to x.

Proof.

Letbe a real number. If (xn) converges to x then there exist an integer m such that for all . This means that for all since for all k. The converse is obvious.

The sufficiency condition for the convergence of a sequence in the above theorem can be replaced as in the following theorem.

Theorem

Let (X,d) be a metric space. A sequences (xn) in X converges to a point x in X if and only if every subsequence of (xn) has a subsequence that converges to x in X.

Proof.

Since a subsequence of a subsequence of a sequence (xn) is a subsequence of (xn), the necessity part follows by the previous theorem.

Conversely, assume that every subsequence of (xn) has a subsequence that converges to x in X. Now suppose on the contrary, that (xn) does not converge to x. So there exist a real numbersuch that for each integer m, there is a positive integer n>m such that . But then we get a subsequence of (xn) which has no subsequence that converges to x.

Example

Let (X,d) be a metric space, and (xn) be a sequence in X that converges to x in X. Letbe any permutation (a bijective mapping) of N. Then the sequencealso converges to x. To see this, letbe a real number. Then there exist an integer m such that for all . Now let m0 be the largest positive integer such that . Such an integer exists since the pointsare finite in number. Now it follows that for all .

Module II- Cluster point of a sequence

Definition

Let (X,d) be a metric space, and (xn) be a sequences in X. A point x in X is said to be a cluster point (or limit point) of (xn) if for each real numberand each integer m, there exist an integer n such that n>m, and .

Example

Let (X,d) be a metric space, then the sequence(x,y,x,y,x,y,x,…) in which x and y alternate, has two cluster points x and y.

The following theorems are immediate consequences of the above definition of a cluster point of a sequence.

Theorem

Let (X,d) be a metric space. A point x in X is a cluster point of a sequence (xn) if and only if for each neighbourhood N of x and for each integer m, there exist an integer n such that n>m, and.

In view of the above theorem, sometimes, we say that a point x is a cluster point of a sequence if the points of the sequence are frequently in every neighbourhood of x.

Theorem

The limit of a convergent sequence is a cluster point of the sequence.

Proof.

Let (X,d) be a metric space, and (xn) sequences in X that converges to x in X. Then for each real numberthere is an integer m such that for all. Now for each integer p, select n>max{p,m}. Then , and so x is a cluster point of the sequence.

Theorem

A convergent sequence in a metric space has a unique cluster point.

Proof.

Let (X,d) be a metric space, and (xn) be a sequence in X that converges to x in X. by the above theorem (xn), x is a cluster point of (xn). Let y be another cluster point of (xn) different from x. Let. Since x is the limit point of (xn), there exist an integer m such that for. Since y is a cluster point, there is an integer n such that n>m and. Then

which is a contradiction.

The converse of the above theorem is not true, that is, if a sequence has a unique cluster point, then it may not be true that the sequence is convergent. Consider the sequence (xn) in R2 defined by

It is easily seen that (0, 0) is a cluster point of (xn), but (xn) does not converge to any point in R2.

From the definition of a bounded sequence of real numbers, it is immediate that a sequence (xn) of real numbers is bounded if and only if the set of points of the sequence {xn: n is an integer} is a bounded set. This result is made the definition of a bounded sequence in metric spaces.

Module III- Bounded sequences

Definition

Let (X,d) be a metric space. A sequence (xn) in X is said to be bounded if the set of points of the sequence {xn: nis an integer} is a bounded set.

Example

In Rn, for each integer m, let ym be a point in the set . Then the sequence (ym) is a bounded sequence in Rn.

In a discrete metric space, every sequence is bounded sequence.

If a metric space (X,d) has the bounded metric d, then every sequence in it is a bounded sequence.

Theorem (Bolzano- Weistrass theorem)

Every bounded sequence in Rn has a cluster point.

Proof.

Let (xk) be a bounded sequence in Rn, and A be the set of points of this sequence, that is, A={xk: k is an integer}. If A is finite, then for some point x in Rn, xk=x for infinitely many values of k. Consequently, ifis a real number, and p is an integer, then there exist an m>p such that xm=x, and so d(xm,x)=0<, where d is the Euclidean metric on Rn. Thus, x is a cluster point in this case. So we assume that A is infinite. Since A is bounded an infinite it has a limit point y in Rn. Therefore for each real number, the neighbourhoodof x contains infinitely many points of A. Hence for each integer p, there exist an integer m such that m>p, and. Thus, x is a cluster point of (xk).

In general, a bounded sequence in a metric space may not have any cluster point. For example, let (X,d) be an infinite discrete metric space. Then any sequence (xn) in X such thatfor is a bounded sequence which has no cluster point.

Theorem

Let (X,d) be a metric space. A point x in X is a cluster point of a sequence (xn) if and only if there exists a subsequence of (xn) that converges to x.

Proof.

First assume that x is a cluster point of (xn). Then there exists an integer n1 such that. Again there exists n2 such that n2>n1, and . By induction there exists an integer nk such that nk>nk-1, and . Thus is a subsequence of (xn) that converges to x.

Conversely, assume that the subsequenceof (xn) converges to x in X. Letbe a real number, and m is an integer. Sinceconverges to x, there exists an integer p such thatfor all . Then for r>max{p,m}, nr>m, and . Thus x is a cluster point of (xn).

Assignment Questions

Let (xn) be a convergent sequence in a metric space (X,d), and i be a fixed positive integer. Show that the sequenceis a bounded sequence in the real line R.
Let (xn) be a sequence in a metric space. Show that if the three subsequences (x2n), (x2n+1) and (x3n) are convergent, (xn) is convergent.
Show that every bounded sequence in Rn has a convergent subsequence.
Prove that the set of all cluster points of a sequence in a metric space is a closed set.

Quiz Questions

In R2, the sequenceis a convergent sequence which converges to

(a)(0,0)(b) (1,1)(c) (1/2,1/2)

Let (X,d) be a metric space, then the sequence (x,y,x,y,x,y,x,…)

(a) has one cluster point(b) has two cluster points(c) has no cluster points

Bolzano- Weistrass theorem states that

(a)Every unbounded sequence in Rn has a cluster point

(b)Every bounded sequence in Rn has not a cluster point

(c)Every bounded sequence in Rn has a cluster point

Answers

(a)2. (b)3. (c)

Glossary

Set: A collection of well defined objects is called a set

Function: A relation from a set X to another set Y which associates every element of X with a unique element of Y is called a function.

Metric space: For a set X (more than 1 element), suppose there is a function d, which maps XXto , and satisfy the following three properties:

For x,y and zX,

(i)d(x,y) 0 and d(x,y) =0 if and only if x=y;

(ii)d(x,y)=d(y,x);

(iii)d(x,y)d(x,z)+d(y,z).(triangle inequality)

Then the pair (X,d) is called a metric space, and d is called distance function, or metric.

Sequence: It is a function whose domain is N, the set of positive integers.

Summary

A sequence (xn) in a metric space (X,d) is said to converge to a point if for each real number , there exist an integer m such that d(xn,x)<for all .

A sequence (xn) in a metric space (X,d) converges to a point x in X if and only if for each neighbourhood N of x, there exist an integer m such that for all .

A sequence in a metric space cannot converge to more than one point.

If (xn) and (yn) are two convergent sequences converging to x and y in X, respectively. Then the sequenceconverges in the real line R.

A sequences (xn) in X converges to a point x in X if and only if every subsequence of (xn) has a subsequence that converges to x in X.

If (xn) is a sequences in X. A point x in X is said to be a cluster point (or limit point) of (xn) if for each real numberand each integer m, there exist an integer n such that n>m, and .

Let (X,d) be a metric space. A point x in X is a cluster point of a sequence (xn) if and only if for each neighbourhood N of x and for each integer m, there exist an integer n such that n>m, and.

The limit of a convergent sequence is a cluster point of the sequence.

A convergent sequence in a metric space has a unique cluster point.

Let (X,d) be a metric space. A sequence (xn) in X is said to be bounded if the set of points of the sequence {xn: nis an integer} is a bounded set.

Every bounded sequence in Rn has a cluster point.

Frequently Asked Questions (FAQs)

Show that a sequence in a metric space cannot converge to more than one point.

Answer:

for all .

Thus d(x,y)<2. Since is arbitrary, d(x,y)=0 which means that x=y, a contradiction.

Prove that the limit of a convergent sequence is a cluster point of the sequence.

Answer:

Prove that every bounded sequence in Rn has a cluster point.

Answer:

Let (xk) be a bounded sequence in Rn, and A be the set of points of this sequence, that is, A={xk: k is an integer}. If A is finite, then for some point x in Rn, xk=x for infinitely many values of k. Consequently, ifis a real number, and p is an integer, then there exist an m>p such that xm=x, and so d(xm,x)=0<, where d is the Euclidean metric on Rn. Thus, x is a cluster point in this case. So we assume that A is infinite. Since A is bounded an infinite it has a limit point y in Rn. Therefore for each real number, the neighbourhoodof x contains infinitely many points of A. Hence for each integer p, there exist an integer m such that m>p, and. Thus, x is a cluster point of (xk).

Books

W. Rudin, Principles of Mathematical analysis, McGraw Hill book company, Auckland (1985).
P. K. Jain and K. Ahmad, Metric Spaces, Narosa, New Delhi (1993).
J. R. Munkers, Topology, Prentice hall of India (1991).
G. F. Simmons, Introduction to topology and Modern Analysis, McGraw Hill (1963).
B. K. Tyagi, Metric Spaces, Cambridge University Press, new Delhi (2010).

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