2.5 Subsequences and cluster points  (Page 5/4)

Let $\left\{a_n\right\}$ be a sequence of real or complex numbers. A subsequence of $\left\{a_n\right\}$ is a sequence $\left\{b_k\right\}$ that is determined by the sequence $\left\{a_n\right\}$ together with a strictly increasing sequence $\left\{n_k\right\}$ of natural numbers. The sequence $\left\{b_k\right\}$ is defined by $b_k=a_{n_k}.$ That is, the $k$ th term of the sequence $\left\{b_k\right\}$ is the $n_k$ th term of the original sequence $\left\{a_n\right\}.$

Here is an interesting generalization of the notion of the limit of a sequence.

Let $\left\{a_n\right\}$ be a sequence of real or complex numbers. A number $x$ is called a cluster point of the sequence $\left\{a_n\right\}$ if there exists a subsequence $\left\{b_k\right\}$ of $\left\{a_n\right\}$ such that $x=lim{b}_k.$ The set of all cluster points of a sequence $\left\{a_n\right\}$ is called the cluster set of the sequence.

Suppose $\left\{a_n\right\}$ is a sequence of real or complex numbers.

1. (Uniqueness of limits) Suppose $lim{a}_n=L,$ and $lim{a}_n=M.$ Then $L=M.$ That is, if the limit of a sequence exists, it is unique.
2. If $L=lim{a}_n,$ and if $\left\{b_k\right\}$ is a subsequence of $\left\{a_n\right\},$ then the sequence $\left\{b_k\right\}$ is convergent, and $lim{b}_k=L.$ That is, if a sequence has a limit, then every subsequence is convergent and converges to that same limit.

Suppose $lim{a}_n=L\text{and}lim{a}_n=M.$ Let $ϵ$ be a positive number, and choose $N_1$ so that $|a_n-L|<ϵ/2$ if $n\ge N_1,$ and choose $N_2$ so that $|a_n-M|<ϵ/2$ if $n\ge N_2.$ Choose an $n$ larger than both $N_1\text{and}{N}_2.$ Then

Therefore, since $|L-M|<ϵ$ for every positive $ϵ,$ it follows that $L-M=0$ or $L=M.$ This proves part (1).

Next, suppose $lim{a}_n=L$ and let $\left\{b_k\right\}$ be a subsequence of $\left\{a_n\right\}.$ We wish to show that $lim{b}_k=L.$ Let $ϵ>0$ be given, and choose an $N$ such that $|a_n-L|<ϵ$ if $n\ge N.$ Choose a $K$ so that $n_K\ge N$ . (How?) Then, if $k\ge K,$ we have $n_k\ge n_K\ge N,$ whence $|b_k-L|=|a_{n_k}-L|<ϵ,$ which shows that $lim{b}_k=L.$ This proves part (2).

REMARK The preceding theorem has the following interpretation. It says that if a sequence converges to a number $L,$ then the cluster set of the sequence contains only one number, and that number is $L.$ Indeed, if $x$ is a cluster point of the sequence, then there must be some subsequence that converges to $x.$ But, by part (2), every subsequence converges to $L.$ Then, by part (1), $x=L.$ Part (g) of [link] shows that the converse of this theorem is not valid. that is, the cluster set may contain only one point, and yet the sequence is not convergent.