20 - Subsequence

Subsequence

Let $(a_n{)}_{n\ge 1}$ be a sequence
A subsequence $(b_r{)}_{r\ge 1}$ of $(a_n{)}_{n\ge 1}$ is defined by a function $f$ such that

1. $f$ is strictly increasing (if $p<q$ then $f(p)<f(q)$)
2. $b_r=a_{f(r)}$

$f(r)$ is often written as $n_r$ so $n_1<n_2<n_3<\cdots$ is strictly increasing sequence of $\mathbb{N}$

Sequence $(b_r)$ has terms of ${\alpha }_{n_1},{\alpha }_{n_2},{\alpha }_{n_3},\cdots$

Note that the subscript variables are just dummy - it can be whatever you want.
Just avoid reusing the same letters especially as $n$ is used for the original sequence

Useful property of f: $n_r\ge r$ for $r\ge 1$

Subsequences of a convergent sequence

Let $(a_n)$ be a sequence
If $(a_n)$ converge, then every subsequence $(a_{n_r})$ of $(a_n)$ converges

Moreover if $a_n\to L$ as $n\to \infty$ then every subsequence also converges to $L$

Note that if two subsequences have different limits then $(a_n)$ doesn’t converge (useful)

Proof

Assume that $(a_n)$ converges to $L$

Let $(a_{n_r})$ be a subsequence of $(a_n)$
Take $ϵ>0$
Since $a_n\to L$, there is $N$ such that if $n\ge N$ then

$$a_n-L<ϵ$$

If $r\ge N$ then $n_r\ge r\ge N$ hence

$$|a_{n_r}-L|<ϵ$$

Hence

$${\alpha }_{n_r}\to L\text{ as }r\to \infty$$