Rudin's proof of every bounded sequence has a convergent subsequence.

Question

This may be really simple, but I couldn't prove the last detail needed in the proof, see below.

Theorem

Every bounded sequence $s_n$ contains a convergent subsequence.

Let $E=\{s_n:n\in\Bbb N\}$. Then $E$ is bounded. If $E$ is infinite, it has a limit (by a previous theorem). And the conclusion follows from theorem 3.6 $[\dots]$

Okay, theorem $3.6$ says that every set of complex numbers with a limit point $z$ has a sequence $z_n\in E\,(n=1,2,\dots)$ such that $z_n\to z$.

The thing is that just having a sequence of $z_{n_k}\to z$ doesn't necessarily imply that $z_{n_k}$ is a subsequence of $s_n$ ($z_{n_k}$ could be $(s_5,s_2,s_1,s_3,\cdots)$).

How can we construct a subsequence of $s_n$ from such sequence $z_{n_k}$?

Answer 1

Suppose that $\langle n_k:k\in\Bbb Z^+\rangle$ is any sequence of distinct positive integers; then it has a strictly increasing subsequence $\langle n_{i_j}:j\in\Bbb Z^+\rangle$. We simply construct the sequence $\langle i_j:j\in\Bbb Z^+\rangle$ recursively, so that both it and $\langle n_{i_j}:j\in\Bbb Z^+\rangle$ are strictly increasing.

Let $i_1=1$. Given $i_j$ for some $j\in\Bbb Z^+$, let

$$i_{j+1}=\min\{k\in\Bbb Z^+:k>i_j\text{ and }n_k>n_{i_j}\}\;;$$

it’s clear that this is always possible, since only finitely many terms $n_k$ are less than or equal to $n_{i_j}$. Thus, the construction goes through, and it’s clear that $\langle n_{i_j}:j\in\Bbb Z^+\rangle$ is as desired.