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


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 $z_{n_k}$?

  • $\begingroup$ That the limit is z means that for every neighborhood of z there exists an N such that for all n > N the z_n are in that neighborhood of z. $\endgroup$ Apr 2 '16 at 23:55

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.

  • $\begingroup$ What if $n_k$ is not a sequence of distinct positive integers (as I believe this hypothesis is not guaranteed from what I wrote). $\endgroup$ Apr 3 '16 at 0:09
  • $\begingroup$ @YoTengoUnLCD: Either it has an infinite subsequence of distinct integers, in which case you work with that, or it has a constant subsequence, in which case the original sequence has a constant subsequence. $\endgroup$ Apr 3 '16 at 0:13
  • $\begingroup$ Awesome! Thank you Brian. I just realized that in the proof of theorem $3.6$, rudin constructs the sequence $z_k$ such that all the $z_k$ are distinct, thus $n_k$ is a sequence of distinct integers (so my question wasn't really necessary), anyway, it's nice to know what can be done if that detail is not remembered. $\endgroup$ Apr 3 '16 at 0:20
  • $\begingroup$ @YoTengoUnLCD: You’re welcome. I should perhaps have said a little more, but you’re generally pretty well on top of things, so I let myself be a little lazy. :-) $\endgroup$ Apr 3 '16 at 0:21

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