Infinite Sequence with an infinite amount of Subsequences

[Supernova93]

Hi, I'm new here and I don't know exactly where I should ask this question so if it's in the wrong area feel free to move it, thanks!

Suppose I have an infinite sequence that seems randomly ordered, seq. An consisting of:

{A1, A2, A3, ... An} to infinity. Ex: {2,5,9,13,18,20,25,33,38,42...}
where each term increases and the sequence diverges.

Now I want to obtain my subsequence by filtering out the numbers which are in the position of the numbers of the sequence..
Ex: of what I mean:

{2,5,9,13,18,20,25,33,38,42...}

I filter out the number in position 2, the number in position 5, the number in position 9 which are: 5 (position 2), 18 (position 5), 38 (position 9).
This can go on to infinity right?

So I get this subsequence after filtering once: {2,9,13,20,25,33,42...}

Now I want to take this subsequence and apply the same process infinitely getting:
{2,13,20,33,42...}
then
{2,20,33...}
then
{2,33...} etc. on to infinity.

My question is how would I denote all this and describe it? Thanks!
How would I describe the fact that my original sequence has infinite subsequences, and describe the operation that I'm applying at each iteration to obtain the next subsequence?

 

[pka]

Suppose I have an infinite sequence that seems randomly ordered, seq. An consisting of: {A1, A2, A3, ... An} to infinity. Ex: {2,5,9,13,18,20,25,33,38,42...}
where each term increases and the sequence diverges.
My question is how would I denote all this and describe it? How would I describe the fact that my original sequence has infinite subsequences, and describe the operation that I'm applying at each iteration to obtain the next subsequence?

There is a very well-known theorem: Every sequence contains a monotone subsequence.

Given the sequence \(\displaystyle \bf{(x_n)}\) define a set \(\displaystyle \mathcal{S}=\{n: (\forall j>n)[x_j>x_n]\}\)
\(\displaystyle \begin{align*} n\in\mathcal{S}&\iff\forall j>n,~x_j>x_n\\ t\notin\mathcal{S}&\iff \exists k>t,~x\le x_t\end{align*}\)

Now to prove the theorem there are two cases for \(\displaystyle \mathcal{S} \) i) \(\displaystyle \mathcal{S} \) is infinite and ii) \(\displaystyle \mathcal{S} \) is finite.

That tells you that it is possible to find your subsequence. BUT it does not tell you how,