Well, I know that I'm going to have some vote-down, because this should be a very simple property but the real story is that... I am not able to prove it!


For any fixed $n\in\mathbb N$, I have a finite partition of the natural numbers


$$
\mathbb N=A\_1^n\cup...\cup A\_{k(n)}^n
$$


(every set of the partition is infinite). Suppose that the sequence $k(n)$ is bounded. I want to make a choice of a set $A\_{i\_n}^n$ for each partition such that $A=\bigcap\_{n=1}^\infty A\_{i\_n}^n$ is infinite.


Intuitively this should be possible, and I have a sketch of the proof making use of the upper density on natural numbers, but I don't think it is completely right and anyway it sounds a bit too complicated.


Motivation: I am basically extracting infinite subsequences from a sequence and I want to be sure to find, at the end of the story, a subsequence.



Simple set-theoretic property to extract a subsequence in infinitely many steps
As requested, here's the comment posted as an answer:


What if the $n$th partition splits the naturals into ten pieces based on the $n$th digit of the decimal expansion?


(Of course ten is chosen for familiarity and not for optimality.)




---


Apologies for such an extensive edit, but I couldn't visualize this either, and this doesn't fit into a comment. Using binary (and indexing from 0) instead of base 10 gives:
 $$A\_0^0 = evens, A\_1^0 = odds$$
 $$A\_0^1 = (0\bmod 4)\cup(1\bmod 4), A\_2^1 = (2\bmod 4)\cup(3\bmod 4)$$
 $$A\_0^n = \{ n \in {\mathbb N} \colon 0\leq n\bmod 2^{n+1} <2^n\},
 A\_1^n = \{ n \in {\mathbb N} \colon 2^n\leq n\bmod 2^{n+1} <2^{n+1}\},$$ 
But now $$\bigcap\_{i=0}^n A\_{i\_n}^n$$ is a single congruence class modulo $2^{n+1}$ for any choice of $i\_n$, and the second positive element of a congruence class must be at least as large as the modulus, and the modulus goes to infinity, so the infinite intersection cannot have two elements.


