For Infinites $A,B\subset\mathbb{N}$ s.t exists $n \in A$ and $r,l < n$ s.t the sets $B\cap(n\mathbb{N}+r)$ and $B\cap(n\mathbb{N}+l)$ are infinite.

Question

We note that for every subset $B$ of $\mathbb{N}$ and for every $n \in \mathbb{N}$ we have the following

$$ B = \bigcup_{0 \leq r < n} (n\mathbb{N} + r) \cap B, $$

where $n\mathbb{N} + r = \{ n \cdot k + r \colon k \in \mathbb{N} \}$.

My question is, for every two infinite subsets $A,B$ of $\mathbb{N}$ exists $n_0 \in A$ and $r_1, r_2 < n_0$ with $r_1 \neq r_2$ s.t the sets

$$ B\cap(n_0 \mathbb{N} + r_1) \text{ and } B \cap (n_0 \mathbb{N} + r_2) $$ are infinites?

I think that this is true, but I don't know how prove this proposition.

Could someone help me prove or disprove this?

Sorry for my english, I'm still practicing.

Answer 1

The set $\ B:= \{n!: n\in\mathbb{N}\}\ $ is a counter-example.

For any $\ n_0\in\mathbb{N},\ (n_0\mathbb{N}+r)\cap B\ $ is infinite if $\ r=0,\ $ else if $\ r\neq 0,\ $ i.e. $\ r\in \{1,2,\ldots, n-1\},\ $ then $\ \left\lvert (n_0\mathbb{N}+r)\cap B \right\rvert \leq n_0-1,\ $ and so $\ (n_0\mathbb{N}+r)\cap B\ $ is finite.