If $X\subset\mathbb{N}$ such that $\vert\mathbb{N}\setminus X\vert=\aleph_0,$ does $X$ or $\mathbb{N}\setminus X$ contain an arithmetic sequence?

Question

Suppose $\ X \subset \mathbb{N},\ $ such that $\ \vert X \vert = \vert \mathbb{N} \setminus X \vert = \aleph_0.\ $ Does either $\ X\ $ or $\ \mathbb{N} \setminus X\ $ contain an (infinite) arithmetic sequence?

Making one of the sets have no arithmetic subsequence is easy: $\ 1,\ 2,\ 4,\ 8,\ \ldots.\ $ But I don't see how to construct sets so that both $\ X\ $ and $\ \mathbb{N} \setminus X\ $ don't have an arithmetic sequence.

Answer 1

Let us denote $Y=\mathbb{N}-X$. We will construct $X$ (and so $Y$) as follows: Add $1$ to $X$, $2, 3$ to $Y$, $4, 5, 6, 7$ to $X$, $8, \dots, 15$ to $Y$, and so on. Note that $X$ and $Y$ are both infinite by construction. Note that both $X$ and $Y$ have arbitrary big gaps, and so they do not contain any infinite arithmetic subseqeunce.