Is there a set of positive integers of density 1 which contains no infinite arithmetic progression? Let $V$ be a set of positive integers whose natural density is 1. Is it necessarily true that $V$ contains an infinite arithmetic progression?—i.e., that there are non-negative integers $a,b,\nu$ with $0\leq b\leq a-1$ so that: $$\left\{ an+b:n\geq\nu\right\} \subseteq V$$ 
In addition to an answer, any references on the matter would be most appreciated.
 A: Here is a concrete counterexample (where $\mathbb{N}$ is the set of positive integers):
$$V:=\mathbb{N}\setminus\{a^2b^2+b:a,b\in\mathbb{N}\}.$$
It is straightforward to see that $V$ has density $1$, but it does not contain an infinite arithmetic progression.
A: Yet another concrete counterexample:
  $$ \bigcup_{n=1}^\infty [n^3+n,(n+1)^3]. $$
More generally, any set containing arbitrarily long gaps is free of infinite arithmetic progressions, and has natural density $1$ if the gaps are properly spaced. 
Incidentally, an incomparably subtler question is whether any set of positive natural density contains arbitrarily long arithmetic progressions. This was conjecture by Erdős and Turán in 1936 and famously proved by Szemerédi almost 40 years later.
A: Another construction is to let $n \notin V$ if and only if $n$ begins with at least $\sqrt{\log{n}}$ consecutive '9's when written in decimal. This satisfies the stronger property that there is no non-constant polynomial $f : \mathbb{N} \rightarrow \mathbb{N}$ whose image is contained entirely in $V$.
