For $S \subseteq \Bbb{N}$, the usual definition of density - natural density, or asymptotic density - goes like this:
$$d_{\text{nat}}(S) = \lim_{n\to\infty} {\#(0\leq k < n \ : \ k\in S) \over n}$$
But the notion of density I want to use is this:
$$d_{\text{arith}}(S) = \sup_{a\in \Bbb{N}} {\#(0\leq b < a\ : \ a\Bbb{N} + b \subseteq S) \over a}$$
It measures the proportion of arithmetic progressions of a given increment-rate that are contained in $S$. As pointed out in an earlier post of mine, this density for a set will very often be $0$, since few sets contain whole arithmetic progressions.
So it is a stronger property to have $d_\text{arith}(S) = 1$ than to have $d_\text{nat}(S)=1$, which is what I am hoping for.
Question:
Is there a necessary and sufficient condition for $d_\text{arith}(S) = 1$, and is it equivalent to "$S^c$ is finite"?
One direction is trivial:
If $S^c$ is finite, then $n \in S$ for all large enough $n > n_0$, in which case $a\Bbb{N} + b \subseteq(S)$ for all $b = 1, 2, \ldots a-1$ whenever $a > n_0$, and so we have $d_\text{arith}(S) = \sup_a {a-1\over a} = 1$
Here is the reason why I suspect the converse may be true: if $d(S) = 1$, then the proportion of congruence classes $[\ b\ ]_{\mod a} \subseteq S$ must approach $1$, in which case we ought to have
$$m_a = \min_{0\leq b < a} (b \ : \ a\Bbb{N} + b \not\subseteq S) \to \infty$$
as $a\to \infty$. If it were not so, then a bounded monotone sequence of integers has a limit - say, by $b_0$, ie $a\Bbb{N} + b_0 \not\subseteq S$ for all sufficiently large $a$.
That is, $\forall a$-big $\exists k$ s.t. $b_0 + ak \not\in S$, in particular $\exists k_0$ s.t. $b_0 + b_0k_0 \not\in S \implies b_0(1+k_0) \not\in S \implies b_0\Bbb{N} \not\subseteq S \implies d(S) \leq 1 - 1/b_0 < 1$?
Or maybe that argument is flawed. Because if not, then for any $n$ we can choose $m_a > n$ to show $a\Bbb{N} + n \subset S \implies n \in S$, or something like that. I'm curious to see what sort of pathological sets can have density $1$ under this definition