Converse of the Erdős Conjecture on Arithmetic Progressions

Question

Clearly, the question of whether large sets always contain arbitrarily long arithmetic progressions is an open question. So my question is not about this conjecture. Instead, it is about the converse. Is the converse known to be true, namely that if a set contains arbitrarily long arithmetic progressions, then it is large?

Answer 1

Certainly not. Just glue together short arithmetic progressions that are really far apart, such as $$\bigcup_{k\ge 1} \{2^{2^k} + i\}_{i=1}^k. $$

It is easy to see that one can make such sets as thin as you want (unless you want them to have finite cardinality).

Answer 2

The set can be arbitrarily sparse. After we have chosen some integers, skip an arbitrarily large amount, then choose $k$ consecutive integers, then skip a huge amount, then choose $k+1$ consecutive integers, then skip as many as you feel like, and so on.