I cannot find the exact same question asked anywhere in this site. I know the related Green-Tao theorem but the gaps between consecutive primes can grow unbounded so it does not seem helpful to answer this question.
What I have tried: assume the largest gap is D and without loss of generality it appears infinitely often. Then I try to apply the pigeonhole principle but don't know how.
Thanks in advance.


Added thought: will there always be a subsequence forming an infinitely long arithmetic progression in the original sequence? I think the answer is NO.



Are there arbitrarily long arithmetic progressions in every increasing sequence of positive integers with bounded gaps between consecutive terms?
You may of course use Szemeredi theorem, as suggested by Alexander Kalmynin.


If you need a more elementary argument, you may apply Van der Waerden theorem as follows: assuming that the gaps are bounded by $T$, color every positive integer $n$ to the color $i\in \{1,\dots,T\}$ if $nT+i$ belongs to your set (so each large enough integer gets at least one color), and find a large monochromatic arithmetic progression. It corresonds to a large progression in the initial set.


