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Mar 2, 2023 at 19:30 vote accept Marcel K. Goh
Mar 1, 2023 at 18:32 answer added Terry Tao timeline score: 4
Feb 28, 2023 at 23:27 history edited Marcel K. Goh CC BY-SA 4.0
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Feb 27, 2023 at 22:38 comment added Marcel K. Goh @TerryTao thanks for the reference, this was exactly what I was looking for! I followed your sketch and everything seems to work well, i.e., intersectivity of $B$ is a sufficient condition for the property I want
Feb 27, 2023 at 20:57 comment added Terry Tao The literature on intersective sets will most likely be relevant home.olemiss.edu/~leth/papers/problems_on_intersective_sets.pdf . Roughly speaking I would expect $(A + B) \cap A$ to have the same density as $A$ when $B$ is intersective, since otherwise $A \backslash (A+B)$ would be a counterexample to the intersectivity of $B$ (or maybe $-B$).
Feb 27, 2023 at 18:03 comment added Marcel K. Goh @mathworker21 Thanks for this idea, i'll think about it and update the post if I figure anything out
Feb 27, 2023 at 17:57 comment added mathworker21 can't you greedily choose $A$ so that $A \cap B = \emptyset$? What happens then? E.g. for $B$ being factorials, if you include $1$ in $A$, then you can't include $2,3,7,25,121,\dots$, so you then include $4 \in A$, etc.. My guess is that the resulting $A$ will have positive density (and maybe this is trivial to prove via a counting argument, but I'm too lazy rn),
Feb 27, 2023 at 17:54 comment added Marcel K. Goh @mathworker21 You're right! I guess I should say nonempty $B$ :))
Feb 27, 2023 at 17:54 history edited Marcel K. Goh CC BY-SA 4.0
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Feb 27, 2023 at 17:53 comment added mathworker21 $B = \emptyset$
Feb 27, 2023 at 17:44 history asked Marcel K. Goh CC BY-SA 4.0