Timeline for Are there infinite sets $A$ and $B$ such that all numbers in $A+B$ are primes?
Current License: CC BY-SA 4.0
8 events
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| Jan 28, 2023 at 19:27 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
corrected DOI link
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| Jan 28, 2023 at 16:17 | comment | added | Will Sawin | @SamHopkins The relation is described in that preprint by the sentences: "We remark that it was previously shown in [8] that Conjecture 1.2 implied that the primes contained the sumset A + B of two infinite [sets] A, B; Theorem 1.3 provides a new proof of this claim." They could have cited this MO answer instead of the reference [8] (but shouldn't have, as [8] was much earlier.) So the relation is that one of their theorems implies the result of this answer, and another of their theorems is separate. | |
| Jan 28, 2023 at 13:40 | comment | added | Sam Hopkins | How does this relate to the recent preprint of Tao and Ziegler: arxiv.org/abs/2301.10303 ? | |
| Jan 28, 2023 at 13:29 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 194 characters in body
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| Nov 18, 2021 at 16:40 | vote | accept | LMP | ||
| Nov 18, 2021 at 13:03 | comment | added | Will Sawin | i@LMP In addition to what fedja said, even the case $|A|= \infty$, $|B|=n$ is known unconditionally by work of Maynard, as Zach Hunter pointed out. | |
| Nov 18, 2021 at 6:37 | comment | added | fedja | @LMP The answer to your last question is "Yes, and that is elementary". You can show more: if a set $P$ does not contain a "recursive interval" $I(N;n_1,\dots,n_m)=\{N+\sum_k\delta_kn_k:\delta_k\in\{0,1\}\}$, then $|P\cap[1,M]|\le C_mM^{q_m}$ with some $q_m<1$, so the primes are just too many. | |
| Nov 17, 2021 at 18:21 | history | answered | Will Sawin | CC BY-SA 4.0 |