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Jun 10, 2024 at 15:32 comment added Emil Jeřábek @PietroMajer 1001's example has $|P\cap[n+1]^2|=n$, and a simple counting argument shows that any bijection-proof $P$ has $|P\cap[n+1]^2|\ge n/\sqrt2$, lest a random permutation on $[n+1]$ give a counterexample. Thus, the interesting choice is $\alpha=1/2$.
Jun 10, 2024 at 11:17 vote accept Dominic van der Zypen
Jun 10, 2024 at 10:35 comment added Pietro Majer You may consider more generally $$\mu_\alpha(P):=\liminf_{n\to\infty}\frac{\text{card}(P\cap[\mathbb N]^2)}{n^{2\alpha}}$$ for $\alpha>0$
Jun 10, 2024 at 10:19 answer added 1001 timeline score: 6
Jun 10, 2024 at 8:45 comment added Daniel Weber I suspect randomly choosing pairs with any nonzero probability works, so the infimum is 0, but I'm not sure how to prove this
Jun 10, 2024 at 8:07 history asked Dominic van der Zypen CC BY-SA 4.0