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Jul 14, 2019 at 22:19 comment added Gabe Conant It seems that the classification is: a subset of $\mathbb{N}$ is a rocket set if and only if it is either empty, or infinite and co-infinite.
Jul 14, 2019 at 20:00 comment added Greg Martin @LSpice Aha, well spotted! And indeed (on further thought) no final segment $\{n,n+1,n+2,\dots\}$ with $n>1$ can be the rocket set of a bijection, because the preimage of $n$ would also be in the rocket set.
Jul 14, 2019 at 19:48 comment added LSpice @GregMartin, I think that this doesn't work if $B$ is a final segment.
Jul 14, 2019 at 19:23 comment added Greg Martin Just to tie off the loose end, here's a modification of the construction: instead of listing $A$ in increasing order, choose any way of writing $A=\{a_1,a_2,\dots\}$ as a sequence, subject only to the restriction $a_1>b_1$. Then the bijection given in the construction definitely has Roc$(f)=B$.
Jul 14, 2019 at 18:36 comment added Martin Sleziak @GregMartin Thanks for the comment. I missed the fact that $a_1$ might be in $\operatorname{Roc}(f)$.
Jul 14, 2019 at 18:35 history edited Martin Sleziak CC BY-SA 4.0
correction pointed out by Greg Martin
Jul 14, 2019 at 18:27 comment added Greg Martin Indeed, this argument shows that any infinite subset of $\Bbb N$ is Roc$(f)$ for some bijection $f$ (and of course, it's impossible for a nonempty finite subset to be Roc$(f)$ for any function $f$). Although we need to be a little careful, because $a_1$ might be in Roc$(f)$ with this construction.
Jul 14, 2019 at 17:00 comment added Pietro Majer The elevators in the Hilbert Hotel
Jul 14, 2019 at 16:18 comment added Dominic van der Zypen Very nice construction, and nicely explained, thanks Martin!
Jul 14, 2019 at 16:17 vote accept Dominic van der Zypen
Jul 14, 2019 at 16:05 history edited Martin Sleziak CC BY-SA 4.0
deleted 3 characters in body
Jul 14, 2019 at 15:45 history answered Martin Sleziak CC BY-SA 4.0