Timeline for answer to Finite $k$-set-respecting splitting of $\mathbb{N}$ by bof
Current License: CC BY-SA 4.0
Post Revisions
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2023 at 2:15 | history | edited | bof | CC BY-SA 4.0 |
edited body
|
| Nov 20, 2023 at 22:41 | history | edited | bof | CC BY-SA 4.0 |
edited body
|
| Nov 20, 2023 at 21:00 | history | edited | bof | CC BY-SA 4.0 |
added 10 characters in body
|
| Nov 20, 2023 at 20:55 | comment | added | Dominic van der Zypen | @SamHopkins I think the "cofinite" bit is a little bug that crept in -- but the elegance of this archetypical bof-solution lies in the following bit (I think): if $F$ has $n$ elements and $S_0,\ldots S_n$ are $n+1$ sets partitioning $\mathbb{N}$, then $F$ cannot intersect all of the $S_i$. So suppose it doesn't intersect $S_j$. Then $F\subseteq (\mathbb{N}\setminus S_j)$. (Apologies for this laudatory speech on bof, but I have seen a lot of answers by him, many of them so simple and to the point like a simple and elegant chess move that still most chess players don't find.) | |
| Nov 20, 2023 at 20:49 | vote | accept | Dominic van der Zypen | ||
| Nov 20, 2023 at 20:17 | history | answered | bof | CC BY-SA 4.0 |