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$\begingroup$ @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.) $\endgroup$

Dominic van der Zypen
– Dominic van der Zypen

2023-11-20 20:55:32 +00:00
Commented Nov 20, 2023 at 20:55

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