Motivation. My sons participated in a large football tournament recently; everyone wanted to be in a team with everyone else at least once. Tricky!

Formulation of the question. For any positive integer $k\in\mathbb{N}$, let $\ \binom {\mathbb N} k\ $ denote the collection of $k$-element subsets of $\mathbb{N}.\ $ Let ${\cal I}$ be the collection of subsets of $\mathbb{N}$ that are both infinite and co-infinite (the complement is infinite).

Fix an integer $k\geq 2$. Is there a finite family $\ {\cal S}\subseteq {\cal I}\ $ such that for every $\ F\in\binom {\mathbb N} k\ $ there is $T\in{\cal S}\ $ with $F\subseteq T$ or $F\subseteq (\mathbb{N}\setminus T)$?   If yes, what is the smallest size such a family ${\cal S}$ can have, in terms of $k$?

Note. I originally posted this question for $k=2$ and arbitrary subsets of $\mathbb{N}$, not just infinite/co-infinite ones. Zach Teitler quickly gave an elegant solution and suggested this generalization. First it seemed $k=2$ was solved, but it turns out $k=2$ is still unresolved.

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    $\begingroup$ I think $S=\{\{1\},\{2\},\{3\}\}$ works. $\endgroup$ Nov 20 at 13:28
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    $\begingroup$ Right @ZachTeitler - sorry. Can it be done with sets that are infinite and co-infinite? Will change the question accordingly $\endgroup$ Nov 20 at 13:30
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    $\begingroup$ Size $|S|=2$ is clearly impossible. How about $|S|=3$ like this: let each $S_i$, $i=1,2,3$, be the set of integers congruent to $0$ or $i$ modulo $4$. $\endgroup$ Nov 20 at 13:36
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    $\begingroup$ This is a fun question. How far can we generalize, eg, to $k$-team games? $\endgroup$ Nov 20 at 13:40
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    $\begingroup$ @WlodAA: see meta.mathoverflow.net/questions/5508/… for a fix. $\endgroup$ Nov 20 at 20:51

1 Answer 1


For $k\ge2$, if we partition $\mathbb N$ into $k+1$ infinite sets $S_0,S_1,\dots,S_k$, then each $S_i$ is coinfinite, and every $k$-element subset of $\mathbb N$ is contained in the complement of some $S_i$.

This is best possible: for $k\ge2$, given $k$ nonempty proper subsets of $\mathbb N$, there is a $k$-element set which meets all of those sets and their complements. The proof is by induction.

For $k=2$, suppose $S_1$ and $S_2$ are nonempty proper subsets of $\mathbb N$. If neither is contained in the other, choose $x_1\in S_1\setminus S_2$ and $x_2\in S_2\setminus S_1$; if one is contained in the other, say $S_1\subseteq S_2$, choose $x_1\in S_1$ and $x_2\in\mathbb N\setminus S_2$. In either case the set $\{x_1,x_2\}$ meets each of the sets $S_1,S_2,\mathbb N\setminus S_1,\mathbb N\setminus S_2$.

For the induction step, let $S_1,\dots,S_k,S_{k+1}$ be nonempty proper subsets of $\mathbb N$, and let $F=\{x_1,\dots,x_k\}$ be a $k$-element set which meets $S_i$ and $\mathbb N\setminus S_i$ for $1\le i\le k$. Choose $x_{k+1}\in\mathbb N$ so that $x_{k+1}\in S_{k+1}\iff x_1\notin S_{k+1}$. Then the set $F\cup\{x_{k+1}\}$ has at most $k+1$ elements and meets $S_i$ and $\mathbb N\setminus S_i$ for $1\le i\le k+1$.

  • $\begingroup$ @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.) $\endgroup$ Nov 20 at 20:55

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