A *clutter* is a pair $C=(V,E)$ where $V\neq\emptyset$ is a set, and $E\subseteq {\cal P}(V)$ such that no member of $E$ is included in another member of $E$. A *matching* in $C$ is a collection of pairwise disjoint members of $E$. Zorn's Lemma implies that every matching is contained in a maximal matching (with respect to $\subseteq$).


Is it possible to find a clutter $C=(V,E)$ with $E\neq\emptyset$ and for every matching $M\subseteq E$ there is a matching $M'\subseteq E$ with $|M|<|M'|$?



Clutters with no maximum-size matchings
Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A\_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A\_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A\_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has arbitrarily large finite matchings. On the other hand, no matching of $\mathcal C$ is infinite. To see this, it suffices to prove that if $A\_{c\_1,p\_1}$ and $A\_{c\_2,p\_2}$ are disjoint, then $p\_1=p\_2$. Suppose $p\_1 \neq p\_2$. By shifting both sequences by $c\_1$, we may assume that $c\_1=0$. Choosing $k \equiv -c\_2 p\_2^{-1} \pmod{p\_1}$, we have $c\_2+kp\_2 \equiv 0 \pmod{p\_1}$, and so $A\_{c\_1,p\_1} \cap A\_{c\_2,p\_2}$ is non-empty.


