 CAC can be derived from ZL in a way echoing the “combinatorial” justification of CAC sketched above. Accordingly suppose we are given a family \(\sH\) of mutually disjoint nonempty sets; call a subset \(S \subseteq \bigcup \sH\) a sampling for \(\sH\) if, for any \(X\in \sH\), either  \(X \subseteq S\) or \(S \cap X\) is nonempty and finite. Minimal samplings are precisely transversals for  \(\sH\);[11]  and the collection \(\sT\) of samplings is clearly nonempty since it contains \(\bigcup \sH\).  So if it can be shown that \(\sT\) is strongly   reductive,[12]   Zorn’s lemma will yield a minimal element of \(\sT\) and so a transversal for \(\sH\).  The strong reductiveness of \(\sT\) may be seen as follows: suppose that \(\{S_{i}: i\in I\}\) is a nest of samplings; let \(S = {\bigcap}_{i\in I} S_{i}\).  We need to show that \(S\) is itself a sampling; to this end let \(X \in \sH\) and suppose \(\neg(X \subseteq S)\). Then there is \(i \in I\) for which  \(\neg(X \subseteq S_{i})\); since \(S_{i}\) is a sampling, \(S_{i}\cap X\) is finite and nonempty, say \(S_{i}\cap X = \{x_{1}, \ldots, x_{n}\}\). Clearly  \(S\cap X\) is then finite; suppose for the sake of contradiction that \(S\cap X = \varnothing\). Then for each \(k = 1,\ldots,n\) there is  \(i_k \in I\) for which \(\neg(x_k \in S_{i_k})\). It follows that \(\neg(S_{i} \subseteq S_{i_k})\), for \(k = 1, \ldots, n\).   So, since the \(S_{i}\) form a chain, each \(S_{i_k}\) is a subset of \(S_{i}\). Let \(S_{j}\) be the least of \(S_{i_1},\ldots, S_{i_k}\); then \(S_{j}\subseteq S_{i}\). But since \(\neg(x_{k} \in S_{j})\), for \(k = 1, \ldots, n\), it now follows that \(S_{j} \cap X = \varnothing\), contradicting the fact that \(S_{j}\) is a sampling. Therefore \(S \cap X\ne \varnothing\); and \(S\) is a sampling as claimed.