Corollary 1

Suppose that $L \neq \emptyset$ and that $J_\lambda \neq \emptyset$ for each $\lambda \in L$. If for e achindex $\lambda \in L$ the family $(X_{\lambda, \iota})_{\iota \in J_\lambda}$ is a partition of $X_\lambda = \bigcup_{\iota \in J_\lambda} X_{\lambda, \iota}$, then the family $\left( \prod_{\lambda \in L} X_{\lambda, f(\lambda)} \right)_{f \in I}$ is a partition of $\prod_{\lambda \in L} X_\lambda$.

If we set

then, by virtue of the first formula of Proposition 9, it is sufficient to show that $\mathrm{P}_f \neq \emptyset$ for all $f \in I$ and that $\mathrm{P}_f \cap \mathrm{P}_g = \emptyset$ whenever $f$ and $g$ are distinct elements of I. The first point follows from Proposition 5, Corollary 2. As to the second, if $f \neq g$, there exists $\lambda \in L$ such that

and therefore, by virtue of the hypothesis, $X_{\lambda, f(\lambda)} \cap X_{\lambda, g(\lambda)} = \emptyset$. It follows that there is no graph belonging to $\mathrm{P}_f \cap \mathrm{P}_g$; for if G were such a graph, we would have $\mathrm{G}(\lambda) \in X_{\lambda, f(\lambda)} \cap X_{\lambda, g(\lambda)} = \emptyset$, which is absurd.