Proposition 8

Let $((X_{\lambda,\iota})_{\iota \in J_\lambda})_{\lambda \in L}$ be a family (with index set L*) of families of sets. Suppose that* $L \neq \emptyset$ and that $J_\lambda \neq \emptyset$ for each $\lambda \in L$. Let

Then we have

("distributivity" of union over intersection, and of intersection over union).

Let $x$ be an element of $\bigcup_{\lambda \in L} \left( \bigcap_{\iota \in J_\lambda} X_{\lambda, \iota} \right)$ and let $f$ be any element of I. There exists an index $\lambda$ such that $x \in \bigcap_{\iota \in J_\lambda} X_{\lambda, \iota}$; consequently $x \in X_{\lambda, f(\lambda)}$ and hence

Since this is true for each $f \in I$, we have

Conversely, let $x$ be an object which does not belong to the set

Then for each $\lambda \in L$ we have $x \notin \bigcap_{\iota \in J_\lambda} X_{\lambda, \iota}$, which means that the set $J'_\lambda$ of indices $\iota \in J_\lambda$ such that $x \notin X_{\lambda, \iota}$ is not empty for each $\lambda \in L$. By Proposition 5, Corollary 2, there exists a functional graph $f$ with domain L such that for each $\lambda \in L$ we have $f(\lambda) \in J'_\lambda$. Consequently $f \in I$ and $x \notin X_{\lambda, f(\lambda)}$ for each $\lambda \in L$; hence

and thus

This completes the proof of the first formula. The second formula follows by applying the first to the family $((\complement_A X_{\lambda, \iota})_{\iota \in J_\lambda})_{\lambda \in L}$, where A denotes the union $\bigcap_{\lambda \in L} \left( \bigcap_{\iota \in J_\lambda} X_\lambda \right)$.