Proposition 2

Let $(X_\iota)_{\iota \in I}$ be a family of sets whose index set I is the union of a family $(J_\lambda)_{\lambda \in L}$ of sets. Then

and (if $L \neq \emptyset$ and $J_\lambda \neq \emptyset$ for each $\lambda \in L$)

("associativity" of union and intersection).

Let $x$ be an element of $\bigcup_{\iota \in I} X_\iota$. There exists an index $\iota \in I$ such that $x \in X_\iota$. Since I is the union of the family $(J_\lambda)_{\lambda \in L}$, there exists an index $\lambda \in L$ such that $\iota \in J_\lambda$, whence $x \in \bigcup_{\iota \in J_\lambda} X_\iota$, and consequently

Conversely, let $x$ be an element of this set. There exists an index $\lambda \in L$ such that $x \in \bigcup_{\iota \in J_\lambda} X_\iota$, hence there exists an index $\iota \in J_\lambda$ (and therefore $\iota \in I$) such that $x \in X_\iota$; it follows that

Now suppose that $L \neq \emptyset$ and $J_\lambda \neq \emptyset$ for each $\lambda \in L$. Then $I \neq \emptyset$. Let $x$ be an element of $\bigcap_{\iota \in I} X_\iota$. If $\lambda \in L$, we have $x \in X_\iota$ for each $\iota \in J_\lambda$ (since $J_\lambda \subset I$), whence $x \in \bigcap_{\iota \in J_\lambda} X_\iota$. Since the last inclusion is true for all $\lambda \in L$, $x$ belongs to $\bigcap_{\lambda \in L} \left( \bigcap_{\iota \in J_\lambda} X_\iota \right)$. Conversely, let $x$ be an element of this latter set, and let $\iota$ be any element of I. There exists $\lambda \in L$ such that $\iota \in J_\lambda$; since $x \in \bigcap_{\iota \in J_\lambda} X_\iota$, we have $x \in X_\iota$, which is true for all $\iota \in I$. Hence $x \in \bigcap_{\iota \in I} X_\iota$. This completes the proof.

For families of subsets of a given set the second part of Proposition 2 remains valid without restriction on L and $J_\lambda$.