Proposition 1

Let $(X_\iota)_{\iota \in I}$ be a family of sets, and let $f$ be a mapping of a set K onto I. Then

and, if $I \neq \emptyset$,

Let $x$ be an element of $\bigcap\limits_{\iota \in I} X_\iota$. There exists an index $\iota \in I$ such that $x \in X_\iota$. Since $f\langle K \rangle = I$, there exists an index $\varkappa \in K$ such that $\iota = f(\varkappa)$, whence $x \in X_{f(\varkappa)}$ and consequently

Conversely, if $x \in \bigcup\limits_{\varkappa \in K} X_{f(\varkappa)}$, there exists $\varkappa \in K$ such that $x \in X_{f(\varkappa)}$, and therefore, since $f(\varkappa) \in I$, $x \in \bigcup\limits_{\iota \in I} X_\iota$. Hence

¶ Now suppose that $I \neq \emptyset$, and let $x$ be an element of $\bigcap\limits_{\iota \in I} X_\iota$. For each element $\varkappa$ of K we have $f(\varkappa) \in I$, hence $x \in X_{f(\varkappa)}$, and therefore

Conversely, let $x$ be an element of $\bigcap\limits_{\varkappa \in K} X_{f(\varkappa)}$. If $\iota$ is any element of I, there exists an element $\varkappa$ of K such that $\iota = f(\varkappa)$, whence $x \in X_\iota$ and consequently $x \in \bigcap_{\iota \in I} X_\iota$. Hence

For families of subsets of a given set, it is clear that the second part of Proposition 1 remains valid without the restriction $I \neq \emptyset$.