Proposition 5

For every family $(X_\iota)_{\iota \in I}$ of subsets of a set E, we have

Let $x \in \complement_E\left(\bigcup_{\iota \in I} X_\iota\right)$. Then $x \in E$ and, for every $\iota \in I$, $x \notin X_\iota$, so that $x \in \complement_E X_\iota$; consequently

Conversely, let $x$ be an element of $\bigcap_{\iota \in I} (\complement_E X_\iota)$; by the definition of intersection we have $x \in E$. Furthermore, if we had $x \in \bigcup_{\iota \in I} X_\iota$, there would exist an index $\varkappa \in I$ such that $x \in X_\varkappa$, which contradicts the hypothesi $x \in \bigcap_{\iota \in I} (\complement_E X_\iota)$; hence

This proves the first formula; the second one is an immediate consequence, in view of the relation $\complement_E(\complement_E X) = X$ for every subset X of E.