TITLE: Infinite DeMorgan laws
QUESTION [7 upvotes]: Let $X$ be a set and $\{Y_\alpha\}$ is infinite system of some subsets of $X$.
Is it true that:
$$\bigcup_\alpha(X\setminus Y_\alpha)=X\setminus\bigcap_\alpha Y_\alpha,$$
$$\bigcap_\alpha(X\setminus Y_\alpha)=X\setminus\bigcup_\alpha Y_\alpha.$$
(infinite DeMorgan laws)
Thanks a lot!

REPLY [2 votes]: (i) The following statements are equivalent:

$ y \in X \setminus \bigcap\limits_{\alpha \in I} A_{\alpha} $
$ y \in X \wedge y \notin \bigcap\limits_{\alpha \in I} A_{\alpha} $
$ y \in X \wedge \neg((\forall \alpha \in I)\: y \in A_{\alpha}) $
$ y \in X \wedge (\exists \alpha \in I)\: y \notin A_{\alpha} $
$ (\exists \alpha \in I)(y \in X \wedge y \notin A_{\alpha}) $
$ (\exists \alpha \in I)(y \in X \setminus A_{\alpha}) $
$ y \in \bigcup\limits_{\alpha \in I}(X \setminus A_{\alpha}) $

(ii) The following statements are equivalent:

$ y \in X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} $
$ y \in X \wedge y \notin \bigcup\limits_{\alpha \in I} A_{\alpha} $
$ y \in X \wedge \neg((\exists \alpha \in I)\: y \in A_{\alpha}) $
$ y \in X \wedge (\forall \alpha \in I)\: y \notin A_{\alpha} $
$ (\forall \alpha \in I)(y \in X \wedge y \notin A_{\alpha}) $
$ (\forall \alpha \in I)(y \in X \setminus A_{\alpha}) $
$ y \in \bigcap\limits_{\alpha \in I}(X \setminus A_{\alpha}) $

Since the first and the last statements are equivalent for all $y$, we have
$$ \underbrace{X \setminus \bigcap\limits_{\alpha \in I} A_{\alpha} = \bigcup\limits_{\alpha \in I}(X \setminus A_{\alpha})}_{(i)} \wedge \underbrace{X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} = \bigcap\limits_{\alpha \in I}(X \setminus A_{\alpha})}_{(ii)}. $$ $\Box$