Theorem. Let $X$ be a set, let $I$ be a non-empty set, and for all $\alpha \in I$ let $A_{\alpha}$ be a subset of $X$. Show that $$X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} = \bigcap\limits_{\alpha \in I} \left( X \setminus A_{\alpha} \right)$$ and $$X \setminus \bigcap\limits_{\alpha \in I}A_{\alpha} = \bigcup\limits_{\alpha \in I} \left( X \setminus A_{\alpha} \right).$$
Proof.

(i) $$y \in \left( X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \leftrightarrow y \in X \wedge y \notin \bigcup\limits_{\alpha \in I} A_{\alpha} \leftrightarrow y \in X \wedge \neg((\exists \alpha \in I)\: y \in A_{\alpha}) \leftrightarrow y \in X \wedge ((\forall \alpha \in I)\: y \notin A_{\alpha}) \leftrightarrow (\forall \alpha \in I)(y \in X \wedge y \notin A_{\alpha}) \leftrightarrow (\forall \alpha \in I)(y \in X \setminus A_{\alpha}) \leftrightarrow y \in \bigcap\limits_{\alpha \in I}(X \setminus A_{\alpha}).$$ Two sets are equal iff every element of one set is element of the other set and vice versa: $$(\forall y) \left( y \in \left( X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \leftrightarrow y \in \bigcup\limits_{\alpha \in I}(X \setminus A_{\alpha}) \right) \leftrightarrow \left( X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} = \bigcap\limits_{\alpha \in I} \left( X \setminus A_{\alpha} \right) \right).$$ (ii) $$y \in \left( X \setminus \bigcap\limits_{\alpha \in I} A_{\alpha} \right) \leftrightarrow y \in X \wedge y \notin \bigcap\limits_{\alpha \in I} A_{\alpha} \leftrightarrow y \in X \wedge \neg((\forall \alpha \in I)\: y \in A_{\alpha}) \leftrightarrow y \in X \wedge ((\exists \alpha \in I)\: y \notin A_{\alpha}) \leftrightarrow (\exists \alpha \in I)(y \in X \wedge y \notin A_{\alpha}) \leftrightarrow (\exists \alpha \in I)(y \in X \setminus A_{\alpha}) \leftrightarrow y \in \bigcup\limits_{\alpha \in I}(X \setminus A_{\alpha}).$$ Two sets are equal iff every element of one set is element of the other set and vice versa: $$(\forall y) \left( y \in \left( X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} \right) \leftrightarrow y \in \bigcup\limits_{\alpha \in I}(X \setminus A_{\alpha}) \right) \leftrightarrow \left( X \setminus \bigcup\limits_{\alpha \in I} A_{\alpha} = \bigcap\limits_{\alpha \in I} \left( X \setminus A_{\alpha} \right) \right).$$
2015 FEB 02 (v1.0)
Contact me: m.herrmann followed by an -at- followed by blaetterundsterne.org.

This document created with the help of MathJax (Thank you guys!)