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.
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