# Example of $(\bigcup\limits_{r\in\Gamma}A_r)^c=\bigcap\limits_{r\in\Gamma}A_r^c$?

I learned the generalized De morgan's Theorem about family of sets, but I can't find any examples although I know how they are proved. Can you give me an example of sets that belong to $(\bigcup\limits_{r\in\Gamma}A_r)^c=\bigcap\limits_{r\in\Gamma}A_r^c$?

"Theorem 8 The Generalized De Morgan's Theorem

Let {$A_r$|$r\in\Gamma$} be an arbitrary family of sets. Then

(a) $(\bigcup\limits_{r\in\Gamma}A_r)^c=\bigcap\limits_{r\in\Gamma}A_r^c$ (b) $(\bigcap\limits_{r\in\Gamma}A_r)^c=\bigcup\limits_{r\in\Gamma}A_r^c$"

" Theorem 6 De Morgan's Theorem

For any two sets A and B,

(a) $(AUB)^c$= $A^c \bigcap B^c$ (b) $(A∩B)^c$ = $A^c \bigcup B^c$ " Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin.

[EDIT] I'm finding an example of Theorem 8, not of Theorem 6.

• What do you mean you can't find any examples? By the theorem, every indexed collection $\{ A_r \mid r \in \Gamma \}$ of sets (to be more precise, subsets of some "universal set") will work. As long as you understand the concepts in that last sentence, you really can't go wrong. – epimorphic Jan 9 '16 at 15:16
• @epimorphic I was finding more concrete example of sets, rather than abstract definition. – buzzee Jan 9 '16 at 15:25

Let $X = \{1,2,3,4,5,6\}$.
Now, consider the subsets of $X$: $A = \{1,4,5\}$ and $B = \{2,3,5\}$.
Clearly, $A \cup B = \{1, 2, 3, 4, 5\}$. That means $(A \cup B)^{c} = \{6\}$.
But $A^{c} = \{2,3,6\}$ and $B^{c} = \{1,4,6\}$. So $A^{c} \cap B^{c} = \{6 \}$.
That shows you at example, at least for the case of two sets, where $(A \cup B)^{c} = A^{c} \cap B^{c}$.