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.

  • 1
    $\begingroup$ 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. $\endgroup$ – epimorphic Jan 9 '16 at 15:16
  • 1
    $\begingroup$ @epimorphic I was finding more concrete example of sets, rather than abstract definition. $\endgroup$ – 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 $(A \cup B)^{c} = A^{c} \cap B^{c}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.