Let {$A_i|i\in I$} be a collection of sets. Show that

$( \bigcap_{i\in I} A_i )^c = \bigcup_{i\in I}(A_i^c)$

here is my attempt using De Morgans Law.


Let B be Universal set

$B \smallsetminus (\bigcap_{i\in I} A_i)$

$= B \bigcap (\bigcap_{i\in I} A_i)^c$

$= (\bigcap_{i\in I} A_i)^c$

Also, $B \smallsetminus (\bigcap_{i\in I} A_i)$

$= \bigcup_{i\in I} B \smallsetminus A_i$

$= \bigcup_{i\in I}(A_i^c)$

Thus, $( \bigcap_{i\in I} A_i )^c = \bigcup_{i\in I}(A_i^c)$

  • $\begingroup$ You seem to be using what you have to prove to prove what you have to prove. The step $B\setminus(\bigcap_{i\in I} A_i) = \bigcup_{i\in I} B\setminus A_i$ needs justification. $\endgroup$ – Graham Kemp Sep 8 '15 at 7:18

In your proof, you don't use De Morgan's law; in fact, the statement you want to proof is called De Morgan's law. However, to prove that the two sets are equal you have to show that \begin{equation} \left(\bigcap_{i \in I} A_i \right)^{c} \subset \bigcup_{i \in I} \big( A_i \big)^c \qquad \text{ and } \qquad\bigcup_{i \in I} \big( A_i \big)^c \subset \left( \bigcap_{i \in I} A_i \right)^c. \end{equation} Let me show you the first inclusion. An element $x$ of the set $\big( \bigcap_{i \in I} A_i \big)^c$ is no element of the set $\bigcap_{i \in I} A_i$. Consequently, there is at least one set $A_i$ such that $x \notin A_i$. That is, $x \in (A_i)^c$. Hence, $x$ is an element of $\bigcup_{i \in I} (A_i)^c$.

The proof of the second inclusion is similar.


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