Prove the intersection of a set complement equals the complement of the union of a set.
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.
Proof:
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, $( \bigcap_{i\in I} A_i )^c = \bigcup_{i\in I}(A_i^c)$
$\endgroup$ 11 Answer
$\begingroup$In your proof, you don't use De Morgan's law; in fact, the statement you want to prove 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 not an 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 for the second inclusion is similar.
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