# De Morgan's Law - Proof of $(\cup_{i} A_{i})^c = (\cap_{i} A_{i}^c)$

I would appreciate it if I could get some feedback on my attempt to proof De Morgan's Law. That is, if it is correct or if there is a better way to do this proof. I am self-studying probability theory from Grimmett and Stirzaker's book Probability and Random Processes and am going through the exercises. Thanks in advance.

Let ${A_i : i \in I}$ be a collection of sets.

$(\cup_{i} A_{i})^c = (A_1 \cup A_2 \cup A_3 \cup A_4 \ldots)^c$ $= (A_1 \cup A_2)^c \cup (A_3 \cup A_4)^c \cup \dots$ $= (A_{1}^c \cap A_{2}^c) \cup (A_{3}^c \cap A_{4}^c)\ldots$ $=(A_{1}^c \cap A_{2}^c \cap A_{3}^c \cap A_{4}^c \ldots)$ $=\cap_{i} A_{i}^c$

• How'd you prove that $$\left(A_1\cup A_2\cup B\right)^c=(A_1\cup A_2)^c\cup B^c\;\;??$$ – DonAntonio Feb 2 '14 at 13:31
• It's not correct. Your second equality is invalid. Prove by showing each side is a subset of the other. (In one direction: "Let $x$ be an element of the LHS. Do stuff using the definitions... So $x$ is an element of the RHS") – David Mitra Feb 2 '14 at 13:31

$$x\in\left(\bigcup_i A_i\right)^c\iff x\notin\bigcup_i Ai\iff\;\forall\,i\;,\;x\notin A_i\iff$$
$$\iff \;\forall\,i\,,\;x\in A_i^c\iff x\in\bigcap_i A_i^c$$