prove a statement (complements of unions)
I want to prove this statement:
$$(A_1 \cup A_2)^c = {A_1}^c \cup {A_2}^c$$
where the $c$ means the complement.
Any help would be greatly appreciated.
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$\begingroup$You will have trouble proving that $(A_1\cup A_2)^c=A_1^c\cup A_2^c$, since it is not true, in general. (In fact, it holds precisely when $A_1=A_2$.) However, the following are true in general:
- $(A_1\cup A_2)^c=A_1^c\cap A_2^c$
- $(A_1\cap A_2)^c=A_1^c\cup A_2^c$
Edit: In response to your (temporary) correction, let me say that you are on the right track. Since $x\notin A_1,$ then by definition, $x\in\:???$ Since $x\notin A_2,$ then $x\in\:???$ Consequently, what can we say? For the other inclusion, you'll basically be doing the same thing, but in reverse.
$\endgroup$ $\begingroup$This is one of De Morgan's Laws. We want to prove that $(A\cup B)^c=A^c \cap B^c$.
Let $x\in(A\cup B)^c$. Then $x\notin A\cup B$. So $x\notin A$ and $x\notin B$. Therefore, $x\in A^c$ and $x\in B^c$. It follows that $x\in A^c \cap B^c$. Thus $(A\cup B)^c\subseteq A^c \cap B^c$. Now, let $x\in A^c \cap B^c$. Then $x\in A^c$ and $x\in B^c$. So $x\notin A$ and $x\notin B$. Therefore $x\notin A\cup B$. It follows that $x\in (A \cup B)^c$. Thus $A^c \cap B^c \subseteq (A\cap B)^c$.
$\endgroup$ $\begingroup$This is false in the general case. For example, if $A_1$ is the set of integers ($\{\ldots,-2,-1,0,1,2,\ldots\}$) and $A_2$ is the set of positive real numbers, and the universe is the set of all real numbers, then $(A_1 \cup A_2)^c$ doesn't contain $-1$ or $\frac12$, but $A_1^c \cup A_2^c$ contains them.
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