So how do you prove with an example that
$$(A\times B)^c \not\subset A^c\times B^c$$
Let's say I define the Universal set to be $\{1, 2, 3, 4\}$, $A = \{2,4\}$, $B = \{3,4\}$, $A^c = \{1,3\}$, and $B^c = \{1,2\}$.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityYou are basically done. Just compute the cartesian products and observe that they are not equal:
\begin{align*} (A\times B)^c &= \{(2,3),(2,4),(4,3),(4,4)\}^C \\ &= \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2)\} \end{align*} and $$ A^c\times B^c=\{(1,1),(1,2),(3,1),(3,2)\}. $$
One thing to note is that it is assumed the universal set to which the complement $(A\times B)^c$ is taken is $\{1,2,3,4\}\times\{1,2,3,4\}$.
Also, it just so happened that $A^c\times B^c\subseteq (A\times B)^c$. Is this true in general? It is! To see this, suppose $(x,y)\in A^c\times B^c$. Then $x\in A^c$ and $y\in B^c$, so we cannot have $(x,y)\in A\times B$. Hence $(x,y)\in (A\times B)^c$.