What is the Cartesian Product of two nonempty sets, one of which includes Ø as an element? Let $A=\{1,2,3\}$ and $B=\{a, b, \{\emptyset\}\}$. Does $A \times B = \{\emptyset\}$ or is it something more like $A \times B = \{ (1, a), (1, b), \{\emptyset\}, (2, a), (2, b),\{\emptyset\}, (3, a), (3, b), \{\emptyset\} \}$?
 A: Neither are correct, but the second is closer. It should be
$$A \times B = \left\{ (1, a), (1, b), (1, \{\emptyset\}), (2, a), (2, b), (2,\{\emptyset\}), (3, a), (3, b), (3,\{\emptyset\}) \right\}.$$
After all, $\{\emptyset\}$ is an element just like any another.
Also, $\{\emptyset\}$ is not the empty set. It is a set that contains only the empty set. Make sure you are familiar with the distinction.
A: 
$$(A \times B):= \{({\color{Red}y},{\color{green}z})| {\color{Red}y}
 \in A \wedge {\color{green}z} \in B\}$$

By definition we have also following: 
$$\begin{array}{}
\begin{array}{c|c|c}
(A \times B)&{\color{green}a} \in B&  {\color{green}b} \in B& {\color{green}\{{\color{green}\emptyset{\color{green}\}}}} \in B\\ \hline
{\color{Red}1} \in A&({\color{Red}1},{\color{green}a})&({\color{Red}1},{\color{green}b})&({\color{Red}1},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} )\\
{\color{Red}2} \in A&({\color{Red}2},{\color{green}a})&({\color{Red}2},{\color{green}b})&({\color{Red}2},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} )\\
{\color{Red}3} \in A&({\color{Red}3},{\color{green}a})&({\color{Red}3},{\color{green}b})&({\color{Red}3},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} )\\
\end{array}&&&
\end{array}$$
 $$(A \times B)=\{({\color{Red}1},{\color{green}a}),({\color{Red}1},{\color{green}b}),({\color{Red}1},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} ),({\color{Red}2},{\color{green}a}),({\color{Red}2},{\color{green}b}),({\color{Red}2},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} ),({\color{Red}3},{\color{green}a}),({\color{Red}3},{\color{green}b}),({\color{Red}3},{\color{green}\{{\color{green}\emptyset{\color{green}\}}}} )\}$$
