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Let $A=\{1,2,3\}$ and $B=\{a, b, \{\emptyset\}\}$. Does $A \times B = \{\emptyset\}$ or is it something more like $A \times B = \{ (1, a), (1, b), \{\emptyset\}, (2, a), (2, b),\{\emptyset\}, (3, a), (3, b), \{\emptyset\} \}$?

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    The set containing the empty set and the empty set are different sets. $\{ \emptyset \} \not = \emptyset $ – 2017-01-30
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    @ΘΣΦ GenSan I think your edit removed the confusion that the asker was having and changed the meaning of the question. – 2017-01-30
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    @Q the Platypus Yeah, my bad. I returned it to the original. – 2017-01-30
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    White of those two sets do you think includes $\emptyset$ as an element? Certainly $A$ doesn't have $\emptyset$ as an element, assuming that natural numbers are realized as sets in the usual way. But $\emptyset$ might be an element of $B,$ because for all we know maybe $a=\emptyset$ or $b=\emptyset.$ – 2017-01-30

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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.

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    Does it make a difference if the question is written as B = { Ø, 1, 2, 3} instead of B = { {Ø}, 1, 2, 3} ? – 2017-01-31
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    Yes, it does. You would then get $$A\times B = \left\{(1,a),(1,b),(1,\emptyset),(2,a),(2,b),(2,\emptyset),(3,a),(3,b),(3,\emptyset)\right\}$$ You have to convince yourself that $\emptyset$ or $\{\emptyset\}$ are not special in this context; they are elements just like anything else. – 2017-01-31
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$$(A \times B):= \{({\color{Red}y},{\color{green}z})| {\color{Red}y} \in A \wedge {\color{green}z} \in B\}$$

By definition we have also following:

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