# Cartesian product of two sets where each set contains the empty element?

What is the Cartesian product of these two sets: $$A = \{\emptyset, 2\}\\ B = \{\emptyset, 3\}$$ I am guessing it is $$\{\emptyset, 2, 3, \{\emptyset, \emptyset\}, \{\emptyset, 3\}, \{2, \emptyset\}, \{2, 3\}\}$$

but the cardinality of a Cartesian product is always $$2^n$$, but there are 7 elements, is this because there exists two empty sets, one from each set A and B but they are the same element, hence the removal of the redundant one?

Edit: What I did to get the sets A and B was to take the power set of $$A = {2} \\ B = {3} \\ P(A) = \{\emptyset, 2\} \\ P(B) = \{\emptyset, 3\} \\ P(A) \times P(B) = \{\emptyset, 2\} \times\{\emptyset, 3\}$$

• Where does $\emptyset$ come from: is that an ordered pair? Sep 16, 2020 at 6:37
• $\emptyset$, $2$, and $3$ are not elements of $A\times B$. Sep 16, 2020 at 6:41
• Empty set, $2$ and $3$ do not belong to the cartesian product .There are only four elements. Sep 16, 2020 at 6:42
• The elements of a Cartesian product is always ordered pairs. None of the elements in your set is an ordered pair. Sep 16, 2020 at 6:42
• Look at my edit, I added why I have got the empty set in both sets. "If S is a finite set with $|S| = n$ elements, then the number of subsets of S is $|P(S)| = 2^n$" quote from wikipedia about power sets Sep 16, 2020 at 6:59

For a set $$A$$ and a set $$B$$, every element of $$A\times B$$ must be of the form $$(a,b)$$ where $$a$$ is an element of $$A$$ and $$b$$ is an element of $$B$$.

In the case of $$\emptyset, 2, 3$$, this is not true. It is not true that $$\emptyset=(a,b)$$ for any pair of values $$a\in A, b\in B$$.

This is also not true for other elements of your solution. $$\{\emptyset,\emptyset\}$$ is not an element of $$A\times B$$, because it is not an ordered pair of two elements. In fact, $$\{\emptyset,\emptyset\}=\{\emptyset\}.$$

Also:

but the cardinality of a Cartesian product is always $$2^n$$

Not true. The cardinality of a Cartesian product of two finite sets $$A$$ and $$B$$ is $$|A|\cdot |B|$$.

• What is the cartesian product of two tuples where they have a common element then? e.g: $A=\{1, 2\},B=\{1, 3\}$ Sep 16, 2020 at 7:06
• @linker $A$ and $B$ are not tuples. They are sets. And the cartesian product is still the same, i.e. the set of all ordered pairs $(a,b)$ such that $a\in A$ and $b\in B$. This definition says nothing about whether $a$ and $b$ can or can't be the same.
– 5xum
Sep 16, 2020 at 7:08
• So $A\times B=\{\{1, 1\}, \{1, 3\}, \{2, 1\}, \{2, 3\}\} = \{\{1\}, \{1, 3\}, \{2, 1\}, \{2, 3\}\}$? Sep 16, 2020 at 7:10
• @linker No. $\{1,1\}$ is not an ordered pair. $(1,1)$ is an ordered pair. Check your textbook or notes for the definition of "ordered pair". Note that there is a difference between $(a,a)$ and $\{a,a\}$.
– 5xum
Sep 16, 2020 at 7:10
• Oh ok so $A \times B = \{(1, 1), (1, 3), (2, 1), (2, 3)}$ but what about $A = \{\emptyset, 1\}, B = \{\emptyset, 3\}$ Sep 16, 2020 at 7:20