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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 $$ A = {2} \\ B = {3} \\ P(A) = \{\emptyset, 2\} \\ P(B) = \{\emptyset, 3\} \\ P(A) \times P(B) = \{\emptyset, 2\} \times\{\emptyset, 3\} $$

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  • $\begingroup$ Where does $\emptyset$ come from: is that an ordered pair? $\endgroup$ – Angina Seng Sep 16 at 6:37
  • $\begingroup$ $\emptyset$, $2$, and $3$ are not elements of $A\times B$. $\endgroup$ – Chrystomath Sep 16 at 6:41
  • $\begingroup$ Empty set, $2$ and $3$ do not belong to the cartesian product .There are only four elements. $\endgroup$ – Kavi Rama Murthy Sep 16 at 6:42
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    $\begingroup$ The elements of a Cartesian product is always ordered pairs. None of the elements in your set is an ordered pair. $\endgroup$ – Arthur Sep 16 at 6:42
  • $\begingroup$ 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 $\endgroup$ – linker Sep 16 at 6:59
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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 $|A|\cdot |B|$.

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  • $\begingroup$ What is the cartesian product of two tuples where they have a common element then? e.g: $A=\{1, 2\},B=\{1, 3\}$ $\endgroup$ – linker Sep 16 at 7:06
  • $\begingroup$ @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. $\endgroup$ – 5xum Sep 16 at 7:08
  • $\begingroup$ So $A\times B=\{\{1, 1\}, \{1, 3\}, \{2, 1\}, \{2, 3\}\} = \{\{1\}, \{1, 3\}, \{2, 1\}, \{2, 3\}\}$? $\endgroup$ – linker Sep 16 at 7:10
  • $\begingroup$ @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 $\{a,a\}$. $\endgroup$ – 5xum Sep 16 at 7:10
  • $\begingroup$ Oh ok so $A \times B = \{(1, 1), (1, 3), (2, 1), (2, 3)}$ but what about $A = \{\emptyset, 1\}, B = \{\emptyset, 3\}$ $\endgroup$ – linker Sep 16 at 7:20

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