# The cross product of two sets

Just reading about the cross product. It says here: $$A\times B= \{(x,y)|x\in A, y\in B\}$$ Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?

Edit: Quoting the free textbook I'm looking at online,

If we are given that A is a set and no other information about A, then there is no ordering to the elements of A. Thus, we cannot speak of “the second element of the set A” unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.

Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated, but it is not okay to have repeated elements in an ordering of S? When we say S = {a, b, c, a}, we know that S contains just the three elements a, b and c. If we were to talk about the ordered set (a, b, c, a) it would not make sense because it would say that the element a is in two places at once: the first position and the last position.

• Yes, $A\times B$ is a set of ordered pairs. There is no problem if there is some $x\in A\cap B$: $\langle x,x\rangle$ is a perfectly good ordered pair. Commented Nov 4, 2012 at 19:29
• An ordered pair can included duplicates. Why do you think otherwise? Commented Nov 4, 2012 at 19:29
• What book are you looking at? I've never heard this called the cross product. Usually it is called the Cartesian product after Rene Descartes. Commented Nov 4, 2012 at 19:33
• math.ucsd.edu/~ebender/DiscreteText1/SF.pdf says " The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set. Commented Nov 4, 2012 at 19:34
• Bender is being unnecessarily confusing. If he wants to write $(a_1,\dots,a_n)$ for an ordered list of distinct objects, that’s fine, but since this notation is more commonly used for $n$-tuples of objects that are not necessarily distinct, he should make it much clearer that his $(a_1,\dots,a_n)$ is not the same as the more usual $(a_1,\dots,a_n)$. Commented Nov 4, 2012 at 19:40

Yes, $A\times B$ is the (new) set of all ordered pairs $(x,y)$ with $x\in A$ and $y\in B$. There will be no problem if $A$ and $B$ share common elements since an ordered pair is precisely such that it can contain duplicates (an unordered pair, i.e. a set $\{a,b\}$ can not). So, for example, $A\times A$ is just the set of all ordered pairs of elements of $A$. For example, $\{1,2\}\times\{1,2\}=\{(1,1),(1,2),(2,1),(2,2)\}$.

EDIT: Ordered means that the order of the elements matters, that is, $(1,2)\ne(2,1)$ whereas $\{1,2\}=\{2,1\}$.

Yes. $A \times B$ is a set of ordered pairs of elements from $A$ and $B$. If $x \in A$ and $x \in B$, then the ordered pair generated will be $(x,x)$ which is a valid ordered pair and is an element of $A \times B$.

For instance, if $A = \{0,1\}$ and $B = \{1,2\}$, then $A \times B = \{(0,1),(0,2),(1,1),(1,2)\}$. Hope this clarifies.

• This book that I'm reading though, says that {} denotes an unordered set while () denotes an ordered set. And ordered sets cannot include duplicates as it is ordered. Does that apply in this case for the set of ordered pairs? Commented Nov 4, 2012 at 19:29
• Your book seems to have strange terminology. Certainly (1,1) is a fine ordered pair. Commented Nov 4, 2012 at 19:31
• @agent154: In the $n$-tuple $(x_1,x_2,\dots,x_n)$ the order matters, but $x_1,x_2,\dots,x_n$ are not required to be distinct. Commented Nov 4, 2012 at 19:35

look at it this way $A\times B =\cup \{ (x,y)|x\in A,\space y\in B\}$, there can never be a duplicate

No,This is no the real multiplication of sets.The real multiplication of sets is: f(x) multiplicity of x A*B={x+y:f(x)f(y)|x:f(x)∈A,y:f(y)∈B} when f(x)=f(y)=1 AB={x+y|x∈A,y∈B} why?because: P({a,b})={0,{a},{b},{a,b}} P({c,d})={0,{c},{d},{c,d}} P({a,b,c,d})={0,{c},{d},{c,d},{a},{a,c},{a,d},{a,c,d},{b},{b,c},{b,d},{b,c,d},{a,b},{a,b,c},{a,b,d},{a,b,c,d}} P(A+B)=P(A)*P(B) People do not know the real multiplication of sets.