Cartesian Products
We can notationally relate pairs of elements from different sets simply by grouping them together in parentheses, separated by a comma.
Occasionally, some mathematicians will use the notation \(A\times B\) to denote Cartesian products.
One can also define Cartesian products of multiple sets in much the same way. For example, if I had sets \(A_1,A_2,A_3,A_4,A_5\), I could define the Cartesian product \(A_1\oplus A_2 \oplus A_3 \oplus A_4 \oplus A_5\) to be the set of all ordered 5-tuples \((a_1,a_2,a_3,a_4,a_5)\) where \(a_1\in A_1, a_2\in A_2, a_3\in A_3, a_4\in A_4, a_5\in A_5\).
Try finding a few Cartesian products in the examples down below!
Try the following problems!
\(A\oplus B= \{(2,0), (2,2), (2, 3),(4,0),(4,2),(4,3),(5,0),(5,2),(5,3)\}\)
\(B\oplus A =\{(0,2),(0,4),(0,5),(2,2),(2,4),(2,5),(3,2),(3,4),(3,5)\}\)
These sets are not the same. Each ordered pair is a distinct element (or ‘object’, if you like), so we need to compare pairs to pairs. In \(A \oplus B\), we have the element \((2,0)\) but this pair cannot be found in \(B\oplus A\). Thus, the sets aren’t the same.
\(A\oplus B=\{(0,1),(0,3),(2,1),(2,3)\}\)
\(A\oplus B\oplus C=\{(0,1,4),(0,1,5),(0,3,4),(0,3,5),(2,1,4),(2,1,5),(2,3,4),(2,3,5)\}\)
For \(A\oplus B\), we are looking for ALL pairs \((a,b)\) where \(a\in A\); i.e. is either 0 or 2, and \(b\in B\); i.e. \(b\) is either 1 or 3. For \(A\oplus B \oplus C\), we are looking for all possible triples \((a,b,c)\) where \(a\in A\), i.e. the first entry is 0 or 2, \(b\in B\), i.e. the second entry is either 1 or 3, and \(c\in C\); i.e. the third entry is either a 4 or 5.
\(A^2=\{(0,0),(0,1),(1,0),(1,1)\}\)
\(A^3=\{(0,0,0),(0,0,1), (0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}\)
For \(A^2\) we are looking for ALL ordered pairs \((a,b)\) where \(a,b\) are either 0 or 1. For \(A^3\) we are looking for ALL ordered 3-tuples \((a,b,c)\) where \(a,b\) or \(c\) are either 0 or 1.