# Cartesian product of sets with different cardinality

What is the correct way of performing a cartesian product over finite sets with different cardinality? For instance:

Let

$$A = \{1, 2, 3\},\quad B = \{4, 5\},\quad C = \{6, 7\}$$

What would be the result of

$$A \times B \times C$$

What I thought it would correct would be to try combining all elements of the sets, but that seems to be incorrect.

• do you know of tuples ? – user451844 Sep 30 '17 at 13:51

By definition, $$A \times B := \{(a,b): a \in A, b \in B\}$$

$A \times B$ $=$ {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}, then $(A \times B)$ $\times$ $C$ $=$ ${(1, 4, 6), (1, 4, 7), (1, 5, 6),(1, 5, 7), (2, 4, 6), (2, 4, 7), (2, 5, 6), (2, 5, 7), (3, 4, 6), (3, 4, 7), (3, 5, 6), (3, 5, 7)}$.

Product of cardinalities of the all the sets is $|3|\times|2|\times|2|$ $=$ $12$

ordering matters it would include:

$(1,4,6),(1,4,7),(1,5,6),(1,5,7),(2,4,6),(2,4,7),(2,5,6),(2,5,7),(3,4,6),(3,4,7),(3,5,6)$ ,and $(3,5,7)$

whereas the Cartesian product $B\times A\times C$ would have:

$(4,1,6),(4,1,7),(4,2,6),(4,2,7),(4,3,6),(4,3,7),(5,1,6),(5,1,7),(5,2,6),(5,2,7),(5,3,6)$ , and $(5,3,7)$ or to put it in set notation:

$$\{(x,y,z):x\in A,y\in B, z\in C\}$$ is the product $A\times B \times C$ whereas $$\{(x,y,z):x\in B,y\in A, z\in C\}$$ is the product $B\times A \times C$