# Cartesian product set

I'm preparing a task on the Cartesian product of two sets and I have run into the following confusion:

I understand that the Cartesian product is not a commutative operation. Generally speaking, AxB does not equal BxA unless A=B or A or B is the empty set. This is usually easy to explain to students because in the definition of a cartesian product, we define it as an ordered pair, meaning order would matter. However, once we move on from this idea to explain what the product set represents, things get a bit fuzzy. For example, If we think of the 52 cards in a standard deck as a product set, we can define set A as the ranks and set B as the suits. How can we explain to students that while AxB and BxA both represent the 52 cards in a standard deck, the sets AxB and BxA are distinct and disjoint sets?

## 2 Answers

They are isomorphic. While $A\times B \neq B \times A$ for any arbitrary distinct sets $A$ and $B$, by defining the map $\phi:A\times B \to B \times A$ by $\phi(a,b)=(b,a)$, we can show that algebraically (or set theoretically) that they behave the same. Just a quick check of the set theoretic parts:

• $\phi(a,b) = \phi(c,d)$ iff $a=c$ and $b=d$. Hence $\phi$ is injective.
• $\phi$ is clearly surjective.

One way to explain this to the students is to say that you just need to be consistent in the order of the sets throughout (i.e., if you choose $A \times B$, always use that ordering), but your initial choice doesn't matter. You can switch to the other ordering by just flipping the order of everything everywhere.

To show they are distinct, I would say say something along the lines that if $A=\{a,b,c\}$ and $B=\{1,2,3\}$, then $(a,1) \in A \times B$ but $(a,1)$ is not in $B \times A$.

I like Chris C's answer; I will offer another point of view.

Perhaps the difficulty lies in the example that has been chosen: a deck of cards isn't naturally a cartesian product for exactly the reason that there is no natural ordering of rank and suit.

Instead, you might look at a (simplified) menu, and consider all of the meals that consist of a main course and a dessert, and represent that as a cartesian product. Here, there is natural ordering. (I'm sure there are better examples out there, this is just the first thing that came to mind.)

• Good example. Of course there is a natural ordering for the cards: you say "ace of spades" and not "spade of aces" Nov 18, 2014 at 22:28
• I had not thought of the natural verbal ordering! That takes care of that. Nov 18, 2014 at 22:32
• Or just A = {first names} and B = {last names} for a class of students; the natural ordering (alphabetically) looks different for AxB and BxA.... Nov 18, 2014 at 22:56