# Thread: Cardinality Of A Cartesian Product

1. ## Cardinality Of A Cartesian Product

Hello,

I am having difficulty understanding why the cardinality of a Cartesian Product is simply the product of the cardinality of the individual sets involved in the Cartesian Product. As an attempt in trying to understand this, I supposed the set A x A, where A has n elements: I understand that 1 goes to 1, 1 goes to 2,..., and 1 goes to n, which can be stated as (1,1),(1,2),...,(1,n); furthermore, 2 goes to 1, 2 goes to 2,..., and 2 goes to n, which can be stated as (2,1), (2,2),...,(2,n). However, this train of thinking did not do an adequate job in elucidating this concept. Could someone please help?

2. ## Re: Cardinality Of A Cartesian Product

Originally Posted by Bashyboy
I am having difficulty understanding why the cardinality of a Cartesian Product is simply the product of the cardinality of the individual sets involved in the Cartesian Product.
First, I will assume that you are considering only finite sets.

Look at the definition: $A\times B=\{(x,y):\;x\in A~\&~y\in B\}$.
How many pairs are possible?
If the cardinality of $A$ is $|A|=n$ then there are only ways $n$ to pick the $x$ value of the pairs.
Likewise, there are only $|B|=m$ ways to pick the $y$ values in the pairs.

Thus $|A\times B|=|A|\cdot|B|=nm$.

Example: $\{a,s,d,f,g\}\times\{1,2,3\}$ contains $15$ pairs.

Now for infinite sets, things are more complicated.

3. ## Re: Cardinality Of A Cartesian Product

If you associate a number with each element of A you can lay out the elements of AXA as:

11, 12, 13, 14, …..
21, 22, 23, 24,…
…………

And then start counting on a diagonal starting from upper left: 11,21,12,31,22,13…. (1,2,,3,…)

Just something I remembered from an analysis text. Not sure I’m convinced.

EDIT: The reason I am not convinced is that if you started counting by row, you would never get past the first row.