# Cardinality Of A Cartesian Product

#### Bashyboy

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?

#### Plato

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

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

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

Now for infinite sets, things are more complicated.

#### Hartlw

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.

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