How to express the cardinality of $∏_{1≤i≤n} A_i$ in terms of cardinalities $|A_1|, |A_2|, . . . , |A_n|$

Question

I was given the problem:

For finite sets $A_1, A_2,\dotsc , A_n$ define their Cartesian product $\prod_{i=1}^n A_i$ as the set of all $n$-sequences $(x_1, x_2,\dotsc, x_n)$, where $x_i \in A_i$ for every $i = 1, 2, \dotsc, n$. Find a formula expressing the cardinality of $\prod_{i=1}^n A_i$ in terms of cardinalities $|A_1|, |A_2|,\dotsc , |A_n|$.

And I am struggling to understand what it is actually asking for, could someone explain it to me please, thanks. :)

Answer 1

$\prod_{1\le i\le n}A_i$ is the cartesian product, that is, all finite sequences $(a_1,\ldots,a_n)$ such that $a_i \in A_i$ for each $i=1,\ldots,n$. How many such sequences can you choose? $|A_1|$ choices for $a_1$, ..., $|A_n|$ choices for $a_n$. Therefore $$ \left|\prod_{1\le i\le n}A_i\right|=\prod_{1\le i\le n}|A_i|. $$

Answer 2

We know that $$|A\times B|=|A|\times|B|\qquad(1)$$.

We want to show $$\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$$ is true for any natural number $n$, where $\prod_{i=1}^n|A_i|=|A_1|\times|A_2|\times\dotsc\times|A_n|$. So, we have use induction.

The base case $n=1$ ($|A_1| = |A_1|$) is trivial. Now suppose inductively that $\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$. We want to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\prod_{i=1}^{n+1}|A_i|.$$ Now we need to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|\qquad(2),$$ i.e., the cardinality of the set $\prod_{i=1}^{n+1}A_i$ is equal to the cardinality of the set $\left(\prod_{i=1}^nA_i\right)\times A_{n+1}$. So $$\begin{aligned}\left|\prod_{i=1}^{n+1}A_i\right|&=&\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|&\qquad\text{by }(2)\\&=&\left|\prod_{i=1}^nA_i\right|\times |A_{n+1}|&\qquad\text{by }(1)\\&=&|A_1|\times|A_2|\times\dotsc\times|A_n|\times|A_{n+1}|&\qquad\text{by induction hypothesis}\\&=&\prod_{i=1}^{n+1}|A_i|.\end{aligned}$$