$\prod_{j=1}^\infty A_j$ is uncountable?

Question

Let $A_j,j\in\mathbb Z^+$ be finite sets with at least two distinct elements.

How to show that $\prod_{j=1}^\infty A_j$ is uncountable?

Answer 1

Identify a subset of each set with {0,1} using AC, and map the corresponding subset of the product onto binary expansions of reals in the interval [0,1], hence uncountable.

Answer 2

Hint:

Replace each $A_j$ by $B_j=\{0,\ldots,n-1\}$, where $n=|A_j|$, using the axiom of choice. Now we have: $$\left|\prod_{j\in\Bbb N}\{0,1\}\right|\leq\left|\prod_{j\in\Bbb N}B_j\right|=\left|\prod_{j\in\Bbb N}A_j\right|.$$

You can bound this from the right by the product of copies of $\Bbb N$ and calculate the exact cardinality too. I am leaving the details for you.

Do note, the axiom of choice is essential and it is consistent that there are families of finite sets whose product is empty, and thus countable.