Prove that any countable cartesian product of countable sets is countable

Question

I want to prove that infinite (yet countable) cartesian product of countable sets is countable.

Here's what I tried:

Step 1:

I proved that for 2 countable sets $ A_1,A_2 $ , the product $ A_{1}\times A_{2} $ is countable.

Step 2:

I proved by induction that for any $n\in \mathbb{N} $ if $ A_{1},...,A_{n} $ are countable sets, then

$ A_{1}\times A_{2}\times,...,\times A_{n} $ is countable.

Now, I want to show that any countable infinite cartesian product would be countable.

How do i show that $ A_{1}\times,....\times A_{\aleph_{0}} $ is countable?

Thanks in advance.

Answer 1

I guess you know that $[0,1)$ is uncountable. If your statement was true, then $\mathbb{N}^\mathbb{N}$ would be countable. But there is a surjection of $\mathbb{N}^\mathbb{N}$ in $[0,1)$ since every $x\in[0,1)$ can be represented by a countable string of digits (writing it's decimal part in base 10 for instance). Then $[0,1)$ would be countable, which is absurd. So a countable cartesian product in general is not countable.

Answer 2

Here are some links that should answer your question:

Infinite Cartesian product of countable sets is uncountable

Countable collection of countable sets and Axiom of choice

Answer 3

You can't prove it because it's false.

The countable product of two element sets is (essentially) the uncountable real unit interval when you think of elements of the product set as binary expansions.

Infinite Cartesian product of countable sets is uncountable