Is the cartesian product of a finite amount of countable sets, countable?

Question

I would assume so. I know that the product of two countable sets is countable, but what about a finite number?

Answer 1

Claim: The cartesion product of $n$ countable set is countable for any finite/natural n.

Proof by induction:

Base cases: $n = 1$. Then $C_1$ is countable. Nothing more needs saying.

$n = 2$. $C_1 \times C_2$ is countable. Why? well, you know why, its the old diagonal list argument. $(a_1,b_1), (a_2,b_1), (a_1,b_2),(a_3, b_1) ....etc.$

I'll assume you know it.

Inductive step:

Assume $\prod_{1}^n C_i$ is countable.

Then $\prod_1^{n+1} C_i = [\prod_1^n C_i] \times C_{n+1}$.

Now $[\prod_1^n C_i]$ is ONE countable set. $C_{n+1}$ is another. So $[\prod_1^n C_i]\times C_{n+1}$ is the cartesian product of TWO countable sets. And is thus countable.

I know, it feels like the rug being pulled from under you. But is works. If cartesian product of two countable sets are countable, just cluster the cartesian products of multiple sets into cluster of two.