The Cartesian product of a finite number of countable sets is countable 
The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$
Let $I_n=\{i \in \mathbb N \mid i \le n\}$ for $n\in \Bbb N$ and $A_i$ be countable for all $i\in I_n$. Then $\prod\limits_{i\in I_n}A_i$ is countable.


My attempt:
Lemma 1: $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ and $\prod\limits_{i\in I_{n+1}}A_i$ are equinumerous for all $n \in \mathbb N$. (I presented a proof here)
Lemma 2: If $A$ and $B$ are countable, then $A \times B$ is countable. (I presented a proof here)
I will prove the theorem by induction on $n$. The theorem is trivially true for $n=0$. Assume that the theorem is true for $n=k$, then $\prod\limits_{i\in I_k}A_i$ is countable. Since $\prod\limits_{i\in I_k}A_i$ is countable (by inductive hypothesis) and $A_{k+1}$ is countable, then by Lemma 2 $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable. Furthermore, by Lemma 1 $(\prod\limits_{i\in I_k}A_i) \times A_{k+1}$ and $\prod\limits_{i\in I_{k+1}}A_i$ are equinumerous, then $\prod\limits_{i\in I_{k+1}}A_i$ is countable too. Thus the theorem is true for $n=k+1$. This completes the proof.


Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!


Let $A$ and $B$ be countable, i.e. we have the enumerations
$$a_1,a_2,a_3,\cdots$$ and $$b_1,b_2,b_3,\cdots$$
Then we can enumerate $A\times B$ as
$$(a_1,b_1),(a_2,b_1),(a_1,b_2),(a_3,b_1),(a_2,b_2),(a_1,b_3),\cdots$$
(notice the sums of the indexes, $2,3,3,4,4,4,\cdots$). You can easily check that all pairs $(a_n,b_m)$ are cited.
Then by induction,
$$\prod_{i=1}^n A_i=\left(\prod_{i=1}^{n-1} A_i\right)\times A_n.$$ is countable.
