(related to Corollary: Cartesian Products of Countable Sets Is Countable)

- By hypothesis, $A_1,\ldots,A_n$ are countable sets and $A=A_1\times\ldots\times A_n$ is their Cartesian product.
- We prove by induction that $A$ is countable.
- Base Case $n=2$
- ... is correct, since $$A_1\times A_2=\bigcup_{a\in A_2}(A_1\times a)$$ and the union of countably many countable sets is countable.

- Induction step $n\to n+1$
- Assume $A_{n+1}$ is countable.
- Then, $$A_1\times \ldots\times A_n\times A_{n+1}=\left(\bigcup_{a\in A_{n+1}}(A_1\times \ldots\times A_n)\times a\right).$$
- By induction hypothesis, $A_1\times \ldots\times A_n$ is countable.
- Apply union of countably many countable sets is countable once again.∎

**Modler, F.; Kreh, M.**: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition