Proving that $A_1\times A_2 \times \cdots \times A_n$ is countable

Question

If $A_1$,$A_2$,...,$A_n$ are countably infinite sets, is $A_1\times A_2\times...\times A_n$ countable?

it seems that an infinite Cartesian product of countable sets may still be uncountable for every x∈[0,1) in a countable set.

Answer 1

Let $A_i=\{ a_{i,j} \}_{j\in\mathbb{N}}$ for $i=1,\dots,n$. Let $p_1,\dots,p_n$ the first $n$ primes.

The function $\varphi: A_1\times\dots\times A_n\rightarrow\mathbb{N}$ defined by $$ \varphi(a_{1,j_1},\dots,a_{n,j_n}):=p_1^{j_1}\cdots p_n^{j_n} $$ is an injection. This shows that $|A_1\times\dots\times A_n|\le \aleph_0=|\mathbb{N}|$. On the other hand, $A_1\times\dots\times A_n$ is an infinite set, so $$ |A_1\times\dots A_n|=\aleph_0=|\mathbb{N}|. $$

Answer 2

You can count the elements of $A_1\times A_2$ in the order

$$11,\ \ 21,12,\ \ 31,22,12,\ \ 41,32,23,14,\cdots$$ where the two indexes relate to the two sets. Note that we formed groups of constant sum, and the size of a group equals its sum minus one.

Countability of the Cartesian product of $n$ sets follows by induction.

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1. $A_1$ is countable.
2. if $A_1\times A_2\times\cdots A_n$ is countable, so is $(A_1\times A_2\times \cdots A_n)\times A_{n+1}$.