$A_1 \times ... \times A_n$ is countable if $A_1, ..., A_n$ are countable Suppose that $A_1, ..., A_n$ are countable sets. Show that the cartesian product $A := A_1 \times ... \times A_n$ is countable.
My attempt:
Sets are said to be countable if they are finite or if they have the same cardinality as some subset of $\mathbb{N}$ (i.e. we can find some bijection $f: A \rightarrow S$ or $f: S \rightarrow A$ where $S \subset \mathbb{N}$).
Assume that $A_1, ..., A_n$ are countable sets. Then, there exists bijections $fi: \mathbb{N} \rightarrow A_i$ for $i = 1, ..., n$.
Define $g: \mathbb{N} \rightarrow A$ as follows
My issue arises here in finding such a bijective function without it being too complicated. How would I go about finding one? I am also open to any suggestions. Any assistance is welcomed.
 A: It is sufficient to prove this in the case where $A_i=\mathbb{N}$ for all $i$. So we have to show that $\mathbb{N^n}$ is a countable set. Let $p_1,p_2,...,p_n$ be the first $n$ prime numbers, and define a map $f:\mathbb{N^n}\to\mathbb{N}$ by:
$f(m_1,m_2,...,m_n)=p_1^{m_1}p_2^{m_2}...p_n^{m_n}$
It is well known that prime decomposition is unique, and so $f$ is injective. So we showed that there is an injection from $\mathbb{N^n}$ to $\mathbb{N}$, and so $\mathbb{N^n}$ is countable. (it is in bijection with the image of $f$, which is a subset of $\mathbb{N}$)
A: Do it by induction.
Prove that if $A,B$ are countable then $A\times B$ are countable (use the "diagonal line" argument).
$1\mapsto (a_1,b_1); 2 \mapsto (a_2,b_1); 3\mapsto (a_1, b_2); 4\mapsto (a_3, b_1); 5\mapsto (a_2,b_2); 6\mapsto (a_1,b_3);7\mapsto (a_4, b_1).... etc. $
Then if we assume $A_1 \times A_2..... \times A_k$ is countable then $A_1\times A_2\times..... \times A_k \times A_{k+1} = (A_1\times A_2\times..... \times A_k) \times A_{k+1}$ is the product of TWO countable sets and is therefore countable.
So by indcuction $A_1 \times A_2..... \times A_n$ is countable for all $n$.
A: It suffices to show that the cartesian product of two countable sets is countable, because an inductive argument will yield the general result.
According to your definition of countable sets, if at least one set is finite, the result is immediate. So the only interesting case is when both sets are countably infinite.
In particular, it is enough to prove that $\mathbb{N}\times\mathbb{N}$ is countable. Define a function $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ as follows. If $n\in\mathbb{N}$ then $n$ can be written uniquely as $2^{m-1}\cdot(2k-1)$, where $m,k\in\mathbb{N}$. Define $f(n)=(m,k)$. By the uniqueness of such expressions, we see that $f$ is one-to-one. On the other hand, if $(m,k)$ is any pair of $\mathbb{N}\times\mathbb{N}$ then setting $n=2^{m-1}\cdot(2k-1)$ yields an integer so that $f(n)=(m,k)$, so $f$ is onto. Thus $f$ is a bijection, and this shows that $\mathbb{N}\times\mathbb{N}$ has the same cardinality as $\mathbb{N}$, i.e. it is countable.
A: You could use an approach similar to the Cantor proof that the algebraic numbers are countable.  Order the $A_k=(a_k(j))$  For an element of $A$,let $B=(a_1(j_1),....a_n(j_n))$. Define $b=\sum_{k=1}^n j_k$.  For each $b$ there are only a finite number of $B's$ with $b$ as the sum.  These can be ordered in any way.  The totality for all (integer) values of $b$ is countable and includes all members of $A$.
