Rudin: set of $n$ tuples are countable I am trying to work through Theorem 2.13 in Rudin, which states; 

Let $A$ be a countable set, and let $B_n$ be the set of all $n$-tuples $(a_1, \ldots, a_n)$ where $a_k \in A$) ($k = 1, \ldots, n$), and the elements $a_1, \ldots, a_n$ need not be distinct. Then $B_n$ is countable.

Rudin proves it by induction on $n$. The base case is rather obvious since $B_1 = A$ and $A$ is countable by assumption. Supposing $B_{n-1}$ is countable, where
$$B_{n-1} := \{(a_1, \ldots, a_{n-1} \mid a_i \in A\},$$
we consider $B_n$, where 
$$B_n := \{(a_1, \ldots, a_{n-1}, a_n) \mid a_i \in A\}.$$
This is where my confusion comes in, and it is mostly a notational issue. Rudin instead writes $B_n$ as the set of order pairs $(b,a)$ where $b \in B_{n-1}$ and $a \in A$. This is, in effect, an $(n-1)$-tuple within an $n$-tuple,i.e., 
$$((a_1, \ldots, a_{n-1}), a)$$
if in fact we were considering an element of $B_{n-1}$. 
My question is: why can he call it $(b,a)$? Is this just an abuse of notation? The argument for why this establishes that $B_n$ is countable makes sense to me, and even if the solution is just that this is a reasonable way to list it, I would be fine with that. But the notion of calling this an ordered pair with element $b \in B_{n-1}$ is what is tripping me up. 

Yes it is an abuse of notation. But this can be justified since there exists a bijection $((a_1,\cdots,a_{n-1}),a)\mapsto (a_1,\cdots,a_{n-1},a)$. This step is just omitted since it is too trivial.
