The context for this question is a specific theorem in Rudin's book. I'm fine with the proof, but I think there are some slight notational abuses which, while ultimately not a huge deal, I'm having trouble wrapping my mind around. The theorem is:
Let $A$ be a countable set and $B_n = \{(a_1, \ldots, a_n) \mid a_i \in A\}$. Then $B_n$ is countable.
Rudin proceeds by induction and claims $B_1 = A$. I'm not certain that's the case, since I don't know if a "one-tuple" is the same as an element of $A$, which is likely a scalar. It's surely a natural identification, sending, say, $a \mapsto (a)$, but are these, truly, "the same thing?"
Similarly, when he goes on to prove the inductive step, after assuming inductively that $B_{n-1}$ is countable, he argues $B_n = \{(b, a) \mid b \in B_{n-1}, a \in A\}$. That doesn't seem completely right to me, as this isn't an $n$-tuple, but a two-tuple containing an $(n-1)$-tuple and the element $a$. Surely it's in bijection with $B_n$: I can just "flatten out" the tuple into an $n$-tuple and get a bijection. But it seems to me these aren't exactly the same thing.
Am I correct to think this is an abuse of notation, akin to saying $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$?
You are right to question that line of reasoning from Rudin, it is indeed an abuse of notation. It would have been more rigorous to give a definition of an $n$-tuple and to discuss the difference (or the sameness) between $a$ and $(a)$ as well as between $A^k\times A$ and $A^{k+1}$.
This is how Tao has done in his Analysis book: