I'm having trouble understanding the last part of the proof of this theorem (2.13) in the Rudin (blue) book:
Let $A$ be a countable set, and let $B_n$ be the set of all $n$-tuples $(a_1,\ldots, a_n)$, where $n\in \mathbb{N}$ and $a_k\in A$ ($k=1,\ldots, n$), and the elements $a_1,\ldots, a_n$ need not be distinct. Then $B_n$ is countable.
I'm not understanding how $B_n$ is the union of a countable set of countable sets. If someone could clarify this or provide a more in depth proof, I would appreciate it! Thank you.
Best Answer
For $a\in A$, let $B_{n,a}=\{(a_1,\ldots,a_n)\in B_n\mid a_n=a\}$. Then we have an obvious bijection $B_{n,a}\leftrightarrow B_{n-1}$ by droping the last term, so by induction we may assume that all $B_{n,a}$ are countable. Then $B_n=\bigcup_{a\in A}B_{n,a}$ shows that $B_n$ is the countable union of countable sets.