Theorem Let $A$ be a countable set, and let $B_{n}$ be the set of all n-tuples $(a_{1}, …, a_{n})$, where $a_{k} \in A (k = 1, …, n)$, and the elements $a_{1}, …, a_{n}$ need not to be distinct. Then $B_{n}$ is countable.
BlockquoteProof That $B_{1}$ is countable is evident, since $B_{1} = A$.(stop here)
Do I misunderstanding the definition of $B_{n}$?
In my opinion, set $B_{1} =$ {$(a_{1}, …, a_{n})$} , which means that there is merely one element in set $B_{1}$ which is an n-tuple $(a_{1}, …, a_{n})$. It is obvious that $B_{1} \neq A$. Would you please show some correct examples of $B_{n}$.
(continue) Suppose $B_{n-1}$ is countable $(n = 2, 3, 4, …)$. The elements of $B_{n}$ are of the form
$$
(b,a) \qquad (b \in B_{n-1}, a \in A).
$$For every fixed $b$, the set of pairs $(b, a)$ is equivalent to $A$, and hence countable. Thus $B_{n}$ is the union of a countable set of countable sets. By Theorem 2.12, $B_{n}$ is countable.
And does $B_{2}$ = ({$a_{1}, …, a_{n}), a_{0})$ , $a_{0} \in A$ ?
Thanks in advance!
Best Answer
$B_n$ is the set of all $n$-tuples of elements in $A$. Note that these are the same $n$.
So when $n=1$, $B_1$ is the set of all $1$-tuples of elements of $A$, ($B_1 = \{ (a_1) : a_1 \in A\}$) which we can identify with $A$.
When $n=2$ we have $B_2 = \{(a_1,a_2) : a_1, a_2 \in A \}$, it's the set of all pairs of elements in $A$.