The union of denumerably many denumerable sets is denumerable. Problem : If for each ${k\in\mathbb N},B_k$ is a denumerable set
, then $\bigcup\limits_{k\in \mathbb N}B_k$ is denumerable.
I know that  $B_k\sim \mathbb{N}$, $\mathbb{N}\sim\mathbb N \times\left \{k\right \}$ for each $k\in \mathbb{N}$.
And I defined the function $f_k:B_k\sim\mathbb N \times\left \{k\right \}$ for each $k \in \mathbb N$
And then I defined the function $f:\bigcup\limits_{k\in \mathbb N} B_k\to\bigcup\limits_{k\in \mathbb N}\mathbb N \times\left \{k\right \}$ i.e. $f:\bigcup\limits_{k\in \mathbb N}B_k\to\mathbb N \times \mathbb N$ by
$$f(x)=\begin{cases}
f_1(x) & \text{if } x\in(B_1-B_2) \\ 
& \text{if } x\in(B_k\cap\ B_{k+1}) \\ 
f_{k+1}(x) & \text{if } x\in(B_{k+1}-(B_k\cup\ B_{k+2})) 
\end{cases}$$
I don't know when $x\in(B_k\cap\ B_{k+1})$ Is this right progress?Please help me.
 A: Here's an easier approach. Notice that if $f : \mathbb{N} \rightarrow X$ is unto (i.e. $(\forall x \in X)(\exists n \in \mathbb{N}) (f(n = x)) $) then for sure $\text{card}(\mathbb{N} )\geq \text{card}( X)$; in otherwords denumerable. 
Notice that each $B_i$ is denumerable by definition therefore there exists a collection $f_i : \mathbb{N} \rightarrow B_i$ that are 1-to-1 and unto. Therefore we can define the following function 
\begin{equation}
f:\mathbb{N}^2 \rightarrow \bigcup_{i \in \mathbb{N}}B_i
\end{equation}
by 
\begin{equation}
f(k,n) = f_{k}(n).
\end{equation}
Let $b \in \bigcup_{i \in \mathbb{N}}B_i$ then by definition there exists 


*

*$\exists k \in \mathbb{N}$ such that $b \in B_k $

*$\exists n \in \mathbb{N}$ such that $b =  f_k(n) $
so that $f$ is unto. If you insist on a 1-to-1 function then you can define 
\begin{equation}
g:\bigcup_{i \in \mathbb{N}}B_i \rightarrow \mathbb{N}^2 
\end{equation}
by 
\begin{equation}
g(b) = \min_{k+n}\{(k,n) \ | \ f(k,n) = b\}
\end{equation}
(minimize over the sum $k+n$ so that we can apply recusrion/induction/well-ordering or whatever you want to call it).
A: Given $f_k\colon B_k\to\Bbb N$, first, you have
$$\begin{align} \mu\colon\bigcup_{k\in\Bbb N}B_k&\to\Bbb N\\
b&\mapsto \min\{\,k\in\Bbb N\mid b\in B_k\,\} .\end{align}$$
Then map
$$ \begin{align}\bigcup_{k\in\Bbb N}B_k&\to\Bbb N\times \Bbb N\\
b&\mapsto \langle f_{\mu(b)}(b),\mu(b)\rangle\end{align}$$
