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.
Best Answer
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
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).