This is the statement I am trying to prove:
Prove that if A and B are denumerable disjoint sets, then $A∪B$ is denumerable.
This is my attempt:
Since A is denumerable, let $f$ be a bijection $f:Z^+→A$ so $A=\left\{ a_{1},a_{2},a_{3},...\right\} $ and since B is denumerable, let $h$ be a bijection $h:Z^+→B$ so $B=\left\{ b_{1},b_{2},b_{3},...\right\} $. Now $A∪B=\left\{ a_{1},b_{1},a_{2},b_{2},a_{3},b_{3},...\right\} $ and is defined by $g(x)=\left\{ \begin{align} & 2k+1, & \text{ if } x=a_k\\ & 2k, & \text{ if } x=b_k, \end{align} \right.$
This is where I don't know how to finish the proof. If $g(x)$ was bijective then I could say $A∪B$ is denumerable, but I know that $g(x)$ is not bijective, though it is surjective. Here's a little more about what I know:
I know that if a function is both injective and surjective that it is bijective or if a function has an inverse that it is bijective. I also know by the Cantor-Berstain theorem that if two functions are injective that there is a bijection.
Based on the general info I just said, I don't know how to proceed. Is there additional theorems I should be using and is there something else I am missing? Could someone help with how I should finish this proof and point out is there were any other errors that I made?
Find bijections f:A -> 2N, g:B -> 2N + 1.
Show f $\cup$ g:A $\cup$ B -> N is a bijection.