Theorem: If $A$ and $B$ are both countable sets, then their union $A\cup B$ is also countable.
I am trying to prove this theorem in the following manner:
Since $A$ is a countable set, there exists a bijective function such that $f:\mathbb{N}\to A$. Similarly, there exists a bijective function $g:\mathbb{N}\to B$. Now define $h:\mathbb{N}\to A\cup B$ such that: $$h(n)=\begin{cases}
f(\frac{n+1}{2})&\text{, n is odd}\\
g(n/2) & \text{, n is even} \\
\end{cases}$$
So in essence, $h(1)=f(1)$, $h(2)=g(1)$, $h(3)=f(2)$ and so on. Now we have to show that h is a bijection.
h(n) is one-one:
Proof: If $h(n_1)=h(n_2)$ then, if $n_1$ and $n_2$ are both either odd or even, we get $n_1=n_2$. But if, suppose $n_1$ is odd and $n_2$ is even, this implies that: $$f\left(\frac{n_1+1}{2}\right)=g\left(\frac{n_2}{2}\right)$$ How can one deduce from this equality that $n_1=n_2$?
I tried to think about this and realized that if $A\cap B=\phi$ then this case is impossible as it would imply that there is a common element in both sets. On the other hand, if we assume that $A\cap B\neq \phi$, then either $f\left(\frac{n_1+1}{2}\right)\in A\cup B$ or $g\left(\frac{n_2}{2}\right)\in A\cup B$….Beyond this I'm clueless.
Edit: Solution by the author-
Best Answer
A set $S$ is countable iff its elements can be enumerated.
Since $A$ is countable, you can enumerate $A=\{a_1,a_2,a_3,...\}$.
Since $B$ is countable you can enumerate $B=\{b_1,b_2,...\}$
Enumerate the elements of $A\cup B$ as $\{a_1,b_1,a_2,b_2,...\}$ and thus $A\cup B$ is countable.