If $A_1$ and $A_2$ are countably infinite, and that $A_1\cap A_2=\phi$. Then prove that $A_1 \cup A_2$ is countably infinite.

Question

If $A_1$ and $A_2$ are countably infinite, and that $A_1\cap A_2=\phi$. Then prove that $A_1 \cup A_2$ is countably infinite.

My work:
As $A_1$ and $A_2$ are countably infinite, there exists a bijection $\theta_1: \mathbb{N}\to A_1$ and $\theta_2: \mathbb{N}\to A_2$. I need to prove now, that there exists a bijection $\theta : \mathbb{N}\to A_1\cup A_2$ which I am not able to find. Please help.

My knowledge about this topic is very limited and I have just started studying about all this.

Answer 1

Define $\phi(2n) = \phi_1(n)$, $\phi(2n-1) = \phi_2(n)$.

Answer 2

Let's look at a trivial example for a moment. Suppose $A = \{a,b,c\}$ and $B = \{d,e\}$. One way of seeing that $A\cup B$ is countable is to right a list containing all the elements. One such list is $a,b,c,d,e$. But this doesn't generalize nicely to countably infinite lists, since we'd never finish $A$.

On the other hand, the list $a,d,b,e,c$, choosing an element from $A$, then an element from $B$, and vice versa, works very well.

In other words, you can explicitly create a list from the two lists you already have.