Showing $ A \cup B \cup C $ is countable if $ A, B $ are countable and $C$ is finite. How does one go about showing $ A \cup B \cup C $ is countable if $ A, B $ are countable and $C$ is finite?
I understand most of the confusion for resolving set theory questions online seem to be the definition. For my course we consider the following definitions:
countable: Finite or $A \sim\mathbb{N}$
uncountable: not countable
finite: The empty set or $A \sim J_n$ where $n \in \mathbb{N}$
infinite: not finite
So I'm not sure if I have this right but looking at the above definitions the way I'm looking to approach this is to consider 6 different cases.


*

*Where C is the empty set and A, B are both finite sets

*Where C is the empty set and A is finite and B is countably infinite

*Where C is the empty set and both A, B are countably infinite


4 - 6. Repeated above but with C being a non-empty finite set
This seems like quite a round about way but intuitively it seems to me like the only way to cover all bases according to the definitions. I'm hoping I might be absolutely wrong on this. Is there a simpler way to prove this?
 A: If $A$ and $B$ are countable, then they can be enumerated, respectively, as:
$a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$
Since $C$ is finite, its elements can be listed as: $c_1, c_2, \ldots, c_n$.
To prove the union of all three is countable, it will suffice to list them in some order $d_1, d_2, d_3, \ldots$, since there is then a natural bijection with $\mathbb{N}$ in which we match $d_k$ with $k \in \mathbb{N}$.
As to the listing for this union, you could list all of $C$, then interweave enumerations of $A$ and $B$:
$c_1, c_2, \ldots, c_n, a_1, b_1, a_2, b_2, a_3, b_3, \ldots $
(You may also specify that no element is listed more than once.)
If either of $A$ or $B$ is finite (since your definition of countable allows for this) then the listing can be tweaked appropriately, e.g., if just $A$ is finite, then list all elements of $C$, then of $A$, and then enumerate $B$.
A: Sets are countable if and only if their elements are 'listable'.  
Now, $C$ is finite, so say it contains elements $c_1, c_2, ..., c_n$ for some $n$
$A$ is countable, so we can create a list:
$a_1, a_2, ... $
Same for $B$:
$b_1, b_2, ...$
So, I would create the following list:
$c_1, c_2, ..., c_n, a_1, b_1, a_2, b_2, ...$
A: Let $j_a: A \rightarrow \mathbb N$ be an injection.  We know that is possible because $A$ is countable.
Let $j_b: B \rightarrow \mathbb N$ be an injection.  We know that is possible because $B$ is countable.
Let $j_c: B \rightarrow \mathbb N_j$ be an bijection.  We know that is possible because $C$ is finite.
Let $k:A\cup B\cup C \rightarrow \mathbb N$ via $k(x) = 3*j_a(x)$ if $x \in A$. If $x \in B$ but $x \not \in A$ let $k(x) = 3*j_b(x) + 1$.  And if $x \in C$ but $x \not \in A$ and $x \not \in B$ let $k(x) = 3*j_c(x) + 2$.
Show that $k$ is an injection and that shows $A\cup B \cup C$ is countable.
Another way to think of it is to start by picking out the first element of $A$ then pick the first element of $B$, then of $C$, then pick the second element, then the third elements and so on.  That's a list of all the elements.  Since they can be listed one after another they are countable.
