Q:

Proof about Countability and Induction

Let n be a positive integer. Using the fact that the union of two countable sets is countable, use induction to prove that the union of n countable sets is countable.
I did the following.
**We begin by induction on n.
Suppose n = 1.
Then we have 1 set and by above, it is countable.
Suppose the union of k sets is countable, for some integer k.
We must show that the union of k+1 sets is also countable.**
From here I am not sure how to "Prove" this without just flat out saying that these would also be countable  by the theorem.

A:

You're right in that it's pretty straightforward. Nevertheless, you have to be careful that you don't make illogical jumps.
You have the base case right. You have 1 countable set and so it is trivially countable.
Now consider the union of $k$ countable sets $S_1, S_2, \ldots S_k$ and assume this union is countable. Let us call this union $A$ i.e. 
$$A = \bigcup_{i=1}^{k}{S_i}$$.
Now consider another countable set $S_{k+1}$. We want to prove that $\bigcup_{i=1}^{k+1}S_i$ is also countable. Well we have:
$$
\bigcup_{i=1}^{k+1}S_i = \left(\bigcup_{i=1}^{k}S_i \right) \cup S_{k+1} \\
= A \cup S_{k+1}
$$
From the induction hypothesis, we know $A$ is countable and $S_{k+1}$ is countable by assumption, and by our rule, we know the union of two countable sets if countable. Since the union of $S_1, S_2, \dots S_{k+1}$ is equal to the union of the countable set $A$ and the countable set $S_{k+1}$, we can conclude that $\bigcup_{i=1}^{k+1}S_i$ is countable.
Note we had to be careful in that we cant just say, the $k+1$ sets are countable because the union of the first $k$ are. Your rule says only the union of two countable sets is countable so you need to make that step of calling the union of the first $k$ a set $A$ and then taking its union with $S_{K+1}$

