Let $A_1, A_2,\dots$ be a sequence of disjoint, finite subsets of $\mathbb{N}$. How can $\bigcup_{n=1}^\infty A_n$ be either finite or infinite? Let $A_n$ be finite subsets of $\mathbb{N}$ that are not $\emptyset$, and $\forall i,j, i\not = j$, $A_i, A_j$ are disjoint, then must 
$$\bigcup_{n=1}^\infty A_n = \mathbb{N}$$
and thus be (countably) infinite?
So basically I am looking for an example of $A_1,A_2,\dots$ such that
$$\bigcup_{n=1}^\infty A_n$$
is finite.
Or, is the only way for the infinite union to be finite for $A_i =A_j = \emptyset$?
Thanks.
Edit: A comment has answered how $\bigcup_{n=1}^\infty A_n \not = \mathbb{N}$ is possible. I am not sure if it must be infinite, though, when $A_n \not = \emptyset$

It must be countable (i.e. in bijection with $\Bbb N$), but not necessarily $=\Bbb N$. 
Let $f: \Bbb N \to \bigcup_n A_n$ be given by $f(n) = \min A_n$. It is clear that $f$ is injective, so $\bigcup_n A_n$ is at least as large as $\Bbb N$. Now since the countable union of finite sets is at most countable, it follows that $\cup_n A_n$ is countable.
As suggested in the comments, the sequence $(A_n)$ where $A_n := \{3n\}$ satisfies the requirement, yet the union is not $\Bbb N$.
