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 $A_n \not = \emptyset$

If the $A_i$ are nonempty and disjoint, each must contain a distinct integer. Therefore any finite subset of $\mathbb{N}$ can only contain a finite number of the $A_i$, since $\bigcup_{i=0}^N A_i$ contains at least $N$ distinct integers.
In conclusion, yes, the union must be infinite under these conditions, but as pointed out in the comments, it need not be all of $\mathbb{N}$.
