Cantor Intersection Theorem in $R^n$ I am looking at the Cantor Intersection Theorem from Apostol's Mathematical analysis.
Let {$Q_1, Q_2, $...} be a countable collection of nonempty sets in $R^n$ such that:
1) $Q_{k+1} \subset Q_k$
2) Each set $Q_k$ is closed and $Q_1$ is bounded.
Then the intersection $\bigcap_{k=1}^\infty Q_k$ is closed and nonempty.
Proof. Let $S$ = $\bigcap_{k=1}^\infty Q_k$. Then $S$ is closed since it is a countable union of closed sets. To show that $S$ is nonempty, we exhibit a point x in $S$. We can assume that each $Q_k$ contains infinitely many points; otherwise the proof is trivial. 
The rest of the proof proceeds by applying Bolzano Weierstrass Thm and I understand it but it is the above part I don't get. Why is the proof trivial if we assume that some $Q_k$ contains only finite points. And if the theorem is obvious when some $Q_k$ contains only finite points obvious, isn't it even more obvious if every set is infinite?
Can anyone point out the problem with my understanding? Thanks.
 A: If $Q_k$ has only finitely many points, then we have that $|Q_k|\geq|Q_{k+1}|\geq\ldots$, this is a decreasing sequence of positive integers, so it must stabilize. This means that for some $n$, every $m>n$ satisfies $Q_m=Q_n$, so clearly the intersection is non-empty, since it is equal to $Q_n$ as well.
The problem when each $Q_k$ is infinite is that infinite sets allow us to push things around, and since removing points from infinite sets might not change it is possible that at every step of the way $|Q_k|=|Q_n|$, but their intersection is suddenly empty after all. Think about the intersection of $A_n=\{k\in\Bbb N\mid k>n\}$ -- all these sets are infinite, but their intersection is empty!
A: If some $Q_k$ is finite, then write $Q_k = \{x_1,x_2,\ldots, x_n\}$. Suppose $\cap Q_k = \emptyset$, then for each $1\leq i\leq n, \exists j = j_i$ such that $x_i \notin Q_{j_i}$. Since $Q_{j_i}$ contains $Q_n$ for all $n\geq j_i$, we may assume that $j_i\geq k$. 
This is true for each $i$, hence if we choose $m = \max\{j_1, j_2, \ldots , j_n\}$, then
$$
x_i \notin Q_m \quad\forall 1\leq i\leq n
$$
Since $Q_m \subset Q_n$ it follows that $Q_m = \emptyset$ which is a contradiction.
Note how crucially we used the fact that $Q_k$ was finite - if it weren't, this process of choosing $j_i$'s would never end. For instance, consider $Q_k = (k,\infty) \subset \mathbb{R}$
