Let $k_i = \{x \in \mathbb N, m\in \mathbb N: x=im\}$ for $i \in \mathbb N$ and $i>1$. Is the $\cap_{i=1}^{\infty} k_i$ empty? I want to give a counter example to the statement saying that: 
Given a collection of closed (not necessarily bounded) sets where the finite intersection of any of these sets is nonempty, the infinite intersection is also nonempty.
The counter example is:
$k_i = \{x \in \mathbb N, m\in \mathbb N: x=im\}$ for $i \in \mathbb N$ and $i>1$.
For $i=2$, $k_2 = \{2,4,6,...\}$.
For $i=3$, $k_3 = \{3,6,9,...\}$.
.....
There may exist other examples to show this, but this came to my mind and I want to see if it works.
I don't know number theory yet.
Thank you.

Yes it works. But $m$ belongs to $k_1,\ldots,k_n$ then all the primes $p_1,\ldots,p_r \le n$ divides $m$. In particular, $m\ge p_1\cdots p_r$. But $r$ is not finite hence such $m$ cannot exist.
