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?

Question

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.

Answer 1

There is no number that is a multiple of every integer. For example, $n$ can not be a multiple of $n+1$.

Answer 2

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.