Find $\bigcap_{k=0}^{\infty} \bigcup_{n=k}^{\infty}A_n$ Let $A_n = \{ m \in \mathbb{N} : \exists k \in \mathbb{N} \ \ km=n$} for $n \in \mathbb{N}$. I am trying to find
$$\bigcap_{k=0}^{\infty} \bigcup_{n=k}^{\infty}A_n$$
In general, this not a very difficult task, however, in this case it bothers me to see $k$ both in the definition of the set with $\exists$ and in the sum index. After all, we want $k$ to exist, not to choose it.
Answer based on user suggestions
$$\bigcap_{k=0}^{\infty} \bigcup_{n=k}^{\infty}A_n = \left( A_0 \cup A_1 \cup \dots \right) \cap \left( A_1 \cup A_2 \cup \dots \right) \cap \left( A_2 \cup A_3 \cup \dots \right) \cap \dots $$
Since $A_n$ is a set of divisors of $n$, $A_0 = \mathbb{N}, A_1 = \{1\}, A_2 = \{1, 2\}, A_3 = \{1, 3\}$ and so on. Moreover, $a \in A_ab$ for $a,b \in \mathbb{N}$. So $\bigcup_{n=k}^{\infty}A_n = \mathbb{N}$ and we have
$$\bigcap_{k=0}^{\infty} \bigcup_{n=k}^{\infty}A_n = \left( A_0 \cup A_1 \cup \dots \right) \cap \left( A_1 \cup A_2 \cup \dots \right) \cap \left( A_2 \cup A_3 \cup \dots \right) \cap \dots = \mathbb{N} \cap \mathbb{N} \cap \mathbb{N} \cap \dots = \mathbb{N}$$
The notation is ambiguous - the $k$ in the definition of $A_n$ has nothing to do with the $k$ in the indexes of the unions and intersections.
 A: It is best that we rephrase the question in layman terms, and then attempt to answer it.
$A_n$, as you can observe,is the set of divisors of $n$.
Suppose that $x \in \displaystyle\bigcap_{k=0}^\infty \displaystyle\bigcup_{n \geq k} A_n$.
This means, that for all $k$, $x \in \displaystyle\bigcup_{n \geq k} A_n$.
This means, that for all $k$, there is an $n$ larger than $k$, such that $x \in A_n$.
This means that for all $k$, there is $n$ larger than $k$, such that $x$ is a divisor of $n$.
Now, having done this, you know that the above statement is true for every $x$, because given $k$, you can take the first multiple of $x$ that comes after $k$, call it $n$, then $x \in A_n$.
Therefore, what you have to attempt to prove is that $\displaystyle\bigcap_{k=0}^\infty \displaystyle\bigcup_{n \geq k} A_n = \mathbb N$ !
We will now rigorously prove this.
How would you do this? One inclusion is obvious: Of course $\displaystyle\bigcap_{k=0}^\infty \displaystyle\bigcup_{n \geq k} A_n$ can consist only of natural numbers, because $A_n$ themselves contain only natural numbers.
The other way, let $x \in \mathbb N$.  Let $k \in \mathbb  N$. Let $l = \lceil \frac kx \rceil$,then note that $n=xl \geq k$ and $n=xl \implies x \in A_n$. Therefore, $x\in \displaystyle\bigcap_{k=0}^\infty \displaystyle\bigcup_{n \geq k} A_n$.
That completes the argument.

Actually, it turns out that $\cap_{k=0}^\infty \cup_{n \geq k} A_n$ has a name : the limit superior of $A_n$. This quantity may also be defined for sequences, using the fact that their images are sets.
Actually, it is easy to see from this definition that an element $x$ is in the limit superior of $A_n$ if and only if $x$ belongs to infinitely many $A_n$. This translates to : $x$ is a divisor of $n$ for infinitely many $n$, since $x\in A_n$ means it is a divisor of $n$. This is true for all $x$, so the claim above follows easily.
There is another quantity, the limit inferior, which is obtained by switching the $\cap$ and $\cup$ in the limsup definition. The limit inferior of the $A_n$ can be obtained using the dual fact that $x$ is in the liminf if and only if $x \in A_n$ eventually i.e. $x \notin A_m$ for only finitely many $m$. It is not difficult to see that the liminf of these $A_n$ is the set $\{1\}$. 
A: Hint. $A_n$ is the set of divisors of $n$ and
$$S:=\bigcap_{k=0}^{\infty} \bigcup_{n=k}^{\infty}A_n$$
consists of integers which belong to $A_n$ for infinitely many $n$.
$S$ is not empty because  $1\in A_n$ for all $n$ (note that $A_0$ is not empty!). 
Moreover if $m$ is a positive integer then $m\in A_{md}$ for all $d\geq 1$. Hence
$$\bigcup_{n=k}^{\infty}A_n\supset \bigcup_{m=1}^{\infty}\bigcup_{d=k}^{\infty}A_{md}\supset \bigcup_{m=1}^{\infty} \{m\}=\mathbb{N}$$
What may we conclude?
