$\mathbb{N} ⊇ A_1 \supset A_2 \supset A_3 \supset \cdots$ but $\bigcap_{n=1}^∞ A_n$ is infinite? What is an example of an infinite intersection of infinite sets is infinite?
I know that the intersection of infinite sets does not need to be infinite. However, I am seeking for an explicit example where it is infinite
I was thinking A n = { [n,infinity) }
 A: Trivially, you could intersect an infinite set with itself. For instance, consider the set of real numbers $\mathbb{R}$. Then
$$\bigcap_{i=1}^{\infty} \mathbb{R} = \mathbb{R}.$$
If you want something nontrivial, consider
$$\bigcap_{i=1}^{\infty} (-\frac{1}{i}-1,\frac{1}{i}+1) = (-1,1).$$
In general, say you have an infinite collection of sets $A=\{A_1,A_2,...\}$ such that $A_1 \subseteq A_2 \subseteq ...$ Then
$$\bigcap_{i=1}^{\infty} A_i = A_1.$$
Update: With the condition $A_n \subset \mathbb{N}$ (proper subset), we can use prime numbers. Let $A_n$ be defined numbers divisible by say $2^n$ or divisible by 3. That is
$$A_n := \{x \in \mathbb{N} : 2^n \mid x \text{ or } 3 \mid x\} = \{x \in \mathbb{N} : 2^n \mid x \} \cup \{x \in \mathbb{N} : 3 \mid x \}.$$
Then (without proof) 
$$... \subset A_3\subset A_2 \subset A_1$$
and
$$\bigcap_{i=1}^{\infty}A_i = \{x \in \mathbb{N} : 3 \mid x \}$$
which is infinite.
A: Let $(B_n)$ be a sequence of sets with empty intersection and $B_{n+1} \subset B_n$, say $B_n =\{n,n+1,...\}$. Now take $A_n =B_n \cup E$. Then  intersection of $A_n$ 's is $E$. You can take $E$ to be any set, in particular a finite set or an infinite set.
A: Let $x_k = 2k$ and define $A_n = \{ \mathbb{N} \setminus \{x_k\}_{k=1}^n \}$.
$$\bigcap_{n=1}^\infty A_n = \text{set of odd natural integers}$$
A: If $A_n$ is $\mathcal{N}$ without the first $n$ prime numbers, then the $A_n$ are nested subsets with an infinite intersection, the non-prime integers.
