How do i go about proving that the set $\bigcap \{A_n: n \in \mathbb{N} \}$ is an empty set?

Question

I am working on the exercises in Introduction to Real Analysis by Bartle, Sherbert, am having trouble on how to make an argument for number 3.

$ \text{For each } n \in \mathbb{N}, A_n = \{(n + 1)k : k \in \mathbb{N}\}$

$ \text{Given } A_n = \{(n + 1)k : k \in \mathbb{N}\}$

1. What is $A_1 \cap A_2?$

$\text{We have}$

$ A_1 = \{ 2k: k \in \mathbb{N}\}$

$ A_2 = \{3k: k \in \mathbb{N} \}$

$ A_1 \cap A_2 = \{ 6k: k \in \mathbb{N} \}$

2. Determine $\bigcup\{A_n: n \in \mathbb{N} \}$ ?

$\text{We have}$

$ A_1 = \{ 2k: k \in \mathbb{N}\}$

$ A_2 = \{3k: k \in \mathbb{N} \}$

$ A_3 = \{4k: k \in \mathbb{N} \}$

$ A_4 = \{5k: k \in \mathbb{N} \}$

$\vdots$

$\bigcup\{A_n: n \in \mathbb{N}\} = A_1 \cup A_2 \cup A_3 \cdots A_n \text{where } n \in \mathbb{N} $

$= \{2k, 3k, 4k \cdots, (n + 1)k: n \in \mathbb{N}\}$

$= \{k: k > 1, k \in \mathbb{N}\}$

$= \mathbb{N}\setminus \{1\}$

3. Determine $\bigcap\{A_n: n \in \mathbb{N}\}$ ?

$\text{Let } \alpha \in \bigcap\{A_n: n \in \mathbb{N}\} $

$\implies \alpha \in A_1, \alpha \in A_2 \cdots \alpha \in A_n$

Answer 1

One simple way to show this is to note $ n \notin A_n $, as $ n < (n + 1) \leq (n + 1)k$ for any $k \in \mathbb{N}$ (here we are excluding $0$ from the naturals). Therefore no natural number can be in all $A_n$.