Proof by contraction for intersection over general index

Question

Use a contradiction argument to justify that

$\bigcap_{n\in\mathbb{N}}[n,n+2] = \emptyset$

is a true statement.

How would I go about proving the negation, that is

$\bigcap_{n\in\mathbb{N}}[n,n+2] \neq \emptyset$

Answer 1

Suppose that $\displaystyle\bigcap_{n\in\mathbb{N}}[n,n+2] \ne \emptyset$. Then $\displaystyle\exists x\in\bigcap_{n\in\mathbb{N}}[n,n+2] $.

This implies that $x\in[1,3]$ and $x\in[4,6]$, which is impossible.

So, $\displaystyle\bigcap_{n\in\mathbb{N}}[n,n+2] = \emptyset$.