Proving that the intersection over a general index is the null set

Question

Prove that the generalized intersection of the interval $[n,n+2]$ is the empty set. I can prove this by contradiction by assuming that there there is an $x$ in the intersection such that $n+2<x.$ I am just not sure where to go from there.

Answer 1

Suppose $X:=\bigcap_{n \in \mathbb Z}[n,n+2]$ is non-empty and let $x\in X.$ Then we have $n \leq x \leq n+2, \forall n \in \mathbb Z.$ In particular, $-3 \leq x \leq -1$ and $1 \leq x \leq 3.$ This gives a contradiction.