The infimum of the set of reciprocals of the natural numbers is zero.

Let $S=\{\frac{1}{n}:n\in \mathbb{N}\}$
The infimum $infS=0$.

Proof

for all $s\in S$, $s>0$, so $0$ a lower bound.

Suppose for the sake of contradiction there is some $w>0$ such that $w$ is the greatest lower bound.
Then $\frac{1}{w}>0$
By the Archimedean property, there is some $y\in \mathbb{N}$ such that $\frac{1}{w}<y$.
Then $w>\frac{1}{y}$.
But $y\in \mathbb{N}$, so $w\in S$
so $w$ not a lower bound (a contradiction)
reductio ad absurdum, QED.