Find the closure of the set $M=\left\{\frac{1}{n}\::\:n\in \mathbb{N}\right\}$ on the line in the Zariski topology.
As I understand it, closed sets in the Zariski topology are all one-point sets, but then the closure of the set $M$ will be the intersection of all closed sets containing the set $M$. That is, the closure of the set $M$ will be the set $M$ itself.
In the affine line $\mathbb{A}^1_k=\operatorname{Spec} k[t]$, the closure of the set $M$ will be $\mathbb{A}^1_k$ in the Zariski topology.
Note the closed subsets in $\mathbb{A}^1_k$ are given by $V(\mathfrak a)$ for an ideal $\mathfrak a$ in $k[t]$. $k[t]$ being PID, any such ideal $\mathfrak a$ will be principal. So $\mathfrak a=(f)$ for some polynomial $f\in k[t].$
But since a non-zero $f$ has atmost finitely many roots, $V(f)$ will have atmost finitely many points. Now $M$ being infinite, closure of $M$ will be entire $\mathbb{A}^1_k$.