Closure is Smallest Closed Successor
Let $(S, \preceq)$ be an ordered set.
Let $f: S \to S$ be a closure operator on $S$.
Let $x \in S$.
Thus $x \preceq f(x)$.
By definition, $f(x)$ is closed.
So $f(x)$ is a closed element that succeeds $x$.
We will now show that it is the smallest such.
Let $k$ be a closed element of $S$ such that $x \preceq k$.
Therefore $f(x) \preceq f(k)$.
Since $k$ is closed, $f(k) = k$.
Thus $f(x) \preceq k$.