Closure is Smallest Closed Successor

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Theorem

Let $(S, \preceq)$ be an ordered set.

Let $f: S \to S$ be a closure operator on $S$.

Let $x \in S$.


Then $f(x)$ is the smallest closed element that succeeds $x$.


Proof

By the definition of closure operator, $f$ is inflationary.

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 $x \preceq k$.

Since $f$ is a closure operator, it is increasing.

Therefore $f(x) \preceq f(k)$.

Since $k$ is closed, $f(k) = k$.

Thus $f(x) \preceq k$.

$\blacksquare$