Closure is Smallest Closed Successor

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Let $\struct {S, \preceq}$ be an ordered set.

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

Let $x \in S$.

Then $\map f x$ is the smallest closed element that succeeds $x$.


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

Thus $x \preceq \map f x$.

By definition, $\map f x$ is closed.

So $\map f x$ 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 $\map f x \preceq \map f k$.

Since $k$ is closed, $\map f k = k$.

Thus $\map f x \preceq k$.