Closure Operator from Closed Elements

Theorem

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

Let $C \subseteq S$.

Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$.

Let $\cl: S \to S$ be defined as follows:

For $x \in S$:

$\map \cl x = \map \min {C \cap x^\succeq}$

where $x^\succeq$ is the upper closure of $x$.

That is, let $\map \cl x$ be the smallest successor of $x$ in $C$.

Then:

$\cl$ is a closure operator on $S$

The closed elements of $\cl$ are precisely the elements of $C$.

Proof

Inflationary

$x$ is a lower bound of $x^\succeq$.

Hence by Lower Bound for Subset, $x$ is also a lower bound of $C \cap x^\succeq$.

By the definition of smallest element, $x \preceq \map \cl x$.

$\Box$

Order-Preserving

Suppose that $x \preceq y$.

Then:

$C \cap y^\succeq \subseteq C \cap x^\succeq$

By Smallest Element of Subset:

$\map \cl x \preceq \map \cl y$

Idempotent

Let $x \in S$.

For each $x \in S$:

$\map \cl x = \map \min {C \cap x^\succeq}$

Thus:

$\map \cl x \in \paren {C \cap x^\succeq} \subseteq C$

That is to say, $\map \cl x$ is its own smallest successor in $C$.

Thus:

$\map \cl x = \map \cl {\map \cl x}$

$\Box$

When $x \in C$, $x$ is the minimum of $C \cap x^\succeq$

Hence, elements of $C$ are closed elements with respect to $\cl$.

Suppose that $x$ is closed with respect to $\cl$.

Then:

$x = \map \min {C \cap x^\succeq}$

so in particular:

$x \in C$

$\blacksquare$