Compact Closure is Directed

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Theorem

Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.

Let $x \in S$.


Then $x^{\mathrm{compact} }$ is directed

where $x^{\mathrm{compact} }$ denotes the compact closure of $x$.


Proof

By Bottom in Compact Closure:

$\bot \in x^{\mathrm{compact} }$

where $\bot$ denotes the smallest element in $L$.

Thus by definition:

$x^{\mathrm{compact} }$ is non-empty.

Let $y, z \in x^{\mathrm{compact} }$

By definition of compact closure:

$y$ and $z$ are compact elements and $y \preceq x$ and $z \preceq x$

By definitions of supremum and upper bound:

$y \vee z \preceq x$

By definition of compact element:

$y \ll y$ and $z \ll z$

where $\ll$ denotes the way below relation.

By Join Succeeds Operands:

$y \preceq y \vee z$ and $z \preceq y \vee z$

By Preceding and Way Below implies Way Below and definition of reflexivity:

$y \ll y \vee z$ and $z \ll y \vee z$

By Join is Way Below if Operands are Way Below:

$y \vee z \ll y \vee z$

By definition:

$y \vee z$ is compact element.

By definition of compact closure:

$y \vee z \in x^{\mathrm{compact} }$

Thus

$\exists v \in x^{\mathrm{compact} }: y \preceq v \land z \preceq v$

$\blacksquare$


Sources