Compact Closure is Subset of Way Below Closure

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Theorem

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

Let $x \in S$.


Then $x^{\mathrm {compact} } \subseteq x^\ll$

where

$x^{\mathrm {compact} }$ denotes the compact closure of $x$,
$x^\ll$ denotes the way below closure of $x$.


Proof

Let $y \in x^{\mathrm {compact} }$.

By definition of compact closure:

$y \preceq x$ and $y$ is compact.

By definition of compact:

$y \ll y$

where $\ll$ denotes the way below relation.

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

$y \ll x$

Thus by definition of way below closure:

$y \in x^\ll$

$\blacksquare$


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