# Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema

Jump to navigation
Jump to search

## Contents

## Theorem

Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.

Let $X$ be a closed under directed suprema subset of $S$.

Then $\complement_S\left({X}\right)$ is inaccessible by directed suprema.

## Proof

Let $D$ be a directed subset of $S$ such that

- $\sup D \in \complement_S\left({X}\right)$

By definition of relative complement:

- $\sup D \notin X$

By definition of closed under directed suprema:

- $D \nsubseteq X$

- $D \nsubseteq \complement_S\left({\complement_S\left({X}\right)}\right)$

Thus by Empty Intersection iff Subset of Relative Complement:

- $D \cap \complement_S\left({X}\right) \ne \varnothing$

$\blacksquare$

## Also See

## Sources

- Mizar article WAYBEL11:funcreg 4