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

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## Contents

## Theorem

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

Let $X$ be an inaccessible by directed suprema subset of $S$.

Then $\complement_S\left({X}\right)$ is closed under directed suprema.

## Proof

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

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

By Empty Intersection iff Subset of Relative Complement:

- $D \cap X = \varnothing$

By definition of inaccessible by directed suprema:

- $\sup D \notin X$

Thus by definition of relative complement:

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

$\blacksquare$

## Also See

## Sources

- Mizar article WAYBEL11:funcreg 3