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

From ProofWiki
Jump to navigation Jump to search

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