Way Below if Between is Compact Set in Ordered Set of Topology

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $L = \struct {\tau, \preceq}$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$

Let $x, y \in \tau$ such that:

$\exists H \subseteq S: x \subseteq H \subseteq y \land H$ is compact


Then:

$x \ll y$


Proof

Let $D$ be a directed subset of $\tau$ such that:

$y \preceq \sup D$

By proof of Topology forms Complete Lattice:

$\ds y \subseteq \bigcup D$

By Subset Relation is Transitive:

$\ds H \subseteq \bigcup D$

By definition:

$D$ is open cover of $H$

By definition of compact:

$H$ has finite subcover $\GG$ of $D$

By Directed iff Finite Subsets have Upper Bounds:

$\exists d \in D: d$ is upper bound for $\GG$

By definitions of upper bound and of $\preceq$:

$\forall z \in \GG: z \subseteq d$

By Union is Smallest Superset/Set of Sets:

$\ds \bigcup \GG \subseteq d$

By definition of cover:

$\ds H \subseteq \bigcup \GG$

By Subset Relation is Transitive:

$x \subseteq d$

Thus by definition of $\preceq$:

$x \preceq d$

Thus by definition of way below relation:

$x \ll y$

$\blacksquare$


Sources