Way Below in Ordered Set of Topology

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Theorem

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

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

Let $x, y \in \tau$.


Then:

$x \ll y$ in $\struct {\tau, \preceq}$

if and only if

for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then:
there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$

where $\ll$ denotes the way below relation.


Proof

Sufficient Condition

Let

$x \ll y$

Let $F$ be a set of open subsets of $S$ such that

$y \subseteq \bigcup F$

By definition of subset:

$F \subseteq \tau$

By proof of Topology forms Complete Lattice:

$\sup F = \bigcup F$

By assumption:

$y \preceq \sup F$

By Topology forms Complete Lattice:

$\struct {\tau, \preceq}$ is complete lattice.

By Way Below in Complete Lattice:

$\exists G \in \map {\it Fin} F: x \preceq \sup G$

where $\map {\it Fin} F$ denotes the set of all finite subsets of $F$.

Thus by proof of Topology forms Complete Lattice:

$x \subseteq \bigcup G$

$\Box$


Necessary Condition

Assume that

for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then
there exists a finite subset $G$ of $F$: $x \subseteq \bigcup G$

Let $I$ be an ideal in $\struct {\tau, \preceq}$ such that

$y \preceq \sup I$

By proof of Topology forms Complete Lattice:

$y \subseteq \bigcup I$

By assumption:

there exists a finite subset $G$ of $I$: $x \subseteq \bigcup G$

By proof of Topology forms Complete Lattice:

$x \preceq \sup I$

By definition of ideal:

$I$ is directed and lower.

By Directed iff Finite Subsets have Upper Bounds:

$\exists u \in I: u$ is upper bound for $G$

By definition of supremum:

$\sup G \preceq u$

By definition of transitivity:

$x \preceq u$

Thus by definition of lower section:

$x \in I$

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:

$x \ll y$

$\blacksquare$


Sources

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