Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 2

Proof
Let $\FF$ be an ultrafilter on $Y$.

For $S \in \FF$, let $\map f S = \inf S$.

Let $p = \sup \map f \FF$.

Then $\FF$ converges to $p$:

Upward rays
Let $a \in X$ with $a \prec p$.

Since $\FF$ is an ultrafilter, either $Y \cap {\uparrow} a \in \FF$ or $Y \cap {\bar \downarrow} a \in \FF$.

that $Y \cap {\bar \downarrow} a \in \FF$.

For each $S \in \FF$:
 * $S \cap {\bar \downarrow} a \in \FF$ because an ultrafilter is a filter.


 * $S \cap {\bar \downarrow} a \ne \O$ because a filter on a set is proper.

By applying the definition of supremum to $p$, it follows that there exists an $S \in \FF$ such that $a \prec \inf S$.

By the definition of infimum:
 * $S \cap {\bar \downarrow} a = \O$

which is a contradiction.

Thus $Y \cap {\uparrow} a \in \FF$.

Downward rays
Let $b \in X$ with $p \prec b$.

Either $Y \cap {\downarrow} b \in \FF$ or $Y \cap {\bar \uparrow} b \in \FF$.

that $Y \cap {\bar \uparrow} b \in \FF$.

Let $b' = \map \inf {Y \cap {\bar \uparrow} b}$.

We have that $b$ is a lower bound of $Y \cap {\bar \uparrow} b$

So by the definition of infimum:
 * $b \preceq b'$

Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity.

By the definition of $b'$ and the definition of $f$:
 * $b' \in \map f \FF$

But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $\map f \FF$.

By the definition of the order topology, the upward and downward rays containing each point form a neighborhood sub-basis for that point.

Thus by the Neighborhood Sub-Basis Criterion for Filter Convergence, $\FF$ converges.

Since every ultrafilter on $Y$ converges, $Y$ is compact by Equivalent Definitions of Compactness.