We will prove that :$x^\gg$ is way below open. Let $z \in x^\gg$. By definition of way above closure: :$x \ll z$ By Way Below has Interpolation Property: :$\exists x' \in S: x \ll x' \land x' \ll z$ Thus by definition of way above closure: :$x' \in x^\gg$ Thus :$x' \ll z$ {{qed|lemma}} Then :$\forall z \in x^\gg: \exists y \in x^\gg: y \ll z$ By Axiom of Choice define a mapping $f: x^\gg \to x^\gg$: :$\forall z \in x^\gg: f\left({z}\right) \ll z$ By definition of way above closure: :$y \in x^\gg$ Define $V := \left\{ {z^\succeq: \exists n \in \N: z = f^n\left({y}\right)}\right\}$ We will prove that :$\forall X, Y \in V: \exists Z \in V: X \cup Y \subseteq Z$ Let $X, Y \in V$. By definition of $V$: :$\exists z_1 \in S: X = z_1^\succeq \land \exists n_1 \in \N: z_1 = f^{n_1}\left({y}\right)$ and :$\exists z_2 \in S: Y = z_2^\succeq \land \exists n_2 \in \N: z_2 = f^{n_2}\left({y}\right)$ We will prove that :$\forall n, k \in \N: f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$ Let $n \in \N$. '''Base Case''': By definition of reflexivity: :$f^{n+0}\left({y}\right) \preceq f^n\left({y}\right)$ '''Induction Hypothesis''': :$f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$ '''Induction Step''': By definition of $f$: :$f^{n+k+1}\left({y}\right) \ll f^{n+k}\left({y}\right)$ By Way Below implies Preceding: :$f^{n+k+1}\left({y}\right) \preceq f^{n+k}\left({y}\right)$ By Induction Hypothesis and definition of transitivity: :$f^{n+k+1}\left({y}\right) \preceq f^{n}\left({y}\right)$ {{qed|lemma}} WLOG: suppose $n_1 \le n_2$ Then :$\exists k \in \N: n_2 = n_1+k$ Then :$z_2 \preceq z_1$ By Preceding iff Meet equals Less Operand: :$z_1 \wedge z_2 = z_2$ By definition of $V$: :$Z := \left({z_1 \wedge z_2}\right)^\succeq \in V$ By Meet Precedes Operands: :$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$ By Upper Closure is Decreasing: :$z_1^\succeq \subseteq Z$ and $z_2^\succeq \subseteq Z$ Thus by Union of Subsets is Subset: :$X \cup Y \subseteq Z$ {{qed|lemma}} Define $F := \bigcup V$. We will prove that :$F$ is way below open. Let $u \in F$. By definition of union: :$\exists Y \in V: u \in Y$ By definition of $V$: :$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$ By definition of $f$: :$z \in x^\gg$ By definition of $f$: :$f\left({z}\right) \ll z$ Then :$z' := f\left({z}\right) = f^{n+1}\left({y}\right)$ By definition of $V$: :${z'}^\succeq \in V$ By definition of reflexivity: :$z' \preceq z'$ By definition of upper closure of element: :$z' \in {z'}^\succeq$ By definition of union: :$z' \in F$ By definition of upper closure of element: :$z \preceq u$ By Preceding and Way Below implies Way Below: :$z' \ll u$ Hence :$\exists g \in F: g \ll u$ {{qed|lemma}} By Upper Closure of Element is Filter: :$\forall X \in V: X$ is a filtered upper set. By Union of Upper Sets is Upper: :$F$ is an upper set. By Union of Filtered Sets is Filtered: :$F$ is filtered. By definition of filter: :$F$ is a way below open filter in $L$. Then :$f^0\left({y}\right) = y$ By definition of $V$: :$y^\succeq \in V$ By definitions of upper closure of element and reflexivity: :$y \in y^\succeq$ By definition of union: :$y \in F$ It remains to prove that :$F \subseteq x^\gg$ Let $u \in F$. By definition of union: :$\exists Y \in V: u \in Y$ By definition of $V$: :$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$ By definition of $f$: :$z \in x^\gg$ By definition of way above closure: :$x \ll z$ By definition of upper closure of element: :$z \preceq u$ and $x \preceq x$ By Preceding and Way Below implies Way Below: :$x \ll u$ Thus by definition of way above closure: :$u \in x^\gg$ {{qed}}