Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.

Let $x, y \in S$ such that
 * $x \ll y$

where $\ll$ denotes the way below relation.

Then there exists a way below open filter in $L$: $y \in F \land F \subseteq x^\gg$

where $x^\gg$ denotes the way above closure of $x$.

Proof
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
 * $x' \ll z$

Then:
 * $\forall z \in x^\gg: \exists y \in x^\gg: y \ll z$

By Axiom of Choice define a mapping $f: x^\gg \to x^\gg$:
 * $\forall z \in x^\gg: \map f z \ll z$

By definition of way above closure:
 * $y \in x^\gg$

Define $V := \set {z^\succeq: \exists n \in \N: z = \map {f^n} y}$

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 $V$:
 * $\exists z_1 \in S: X = z_1^\succeq \land \exists n_1 \in \N: z_1 = \map {f^{n_1} } y$

and
 * $\exists z_2 \in S: Y = z_2^\succeq \land \exists n_2 \in \N: z_2 = \map {f^{n_2} } y$

We will prove that
 * $\forall n, k \in \N: \map {f^{n + k} } y \preceq \map {f^n} y$

Let $n \in \N$.

Base Case:

By definition of reflexivity:
 * $\map {f^{n + 0} } y \preceq \map {f^n} y$

Induction Hypothesis:


 * $\map {f^{n + k} } y \preceq \map {f^n} y$

Induction Step:

By definition of $f$:
 * $\map {f^{n + k + 1} } y \ll \map {f^{n + k} } y$

By Way Below implies Preceding:
 * $\map {f^{n + k + 1} } y \preceq \map {f^{n + k} } y$

By Induction Hypothesis and definition of transitivity:
 * $\map {f^{n + k + 1} } y \preceq \map {f^n} y$

, 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 $V$:
 * $Z := \paren {z_1 \wedge z_2}^\succeq \in V$

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:
 * $X \cup Y \subseteq Z$

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 $V$:
 * $\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = \map {f^n} y$

By definition of $f$:
 * $z \in x^\gg$

By definition of $f$:
 * $\map f z \ll z$

Then
 * $z' := \map f z = \map {f^{n + 1} } y$

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
 * $\exists g \in F: g \ll u$

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:
 * $\map {f^0} y = y$

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 $V$:
 * $\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = \map {f^n} y$

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:
 * $u \in x^\gg$