Way Below in Lattice of Power Set

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Theorem

Let $X$ be a set.

Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be a lattice of power set of $X$ where $\mathord\preceq = \mathord\subseteq \cap \paren {\powerset X \times \powerset X}$

Let $x, y \in \powerset X$.


Then $x \ll y$ if and only if

for every a set $Y$ of subsets of $X$ such that $y \subseteq \bigcup Y$
then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$

where $\ll$ denotes the way below relation.


Proof

Sufficient Condition

Let $x \ll y$

Let $Y$ be a set of subsets of $X$ such that

$y \subseteq \bigcup Y$

By definitions of power set and subset:

$Y \subseteq \powerset X$

By Power Set is Complete Lattice:

$\bigcup Y = \sup Y$

By definition of $\preceq$:

$y \preceq \sup Y$

By Way Below in Complete Lattice:

there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

By the proof of Power Set is Complete Lattice:

$\bigcup Z = \sup Z$

Thus by definition of $\preceq$:

there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$

$\Box$


Necessary Condition

Suppose

for every a set $Y$ of subsets of $X$ such that $y \subseteq \bigcup Y$
then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$

We will prove that

for every a subset $Y$ of $\powerset X$ such that $y \preceq \sup Y$
then there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

Let $Y$ be a subset of $\powerset X$ such that

$y \preceq \sup Y$

By definition of power set:

$Y$ is a set of subsets of $X$.

By Power Set is Complete Lattice:

$\bigcup Y = \sup Y$

By definition of $\preceq$:

$y \subseteq \bigcup Y$

By assumption:

there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$

By the proof of Power Set is Complete Lattice:

$\bigcup Z = \sup Z$

Thus by definition of $\preceq$:

there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$

$\Box$


Thus by Way Below in Complete Lattice:

$x \ll y$

$\blacksquare$


Sources