Power Set is Complete Lattice/Proof 1
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Theorem
Let $S$ be a set.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on the power set $\powerset S$ of $S$ by the relation $\subseteq$.
Then:
- $\struct {\powerset S, \subseteq}$ is a complete lattice
where for every subset $\mathbb S$ of $\powerset S$:
- the infimum of $\mathbb S$ necessarily admitted by $\mathbb S$ is $\bigcap \mathbb S$.
Proof
From Subset Relation on Power Set is Partial Ordering, we have that $\subseteq$ is a partial ordering.
We note in passing that for any set $S$:
- From Supremum of Power Set, $\powerset S$ has a supremum, that is, $S$ itself
- From Infimum of Power Set, $\powerset S$ has an infimum, that is, $\O$.
These are also the maximal and minimal elements of $\powerset S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then from Union is Smallest Superset:
- $\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
and from Intersection is Largest Subset:
- $\paren {\forall X \in \mathbb S: T \subseteq X} \iff T \subseteq \bigcap \mathbb S$
So $\bigcap \mathbb S$ is the infimum and $\bigcup \mathbb S$ is the supremum of $\struct {\mathbb S, \subseteq}$.
Hence by definition $\powerset S$ is a complete lattice.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Example $14.3$