Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice
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Theorem
Let $A$ be a set.
Let $\SS$ be a set of subsets of $A$ such that:
- $A \in \SS$
- for every non-empty subset $\TT$ of $\SS$, $\ds \bigcap \TT \in \SS$, where $\ds \bigcap \TT$ denotes the intersection of $\TT$.
Then:
- the ordered set $\struct {\SS, \subseteq}$ is a complete lattice
where:
- $\ds \bigcap \TT$ is the infimum necessarily admitted by $\TT$.
Proof
From Subset Relation is Ordering, $\struct {\SS, \subseteq}$ is indeed an ordered set.
We have by definition of $\SS$ that:
- $\forall H \in \SS: H \subseteq A$
Hence $A$ is the greatest element of $\SS$ with respect to the ordered set $\struct {A, \subseteq}$.
Then we have that:
- $\forall K \in \TT: \ds \bigcap \TT \subseteq K$
and so $\TT$ admits an infimum, which is $\ds \bigcap \TT$.
Hence the conditions of Ordered Set with Greatest Element whose Subsets have Infimum is Complete Lattice.
The result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(b)}$