Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice

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 $14.11 \ \text{(b)}$