Proof. We need to check that
satisfies the
defining properties.
Property 1: Since every element of the set
contains
, we have
.
Property 2: For every element
of
such that
, it also is the case that
because an intersection of a family of sets is a subset of any member of the family. In other words (or rather, symbols),
hence

. By the first property proven above,

so

. Thus,

.
Property 3: Let
and
be two subsets of
such that
. Then if, for some other subset
of
, we have
, it follows that
. Hence,
so

.
