Intersection of filters
Let and
be filters on a same set
.
Then is also a filter on
.
Proof:
belongs to both
and
, hence it belongs to
, and
.
Since is not an element of
, it is not an element of
.
Let be an element of
and
a superset of
. Then
and
; hence
is an element of
. Similarly
; hence
.
Let and
be elements of
. Then they are both elements of
, hence their intersection belongs to
. Similarly their intersection belongs to
. Hence
.
Note:
The filter is also equal to the set of unions of an element
of
and an element
of
. In effect, such an
is a superset of
, hence belongs to
, and a superset of
, hence belongs to
, and so belongs to
. Conversely, an element of
belongs both to
and to
, and, being the union of itself with itself, is the union of an element of
with an element of
.