## 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 .