Precisely
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Intersection of filters

Let \mathcal F_1 and \mathcal F_2 be filters on a same set E.

Then \mathcal F = \mathcal F_1 \cap \mathcal F_2 is also a filter on E.

Proof:

E belongs to both \mathcal F_1 and \mathcal F_2, hence it belongs to \mathcal F, and \mathcal F \neq \emptyset.

Since \emptyset is not an element of \mathcal F_1, it is not an element of \mathcal F.

Let A be an element of \mathcal F and B \subseteq E a superset of A. Then A \in \mathcal F_1 and A \subseteq B \subseteq E; hence B is an element of \mathcal F_1. Similarly B \in \mathcal F_2; hence B \in \mathcal F.

Let A and B be elements of \mathcal F. Then they are both elements of \mathcal F_1, hence their intersection belongs to \mathcal F_1. Similarly their intersection belongs to \mathcal F_2. Hence A \cap B \in \mathcal F.

Note:

The filter \mathcal F = \mathcal F_1 \cap \mathcal F_2 is also equal to the set of unions of an element X_1 of \mathcal F_1 and an element X_2 of \mathcal F_2. In effect, such an X = X_1 \cup X_2 is a superset of X_1, hence belongs to \mathcal F_1, and a superset of X_2, hence belongs to \mathcal F_2, and so belongs to \mathcal F. Conversely, an element of \mathcal F belongs both to \mathcal F_1 and to \mathcal F_2, and, being the union of itself with itself, is the union of an element of \mathcal F_1 with an element of \mathcal F_2.