Filter Basis Generates Filter

Theorem
Let $$X$$ be a set and $$\mathcal{B} \subset \mathcal{P}(X)$$. Then $$\mathcal{F} := \{ F \subseteq X | \exists U \in \mathcal{B}: U \subseteq V\}$$ is a filter on $$X$$ iff the following conditions hold:
 * 1) For all $$V_1, V_2 \in \mathcal{B}$$ exists $$U \in \mathcal{B}$$ such that $$U \subseteq V_1 \cap V_2$$
 * 2) $$\emptyset \not \in \mathcal{B}$$ and $$\mathcal{B} \ne \emptyset$$

Any such $$\mathcal{B}$$ is called a filter basis. $$\mathcal{F}$$ is said to be generated by $$\mathcal{B}$$.

Proof
Assume first that $$\mathcal{F}$$ is a filter on $$X$$. Then $$X \in \mathcal{F}$$ and thus $$\mathcal{B} \ne \emptyset$$. Because $$\emptyset \not \in \mathcal{F}$$ we have that $$\emptyset \not \in \mathcal{B}$$ since $$\mathcal{B} \subseteq \mathcal{F}$$. Let $$V_1, V_2 \in \mathcal{V}$$, then $$V_1, V_2 \in \mathcal{F}$$. Because $$\mathcal{F}$$ is a filter it follows that $$V := V_1 \cap V_2 \in \mathcal{F}$$. The definition of $$\mathcal{F}$$ implies therefore that there is $$U \in \mathcal{B}$$ such that $$U \subseteq V = V_1 \cap V_2$$.

Assume now that $$\mathcal{B}$$ satisfies conditions (1) and (2). To show that $$\mathcal{F}$$ is a filter note that because $$\mathcal{B} \ne \emptyset$$ there is a $$B \in \mathcal{B}$$. Because $$\mathcal{B} \subset \mathcal{P}(X)$$ we know that $$B \subseteq X$$ and thus $$X \in \mathcal{F}$$ by the definition of $$\mathcal{F}$$. Since the only subset of $$\emptyset$$ is $$\emptyset$$ and since $$\emptyset \not \in \mathcal{B}$$ it follows that $$\emptyset \not \in \mathcal{F}$$. Let $$V_1, V_2 \in \mathcal{F}$$. Then there exist $$U_1, U_2 \in \mathcal{B}$$ such that $$U_1 \subseteq V_1$$ and $$U_2 \subseteq V_2$$. Because $$\mathcal{B}$$ satisfies condition (1) there exists a set $$U \in \mathcal{B}$$ for which $$U \subseteq U_1 \cap U_2$$ holds. Since $$U_1 \cap U_2 \subseteq V_1 \cap V_2$$ this implies that $$U \subseteq V_1 \cap V_2$$ and therefore $$V_1 \cap V_2 \in \mathcal{F}$$. $$\square$$