Filter Basis Generates Filter

Theorem
Let $$X$$ be a set, and $$\mathcal P \left({X}\right)$$ be the power set of $$X$$.

Let $$\mathcal B \subset \mathcal P \left({X}\right)$$.

Then:
 * $$\mathcal F = \left\{{V \subseteq X: \exists U \in \mathcal B: U \subseteq V}\right\}$$ is a filter on $$X$$

iff:
 * 1) $$\forall V_1, V_2 \in \mathcal B: \exists U \in \mathcal B: U \subseteq V_1 \cap V_2$$
 * 2) $$\varnothing \not \in \mathcal B, \mathcal B \ne \varnothing$$

That is, $$\mathcal B$$ is a filter basis of the filter $$\mathcal F$$, which is generated by $\mathcal B$.

Proof
Assume first that $$\mathcal F$$ is a filter on $$X$$.

Then $$X \in \mathcal F$$ and thus $$\mathcal B \ne \varnothing$$.

Because $$\varnothing \not \in \mathcal F$$ we have that $$\varnothing \not \in \mathcal B$$, 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 $$\mathcal B \ne \varnothing$$ there is a $$B \in \mathcal B$$.

Because $$\mathcal B \subset \mathcal P \left({X}\right)$$ we know that $$B \subseteq X$$ and thus $$X \in \mathcal F$$ by the definition of $$\mathcal F$$.

Since the only subset of $$\varnothing$$ is $$\varnothing$$ and since $$\varnothing \not \in \mathcal B$$ it follows that $$\varnothing \not \in \mathcal F$$.

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 $$V_1 \cap V_2 \in \mathcal F$$.