Equivalence of Definitions of Ultrafilter on Set

Definition 1 implies Definition 2
Let $S$ be a set and $\mathcal{F} \subseteq \mathcal{P} (S)$ a filter on $S$ which fulfills the condition:
 * if whenever $\mathcal{G}$ is a filter on $S$ and $\mathcal{F} \subseteq \mathcal{G}$ holds, then $\mathcal{F} = \mathcal{G}$

By definition of Filter, $\mathcal{F}$ satisfies the Finite Intersection Property and is a non-empty set of subsets of $S$.

Let $U \subseteq S$ and suppose $U \not \in \mathcal{F}$. Let $V=U^c$.

Suppose there exists $C \in \mathcal{F}$ such that
 * $C \cap V = \emptyset$

By Empty Intersection iff Subset of Complement, it follows that
 * $C \subset U$

Thus $U \in \mathcal{F}$, a contradiction.

Hence the set $\Omega = \{C \cap V : C \in \mathcal{F}\}$ is a non-empty collection of non-empty sets. Note that $\Omega$ is downward directed, that is, $\Omega$ is a filter basis.

From Filter Basis Generates Filter, $\Omega$ generates a filter $\mathcal{G}$ on $S$ that contains $\mathcal{F}$ and $\{V\}$.

By hypothesis, we have $\mathcal{F}=\mathcal{G}$. Hence $U^c=V \in \mathcal{F}$.

Thus $\mathcal{F}$ is an ultrafilter by Definition 2.

Definition 2 implies Definition 1
Let $S$ be a non-empty set and $\mathcal{F}$ a non-empty set of subsets on $S$ which fulfills the conditions:
 * $\mathcal{F}$ has the finite intersection property
 * For all $U \subseteq S$, either $U \in S$ or $U^c \in S$