Equivalence of Definitions of Ultrafilter on Set

Proof
Let $S$ be a set.

Definition 1 implies Definition 4
Let $\FF \subseteq \powerset S$ a filter on $S$ which fulfills the condition:
 * whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$

By definition of filter, $\FF$ satisfies the Finite Intersection Property and is a non-empty set of subsets of $S$.

Let $U \subseteq S$.

$U \notin \FF$.

Let $V = U^c$, where $U^c$ denotes the relative complement of $U$ in $S$.

Suppose there exists $C \in \FF$ such that:
 * $C \cap V = \O$

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

Thus $U \in \FF$, which contradicts our hypothesis that $U \notin \FF$.

Hence the set $\Omega = \set {C \cap V: C \in \FF}$ is a non-empty set of non-empty set.

Note that $\Omega$ is downward directed, that is, $\Omega$ is a filter basis.

From Filter Basis Generates Filter, $\Omega$ generates a filter $\GG$ on $S$ that contains $\FF$ and $\set V$.

By hypothesis, we have:
 * $\FF =\GG$

Hence $U^c = V \in \FF$.

Thus $\FF$ is an ultrafilter by Definition 2.

Definition 4 implies Definition 1
Let $S$ be a non-empty set and $\FF$ a non-empty set of subsets on $S$ which fulfills the conditions:


 * $\FF$ has the finite intersection property
 * For all $U \subseteq S$, either $U \in \FF$ or $U^c \in \FF$.

Because $\FF$ has the finite intersection property, it follows that $\O \notin \FF$.

As $\O \notin \FF$, it follows that $\O^c = S \in \FF$.

Let $U, V \in \FF$.

Then either $U \cap V \in \FF$ or $\paren {U \cap V}^c \in \FF$.

However we observe that:
 * $\paren {U \cap V}^c \cap U \cap V = \O$

By finite intersection property, it follows that $U \cap V \in \FF$.

Now let $U \in \FF$ and $V \in S$ such that $U \subseteq V \subseteq S$.

Then either $V \in \FF$ or $V^c \in \FF$.

However we observe that:
 * $V^c \cap U \subseteq V^c \cap V = \O$

By finite intersection property, it follows that $V \in \FF$.

Therefore $\FF$ is a filter.

Let $\GG$ be a filter on $S$ such that $\FF \subseteq \GG$.

Assume that $\FF \subsetneq \GG$.

Then there exists $A \in \GG \setminus \FF$.

Since $\O \notin \GG$ this implies that $A^c \notin \GG$.

As $\FF \subsetneq \GG$, it follows that $A^c \notin \FF$.

Therefore neither $A \in \FF$ nor $A^c \in \FF$, a contradiction to our assumption.

Thus $\FF = \GG$, which implies that $\FF$ is an ultrafilter by definition 1.