Equivalence of Definitions of Ultrafilter on Set

Proof
Let $S$ be a set.

Definition 1 implies Definition 4
Let $\mathcal F \subseteq \mathcal P \left({S}\right)$ a filter on $S$ which fulfills the condition:
 * 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 $U \subseteq S$.

$U \notin \mathcal F$.

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

Suppose there exists $C \in \mathcal F$ such that:
 * $C \cap V = \varnothing$

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

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

Hence the set $\Omega = \left\{ {C \cap V: C \in \mathcal F}\right\}$ 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 $\mathcal G$ on $S$ that contains $\mathcal F$ and $\left\{ {V}\right\}$.

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 4 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 $U^c \in S$.

Because $\mathcal F$ has the finite intersection property, it follows that $\varnothing \notin \mathcal F$.

As $\varnothing \notin \mathcal F$, it follows that $\varnothing^c = S \in \mathcal F$.

Let $\mathcal G$ be a filter on $S$ such that $\mathcal F \subseteq \mathcal G$.

Assume that $\mathcal F \subsetneq \mathcal G$.

Then there exists $A \in \mathcal G \setminus \mathcal F$.

Since $\varnothing \notin \mathcal G$ this implies that $\complement_S \left({A}\right) \notin \mathcal G$.

As $\mathcal F \subsetneq \mathcal G$, it follows that $\complement_S \left({A}\right) \notin \mathcal F$.

Therefore neither $A \in \mathcal F$ nor $\complement_S \left({A}\right) \in \mathcal F$, a contradiction to our assumption.

Thus $\mathcal F = \mathcal G$, which implies that $\mathcal F$ is an ultrafilter.