Equivalence of Definitions of Ultrafilter on Set/Equivalence of Definitions 1, 2 and 3

Proof
Let $S$ be a set.

Definition 1 implies Definition 2
Let $\FF$ be an ultrafilter on $S$ by definition 1.

Thus $\FF \subseteq \powerset S$ is a filter on $S$ which fulfills the condition:
 * whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$.

Let $A \subseteq S$ and $B \subseteq S$ such that:
 * $A \cap B = \O$
 * $A \cup B \in \FF$

$A \notin \FF$ and $B \notin \FF$.

Consider the set $\BB := \set {V \cap A: V \in \FF} \cup \set {V \cap B: V \in \FF}$.

This is a basis of a filter $\GG$ on $S$, for which $\FF \subseteq \GG$ holds.

Let $U \in \FF$.

But by definition of a filter:
 * $U, V \in \FF \implies U \cap V \in \FF$

But $\O \notin \FF$.

This contradicts our initial hypothesis.

Hence either:
 * $A \in \FF$

or:
 * $B \in \FF$

and so $\FF$ fulfills the conditions to be an ultrafilter by Definition 2.

Definition 2 implies Definition 3
Let $\FF$ be an ultrafilter on $S$ by definition 2.

That is, $\FF \subseteq \powerset S$ is a filter on $S$ which fulfills the condition:
 * for every $A \subseteq S$ and $B \subseteq S$ such that $A \cap B = \O$ and $A \cup B \in \FF$, either $A \in \FF$ or $B \in \FF$.

Let $A \subseteq S$.

We have:

So $\FF$ fulfills the conditions to be an ultrafilter by Definition 3.

Definition 3 implies Definition 1
Let $\FF$ be an ultrafilter on $S$ by definition 3.

That is, $\FF \subseteq \powerset S$ is a filter on $S$ which fulfills the condition:
 * for any $A \subseteq S$ either $A \in \FF$ or $\relcomp S A \in \FF$ holds.

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

$\FF \subsetneq \GG$.

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

By definition of filter, $\O \notin \GG$.

But from Intersection with Relative Complement is Empty:
 * $A \cap \relcomp S A$

and so:
 * $\relcomp S A \notin \GG$

By hypothesis:
 * $\FF \subsetneq \GG$

and so:
 * $\relcomp S A \notin \FF$

Therefore neither $A \in \FF$ nor $\relcomp S A \in \FF$.

This contradicts our assumption:
 * for any $A \subseteq S$ either $A \in \FF$ or $\relcomp S A \in \FF$ holds.

Thus:
 * $\FF = \GG$

and so $\FF$ fulfills the conditions to be an ultrafilter by Definition 1.