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 $\mathcal F$ be an ultrafilter on $S$ by definition 1.

Thus $\mathcal F \subseteq \mathcal P \left({S}\right)$ is a filter on $S$ which fulfills the condition:
 * whenever $\mathcal G$ is a filter on $S$ and $\mathcal F \subseteq \mathcal G$ holds, then $\mathcal F = \mathcal G$.

Let $A \subseteq S$ and $B \subseteq S$ such that:
 * $A \cap B = \varnothing$
 * $A \cup B \in \mathcal F$

$A \notin \mathcal F$ and $B \notin \mathcal F$.

Consider the set $\mathcal B := \left\{{V \cap A: V \in \mathcal F}\right\} \cup \left\{{V \cap B: V \in \mathcal F}\right\}$.

This is a basis of a filter $\mathcal G$ on $S$, for which $\mathcal F \subseteq \mathcal G$ holds.

Let $U \in \mathcal F$.

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

But $\varnothing \notin \mathcal F$.

This contradicts our initial hypothesis.

Hence either:
 * $A \in \mathcal F$

or:
 * $B \in \mathcal F$

and so $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 2.

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

That is, $\mathcal F \subseteq \mathcal P \left({S}\right)$ is a filter on $S$ which fulfills the condition:
 * for every $A \subseteq S$ and $B \subseteq S$ such that $A \cap B = \varnothing$ and $A \cup B \in \mathcal F$, either $A \in \mathcal F$ or $B \in \mathcal F$.

Let $A \subseteq S$.

We have:

So $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 3.

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

That is, $\mathcal F \subseteq \mathcal P \left({S}\right)$ is a filter on $S$ which fulfills the condition:
 * for any $A \subseteq S$ either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$ holds.

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

$\mathcal F \subsetneq \mathcal G$.

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

By definition of filter, $\varnothing \notin \mathcal G$.

But from Intersection with Relative Complement is Empty:
 * $A \cap \complement_S \left({A}\right)$

and so:
 * $\complement_S \left({A}\right) \notin \mathcal G$

By hypothesis:
 * $\mathcal F \subsetneq \mathcal G$

and so:
 * $\complement_S \left({A}\right) \notin \mathcal F$

Therefore neither $A \in \mathcal F$ nor $\complement_S \left({A}\right) \in \mathcal F$.

This contradicts our assumption:
 * for any $A \subseteq S$ either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$ holds.

Thus:
 * $\mathcal F = \mathcal G$

and so $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 1.