=== 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: :$A \cap B = \varnothing$ :$A \cup B \in \mathcal F$ {{AimForCont}} $A \notin \mathcal F$ 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 $U \in \mathcal F$. {{begin-eqn}} {{eqn | l = A       | o = \notin       | r = \mathcal F       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = U \cap A       | r = \varnothing       | c = {{Defof|Filter on Set/Definition 1|Filter}}: Axiom $(4)$: $U \cap A \subseteq A \implies A \in \mathcal F$ }} {{eqn | ll= \leadsto       | l = U       | o = \subseteq       | r = \complement_S \left({A}\right)       | c = Empty Intersection iff Subset of Complement }} {{eqn | l = B       | o = \notin       | r = \mathcal F       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = U \cap B       | r = \varnothing       | c = {{Defof|Filter on Set/Definition 1|Filter}}: Axiom $(4)$: $U \cap B \subseteq B \implies B \in \mathcal F$ }} {{eqn | ll= \leadsto       | l = U       | o = \subseteq       | r = \complement_S \left({B}\right)       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \leadsto       | l = U       | o = \subseteq       | r = \complement_S \left({A}\right) \cap \complement_S \left({B}\right)       | c = Intersection is Largest Subset }} {{eqn | ll= \leadsto       | l = U       | o = \subseteq       | r = \complement_S \left({A \cup B}\right)       | c = De Morgan's Laws: Relative Complement of Union }} {{eqn | ll= \leadsto       | l = U \cap \left({A \cup B}\right)       | r = \varnothing       | c = Empty Intersection iff Subset of Complement }} {{end-eqn}} 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. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = A \cup \complement_S \left({A}\right)       | r = S       | c = Union with Relative Complement }} {{eqn | ll= \leadsto       | l = A \cup \complement_S \left({A}\right)       | o = \in       | r = \mathcal F       | c = {{Defof|Filter on Set/Definition 1|Filter}}: Axiom $(1)$: $S \in \mathcal F$ }} {{eqn | l = A \cap \complement_S \left({A}\right)       | r = \varnothing       | c = Intersection with Relative Complement is Empty }} {{eqn | ll= \leadsto       | l = \left({A \in \mathcal F}\right)       | o = \lor       | r = \left({\complement_S \left({A}\right) \in \mathcal F}\right)       | c = by hypothesis: {{Defof|Ultrafilter on Set/Definition 2|Ultrafilter by Definition 2}} }} {{end-eqn}} So $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 3. {{qed|lemma}} === 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 $\mathcal F \subseteq \mathcal G$. {{AimForCont}} $\mathcal F \subsetneq \mathcal G$. 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.