# Equivalence of Definitions of Ultrafilter on Set

## Theorem

The following definitions of the concept of ultrafilter on a set $S$ are equivalent:

### Definition 1

Let $S$ be a set.

Let $\mathcal F \subseteq \powerset S$ be a filter on $S$.

Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

there is no filter on $S$ which is strictly finer than $\mathcal F$

or equivalently, if and only if:

whenever $\mathcal G$ is a filter on $S$ and $\mathcal F \subseteq \mathcal G$ holds, then $\mathcal F = \mathcal G$.

### Definition 2

Let $S$ be a set.

Let $\mathcal F \subseteq \mathcal P \left({S}\right)$ be a filter on $S$.

Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

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 $B \in \mathcal F$.

### Definition 3

Let $S$ be a set.

Let $\mathcal F \subseteq \mathcal P \left({S}\right)$ be a filter on $S$.

Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

for every $A \subseteq S$, either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$

where $\complement_S \left({A}\right)$ is the relative complement of $A$ in $S$, that is, $S \setminus A$.

### Definition 4

Let $S$ be a non-empty set.

Let $\mathcal F$ be a non-empty set of subsets of $S$.

Then $\mathcal F$ is an ultrafilter on $S$ if and only if both of the following hold:

$\mathcal F$ has the finite intersection property
For all $U \subseteq S$, either $U \in \mathcal F$ or $U^c \in \mathcal F$

where $U^c$ is the complement of $U$ in $S$.

## 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:

$A \cap B = \varnothing$
$A \cup B \in \mathcal F$

Aiming for a contradiction, suppose $A \notin \mathcal F$ 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 $U \in \mathcal F$.

 $\displaystyle A$ $\notin$ $\displaystyle \mathcal F$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle U \cap A$ $=$ $\displaystyle \varnothing$ Definition of Filter: Axiom $(4)$: $U \cap A \subseteq A \implies A \in \mathcal F$ $\displaystyle \leadsto \ \$ $\displaystyle U$ $\subseteq$ $\displaystyle \complement_S \left({A}\right)$ Empty Intersection iff Subset of Complement $\displaystyle B$ $\notin$ $\displaystyle \mathcal F$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle U \cap B$ $=$ $\displaystyle \varnothing$ Definition of Filter: Axiom $(4)$: $U \cap B \subseteq B \implies B \in \mathcal F$ $\displaystyle \leadsto \ \$ $\displaystyle U$ $\subseteq$ $\displaystyle \complement_S \left({B}\right)$ Empty Intersection iff Subset of Complement $\displaystyle \leadsto \ \$ $\displaystyle U$ $\subseteq$ $\displaystyle \complement_S \left({A}\right) \cap \complement_S \left({B}\right)$ Intersection is Largest Subset $\displaystyle \leadsto \ \$ $\displaystyle U$ $\subseteq$ $\displaystyle \complement_S \left({A \cup B}\right)$ De Morgan's Laws: Relative Complement of Union $\displaystyle \leadsto \ \$ $\displaystyle U \cap \left({A \cup B}\right)$ $=$ $\displaystyle \varnothing$ Empty Intersection iff Subset of Complement

But by definition of a filter:

$U, V \in \mathcal F \implies U \cap V \in \mathcal F$

But $\varnothing \notin \mathcal F$.

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.

$\Box$

### 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:

 $\displaystyle A \cup \complement_S \left({A}\right)$ $=$ $\displaystyle S$ Union with Relative Complement $\displaystyle \leadsto \ \$ $\displaystyle A \cup \complement_S \left({A}\right)$ $\in$ $\displaystyle \mathcal F$ Definition of Filter: Axiom $(1)$: $S \in \mathcal F$ $\displaystyle A \cap \complement_S \left({A}\right)$ $=$ $\displaystyle \varnothing$ Intersection with Relative Complement is Empty $\displaystyle \leadsto \ \$ $\displaystyle \left({A \in \mathcal F}\right)$ $\lor$ $\displaystyle \left({\complement_S \left({A}\right) \in \mathcal F}\right)$ by hypothesis: Definition of Ultrafilter by Definition 2

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

$\Box$

### 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 $\mathcal F \subseteq \mathcal G$.

Aiming for a contradiction, suppose $\mathcal F \subsetneq \mathcal G$.

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

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

$A \cap \complement_S \left({A}\right)$

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 $\complement_S \left({A}\right) \in \mathcal F$.

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.

$\Box$

### 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 $U \subseteq S$.

Aiming for a contradiction, suppose $U \notin \mathcal F$.

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.

$\Box$

### 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.