# 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 $\FF \subseteq \powerset S$ be a filter on $S$.

Then $\FF$ is an **ultrafilter (on $S$)** if and only if:

- there is no filter on $S$ which is strictly finer than $\FF$

or equivalently, if and only if:

- whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$.

### Definition 2

Let $S$ be a set.

Let $\FF \subseteq \powerset S$ be a filter on $S$.

Then $\FF$ is an **ultrafilter (on $S$)** if and only if:

- 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 $B \in \FF$.

### Definition 3

Let $S$ be a set.

Let $\FF \subseteq \powerset S$ be a filter on $S$.

Then $\FF$ is an **ultrafilter (on $S$)** if and only if:

- for every $A \subseteq S$, either $A \in \FF$ or $\relcomp S A \in \FF$

where $\relcomp S A$ is the relative complement of $A$ in $S$, that is, $S \setminus A$.

### Definition 4

Let $S$ be a non-empty set.

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

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

- $\FF$ has the finite intersection property
- For all $U \subseteq S$, either $U \in \FF$ or $U^\complement \in \FF$

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

## Proof

Let $S$ be a set.

### Equivalence of Definitions 1, 2 and 3

### 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:

- $A \cap B = \O$
- $A \cup B \in \FF$

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

\(\ds A\) | \(\notin\) | \(\ds \FF\) | by hypothesis | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds U \cap A\) | \(=\) | \(\ds \O\) | Definition of Filter: Axiom $(4)$: $U \cap A \subseteq A \implies A \in \FF$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \relcomp S A\) | Empty Intersection iff Subset of Complement | ||||||||||

\(\ds B\) | \(\notin\) | \(\ds \FF\) | by hypothesis | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds U \cap B\) | \(=\) | \(\ds \O\) | Definition of Filter: Axiom $(4)$: $U \cap B \subseteq B \implies B \in \FF$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \relcomp S B\) | Empty Intersection iff Subset of Complement | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \relcomp S A \cap \relcomp S B\) | Intersection is Largest Subset | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds U\) | \(\subseteq\) | \(\ds \relcomp S {A \cup B}\) | De Morgan's Laws: Relative Complement of Union | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds U \cap \paren {A \cup B}\) | \(=\) | \(\ds \O\) | Empty Intersection iff Subset of Complement |

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.

$\Box$

### 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:

\(\ds A \cup \relcomp S A\) | \(=\) | \(\ds S\) | Union with Relative Complement | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds A \cup \relcomp S A\) | \(\in\) | \(\ds \FF\) | Definition of Filter: Axiom $(1)$: $S \in \FF$ | ||||||||||

\(\ds A \cap \relcomp S A\) | \(=\) | \(\ds \O\) | Intersection with Relative Complement is Empty | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds A\) | \(\in\) | \(\ds \FF\) | by hypothesis: Definition of Ultrafilter by Definition 2 | ||||||||||

\(\, \ds \lor \, \) | \(\ds \relcomp S A\) | \(\in\) | \(\ds \FF\) |

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

$\Box$

### 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 $\FF \subseteq \GG$.

Aiming for a contradiction, suppose $\FF \subsetneq \GG$.

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.

$\Box$

### Definition 1 implies Definition 4

Let $\FF \subseteq \powerset S$ a filter on $S$ which fulfills the condition:

- whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$

By definition of filter, $\FF$ satisfies the Finite Intersection Property and is a non-empty set of subsets of $S$.

Let $U \subseteq S$.

Aiming for a contradiction, suppose $U \notin \FF$.

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

Suppose there exists $C \in \FF$ such that:

- $C \cap V = \O$

By Empty Intersection iff Subset of Complement, it follows that:

- $C \subset U$

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

Hence the set $\Omega = \set {C \cap V: C \in \FF}$ 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 $\GG$ on $S$ that contains $\FF$ and $\set V$.

By hypothesis, we have:

- $\FF =\GG$

Hence $U^c = V \in \FF$.

Thus $\FF$ is an ultrafilter by Definition 2.

$\Box$

### Definition 4 implies Definition 1

Let $S$ be a non-empty set and $\FF$ a non-empty set of subsets on $S$ which fulfills the conditions:

- $\FF$ has the finite intersection property
- For all $U \subseteq S$, either $U \in \FF$ or $U^c \in \FF$.

Because $\FF$ has the finite intersection property, it follows that $\O \notin \FF$.

As $\O \notin \FF$, it follows that $\O^c = S \in \FF$.

Let $U, V \in \FF$.

Then either $U \cap V \in \FF$ or $\paren {U \cap V}^c \in \FF$.

However we observe that:

- $\paren {U \cap V}^c \cap U \cap V = \O$

By finite intersection property, it follows that $U \cap V \in \FF$.

Now let $U \in \FF$ and $V \in S$ such that $U \subseteq V \subseteq S$.

Then either $V \in \FF$ or $V^c \in \FF$.

However we observe that:

- $V^c \cap U \subseteq V^c \cap V = \O$

By finite intersection property, it follows that $V \in \FF$.

Therefore $\FF$ is a filter.

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

Assume that $\FF \subsetneq \GG$.

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

Since $\O \notin \GG$ this implies that $A^c \notin \GG$.

As $\FF \subsetneq \GG$, it follows that $A^c \notin \FF$.

Therefore neither $A \in \FF$ nor $A^c \in \FF$, a contradiction to our assumption.

Thus $\FF = \GG$, which implies that $\FF$ is an ultrafilter by definition 1.

$\blacksquare$