# Equivalence of Definitions of Ultrafilter on Set/Definition 1 iff Definition 3

## 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 $\FF \subseteq \mathcal G$ holds, then $\FF = \GG$.

### 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, $S \setminus A$.

## Proof

Let $S$ be a set.

### Definition 1 implies Definition 3

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 $A \subseteq S$.

Aiming for a contradiction, suppose $A \notin \mathcal F$ and $\complement_S \left({A}\right) \notin \mathcal F$.

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

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

Let $U \in \mathcal F$.

 $\displaystyle \complement_S \left({A}\right)$ $\notin$ $\displaystyle \mathcal F$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle U$ $\not \subseteq$ $\displaystyle \complement_S \left({A}\right)$ Definition of Filter: Axiom $(4)$: $U \subseteq \complement_S \left({A}\right) \subseteq S \implies \complement_S \left({A}\right) \in \mathcal F$ $\displaystyle \leadsto \ \$ $\displaystyle A \cap U$ $\ne$ $\displaystyle \varnothing$ Empty Intersection iff Subset of Complement $\displaystyle \leadsto \ \$ $\displaystyle A \cap U$ $\in$ $\displaystyle \mathcal G$ Definition of $\mathcal G$ by way of $\mathcal B$ $\displaystyle \leadsto \ \$ $\displaystyle A$ $\in$ $\displaystyle \mathcal G$ Definition of Filter: Axiom $(4)$: $A \cap U \subseteq A \subseteq S \implies A \in \mathcal G$

But we have:

$A \notin \mathcal F$

and so:

$\mathcal F \subsetneq \mathcal G$

Thus $\mathcal F$ is not an ultrafilter by Definition 1.

Hence either:

$A \in \mathcal F$

or:

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

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