Equivalence of Definitions of Ultrafilter on Set/Equivalence of Definitions 1, 2 and 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 $\FF \subseteq \GG$ 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, $S \setminus A$.
Proof
Let $S$ be a set.
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.
$\blacksquare$