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

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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 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, $S \setminus A$.


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.

This contradicts our initial hypothesis.

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, $\varnothing \notin \mathcal G$.

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.

$\blacksquare$