Filter on Set is Proper Filter
Theorem
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the poset defined on $\powerset S$ by the subset relation.
Let $\FF$ be a filter on $S$.
Then $\FF$ is a proper filter on $\struct {\powerset S, \subseteq}$.
Proof
From the general definition of a filter, we have:
A filter on $\struct {S, \preccurlyeq}$ is a subset $\FF \subseteq S$ which satisfies the following conditions:
- $\FF \ne \O$
- $x, y \in \FF \implies \exists z \in \FF: z \preccurlyeq x, z \preccurlyeq y$
- $\forall x \in \FF: \forall y \in S: x \preccurlyeq y \implies y \in \FF$
A filter $\FF$ is proper if it does not equal $S$ itself.
From the definition of a filter on a set, we have:
A filter on $T$ is a set $\FF \subset \powerset T$ which satisfies the following conditions:
- $T \in \FF$
- $\O \notin \FF$
- $U, V \in \FF \implies U \cap V \in \FF$
- $\forall U \in \FF: U \subseteq V \subseteq T \implies V \in \FF$
We can identify:
- $\powerset T$ with $S$
- $\subseteq$ with $\preccurlyeq$.
Filter Not Empty
We have that $T \in \FF$ and so $\FF \ne \O$.
Preceding Elements in Filter
We have that:
- $U, V \in \FF \implies U \cap V \in \FF$
From Intersection is Subset, we have that $U \cap V \subseteq U$ and $U \cap V \subseteq V$.
So identifying $U$ with $x$, $V$ with $y$ and $U \cap V$ with $z$ it is clear that:
- $x, y \in \FF \implies \exists z \in \FF: z \preccurlyeq x, z \preccurlyeq y$
Succeeding Elements in Filter
We have that:
- $\forall U \in \FF: U \subseteq V \subseteq T \implies V \in \FF$
This can be rewritten:
- $\forall U \in \FF, V \in \powerset T: U \subseteq V \implies V \in \FF$
Identifying $U$ with $x$ and $V$ with $y$, this translates as:
- $\forall x \in \FF, y \in S: x \preccurlyeq y \implies y \in \FF$
Proper Filter
For $\FF$ to be a proper filter on $\struct {\powerset T, \subseteq}$, it must not equal $\powerset T$.
This is seen to be satisfied by the axiom $\O \notin \FF$.
All axioms are fulfilled, hence the result.
$\blacksquare$
Note about Axioms
It seems at first glance that the condition $T \in \FF$ is not axiomatic, as it is clear from the third property:
- $U \in \FF: U \subseteq T \subseteq T \implies T \in \FF$
However, one of the properties of a filter is that it is specifically not empty.
Specifying that $T \in \FF$ is therefore equivalent to specifying that $\FF \ne \O$.
Thus it would be possible to cite the first axiom as $\FF \ne \O$ instead, but this is rarely done.