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.