# Filter on Set is Proper Filter

## Contents

## Theorem

Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ denote the power set of $S$.

Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the poset defined on $\mathcal P \left({S}\right)$ by the subset relation.

Let $\mathcal F$ be a filter on $S$.

Then $\mathcal F$ is a proper filter on $\left({\mathcal P \left({S}\right), \subseteq}\right)$.

## Proof

From the general definition of a filter, we have:

A **filter on $\left({S, \preccurlyeq}\right)$** is a subset $\mathcal F \subseteq S$ which satisfies the following conditions:

- $\mathcal F \ne \varnothing$

- $x, y \in \mathcal F \implies \exists z \in \mathcal F: z \preccurlyeq x, z \preccurlyeq y$

- $\forall x \in \mathcal F: \forall y \in S: x \preccurlyeq y \implies y \in \mathcal F$

A filter $\mathcal F$ 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 $\mathcal F \subset \mathcal P \left({T}\right)$ which satisfies the following conditions:

- $T \in \mathcal F$

- $\varnothing \notin \mathcal F$

- $U, V \in \mathcal F \implies U \cap V \in \mathcal F$

- $\forall U \in \mathcal F: U \subseteq V \subseteq T \implies V \in \mathcal F$

We can identify:

- $\mathcal P \left({T}\right)$ with $S$
- $\subseteq$ with $\preccurlyeq$.

### Filter Not Empty

We have that $T \in \mathcal F$ and so $\mathcal F \ne \varnothing$.

### Preceding Elements in Filter

We have that:

- $U, V \in \mathcal F \implies U \cap V \in \mathcal F$

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:

- $x, y \in \mathcal F \implies \exists z \in \mathcal F: z \preccurlyeq x, z \preccurlyeq y$

### Succeeding Elements in Filter

We have that:

- $\forall U \in \mathcal F: U \subseteq V \subseteq T \implies V \in \mathcal F$

This can be rewritten:

- $\forall U \in \mathcal F, V \in \mathcal P \left({T}\right): U \subseteq V \implies V \in \mathcal F$

Identifying $U$ with $x$ and $V$ with $y$, this translates as:

- $\forall x \in \mathcal F, y \in S: x \preccurlyeq y \implies y \in \mathcal F$

### Proper Filter

For $\mathcal F$ to be a proper filter on $\left({\mathcal P \left({T}\right), \subseteq}\right)$, it must not equal $\mathcal P \left({T}\right)$.

This is seen to be satisfied by the axiom $\varnothing \notin \mathcal F$.

All axioms are fulfilled, hence the result.

$\blacksquare$

## Note about axioms

It seems at first glance that the demand $T \in \mathcal F$ is not axiomatic, as it is clear from the third property:

- $U \in \mathcal F: U \subseteq T \subseteq T \implies T \in \mathcal F$

However, one of the properties of a filter is that it is specifically not empty.

Specifying that $T \in \mathcal F$ is therefore equivalent to specifying that $\mathcal F \ne \varnothing$.

Thus it would be possible to cite the first axiom as $\mathcal F \ne \varnothing$ instead, but this is rarely done.