Filter on Set is Proper Filter

Theorem
Let $$X$$ be a set, and $$\mathcal P \left({X}\right)$$ be the power set of $$X$$.

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

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

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

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 $$X$$ is a set $$\mathcal F \subset \mathcal P \left({X}\right)$$ which satisfies the following conditions:


 * $$X \in \mathcal F$$


 * $$\varnothing \notin \mathcal F$$


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


 * $$\forall U \in \mathcal F: U \subseteq V \subseteq X \implies V \in \mathcal F$$

We can identify:
 * $$\mathcal P \left({X}\right)$$ with $$S$$;
 * $$\subseteq$$ with $$\preccurlyeq$$.

Filter Not Empty
We have that $$X \in \mathcal F$$ 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 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:
 * $$\forall U \in \mathcal F: U \subseteq V \subseteq X \implies V \in \mathcal F$$

This can be rewritten:
 * $$\forall U \in \mathcal F, V \in \mathcal P \left({X}\right): U \subseteq V \implies V \in \mathcal F$$

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 $$\left({\mathcal P \left({X}\right), \subseteq}\right)$$, it must not equal $$\mathcal P \left({X}\right)$$.

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

All axioms are fulfilled, hence the result.

Note about axioms
It seems at first glance that the demand $$X \in \mathcal F$$ is not axiomatic, as it is clear from the third property:
 * $$U \in \mathcal F: U \subseteq X \subseteq X \implies X \in \mathcal F$$

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

Specifying that $$X \in \mathcal F$$ 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.