Fréchet Filter is Filter

Theorem
Let $S$ be an infinite set.

Let $\FF$ be the Fréchet filter on $S$.

Then $\FF$ is a filter on $S$.

Proof
By definition, $\FF$ must satisfy the filter on set axioms:

$\paren {\text F 1}$
By Set Difference with Self is Empty Set, we have that $S \setminus S = \empty$.

But by Cardinality of Empty Set, the empty set is finite.

Therefore, $S$ is cofinite in $S$ by definition.

Thus, by definition of Fréchet filter, $S \in \FF$.

$\paren {\text F 2}$
By Set Difference with Empty Set is Self, $S \setminus \empty = S$.

But by assumption, $S$ is infinite.

Therefore, $\empty$ is not cofinite in $S$ by definition.

Thus, by definition of Fréchet filter, $\empty \notin \FF$

$\paren {\text F 3}$
Let $U, V \in \FF$.

Then, by definition of Fréchet filter, $U$ and $V$ are cofinite in $S$.

That is, $S \setminus U$ and $S \setminus V$ are finite.

Thus, by Union of Finite Sets is Finite, $\paren {S \setminus U} \cup \paren {S \setminus V}$ if finite.

But by Difference with Intersection:
 * $S \setminus \paren {U \cap V} = \paren {S \setminus U} \cup \paren {S \setminus V}$

Therefore, $S \setminus \paren {U \cap V}$ is finite.

Thus, by definition, $U \cap V$ is cofinite in $S$.

Then, by definition of Fréchet filter, $U \cap V \in \FF$.

$\paren {\text F 4}$
Let $U \in \FF$.

Let $V$ be a set such that $U \subseteq V \subseteq S$.

Then, by Set Difference with Subset is Superset of Set Difference:
 * $S \setminus V \subseteq S \setminus U$

By definition of Fréchet filter, $U$ is cofinite in $S$.

That is, $S \setminus U$ is finite.

But then, by Subset of Finite Set is Finite, $S \setminus V$ is finite.

Thus, $V$ is cofinite in $S$ by definition.

Therefore, $V \in \FF$, by definition of Fréchet filter.

As $\FF$ satisfies the filter on set axioms, it is a filter on $S$ by definition.