# Equivalence of Definitions of Filter on Set

## Theorem

The following definitions of the concept of Filter on Set are equivalent:

### Definition 1

A set $\FF \subset \powerset S$ is a filter on $S$ (or filter of $S$) if and only if $\FF$ satisfies the filter on set axioms:

 $(\text F 1)$ $:$ $\ds S \in \FF$ $(\text F 2)$ $:$ $\ds \O \notin \FF$ $(\text F 3)$ $:$ $\ds U, V \in \FF \implies U \cap V \in \FF$ $(\text F 4)$ $:$ $\ds \forall U \in \FF: U \subseteq V \subseteq S \implies V \in \FF$

### Definition 2

A set $\FF \subset \powerset S$ is a filter on $S$ (or filter of $S$) if and only if $\FF$ satisfies the filter on set axioms:

 $(\text F 1)$ $:$ $\ds S \in \FF$ $(\text F 2)$ $:$ $\ds \O \notin \FF$ $(\text F 3)$ $:$ $\ds \forall n \in \N: U_1, \ldots, U_n \in \FF \implies \bigcap_{i \mathop = 1}^n U_i \in \FF$ $(\text F 4)$ $:$ $\ds \forall U \in \FF: U \subseteq V \subseteq S \implies V \in \FF$

## Proof

Conditions $(\text F 1)$, $(\text F 2)$ and $(\text F 4)$ are the same for both definitions.

It remains to establish that $(\text F 3)$ in Definition 1 is equivalent to $(\text F 3)$ in Definition 2.

### $(1)$ implies $(2)$

Let $\FF$ be a filter on $S$ by definition $1$.

Then by definition:

$U_1, U_2 \in \FF \implies U_1 \cap U_2 \in \FF$

Suppose that for some $k \in \N$:

$(\text F 3): U_1, \ldots, U_k \in \FF \implies \ds \bigcap_{i \mathop = 1}^k U_i \in \FF$

Then we have:

$\ds \paren {\bigcap_{i \mathop = 1}^k U_i} \cap U_{k + 1} = \bigcap_{i \mathop = 1}^{k + 1} U_i$

by hypothesis.

Hence by induction:

$U_1, \ldots, U_k, U_{k + 1} \in \FF \implies \ds \bigcap_{i \mathop = 1}^{k + 1} U_i \in \FF$

Thus $\FF$ is a filter on $S$ by definition $2$.

$\Box$

### $(2)$ implies $(1)$

Let $\FF$ be a filter on $S$ by definition $2$.

Then by definition:

$(\text F 3): \forall n \in \N: U_1, \ldots, U_n \in \FF \implies \ds \bigcap_{i \mathop = 1}^n U_i \in \FF$

In particular when $n = 2$:

$U_1, U_2 \in \FF \implies U_1 \cap U_2 \in \FF$

Thus $\FF$ is a filter on $S$ by definition $1$.

$\blacksquare$