Equivalence of Definitions of Filter on Set

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 $2$.

$(2)$ 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 $1$.