Filter is Finer iff Sets of Basis are Subsets

Theorem

Let $S$ be a set.

Let $\powerset S$ denote the power set of $S$.

Let $\FF, \FF' \subset \powerset S$ be two filters on $S$.

Let $\FF$ have a basis $\BB$.

Let $\FF'$ have a basis $\BB'$.

$\FF$ is finer than $\FF'$ if and only if for every set of $\BB'$, there is a set of $\BB$ subset to it.

Suppose for every set of $\BB'$, there is a set of $\BB$ subset to it.

Pick any $U \in \FF'$.

Then from definition of a basis:

$\exists V' \in \BB': V' \subseteq U$

By our assumption:

$\exists V \in \BB: V \subseteq V' \subseteq U \subseteq S$

By definition of a filter:

$U \in \FF$.

Hence we have $\FF' \subseteq \FF$.

$\Box$

Suppose there is some set of $\BB'$ with no set of $\BB$ subset to it.

Let $U$ be such a set.

Then $U \in \BB' \subseteq \FF'$.

From our assumption:

$\nexists V \in \BB: V \subseteq U$

From the definition of a basis:

$U \notin \FF$

This gives $\FF' \nsubseteq \FF$.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters