Example

Definition of a topology on a group by means of a set of subgroups. If $\mathfrak{B}$ is a filter base on a group $G$, formed of subgroups of $G$, then it is immediately seen that $\mathfrak{B}$ satisfies axioms (GV_{I}) and (GV_{II}), since $H . H^{-1} = H$ for any subgroup $H$ of $G$. Hence the set $\mathfrak{B}$ will be a fundamental system of neighbourhoods of $e$ in a topology compatible with the group structure of $G$, provided that $\mathfrak{B}$ satisfies (GV_{III}); this will in particular be the case if all the subgroups in $\mathfrak{B}$ are normal, hence always if $G$ is commutative. The topology thus defined is Hausdorff, by Proposition 2, if and only if the intersection of all the subgroups in $\mathfrak{B}$ consists only of $e$. The most interesting cases are those in which the subgroup $\{ e \}$ is not in $\mathfrak{B}$ (otherwise the topology defined by $\mathfrak{B}$ is the discrete topology): if $\{ e \} \notin \mathfrak{B}$, the topology defined by $\mathfrak{B}$ is Hausdorff only if $\mathfrak{B}$ is an infinite set.

Since the intersection of two subgroups is a subgroup, we can define a topology on $G$, compatible with its group structure, starting from any set $\mathfrak{F}$ of subgroups of $G$: let $\mathfrak{G}$ be the set of all subgroups $a.H.a^{-1}$, where $H \in \mathfrak{F}$ and $a \in G$, and let $\mathfrak{B}$ be the set of all finite intersections of subgroups belonging to $\mathfrak{G}$. Then $\mathfrak{B}$ is a filter base and satisfies (GV$_{\text{III}}$).

Consider in particular the additive group of a ring $A$. Every set $\mathfrak{F}$ of ideals of $A$ defines a topology compatible with this additive group structure. This topology is Hausdorff if the intersection of all the ideals of $\mathfrak{F}$ is the zero ideal, and it is not discrete if no finite intersection of the ideals of $\mathfrak{F}$ is the zero ideal. Topologies defined in this way play a large part in the theory of numbers (see the Exercises of §§ 6 and 7 of this chapter).