van Dantzig's theorem

Remarks

The proof has two steps. A locally compact Hausdorff totally disconnected space has a basis of compact open sets, so the given neighbourhood contains a compact open set $V \ni e$. By compactness of $V$ there is a symmetric open neighbourhood $W$ of $e$ with $V W \subseteq V$. Then all products of elements of $W$ lie in $V$, so the subgroup generated by $W$ is open and lies in $V$. An open subgroup is closed, so this subgroup is compact.

Example: in the $p$-adic numbers $\Q_p$, the subgroups $p^k \Z_p$ with $k \ge 0$ are compact and open, and they shrink to $0$. In $\mathrm{GL}_n(\Q_p)$, the congruence subgroups $I + p^k \mathrm{M}_n(\Z_p)$ with $k \ge 1$ play the same role.

Local compactness is necessary. The group $\Q$ with the topology of $\R$ is totally disconnected and not locally compact. An open subgroup of it contains an interval around $0$, so it is all of $\Q$, which is not compact.

Going deeper

For a student

The theorem is the starting point of the structure theory of totally disconnected locally compact groups. It shows that such a group is a union of cosets of a compact open subgroup $U$, with a discrete coset space $G/U$. A totally disconnected compact group is profinite, so these groups are locally profinite. Compact open subgroups are used to define Cayley–Abels graphs and the Hecke algebras of $p$-adic groups.