Proposition 5

Let $G$ be a filtered group whose filtration $(G_n)$ consists of invariant subgroups of $G$. The following conditions are equivalent:

(a) $G$ is a complete topological group.

(b) The associated Hausdorff group $G' = G / (\bigcap_{n \in \mathbf{Z}} G_n)$ is complete.

(c) Every Cauchy sequence in $G$ is convergent.

If $G$ is commutative and written additively, these conditions are also equivalent to the following:

(d) Every family $(x_\lambda)_{\lambda \in L}$ of elements of $G'$ which converges to 0 with respect to the filter $\mathscr{F}$ of complements of finite subsets of $L$ is summable in $G'$.

For a filter on $G$ to be a Cauchy filter (resp. a convergent filter), it is necessary and sufficient that its image under the canonical mapping $G \to G'$ be a Cauchy (resp. convergent) filter (General Topology, Chapter II, § 3, no. 1, Proposition 4); whence first of all the equivalence of (a) and (b); on the other hand, as $G'$ is metrizable, the equivalence of (b) and (c) follows from Proposition 9 of General Topology, Chapter IX, § 2, no. 6.

Suppose now that $G$ is commutative. Suppose that $G'$ is complete and let $(x_\lambda)_{\lambda \in L}$ be a family of elements of $G'$ which converge to 0 with respect to $\mathscr{F}$. For every neighbourhood $V'$ of 0 in $G$ which is a subgroup of $G'$, there exists a finite subset $J$ of $L$ such that the condition $\lambda \in L - J$ implies $x_\lambda \in V'$; then $\sum_{\lambda \in H} x_\lambda \in V'$ for every finite subset $H$ of $L$ not meeting $J$, which shows that the family $(x_\lambda)_{\lambda \in L}$ is summable (General Topology, Chapter III, § 5, no. 2, Theorem 1).

Conversely, suppose that condition (d) holds and let $(x_n)$ be a Cauchy sequence on $G'$; the family $(x_{n+1} - x_n)$ is then summable and in particular the series with general term $x_{n+1} - x_n$ is convergent and hence the sequence $(x_n)$ is convergent.

Let $G$ be a filtered group whose filtration $(G_r)$ consists of normal subgroups of $G$; the quotient groups $G/G_n$ are discrete and hence complete, since the $G_r$ are open in $G$. Let $f_n$ be the canonical mapping $G \to G/G_n$ and for $m \leq n$ let $f_{mn}$ be the canonical mapping $G/G_n \to G/G_m$; $(G/G_n, f_{mn})$ is an inverse system of discrete groups with $\mathbf{Z}$ as indexing set (General Topology, Chapter III, § 7, no. 3). Let $\tilde{G}$ be the topological group the inverse limit of this inverse system and for all $n$ let $g_n : \tilde{G} \to G/G_n$ be the canonical mapping; let $f : G \to \tilde{G}$ be the inverse limit of the inverse system of mappings $(f_n)$ such that $f_n = g_n \circ f$ for all $n$; finally, let $j$ be the canonical mapping of $G$ to its Hausdorff completion $\hat{G}$; as the $G/G_n$ are complete, there exists a unique topological group isomorphism $i : \hat{G} \to \tilde{G}$ such that $f = i \circ j$ (loc. cit., Corollary 1 to Proposition 2); we shall call it the canonical isomorphism of $\hat{G}$ onto $\tilde{G}$.

Let $H$ be another filtered group whose filtration $(H_r)$ consists of normal subgroups of $H$ and let $u : G \to H$ be a homomorphism compatible with the filtrations (no. 4). Set $\tilde{H} = \lim H/H_n$; for all $n$, $u$ defines by taking quotients a homomorphism $u_n : G/G_n \to H/H_n$ and the $u_n$ obviously form an inverse system of mappings; set $\tilde{u} = \lim u_n$. Moreover let $\hat{H}$ be the Hausdorff completion of $H$ and $ti : \hat{G} \to H$ the homomorphism derived from $u$ by passing to the Hausdorff completions (General Topology, Chapter II, § 3, no. 7, Proposition 15). It follows immediately from the definitions that if $\hat{G}$ is identified with $\tilde{G}$ and $\hat{H}$ with $\tilde{H}$ by means of the canonical isomorphisms, $ti$ is identified with $\tilde{u}$. We conclude in particular that, if, for all $n$, $u_n$ is an isomorphism, then $ti$ is an isomorphism of topological groups.

Examples of complete filtered groups and rings