Proposition 4

If G is a metrizable group, then every Hausdorff quotient group G/H of G is metrizable. If G is also complete, then G/H is complete ().*

The first part of the proposition is a consequence of the fact that the identity element of G/H has a countable fundamental system of neighbourhoods in G/H; for if (V_n) is a fundamental system of neighbourhoods of e in G, then the canonical images $\dot{V}_n$ of the sets V_n in G/H form a fundamental system of neighbourhoods of the identity element of G/H (Chapter III, § 2, no. 6, Proposition 17).

To show that G/H is complete if G is complete, it is enough, by § 2, no. 6, Proposition 9, to show that every Cauchy sequence $(\dot{x}_n)$ (with respect to the left uniformity of G/H) is convergent. We may assume, by passing to a subsequence of $(\dot{x}_n)$ if necessary, that for each pair of indices $p, q$ such that $p \geq n$ and $q \geq n$, we have $\dot{x}_p^{-1}\dot{x}_q \in \dot{V}_n$; this means that for each pair of points $y \in \dot{x}_p, z \in \dot{x}_q$, we have $y^{-1}z \in HV_n = V_nH$; and therefore, for each $y \in \dot{x}_p$, the intersection of $\dot{x}_q$ and the neighbourhood $yV_n$ of $y$ is not empty. Suppose then that the sequence (V_n) has been chosen so that $V_{n+1}^2 \subset V_n$, and define inductively a sequence $(x_n)$ of points of G, such that $x_n \in \dot{x}_n$ and $x_{n+1} \in x_n V_n$; this is possible by what has been said. It follows then by induction that for each $p > 0$ we have $x_{n+p} \in x_n V_n V_{n+1} \ldots V_{n+p-1} \subset x_n V_{n-1}$. The sequence $(x_n)$ is therefore a Cauchy sequence in G, so it converges to a point $a$; and it follows immediately that the canonical image $\dot{a}$ of $a$ in G/H is the limit of the sequence $(\dot{x}_n)$.