Remark 2

Let d be a left-invariant metric which defines the topology of a metrizable group G, and let H be a closed normal subgroup of G. If $\dot{x}$ and $\dot{y}$ are any two points of G/H, consider the distance $d(\dot{x}, \dot{y})$ of the two closed subsets $\dot{x}, \dot{y}$ in G (§ 2, no. 2); we shall see that this function is a left-invariant metric on G/H and defines the topology of this quotient group.

Notice first that if $x \in \dot{x}$ and $y \in \dot{y}$ we have $d(\dot{x}, \dot{y}) = d(x, Hy)$; for $d(x, Hy) = \inf_{h \in H} d(x, h y)$, and therefore $d(h'x, Hy) = d(x, Hy)$ for all $h' \in H$, since d is left-invariant; this proves the assertion (§ 2, no. 2). Hence for each $\dot{z} \in G/H$ we have [§ 2, no. 2, formula (2)]

and since this inequality is valid for all $x \in \dot{x}$ and all $y \in \dot{y}$, we have $|d(\dot{x}, \dot{z}) - d(\dot{y}, \dot{z})| \leq d(\dot{x}, \dot{y})$, which shows that $d(\dot{x}, \dot{y})$ is a metric on G/H. Moreover, for any $z \in \dot{z}$, we have

from above; but since $hzy = z(z^{-1}hz) y$ and since $z^{-1}hz$ runs through H as h runs through H (H being normal), the left-invariance of $d(x, y)$ shows that we have $d(zx, Hz y) = d(x, Hy) = d(\dot{x}, \dot{y})$. Finally, if V is a neighbourhood of e in G defined by $d(e, x) < \alpha$, the image V of V in G/H is the set defined by $d(e, \dot{x}) < \alpha$; this completes the proof.