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Let $(X,d)$ be a metric spaces and $G$ a discrete group which acts by isometries on $X$ and the action is properly discontinuous. Then it's possible to define an induced metric $\hat{d}$ on the quotient space $\hat{X}:=X/G$ saying that the distance between two orbits is the infimum of the distance between any pair of representatives.

I've read that the completeness of $(\hat{X},\hat{d})$ implies the completeness of $(X,d)$, but I can't work out the proof.

Suppose $x_n$ is a Cauchy sequence for $d$, then, if $\hat{x}_n$ is the sequence of orbits in $\hat{X}$, $\hat{x}_n$ is Cauchy for $\hat{d}$ in $\hat{X}$. Then we know there exists $\hat{x}\in \hat{X}$ which is the limit point of $\hat{x}_n$ in $\hat{X}$ for $\hat{d}$. But I can't see why this should imply the existence of a limit point $x$ for $x_n$ in $X$ for $d$.

1 Answers1

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Assume WLOG that $\hat{d}(\hat{x}_n,\hat{x})<\frac{1}{2n}$ for all $n$. If not, we can pass to a subsequence and apply the following argument to it, since if a subsequence of a Cauchy sequence has a limit, the whole sequence does. By the definition of $\hat d$ we can find $a_n$ in the orbit of $x_n$ and $b_n$ in the orbit of $\hat x$ such that $d(a_n,b_n)<\frac1n$. We can find $g_n\in G$ such that $g_n a_n=x_n$. Since $G$ acts by isometries we obtain $d(x_n, g_nb_n)<\frac1n$. Since $(x_n)$ is Cauchy, so is $(g_nb_n)$, but the latter is contained in the orbit of $\hat x$, which is discrete, since the action is properly discontinuous (see here). This means $g_nb_n=x$ for some $x\in X$ and large $n$. This $x$ is the sought limit.

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