If $f_n$, f are isometries, then $\underset{n\to\infty}{\lim}\overset{\infty}{\underset{i=0}{\sum}}2^{-i-1}d(f_n(x_i), f(x_i))=0$.

Question

The Problem: Suppose $(X, d)$ is a complete separable metric space, and $\{x_i\}\subseteq X$ is a countable dense subset. Suppose $(f_n)$ is a sequence of isometries of $X$ such that $\underset{n\to\infty}{\lim}f_n(x_i)=f(x_i)$ for all $x_i\in\{x_i\}$, where $f: X\to X$ is also an isometry. Show that $\underset{n\to\infty}{\lim}\overset{\infty}{\underset{i=0}{\sum}}2^{-i-1}\frac{d(f_n(x_i), f(x_i))}{1+d(f_n(x_i), f(x_i))}=0$.

Source: Classical Descriptive Set Theory by Alexander S. Kechris.

The problem is a simplification of a statement from the book. The original statement is:

Let $(X, d)$ be a complete separable metric space. Denote by $Iso(X, d)$ the group of its isometries. Put on $Iso(X, d)$ the topology generated by the functions $f\mapsto f(x)$ for $x\in X$. This is a Polish group with a compatible complete metric given by $\delta(f, g)=\overset{\infty}{\underset{n=0}{\sum}}2^{-n-1}(\frac{d(f(x_n), g(x_n))}{1+d(f(x_n), g(x_n))}+\frac{d(f^{-1}(x_n), g^{-1}(x_n))}{1+d(f^{-1}(x_n), g^{-1}(x_n))})$, where $\{x_n\}$ is dense in $X$.

A couple of things I used to simplify the statement were:

1. The said topology on $Iso(X, d)$ is the same as the pointwise convergence topology.
2. A sequence $(f_n)\in Iso(X, d)$ converges to $f\in Iso(X, d)$ if and only if for all $x\in X, f_n(x)\to f(x)$ in $X$.

My Question: The statement is clearly false without the functions being isometries; but how exactly should I make use of the isometry condition? Any help would be greatly appreciated.

Answer 1

Suppose $\varepsilon>0$. Let $k\in\mathbb{N}$ be so large such that $\overset{\infty}{\underset{i=k}{\sum}}\frac{1}{2^{i+1}}<\frac{\varepsilon}{2}$; then for any $n\in\mathbb{N}$, we have $\overset{\infty}{\underset{i=k}{\sum}}\frac{1}{2^{i+1}}\frac{d(f_n(x_i), f(x_i))}{1+d(f_n(x_i), f(x_i))}<\frac{\varepsilon}{2}$. Now, for each $i\in\{1, 2, \dots, k\}$, pick $N_i\in\mathbb{N}$ such that $n\geq N_i$ implies $d(f_n(x_i), f(x))<\frac{\varepsilon}{4(1-\frac{1}{2^{k+1}})}$ Then $n\geq N:=\max\{N_i|i=1, \dots, k\}$ implies $\overset{k-1}{\underset{i=1}{\sum}}\frac{1}{2^{i+1}}\frac{d(f_n(x_i), f(x_i))}{1+d(f_n(x_i), f(x_i))}<\frac{\varepsilon}{2}$. It then follows that $n\geq N\implies\overset{\infty}{\underset{i=1}{\sum}}\frac{1}{2^{i+1}}\frac{d(f_n(x_i), f(x_i))}{1+d(f_n(x_i), f(x_i))}<\varepsilon$.