about a sequence of isometries' convergency. Let $M$ be a compact metric space, let $(i_n)$ be a sequence of isometries: $M \rightarrow M$. I've already showed that there exists a subsequence $(i_{n_k})$ that converges to $i$ which is also a isometry. So I want to show whether the sequence of $i_{n_k}^{-1}$ converges to $i^{-1}$.
I thought about it in $R^n$ and let $i_n$ be a sequence of rotation. But I don't know whether it holds for any metric space. If it is not true, any counterexample?
 A: Suppose $i_m^{-1}$ does not converge uniformly to $i^{-1}$.  Then there are $\epsilon > 0$ and a sequence $x_m$ in $M$ such that $d(i_m^{-1}(x_m), i^{-1}(x_m)) > \epsilon$.  This means there are $y_m$ and $z_m$ with $i_m(y_m) = x_m$, $i(z_m) = x_m$, and $d(y_m, z_m) > \epsilon$.  Taking subsequences, we may assume $x_m \to x$, $y_m \to y$ and $z_m \to z$, and $d(y, z) = \lim_m d(y_m, z_m) \ge \epsilon$.  Note that $i(z) = \lim_m i(z_m) = \lim_m x_m = x$.  But 
$$ \eqalign{d(y,z) &= d(i(y), i(z))\cr
& <= d(i(y), i_m(y)) + d(i_m(y), i_m(y_m)) + d(i_m(y_m),i(z))\cr
& = d(i(y), i_m(y)) + d(y, y_m) + d(x_m, x)\cr
& \to 0} $$
contradicting the assertion $d(y, z)\ge \epsilon$.
A: I try to show it directly. Not sure it is right... 
Let $y$ be an arbitrary point in $M$. And $i_{n_k}(p) \to i$ uniformly. Isometry is a homeomorphism, it follows that there must be unique $p$ and $q$ such that $i_{n_k}(p)=y$, and $i(q)=y$. Now $$d(i_{n_k}^{-1}(y),i^{-1}(y))=d(p,q)$$ If we can show $d(p,q)<\epsilon $, for every $\epsilon > 0$, for large enough $i_{n_k}$, then since y is arbitrary in $M$, we have $i_{n_k}^{-1} \to i$ uniformly. 
$d(i(p),i(q)) -d(i_{n_k}(p),i(p)) \leq d(i_{n_k}(p),i(q)) =0$. By choosing large enough $n_k$, $d(i_{n_k}(p),i(p)) < \epsilon$, we have $d(p,q)=d(i(p),i(q)) < \epsilon$. 
So, $d(p,q) < \epsilon$. We are done.
