I'm trying to prove the following theorem: let $f_n : M \to N $ a sequence of isometries of Riemannian manifolds that converges uniformly to a function $f:M \to N$: prove that $f$ is an isometry too.
What I did already:
It is enough to prove the theorem for $M,N$ connected, so we can use the fact that $f$ is a Riemannian isometry if and only if it is a metric isometry, with respect to Riemannian distance. Now, $f$ is continuous because it is the limit of a uniformly convergent sequence of continuous maps, and preserves the distance because the distance function is continuous. So, $f$ is also injective, and then it is open by the invariance of domain theorem.
Now, it suffices to show that $f$ is closed or surjective.
Any help?
We show that $f$ is surjective. Suppose that exist $y\notin Im(f)$. Since $f_n$ is diffeomorphism for all $n$, there exist unique $x_n\in M$ such that $f_n(x_n)=y$. By the uniform convergence follows that $\lim f(x_n)=y$. Note that $x_n$ can't have accumulation point ( because $x_{n_k}\to x\in M$ implies that $f(x_{n_k})\to f(x)$, so $f(x)=y$. Contradiction! ). So we can choose a subsequence $x_{n_k}$ such that $d(x_{n_k}, x_{n_{k+1}}) > \delta$ for some $\delta > 0$. Therefore we have $d(f(x_{n_{k}}), f(x_{n_{k+1}}))=d(x_{n_k}, x_{n_{k+1}}) > 0$. Absurd! Since $\lim f(x_{n_k})=y$.