Is the uniform limit of (bijective) metric space isometries also a (bijective) metric space isometry? Suppose $X, Y$ are metric spaces and $\varphi_{i}:X\rightarrow Y$ are bijective isometries that uniformly converge to a map $\varphi:X\rightarrow Y$. Is $\varphi$ a bijective isometry?
According to this post and the answer. There is an analogous result in Riemannian geometry if you assume $X, Y$ are connected Riemannian manifolds and $d_{X}, d_{Y}$ are the distance functions. However, the hint by Jack Lee to show that $\varphi$ is surjective relies on Riemannian geometry.
Does the result that $\varphi$ is surjective hold in general for metric spaces? Or is this result specific to Riemannian manifolds? Is there a counterexample?
Do we need some assumption about Cauchy completeness? 
 A: Assuming Cauchy completeness is a sufficient condition, one should only expand the argument of @WoolierThanThou. Unfortunately, I cannot think of any counterexample.
Let $\{\varphi_n:X\rightarrow Y\}$ be the sequence of uniformly converging bijective isometries and $\varphi:X\rightarrow Y$ its limit. It is clear, that $\varphi$ is an injective isometry, as uniform convergence implies punctual convergence. So we only have to show that $\varphi$ is surjective.
Without loss of generality, by passing to a subsequence if necessary, assume that
$$d_\infty(\varphi_n,\varphi)<1/n$$
where $d_\infty(\psi,\tilde{\psi}):=\sup_{x\in X}d(\psi(x),\tilde{\psi}(x))$. 
Let $y\in Y$. We will construct $x\in X$ such that $\varphi(x)=y$. This will show surjectivity. Let the sequence $\{x_n\}$ be given by
$$x_n:=\varphi_n^{-1}(y).$$
Note that
$$d(\varphi(x_n),y)=d(\varphi(x_n),\varphi_n(x_n))<d_\infty (\varphi,\varphi_n)\leq 1/n$$
and so
$$\lim_{n\to\infty}\varphi(x_n)=y.$$
Hence it is enough to show that $\{x_n\}$ is convergent. Until here I only repeated what @WoolierThanThou said.
Now, for all $k,l\geq 2n$ and $m$,
$$d(x_k,x_l)=d(\varphi_m(x_k),\varphi_m(x_l))\leq d(\varphi_m(x_k),y)+d(y,\varphi_m(x_l))$$
where the first equality follows from the fact that $\varphi_m$ is an isometry and the second inequality is the triangular inequality. Taking the limit $m\to\infty$, we get that for all $k,l\geq 2n$,
$$d(x_k,x_l)\leq d(\varphi(x_k),y)+d(y,\varphi(x_l))< 1/k+1/l\leq 1/(2n)+1/(2n)\leq 1/n,$$
where the second inequality follows from the inequality in the previous paragraph. Hence $\{x_n\}$ is a Cauchy sequence and thus convergent by the initial assumption. We are done.
