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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?

Maximal Ideal
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    For every $\varepsilon>0,$ let $N$ so large that $||\varphi_i-\varphi_j||\leq \varepsilon$ for $i,j\geq N$. Now let $y\in Y$ and $x_{\varepsilon}=\varphi_N^{-1}(y)$. Then, $d_Y(\varphi(x_{\varepsilon}),\varphi_{N}(x))\leq \varepsilon$. Then, necessarily, $\lim_{n\to\infty}\varphi(x_{1/n})=y,$ so the question reduces to whether or not the sequence $x_{1/n}$ converges (or has a convergent subsequence). I think this becomes a question of whether $X$ is locally compact (the $x_{1/n}$ sequence must be bounded). – WoolierThanThou Jul 17 '19 at 06:47

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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.

Josué Tonelli-Cueto
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