A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] \rightarrow X$ is the sup of
$$ d(c(0), c(t\_1)) + d(c(t\_1), d(t\_2)) + \cdots + d(c(t\_{N-1}), c(1)) $$
over all $0 < t\_1 < t\_2\cdots < t\_{N-1} < 1$ and $N > 0$.


A geodesic space is a length space, where for each $x,y \in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.


A Riemannian manifold $M$ and its metric completion $\overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.


But is $\overline{M}$ necessarily a geodesic space? If not, what is a counterexample?


This was motivated by my flawed answer to [Minimizing geodesics in incomplete Riemannian manifolds](https://mathoverflow.net/q/317523/613)


Also, note that if $\overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $\overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.



Is the metric completion of a Riemannian manifold always a geodesic space?
I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $\Sigma:=\{1,\tfrac{1}{2},\tfrac{1}{3},\ldots\}\cup \{-1,-\tfrac{1}{2},-\tfrac{1}{3},\ldots\}$. The set $\Sigma\cup \{0\}$ is closed in $\mathbb{R}$.


Let $(M,g)$ be the complement of $[0,1]\times (\Sigma\cup\{0\})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $\overline{M}$ of $(M,g)$ contains the following "extra points":


* $\{0,1\}\times (\Sigma\cup\{0\})$
* for each $(t,s)\in(0,1)\times\Sigma$, two points $(t,s)\_\pm$, coming from the (two different) directional limits $\lim\_{y\to s^\pm}(t,y)$.


Importantly, as far as I can tell, there is nothing in $\overline{M}$ corresponding to the points in the segment $(0,1)\times\{0\}$.


If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $\overline{M}$ connecting them.


