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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 $0 < t_1 < t_2cdots < t_N-1 < 1$ 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 $overlineM$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.

But is $overlineM$ necessarily a geodesic space? If not, what is a counterexample?

This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds

Also, note that if $overlineM$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overlineM$ is not locally compact is the universal cover of the punctured flat plane. However, this is still 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 $Sigma:=1,tfrac12,tfrac13,ldotscup -1,-tfrac12,-tfrac13,ldots$. The set $Sigmacup 0$ is closed in $mathbbR$.

Let $(M,g)$ be the complement of $[0,1]times (Sigmacup0)$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overlineM$ of $(M,g)$ contains the following "extra points":

• $0,1times (Sigmacup0)$




• for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_yto s^pm(t,y)$.

Importantly, as far as I can tell, there is nothing in $overlineM$ corresponding to the points in the segment $(0,1)times0$.

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 $overlineM$ connecting them.