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It seems that I have the needed example, but I want it to be simple and self-explaining...

Construct a nontrivial complete metric space $X$ with intrinsic metric which has no nontrivial minimizing geodesics.


  • A metric $d$ is called intrinsic if for any two points $x$, $y$ and any $\epsilon>0$ there is an $\epsilon$-midpoint $z$; i.e. $d(x,z),d(x,y)<\tfrac12 d(x,y)+\epsilon$.

  • A minimizing geodesic is nontrivial if it connects two distinct points.

  • A meric space is nontrivial if it contains two distinct points.


  • Clearly, $X$ can not be locally compact.
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Wow. Please show us your example! –  Deane Yang Feb 17 '10 at 16:36
Side comment/question: why is the intrinsic property called like that? What is intrinsic about an intrinsic metric? –  Mariano Suárez-Alvarez Feb 17 '10 at 16:51
I suggest looking at Urysohn's universal complete separable metric space. If any separable example exists, then it isometrically embeds into this space. –  François G. Dorais Feb 17 '10 at 17:00
A couple of editorial suggestions: as stated, the silly "single point space" works. Also you might mention that the example you need cannot be locally compact by Hopf-Rinow –  Igor Belegradek Feb 17 '10 at 17:23
My guess is that you can built such a space inside $L_1$ using the fact that the set of metric midpoints between two points in $L_1$ is huge. The start would be with $0$ and $1_{(0,1)}$ and perturb a sequence $f_n$ of independent random variables with distribution $P[f=1]=P[f=0]=1/2$ to make $f_n$ an $1/(2n)$-approximate midpoint between $0$ and $1_{(0,1)}$; the independence keeps $(f_n)$ well separated in $L_1(0,1)$ so that the set constructed so far is closed. Repeat this construction between every pair of points. After countably many steps you should have an example. (Only a guess.) –  Bill Johnson Feb 17 '10 at 18:46

3 Answers 3

up vote 5 down vote accepted

Well, the unit ball in $c_0$ is almost what you want (there is no unique shortest curve between points). All we need now is to enhance "bypasses" and to give disadvantage to "straight lines". This can easily be done by taking the distance element to be $(2+\sum_n 2^{-n}x_n)^{-1}\|dx\|_\infty$, which is never less than the usual distance element in $c_0$ and never greater than 3 times it in the unit ball. Now, if we have any continuous finite length curve $x(t)$ from $y$ to $z$ parametrized by the arclength, we can easily shorten it by replacing the $m$-th position by the maximum of the actual value of $x_m(t)$ and $y_m+t(z_m-y_m)/d+\frac 12 \min(t,d-t)$, where $d$ is the length of $x(t)$, which will work if $m$ is large enough since $\max_t|x_m(t)|\to 0$ as $m\to\infty$ and both functions change slower than the distance along the original curve.

This is certainly self-explaining (the shortest curve escapes from $c_0$ to $\ell^\infty$) but I do not know if it is simple enough for your purposes.

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There are metric simplicial graphs (each edge has length $1$) even quasi-isometric to the real line $\mathbb{R}$ (and as such Gromov hyperbolic) with no infinite geodesics: Start with the integers $\mathbb{Z}$ and connect any two distinct integers $x, y$ with a simplicial interval of length $|x-y|+1$ and otherwise disjoint from $\mathbb{Z}$. Any infinite geodesic must pass through the concatenation of two such intervals, but no concatenation is a geodesic -- it can be shortcut by a single interval. I hope this helps with the question you have in mind.

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Thanks. It is not what I want, but it is nice construction and it is relevant. –  Anton Petrunin Feb 18 '10 at 2:21
Your example is not quasi-isometric to the real line, correct? –  Sam Nead Mar 28 '10 at 20:47

There is a very simple example of an intrinsic, complete metric space that is not geodesic (read in Ballmann's "Lectures on Spaces of Nonpositive Curvature": it is the graph on two vertices $x,y$, linked by edges $e_n$ of length $1+1/n$.

Of course it does not answer your question, but it may be possible to improve this example to one that does. Call $X_1$ the graph described above, and define $X_{n+1}$ from $X_n$ as follows: $X_n$ has a vertex $x'$ for each vertex $x$ of $X_n$, plus a vertex $v_e$ for each edge $e$ of $X_n$. For each edge $e=(xy)$ of $X_n$ we define edges $f_e^n$ and $g_e^n$ of $X{n+1}$: $f_e^n$ connects $x'$ to $v_e$ and has length $(1+1/n)$ times the original length of $e$, and $g_e^n$ does the same but replacing $x'$ by $y'$.

Now it should be possible to construct the desired example by a limiting process. For example, take all vertices along the construction: the distance between any two of this points is constant as long it is defined, so we get a metric space. Its completion might be what you want (but I a not so sure of that after witting these lines).

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This approach should work, but one has to perform a lot of stupid calculations to make it precise. I think that Fedja's example is much easier. –  Anton Petrunin Mar 28 '10 at 18:58

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