Consider $\mathbb{Z}$ the integers. Is this considered a geodesic metric space under the usual metric? Why? given $n,m$ integers, we would require to have a map $f: [0,c] \rightarrow \mathbb{Z}$ such that $f(0)=n$ and $f(c)=m$ and also $f$ must be a quasi-isometric embedding. But how can this be?
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1No such map $f$ can be continuous, so no. – Qiaochu Yuan Apr 23 '15 at 09:14
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In order to even contemplate geodesics between points $x$ and $y$ of a metric space $M$, there must be some path from $x$ to $y$, i.e., a continuous function $\gamma: [0,1] \rightarrow M$ with $\gamma(0) = x$, $\gamma(1) = y$. In particular a geodesic metric space is certainly path-connected.
The integers under the standard metric form an infinite, discrete topological space, so no two distinct points can be connected by a path. So the answer is no. For that matter, no metric on the integers makes them connected, so there is really no hope here.
Moreover, since no countably infinite metric space can be geodesic, being geodesic is certainly not a quasi-isometry invariant.
Pete L. Clark
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