Complete metric but not uniformly locally compact on the real line Is it possible to give a complete metric over the real line R, which is compatible with the Euclidean topology, but not uniformly locally compact?
Definition: A metric space (X,d) is uniformly locally compact if there exists $\varepsilon >0$ such that the closure of the open ball $B_d(x,\varepsilon)$ is compact for every $x\in X$.
 A: Choose an injection $f:\mathbb N\times\mathbb N\to \mathbb R$ whose image is uniformly discrete, so $(n,m)\neq (n',m')$ implies $|f(n,m)-f(n',m')|>\delta.$ Define a metric on $\mathbb N\times\mathbb N$ by
$$d_{\mathbb N\times\mathbb N}((n,m),(n',m'))=\begin{cases}
\infty&\text{ if $n\neq n',$}\\
1/n&\text{ if $n=n'$ but $m\neq m',$}\\
0&\text{ if $n=n'$ and $m=m'.$}
\end{cases}$$
Then take $d$ to be the largest metric on $\mathbb R$ satisfying $d(x,y)\leq |x-y|$ and $d(f(n,m),f(n',m'))\leq d_{\mathbb N\times\mathbb N}((n,m),(n',m')).$ In other words, it's the path metric where there are edges between $x$ and $y$ of length $|x-y|,$ and also edges between $f(n,m)$ and $f(n,m')$ of length $1/n$ for $m\neq m',$ and the distance between two points is the infimum of the total length of edges, taken over finite paths joining those points.
We need to check that $d(x,y)>0$ for $x\neq y.$ Any path from $x$ to $y$ of length less than $\delta/2$ either goes through points $x',y'$ in the image of $f$ with $d(x,x')<\delta/2$ and $d(y',y)<\delta/2,$ or it does not go through any points in the image of $f.$ In the latter case the path has length at least $|x-y|.$ In the former case there is a unique choice of $x'$ and $y',$ and the path has length at least $d(x,x')+d(x',y')+d(y',y),$ which is positive unless $x=y.$
I claim that the metric is complete. Any Cauchy sequence eventually lies in some $\delta/2$-ball. The distance between $f(n,m)$ and $f(n',m')$ for $n\neq n'$ is at least $\delta.$ So there is at most one $n$ such that this ball contains points of the form $f(n,m).$ So any distances less than $1/n$ are ordinary Euclidean distances, and the Cauchy sequence converges in the usual metric.
But it's not uniformly locally compact, because for each $n$ the ball $B_d(f(n,m),2/n)$ contains the infinite discrete set $\{f(n,m)\mid m\in\mathbb N\}.$
