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 $x\in X$.

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    $\begingroup$ What does "uniformly locally compact" mean? $\endgroup$ – David C. Ullrich Feb 22 '18 at 15:47
  • $\begingroup$ Dear David, I have change my edit to give the definition. $\endgroup$ – ASM Feb 23 '18 at 9:37

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 $\{f(n,m)\mid m\in\mathbb N\}.$

  • $\begingroup$ Is it complete? $\endgroup$ – ASM Feb 23 '18 at 13:28
  • $\begingroup$ What if $x=0$ or $y=0$? usual distance as the max term becomes "infinity"? $\endgroup$ – Henno Brandsma Feb 23 '18 at 13:32
  • $\begingroup$ @ASM: yes, in fact it is locally isometric to the usual metric $\endgroup$ – Dap Feb 23 '18 at 13:55
  • $\begingroup$ @HennoBrandsma: yes, $1/0=\infty$ $\endgroup$ – Dap Feb 23 '18 at 13:55
  • $\begingroup$ @Dap Please Dap, can you be more specific in your proof? I d'ont see that it is a metric or that it is complete, even if it is locally isometric. Thank you a lot. $\endgroup$ – ASM Feb 23 '18 at 15:33

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