0

Can someone provide me with an example as such? A metric space which is not totally bounded with a non locally-compact completion.

User666x
  • 844

1 Answers1

2

Let $B \subset \mathbb{R}^2$ be the closed unit ball, but equipped with the metric $$d(p,q) = \begin{cases} |p| + |q| & \text{if $p$, $q$ are not on the same ray based at the origin} \\ | \, |p| - |q| \, | & \text{otherwise} \end{cases} $$ This is a complete metric space. You can think of this as the "metric tree" obtained from the disjoint union of all the oriented segments $[0,p]$, $p \in S^1$, by identifying the initial points of these segments.

$B$ is not totally bounded, because for any point $p \in S^1 \subset B$, any set of diameter $<2$ containing $p$ misses all of the other points in $S^1$. It therefore takes uncountably many balls of radius $< 1$ to cover $B$.

Also, $B$ is not locally compact, because for any $r > 0$ we can find a sequence with no convergent subsequence staying within the radius $r$ ball of the origin: simply pick a sequence of points at distance $r/2$ from the origin all lying on different rays based at the origin. The distance between any two points on this sequence equals $r$.

Lee Mosher
  • 120,280
  • Thank you. Is there a name formthe interesting metric you pointed out? – User666x Feb 16 '16 at 15:17
  • These metric trees are quite common in geometric group theory, and they fall under a larger class of metric spaces called "real trees" or "$\mathbb{R}$-trees" for short. – Lee Mosher Feb 16 '16 at 21:09