9
$\begingroup$

As far as I know the answer to the question: "Is it true that a completion of a locally compact length space is locally compact?" - Negative.

Does anybody know some metric and/or topological conditions for locally compact length space $(X,d)$ such that its completion $\bar{X}$ is locally compact?

$\endgroup$
9
  • $\begingroup$ I'm intrigued: I've never heard of a "length space", can you give a reference? $\endgroup$ Aug 20, 2010 at 17:13
  • 2
    $\begingroup$ Dmitri Burago; Yuri Burago; and Sergei Ivanov - A Course in Metric Geometry. Chapter 2. (length space = space with intrinsic metric) $\endgroup$ Aug 20, 2010 at 17:31
  • 4
    $\begingroup$ Consider the universal cover of the punctured Euclidean plane with the (incomplete) induced Riemannian metric. Its universal cover is a length space but its metric completion is not locally compact $\endgroup$ Aug 20, 2010 at 23:05
  • 4
    $\begingroup$ @ Paul Siegel: I have only reference to the book Martin R. Bridson, André Haefliger - Metric spaces of non-positive curvature. page 34. "... (4) Prove that there exists a geodesic metric space which is locally compact but whose completion is neither geodesic nor locally compact (Hint: Consider the induced path metric space on the following subset of the Euclidean plane: $(0,1]\times \{0\}\cup(0,1]\times \{1\}\cup \bigcup_{n=1}^{\infty}\{1/n\}\times [0,1]$.) $\endgroup$ Aug 21, 2010 at 7:17
  • 1
    $\begingroup$ @ Pietro Majer: $(X,d)$ - metric space, $d$ is intrinsic metric if for any two points $x,y\in X$ the distance $d(x,y)=\inf_{\gamma}\{L(\gamma)\}$, where $\gamma$ is path connecting $x,y$. (See book Dmitri Burago; Yuri Burago; and Sergei Ivanov - A Course in Metric Geometry.) $\endgroup$ Aug 21, 2010 at 7:59

1 Answer 1

2
$\begingroup$

A necessary and sufficient condition (but I do not feel satisfied with that) for the locally compact length space $X$ to have a locally compact completion is that there exists some $r>0$ such that each ball of radius $r$ in $X$ is totally bounded.

In fact, if the condition holds closed balls of radius $r/2$ in $\overline{X}$ are compact. On the other hand, suppose that $\overline{X}$ is locally compact. Then, as it is a complete length space, it is proper (this is called the Hopf-Rinow Theorem in the book by Bridson and Haefliger). This should imply that balls of any radius in $X$ are totally bounded.

The main reason why I am not satisfied with it is that the proof that the condition is sufficient does not use that $X$ is a length space, so this is not really the answer to what you asked. I thought it might be relevant, anyway...

$\endgroup$
1
  • $\begingroup$ Unfortunately it is not useful. My fruitless search in the Internet give me only new (for me) definition - A metric space is said to be locally precompact space if its completion is locally compact. Article "Detecting Hilbert manifolds among isometrically homogeneous metric space" Taras O.Banakh and Dusan Repovs arxiv.org/pdf/0908.4205 $\endgroup$ Oct 28, 2010 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.