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?

Improve this question
$\endgroup$
9
  • $\begingroup$ I'm intrigued: I've never heard of a "length space", can you give a reference? $\endgroup$
    – Andrew Stacey
    Commented 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) $\endgroup$
    – Ivan Gundyrev
    Commented 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 $\endgroup$
    – Igor Belegradek
    Commented 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: $(0,1]\times \{0\}\cup(0,1]\times \{1\}\cup \bigcup_{n=1}^{\infty}\{1/n\}\times [0,1]$.) $\endgroup$
    – Ivan Gundyrev
    Commented 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.) $\endgroup$
    – Ivan Gundyrev
    Commented Aug 21, 2010 at 7:59

1 Answer 1

Reset to default
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...

Improve this answer
$\endgroup$
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 $\endgroup$
    – Ivan Gundyrev
    Commented Oct 28, 2010 at 17:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .