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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?

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I'm intrigued: I've never heard of a "length space", can you give a reference? –  Andrew Stacey Aug 20 '10 at 17:13
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Dmitri Burago; Yuri Burago; and Sergei Ivanov - A Course in Metric Geometry. Chapter 2. (length space = space with intrinsic metric) –  Ivan Gundyrev Aug 20 '10 at 17:31
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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 –  Igor Belegradek Aug 20 '10 at 23:05
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@ 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: $(0,1]\times \{0\}\cup(0,1]\times \{1\}\cup \bigcup_{n=1}^{\infty}\{1/n\}\times [0,1]$.) –  Ivan Gundyrev Aug 21 '10 at 7:17
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@ 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.) –  Ivan Gundyrev Aug 21 '10 at 7:59
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1 Answer

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...

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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 –  Ivan Gundyrev Oct 28 '10 at 17:09
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