Locally compact, complete, bounded spaces - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-21T09:06:24Z http://mathoverflow.net/feeds/question/73170 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73170/locally-compact-complete-bounded-spaces Locally compact, complete, bounded spaces Pablo Pigeon 2011-08-18T17:40:06Z 2011-08-18T20:13:34Z <p>Is every locally compact, complete, bounded space compact?</p> <p>It seems that this fact is implicitly used in the definition of Gromov-Hausdorff convergence for pointed spaces on wikipedia, but I'm not able to prove it.</p> http://mathoverflow.net/questions/73170/locally-compact-complete-bounded-spaces/73173#73173 Answer by Stefan Geschke for Locally compact, complete, bounded spaces Stefan Geschke 2011-08-18T18:47:46Z 2011-08-18T20:13:34Z <p>A metric space is compact if and only if it is complete and totally bounded. This is the Heine-Borel theorem. Is this what you are looking for?</p> http://mathoverflow.net/questions/73170/locally-compact-complete-bounded-spaces/73174#73174 Answer by Mahdi Majidi-Zolbanin for Locally compact, complete, bounded spaces Mahdi Majidi-Zolbanin 2011-08-18T18:48:50Z 2011-08-18T18:48:50Z <p>I don't believe this is true. For example, take an <em>infinite</em> set and put discrete metric on it, that is, \$d(x,y)=0\$ if \$x=y\$ and \$d(x,y)=1\$ if \$x\not=y\$. Then I believe this is locally compact, complete and bounded, but it is not compact. Did I understand your question correctly?</p>