Is every locally compact, complete, bounded space compact?
It seems that this fact is implicitly used in the definition of GromovHausdorff convergence for pointed spaces on wikipedia, but I'm not able to prove it.
Is every locally compact, complete, bounded space compact? It seems that this fact is implicitly used in the definition of GromovHausdorff convergence for pointed spaces on wikipedia, but I'm not able to prove it. 

closed as too localized by Stefan Geschke, Yemon Choi, Dmitri Pavlov, Andreas Blass, Bill Johnson Aug 19 '11 at 0:01This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


I don't believe this is true. For example, take an infinite 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? 


A metric space is compact if and only if it is complete and totally bounded. This is the HeineBorel theorem. Is this what you are looking for? 

