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Is every locally compact, complete, bounded space compact?

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.

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 Do you mean Metric space? – Mahdi Majidi-Zolbanin Aug 18 2011 at 18:43
You talk about metric spaces, I assume? Otherwise I wouldn't know what completeness and boundedness mean. Moreover, by boundedness, do you just mean that the metric is bounded? Discrete spaces are locally compact and carry complete bounded metrics without being compact. – Stefan Geschke Aug 18 2011 at 18:44
By bounded, do you mean finite diameter, or totally bounded? This affects whether or not your question is true. – Richard Rast Aug 18 2011 at 18:45

closed as too localized by Stefan Geschke, Yemon Choi, Dmitri Pavlov, Andreas Blass, Bill Johnson Aug 19 2011 at 0:01

2 Answers

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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?

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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?

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If every complete and totally bounded space is compact, then a countable union of finite sets is always countable. – Ricky Demer Aug 18 2011 at 18:56
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 Certainly true, the implication that you state, its antecedent, and also the conclusion. Can you elaborate what you want to say in your comment? – Stefan Geschke Aug 18 2011 at 20:08
The implication is provable in ZF, unlike the conclusion. – Ricky Demer Aug 18 2011 at 21:09
Oh, I see. Thanks for the clarification. – Stefan Geschke Aug 18 2011 at 21:13

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