I am havig some trouble with topology. If you have a metric space which is totally bounded en locally compact, is it then compact? At first I though that this was not true. I tried some dicrete metric space and it failed. I tried to take the space of rational numbers with the interval [0,1] and it failed. So I could not find a good couterexample. I also could not prove that it is true... I still believe there is a counterexample. Any tips?

  • $\begingroup$ Compactness of a metric space is equivalent to being complete and totally bounded. Does local compactness imply completeness? $\endgroup$ – Ian Apr 1 '17 at 11:48
  • $\begingroup$ @Ian no, but local compactness plus metrisable does imply being completely metrisable. But you might need to change to an equivalent metric, see $(0,1)$, e.g. $\endgroup$ – Henno Brandsma Apr 1 '17 at 14:08
  • $\begingroup$ @HennoBrandsma It's true, but once you change to the other metric, such a set is not totally bounded anymore. The topology itself can't "tell" which of completeness and total boundedness is the problem with the metric, but one or the other will be. $\endgroup$ – Ian Apr 2 '17 at 16:34

I think $(0,1)$ is totally bounded, locally compact, and not compact.

  • 1
    $\begingroup$ Because it's an open set whose closure is compact. $\endgroup$ – Henno Brandsma Apr 1 '17 at 14:09

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