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A metric space $(X,d)$ is said to be bounded if it is equal to a ball $B(x,r)$ of it. It is said to be totally bounded if for all $\epsilon>0$ there is a finite covering of $X$ by $\epsilon$-balls. Every totally bounded space is a bounded space. For a subset of a Euclidean space they are equivalent, but not in general.

The problem is that the counterexamples that I'm aware of are counterexamples because they are infinite discrete spaces; so a ball small enough can only contain its center. Is every bounded connected metric space then totally bounded?

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    $\begingroup$ Take closed unit ball in any infinite-dimensional Banach space. $\endgroup$
    – Moishe Kohan
    Commented Jul 5 at 21:43
  • $\begingroup$ @MoisheKohan Indeed, thank you so much! $\endgroup$
    – Noiril
    Commented Jul 5 at 21:57
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    $\begingroup$ Take $\Bbb R$ with distance $d(x,y)=\min(\lvert x-y\rvert,1)$. The resulting space is homeomorphic to $\Bbb R$, hence connected, and there is no finite covering by $\frac12$-balls. $\endgroup$
    – Sassatelli Giulio
    Commented Jul 5 at 22:28
  • $\begingroup$ @SassatelliGiulio Thank you so much! Right under my nose. $\endgroup$
    – Noiril
    Commented Jul 5 at 22:36

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