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For a subset of a metric space, quoted from Wikipedia:

Total boundedness implies boundedness. For subsets of $\mathbb{R}^n$ the two are equivalent.

I was wondering what are some more general conditions for a metric space than being $\mathbb{R}^n$, so that boundedness can imply totally boundedness?

Thanks and regards!

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Compactness certainly does the trick. –  JSchlather Apr 19 '11 at 22:14

1 Answer 1

up vote 7 down vote accepted

A metric space is totally bounded if and only if its completion is compact. A subset of a complete metric space is totally bounded if and only if its closure is compact. A metric space $X$ has the property that its bounded subsets are totally bounded if and only if the completion of $X$ has the property that its closed and bounded subsets are compact, a property sometimes called the Heine-Borel property.

Montel spaces are examples of these.

Here's an open access article by Williamson and Janos you may find interesting. For example, Theorem 1 (which they credit to a 1937 paper of Vaughan) says that a metrizable, $\sigma$-compact, locally compact topological space has a compatible metric with the Heine-Borel property.

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