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I'm interested in conditions on a metric space $X$ which imply that boundedness is equivalent to total boundedness (or, assuming that $X$ is complete, that compactness is equivalent to precompactness). If $X$ is a normed space, then we know that this is true if and only if $X$ is finite-dimensional. Is there some suitable concept of dimension for general metric spaces such that boundedness implies total boundedness if and only if the space has finite dimension in this sense? You may assume any reasonable hypotheses on $X$ (e.g. that $X$ is a Polish metric space).

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up vote 4 down vote accepted

If $X$ is locally compact, then it has this Heine-Borel-property. For topological vector spaces locally compactness is equivalent to finite dimension if I remember correctly.

But there are other examples even vector spaces that have the Heine-Borel-property without being locally compact. The space $H(U)$ of holomorphic functions on an open set $U\subseteq\mathbb{C}$ with the topology of locally uniform convergence of all derivatives. This is Montel's theorem and therefore such spaces are called Montel spaces (well a certain additional condition is needed, but that's not the point) Another example is the Schwartz-Space $\mathcal{S}(\mathbb{R}^n)$ of rapidly decreasing functions. Because being Montel is stable under taking strong duals, the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n)$ has the Heine-Borel-property too (but is not metrizable).

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Thanks. In case anyone is interested, I came across the following article on the subject: (thanks to Google). – Mark Aug 28 '10 at 21:17

Doubling metric spaces, which are studied a lot, have this property. A metric space is doubling provided there exists $n$ so that for every $r>0$, a ball of radius $2r$ is covered by $n$ balls of radius $r$. Google "doubling metric spaces"; some of the half million hits should be useful.

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