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This is a follow-up to my question here. A metric has the Heine-Borel property if a set is closed and bounded with respect to the metric if and only if it is compact. Now if a metrizable topology is induced by a metric with the Heine-Borel property, then it is locally compact and separable.

My question is, is the converse true? That is, if a topologicy is separable, locally compact, and metrizable, then is it induced by some metric with the Heine-Borel property?

If not, what is an example of such a topology all of whose metrics fail to have the Heine-Borel property?

Keshav Srinivasan
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1 Answers1

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Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Y\setminus X=\{\infty\}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |\frac{1}{d(x_1,\infty)} - \frac{1}{d(x_2,\infty)}|$ for $x_1,x_2 \in X$.

One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $\infty$", essentially.

Henno Brandsma
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