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Why is Heine-Borel theorem in metric spaces stated without mentioning the metric that is defined on the set $\mathbb{R}^n$?

Partey5
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2 Answers2

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It's just convention: when a space has a "usual" structure, it's understood that we use that structure by default. If a metric other than the usual Euclidean metric was intended here, you would say so explicitly.

Likewise, when we say something like "the set $n\mathbb{Z}$ is a subgroup of the integers $\mathbb{Z}$", it's understood that we are talking about the group of integers under the operation of usual addition, and not some other operation. There are certainly other binary operations on the set of integers that would make it into a group, but if we meant one of them, we would have said so.

Nate Eldredge
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One reason is that $\mathbf{R}^n$ is a finite dimensional vector space and all norms on finite dimensional vector spaces are equivalent. So the metric doesn't matter because equivalent norms define the same topology! And the topology dictates which sequences converge.

Nick Castillo
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  • I’m considering metrics and not norms though. How does this affect this? – Partey5 Mar 14 '20 at 17:01
  • For one, every norm defines a metric. When thinking about the HB property on a set like $\mathbf{R}^n$ we usually take advantage of the extra structure it carries; it's a topological vector space when the topology comes from some norm. – Nick Castillo Mar 14 '20 at 17:15
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    I think Anteater23's point is that there are metrics on $\mathbb{R}^n$ that are not induced by norms, and many of them do not satisfy the Heine-Borel theorem. You can even have a metric on $\mathbb{R}^n$ which induces the same topology as the usual one, yet does not satisfy Heine-Borel (e.g. $\mathbb{R}^n$ is homeomorphic to an open ball, which has closed bounded subsets that are not compact). – Nate Eldredge Mar 14 '20 at 17:39