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I have been reading on the Heine–Borel theorem and Heine–Borel property and their relation to topological vector spaces.

The Heine–Borel theorem states each subset of Euclidean space $\mathbb{R}^n$, is closed and bounded if and only if it is compact.

A topological vector space is said to have the Heine–Borel property if each closed and bounded set in is compact.

From this I understand that not every TVS has the Heine–Borel property. However, what about the converse? i.e, Is each compact subset of TVS, closed and bounded?

gbd
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  • You have to be careful how exactly you define bounded without the use of a norm – Evangelopoulos Foivos May 03 '22 at 07:03
  • @EvangelopoulosF., In our textbook, a subset $A$ of a TVS is bounded if for all nbd $V$ of $0$ there exists $\lambda>0$ such that $A\subseteq \lambda V$ . – gbd May 03 '22 at 07:10

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In a topological vector space, one point sets (which are clearly compact) are closed if and only if the space is Hausdorff (see, for example, Rudin's book on functional analysis). In a Hausdorff space, every compact set is closed.

If $C$ is compact and $V$ an open neighborhood of $0$, the family of sets of the form $nV$ is an open cover of $C$. If a finite subcover exists, there must exist a cover given by a single such $n V$, so every compact set in a topological vector space is bounded.

Michael Greinecker
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  • I understand the second part, but how does every one point set is compact and closed imply every set is every compact set is closed? – gbd May 03 '22 at 07:20
  • @gbd It does not. The point is that in a TVS, every compact set is closed if and only if the TVS has the Hausdorff property. – Michael Greinecker May 03 '22 at 07:23
  • What I mean is if we assume that a TVS has Hausdorff property, then I know that every singleton is closed and compact (we use Rudin's book). However, I don't understand how that implies every compact set is closed. – gbd May 03 '22 at 07:29
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    In a Hausdorff space every compact set is closed – Evangelopoulos Foivos May 03 '22 at 07:49
  • Does every finite-dimensional topological vector space has the Heine–Borel property? – gbd May 03 '22 at 08:32
  • Yes; if they are Hausdorff, they are linearly homeomorphic to $\mathbb{R}^n$ with the usual topology. If they are not Hausdorff, they have a quotient that is. But this is a different question. – Michael Greinecker May 03 '22 at 09:19