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Are there any known characterizations of compactly generated spaces among all topological vector spaces? Certainly all finite-dimensional vector spaces are compactly generated (because they are locally compact). More generally, all first-countable topological vector spaces are compactly generated. I would like to know if there are more, and in particular how to check efficiently if given topological vector space is compactly generated. Results valid only for special clases of TVSes (e.g. locally convex) are certainly welcome.

For reference, topological space $X$ is said to be compactly generated if subset $C \subseteq X$ is closed if and only if $C \cap K$ is closed for every compact subspace $K \subseteq X$. Secondly, I don't include Hausdorffness in the definition of a compact space. However results valid only under this additional assumption are no less interesting.

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    $\begingroup$ what does compactly generated mean? $\endgroup$ – Tim kinsella Feb 3 at 21:01
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    $\begingroup$ Dear @Tim kinsella , I included the definition in my question. $\endgroup$ – Blazej Feb 3 at 21:09
  • $\begingroup$ You can add Cech-complete (Hausdorff) spaces to the classes that also imply being a $k$-space, besides locally compact Hausdorff and first countable. $\endgroup$ – Henno Brandsma Feb 3 at 22:31
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This is only a partial answer.

Each finite-dimensional vector space $V$ has a unique Hausdorff topology making it a TVS. Using this, any vector space can be endowed with the finite topology which is the weak topology determined by the set of all finite-dimensional linear subspaces (with their natural topology). See for example the answer to Generic topology on a vector space?

You will see that this results in a $k$-space topology on $V$. Unfortunately, in general we do not get a TVS by this construction. In fact, if $V$ has a basis of cardinality $\ge 2^{\aleph_0}$, then the addition is not continuous.

However, if $V$ has countable dimension, then $V$ is a TVS. See for example https://arxiv.org/abs/1801.09085.

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