Timeline for Is every Montel locally convex vector space compactly generated?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2014 at 15:26 | answer | added | TaQ | timeline score: 5 | |
| Oct 16, 2014 at 21:32 | vote | accept | Pedro Lauridsen Ribeiro | ||
| Oct 16, 2014 at 11:15 | answer | added | Jochen Wengenroth | timeline score: 6 | |
| Oct 16, 2014 at 10:12 | comment | added | Jochen Wengenroth | A counterexample to the Banach-Dieudonne theorem for non-metrizable spaces was first given by Komura [link.springer.com/article/10.1007%2FBF01361183] | |
| Oct 15, 2014 at 15:52 | comment | added | Pedro Lauridsen Ribeiro | @JochenWengenroth Are there counterexamples to the Banach-Dieudonné theorem if $X$ is no longer the dual of a Fréchet space? Can one not circumvent those if one assumes, say, that $X$ is (semi-)Montel and bornological as in my above comment? | |
| Oct 15, 2014 at 15:30 | comment | added | Pedro Lauridsen Ribeiro | @JohannesHahn of course, you are right, I forgot about this difference. | |
| Oct 15, 2014 at 14:10 | comment | added | Johannes Hahn | ... The question then becomes whether or not this maximum exists, i.e. if there is any topology at all that makes the given maps continuous. For linear maps, this is always true, for arbitrary maps this might be untrue. In our case we have inclusion maps of subspaces so the topology we're starting with demonstrates that the supremum is taken over a non-empty set. | |
| Oct 15, 2014 at 14:09 | comment | added | Johannes Hahn | @PedroLauridsenRibeiro You misunderstand me. The final topology w.r.t. a family of linear maps will in general not be linear. The final linear topology is something different: It is the supremum (w.r.t. inclusion) of all linear topology that makes the given maps continuous. The final locally convex topology is defined in the same way. So one takes a supremum not in the lattice of all topologies, but in the lattice of linear and l.c. topologies respectively... | |
| Oct 15, 2014 at 11:16 | comment | added | Jochen Wengenroth | Kriegl-Michor really mean by $kX$ the finest topology (not necessarily locally convex) making all inclusions of compact subsets continuous. The proof uses the the Banach-Dieudonne theorem for which metrizability is quite essential. | |
| Oct 15, 2014 at 6:24 | history | edited | Stefan Waldmann |
added tag ;)
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| Oct 15, 2014 at 2:08 | comment | added | Pedro Lauridsen Ribeiro | In view of that, if $X$ is bornological (hence quasi-barrelled) and semi-Montel (hence Montel), I'm willing to bet that $X$ is compactly generated. That would actually be enough for the purposes I have in mind... | |
| Oct 15, 2014 at 2:04 | comment | added | Pedro Lauridsen Ribeiro | @JohannesHahn Well, it seems to me that these inclusions must be extended to the inclusions of the vector subspaces generated by each compact subset, otherwise we cannot guarantee that the final topology will be linear. Since we also want a locally convex topology, it suffices to consider absolutely convex compact subsets (for $X$ semi-Montel, these are the bipolars of bounded subsets of $X$). The picture that seems to emerge is that, for $X$ semi-Montel, the LCTVS topology generated by the inclusions of compact subsets is the bornologification of $X$. | |
| Oct 15, 2014 at 0:26 | comment | added | Johannes Hahn | Since we are looking at vector spaces here, each question that about a topology has slight variations asking about vector space topologies (i.e. topologies that turn the vector space into a topological vector space), LCTVS topologies etc. So let me ask: What can we say about the TVC or LCTVS topology generated by inclusions of compact sets? | |
| Oct 15, 2014 at 0:14 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
corrected a small typo
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| Oct 14, 2014 at 20:51 | history | asked | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |