Timeline for Ordered vector space that can be embedded into its bidual
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 23, 2023 at 20:33 | comment | added | Cheuk Ting Li | @TobiasFritz Thanks! Looks like considering the finest locally convex topology should work. Yes this is indeed related to our email exchange. We can talk more about it if you are interested. | |
| May 23, 2023 at 20:28 | comment | added | Cheuk Ting Li | @JochenGlueck Indeed it seems that ordered vector spaces are usually studied together with a topology. The property in this question is intended to be a lemma for a theorem. If the lemma depends on the topology on the vector space, the theorem would have to depend on it as well, so I would like to avoid this for the sake of generality. | |
| May 23, 2023 at 9:28 | comment | added | Tobias Fritz | Oh, I realize that we've recently communicated independently of this, and now I wonder whether this question has to do with our email exchange ;) | |
| May 23, 2023 at 9:26 | comment | added | Tobias Fritz | Every real vector space $V$ has a finest locally convex topology, which is such that all linear functionals are continuous. This is helpful because elements outside of the positive cone of $V$ can be separated by monotone functionals if and only if the cone is closed in that topology. I think this means that your bidual embeddability holds iff the cone is closed in this sense. For certain kinds of cones this can be translated into a simpler condition, e.g. if the cone has an order unit. Let me know if you need more details. | |
| May 22, 2023 at 21:50 | comment | added | Jochen Glueck | (For context, the reason why I find this is surprising is that most situations where I'm used to working with linear functionals on ordered vector spaces occur within some kind of topological framework.) | |
| May 22, 2023 at 21:37 | comment | added | Jochen Glueck | Hmm, this seems a bit surprising to me. Could you elaborate a bit on the setting you're working in? E.g., doesn't your space $V$ carry any topological structure? | |
| May 22, 2023 at 21:32 | comment | added | Cheuk Ting Li | @JochenGlueck Thanks for the comment. Unfortunately, I cannot assume continuity. | |
| May 22, 2023 at 20:50 | comment | added | Jochen Glueck | (By the way, the same is also true if $V$ is only pre-ordered. This indicates that the property you're interested in might not be directly related to the regularity property that you linked on Wikipedia, since the order dual can be very small of $V$ is only pre-ordered.) | |
| May 22, 2023 at 20:48 | comment | added | Jochen Glueck | If $V$ is a normed space (or more generally a locally convex topological vector space, I think) and you only consider continuous linear functionals, then this is true if and only if the cone in $V$ is closed. That's a consequence of the Hahn-Banach separation theorem. | |
| May 22, 2023 at 20:41 | history | asked | Cheuk Ting Li | CC BY-SA 4.0 |