Timeline for Are closed convex subsets of a Banach space weakly closed without the axiom of choice?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2019 at 15:40 | comment | added | Asaf Karagila♦ | @Robert: Since you mentioned Solovay's model, $\ell^\infty/c_0$ has a trivial dual there. | |
| Nov 3, 2019 at 14:29 | comment | added | Robert Furber | Asaf's answer is simple and to the point, but I can't help myself from giving another example. In Solovay's model (with all subsets of $\mathbb{R}$ Lebesgue measurable), the dual space of $\ell^\infty$ is $\ell^1$, so the weak topology on $\ell^\infty$ is the same as the weak-* topology (as the dual of $\ell^1$). Therefore the unit ball of $c_0$ is norm-closed in $\ell^\infty$, but not weakly closed, because it is not weak-* closed (proving these facts does not require choice). | |
| Nov 3, 2019 at 10:52 | vote | accept | Jan | ||
| Nov 3, 2019 at 8:08 | answer | added | Asaf Karagila♦ | timeline score: 3 | |
| Nov 3, 2019 at 8:00 | review | First posts | |||
| Nov 3, 2019 at 12:55 | |||||
| Nov 3, 2019 at 7:58 | history | asked | Jan | CC BY-SA 4.0 |