Timeline for About reflexivity of ultrapower
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2018 at 3:58 | comment | added | MSMalekan | @MatthewDaws You are right, thank you! | |
| Mar 4, 2018 at 3:45 | vote | accept | MSMalekan | ||
| Mar 3, 2018 at 17:43 | comment | added | Matthew Daws | I'm not sure why you believe that a graduate student (myself some years ago) should be trusted to have stated a theorem in maximum generality... | |
| Mar 3, 2018 at 17:42 | answer | added | Matthew Daws | timeline score: 7 | |
| Mar 2, 2018 at 21:05 | history | edited | Martin Sleziak |
added (ultrapowers) tag
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| Mar 2, 2018 at 20:59 | history | edited | MSMalekan | CC BY-SA 3.0 |
edited tags
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| Mar 2, 2018 at 19:14 | comment | added | MSMalekan | @MatthewDaws As you mentioned in your thesis, I think a Banach space $E$ is superreflexive iff $(E)_\mathcal U$ is reflexive for all countably incomplete ultrafilter $\mathcal U$, not only for one ultrafilter! | |
| Mar 2, 2018 at 19:00 | comment | added | Matthew Daws | See Theorem 6.3 in Heinrich's paper, using that reflexivity is separably determined. | |
| Mar 2, 2018 at 18:28 | comment | added | MSMalekan | @MatthewDaws Can you please let me know where I can find the proof of your claim? | |
| Mar 2, 2018 at 10:50 | comment | added | Jochen Glueck | Oh, you're right; it didn't take into account that $\mathcal{U}$ is fixed. But see @Mathew Daws' comment. | |
| Mar 2, 2018 at 8:19 | comment | added | Matthew Daws | A single countably incomplete ultrafilter is enough. | |
| Mar 2, 2018 at 4:30 | comment | added | MSMalekan | @JochenGlueck Is it enough for a Banach space $E$ to be superreflexive, if $(E)_\mathcal U$ is reflexive only for one ultrafilter?!!! I think we have $E$ is superreflexive if and only if $(E)_\mathcal U$ is reflexive for all ultrafilter $\mathcal U$. | |
| Mar 1, 2018 at 20:58 | comment | added | Jochen Glueck | Yes: $(E)_{\mathcal{U}}$ is reflexive if and only if $E$ is super reflexive if and only if $E$ is uniformly convex for an equivalent norm if and only if $\ell^2(E)$ is uniformly convex for an equivalent norm if and only if $\ell^2(E)$ is super reflexive if and only if $(\ell^2(E))_{\mathcal{U}}$ is reflexive. | |
| Mar 1, 2018 at 20:25 | history | asked | MSMalekan | CC BY-SA 3.0 |