Timeline for Hahn-Banach and the "Axiom of Probabilistic Choice"
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2017 at 10:31 | comment | added | user111966 | 1. D. Pincus, The strength of Hahn–Banach's Theorem, in: Victoria Symposium on Non-standard Analysis, Lecture notes in Math. 369, Springer 1974, pp. 203-248. Implies Hahn-Banach theorem is weaker than the ultrafilter lemma | |
| Sep 14, 2017 at 16:56 | comment | added | Asaf Karagila♦ | @Fedor: You might have remembered that HB+SKM (Strong Krein-Milman) imply AC. | |
| Sep 14, 2017 at 16:52 | comment | added | Fedor Petrov | @Wojowu ah, indeed, I remembered not well | |
| Sep 14, 2017 at 16:37 | comment | added | Wojowu | @FedorPetrov Not at all. It is very much weaker. It follows from the ultrafilter lemma (and at the same time doesn't imply it). | |
| Sep 14, 2017 at 16:33 | comment | added | Fedor Petrov | If I remember well, HB is equivalent to AC. | |
| Sep 14, 2017 at 16:31 | history | edited | Alexander Pruss | CC BY-SA 3.0 |
added 183 characters in body
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| Sep 14, 2017 at 15:41 | comment | added | Asaf Karagila♦ | Oh, I wasn't implying it's trivial. I was hoping for someone to complete the proof. :) | |
| Sep 14, 2017 at 15:40 | comment | added | Alexander Pruss | ... sorry, I don't see the "so"... | |
| Sep 14, 2017 at 15:20 | comment | added | Asaf Karagila♦ | Well, using Bartle integrals you can get linear functionals on $\ell^\infty(A_i)$ which do not come from $\ell^1(A_i)$. And HB is equivalent to "Every Banach space has a nontrivial functional" (continuous or otherwise). So... | |
| Sep 14, 2017 at 14:57 | history | asked | Alexander Pruss | CC BY-SA 3.0 |