Timeline for Hahn-Banach without Choice
Current License: CC BY-SA 2.5
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 15, 2010 at 20:19 | vote | accept | Mark Kim-Mulgrew | ||
| Nov 15, 2010 at 20:19 | vote | accept | Mark Kim-Mulgrew | ||
| Nov 15, 2010 at 20:19 | |||||
| Nov 12, 2010 at 22:55 | comment | added | Andrés E. Caicedo | Also, "From the folk theorem stating that the Kreın-Milman theorem (KM) and (PI) effectively imply the axiom of choice (AC), it follows immediately also that (HB) and (VKM) imply (AC)." Since (PI) does not imply (AC), we have that (KM) and (HB) do not suffice to prove (AC), in particular, (VKM) is strictly stronger than (KM). | |
| Nov 12, 2010 at 22:55 | comment | added | Andrés E. Caicedo | To complement Gerald's comment: The paper Gerald mentions is Bell-Jellett. "On the relationship between the Boolean prime ideal theorem and two principles in functional analysis", Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 19 1971 191–-194. From the review by Luxemburg: "The authors’ main purpose is to show that the Hahn-Banach extension principle (HB) and an extended version of the Kreın-Milman theorem (VKM), stating the existence of extreme points for quasicompact convex subsets of a locally convex Hausdorff space, imply the prime ideal theorem for Boolean algebras (PI)." | |
| Nov 12, 2010 at 20:03 | comment | added | Gerald Edgar | All I could find quickly was MR0282186 in MathSciNet. That seems to say that it is PI+KM that proves AC, while HB is weaker than PI. | |
| Nov 12, 2010 at 19:13 | comment | added | Andrés E. Caicedo | Hi Gerald. That's interesting! I don't recall having run into that statement before. Do you know of a reference by any chance? | |
| Nov 12, 2010 at 18:44 | comment | added | Gerald Edgar | A curious remark. Hahn-Banach (HB) is strictly weaker than AC, in the sense that ZF+HB does not prove AC. There is another proposition of functional analysis in a similar state: the Krein-Milman theorem (KM), which states a nonempty compact convex set in a Banach space has an extreme point. It is known that ZF+KM does not prove AC. But (the curious bit) ZF+KM+HB does prove AC. | |
| Nov 12, 2010 at 17:29 | history | answered | Andrés E. Caicedo | CC BY-SA 2.5 |