Timeline for Basis of l^infinity
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2023 at 10:28 | answer | added | Vincent R.B. Blazy | timeline score: -2 | |
| Dec 26, 2009 at 0:54 | vote | accept | Shake Baby | ||
| Dec 26, 2009 at 0:54 | vote | accept | Shake Baby | ||
| Dec 26, 2009 at 0:54 | |||||
| Dec 12, 2009 at 3:29 | comment | added | Joel David Hamkins | Descriptive set theory provides many ways to measure the complexity of such sets, with its various hierarchies, and these gives substance to various notions of what it means to "exhibit" the basis. For example, perhaps there can be no Borel basis, but there can be a projective basis. | |
| Dec 12, 2009 at 3:26 | history | edited | Joel David Hamkins |
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| Dec 12, 2009 at 3:25 | answer | added | Joel David Hamkins | timeline score: 24 | |
| Nov 13, 2009 at 4:57 | history | edited | Greg Kuperberg |
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| Nov 13, 2009 at 4:57 | answer | added | Greg Kuperberg | timeline score: 17 | |
| Nov 13, 2009 at 4:03 | comment | added | Harald Hanche-Olsen | A bit of googling reveals the existence of a theorem stating that it is consistent with ZF set theory (without the axiom of choice) that $\mathbb{R}$ has no Hamel basis over $\mathbb{Q}$. Which does not answer the question, but it lends a small bit of support to a “no” answer. | |
| Nov 13, 2009 at 3:52 | comment | added | Tom Leinster | Darsh, I agree. But then the question becomes: what precisely does "explicit" mean? It's still not at the level of a well-posed mathematical question. Anyway, along with Qiaochu, I suspect that however you formalize the original question ("Is it possible to exhibit a basis?"), the answer is "no". | |
| Nov 13, 2009 at 3:42 | comment | added | Darsh Ranjan | It could also mean "give an explicit description of such a basis" (rather than simply proving that one exists); that's usually how I interpret the verb "exhibit" in mathematics. | |
| Nov 13, 2009 at 3:03 | comment | added | Tom Leinster | To expand on Qiaochu's comment: what precisely do you mean by "exhibit"? If you're asking whether there EXISTS a basis for l^infinity, then presumably you know the answer (assuming the axiom of choice). One way to make your question precise is this: do some specific axioms for set theory, excluding Choice, imply the existence of a basis for l^infinity? | |
| Nov 13, 2009 at 2:48 | comment | added | Qiaochu Yuan | My guess is, "not without the axiom of choice." | |
| Nov 13, 2009 at 2:47 | history | asked | Shake Baby | CC BY-SA 2.5 |