Timeline for Algebraic Dual / Continuous Dual
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| S Apr 29, 2020 at 3:49 | history | suggested | Kei | CC BY-SA 4.0 |
make \* -> \ast
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| Apr 29, 2020 at 2:39 | review | Suggested edits | |||
| S Apr 29, 2020 at 3:49 | |||||
| Feb 5, 2010 at 18:11 | history | edited | Yemon Choi |
added bsp tag
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| Feb 5, 2010 at 2:55 | comment | added | Yemon Choi | @Harry: it depend what you mean by "anything" -- or, I suppose, what you mean by FA :) | |
| Feb 5, 2010 at 1:53 | comment | added | Harry Gindi | I can't see doing anything in FA without CH. | |
| Feb 5, 2010 at 1:41 | history | edited | Ady | CC BY-SA 2.5 |
added 225 characters in body
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| Feb 4, 2010 at 11:51 | answer | added | Pandelis Dodos | timeline score: 10 | |
| Feb 4, 2010 at 1:52 | answer | added | Ilya Grigoriev | timeline score: 0 | |
| Feb 4, 2010 at 0:50 | comment | added | Ady | @Jonas Just to avoid "pathological" sets. Since I'm not believe in them. And because this is a FA question, and I'm liking CH. | |
| Feb 4, 2010 at 0:22 | comment | added | Jonas Meyer | Why CH ? | |
| Feb 4, 2010 at 0:04 | history | edited | Ady | CC BY-SA 2.5 |
added 32 characters in body
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| Feb 3, 2010 at 20:42 | answer | added | Ben | timeline score: 0 | |
| Feb 3, 2010 at 12:08 | comment | added | Pandelis Dodos | I think that in order to test our intuition on this problem we should look at James tree spaces. A tree is a partial order set $(T,<)$ such that for every $t$ in $T$ the set $\{ s\in T: s<t\}$ equiped with the relation $<$ is well-ordered. For every tree $T$ the corresponding James tree space $JT$ is hereditarily $\ell_2$ (in particular, it does not contain a copy of $\ell_1$). Now there are many special trees (Kurepa, Souslin etc). Why for EVERY tree $T$ the cardinality of its topological dual $JT^*$ is strictly less than the cardinality of its algebraic dual $JT'$? I just cannot see. | |
| Feb 2, 2010 at 2:32 | comment | added | Kim Morrison | Ady had replied, in part, "@Yemon E = the space of all bounded sequences. Or, E = the dual of C[0,1]." I've taken the liberty of deleting the rest of the comment, and the ensuing conversation. Please complain on meta :-) | |
| Feb 1, 2010 at 22:37 | comment | added | Yemon Choi | I'm seconding Pete's comment/question. Could you please give an example of an infinite-dimensional Banach space $E$ for which the algebraic and continuous dual spaces are isomorphic as vector spaces? If not, then I think you need to open a separate question to address this... | |
| Feb 1, 2010 at 22:28 | comment | added | Ady | @Mariano By "topologically isomorphic" I mean "homeomorphic" only. | |
| Feb 1, 2010 at 22:26 | comment | added | Ady | But I didn't say they could have different algebraic dimensions, yet being homeomorphic. Still, this is the canonical definition. | |
| Feb 1, 2010 at 22:25 | comment | added | Pete L. Clark | I'm no Banachist, but the hypothesis that the algebraic and topological duals have the same dimension seems very unlikely to me. (At the very least, this cannot happen for a reflexive space.) Do you know an example of an infinite-dimensional Banach space with this property? | |
| Feb 1, 2010 at 22:22 | comment | added | Mariano Suárez-Álvarez | I'm just wondering why you phrased the definition ion your last sentence in the way you did: how could the subspace be topologically isomorphic and not algebraically isomorphic? | |
| Feb 1, 2010 at 22:20 | comment | added | Ady | Well, that's the definition, isn't it ? When you are saying that two normed spaces are isomorphic, are you omitting the linearity ? :-) | |
| Feb 1, 2010 at 22:07 | comment | added | Mariano Suárez-Álvarez | (If the subspace is topologically isomorphic, it will be algebraically isomorphic, no?) | |
| Feb 1, 2010 at 22:04 | history | asked | Ady | CC BY-SA 2.5 |