Timeline for Which vector spaces are duals ?
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Feb 5, 2011 at 2:49 | vote | accept | Georges Elencwajg | ||
| Feb 3, 2011 at 19:08 | comment | added | Amit Kumar Gupta | For some basics of cardinal arithmetic, see chapter 5 in Jech's "Set Theory." Easton showed that for $\kappa$ regular, $2^{\kappa}$ can be anything of cofinality greater than $\kappa$. Chapter 15 in that same book deals with applications of forcing and covers Easton forcing. The function $\gimel (\kappa) = \kappa ^{\mathrm{cf}(\kappa)}$ plays an important role in cardinal arithmetic, and in light of Easton's theorem, it's of most interest in the case of singular $\kappa$. Abraham and Magidor's article "Cardinal Arithmetic" in "The Handbook of Set Theory" is a good reference. | |
| Feb 3, 2011 at 17:43 | comment | added | François Brunault | @Amit : could you be more precise (or give a reference) on consistency results with ZFC? Thanks in advance (I'm by no means expert in set theory). | |
| Feb 3, 2011 at 17:15 | comment | added | Amit Kumar Gupta | Note that if $\mathrm{dim}_K(V) = \lambda \geq \omega$, then a $K$-v.s. $W$ is (isomorphic to) $V^{\ast}$ iff $|W| = |K|^{\lambda}$ iff $\mathrm{dim}_K(W) = |K|^{\lambda}$. So a $K$-v.s. is isomorphic to another infinite dimensional $K$-v.s. iff it's cardinality, or equivalently, its dimension over $K$, is an infinite power of $|K|$. Now you may ask, which cardinals are infinite powers of $|K|$? The short answer is, there's a few basic things we can say about cardinal exponentiation, but aside from those basic restrictions, any other possibility is consistent relative to ZFC. | |
| Feb 3, 2011 at 17:02 | comment | added | Amit Kumar Gupta | Note that because of your useful formula, the condition in your precise question reduces to $\mathrm{dim}_K(V) \geq |K|$ | |
| Feb 3, 2011 at 11:12 | comment | added | François Brunault | The precise question seems indeed to reduce to set theory (see my answer below). | |
| Feb 3, 2011 at 10:52 | answer | added | François Brunault | timeline score: 15 | |
| Feb 3, 2011 at 9:04 | comment | added | Georges Elencwajg | Dear Mariano, we'll see in the answers what this really is (Kaplanski-Erdős involves hard linear algebra), but you are definitely right that I should add the tag "set-theory". I have just done it: thanks for your remark. | |
| Feb 3, 2011 at 9:01 | answer | added | Stephen S | timeline score: 10 | |
| Feb 3, 2011 at 8:53 | history | edited | Georges Elencwajg |
added tag "set-theory"
|
|
| Feb 3, 2011 at 8:47 | comment | added | Mariano Suárez-Álvarez | This is really set theory rather than linear algebra, no? | |
| Feb 3, 2011 at 8:41 | answer | added | Stefan Geschke | timeline score: 3 | |
| Feb 3, 2011 at 8:32 | history | asked | Georges Elencwajg | CC BY-SA 2.5 |