Timeline for "A gentleman never chooses a basis."
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 14 at 8:42 | comment | added | Max Demirdilek | The finite-dimensional vector spaces are precisely the compact objects in the category of vector spaces, i.e. those vector spaces $V$ such that $\operatorname{Hom}(V,-)$ preserves filtered colimits. That is essentially what Todd writes in a comment under his answer, though. | |
| Oct 5, 2018 at 17:01 | comment | added | Kevin Buzzard | Quaochu: here is an answer to your question! You find down the back of your sofa a function which sends a vector space to a cardinal, and a theorem saying that the function sends V to the cardinality of any basis of V. You choose to discard the theorem and then you define a vector space to be finite-dimensional if the cardinal returned by the function is finite. You decide not to investigate how the function was defined. | |
| Jan 18, 2017 at 19:15 | comment | added | Noam D. Elkies | finite spanning sets? | |
| Oct 10, 2016 at 21:48 | comment | added | Qiaochu Yuan | Yeah, that's what I meant, sorry. In my defense it's been 7 years since I could edit that comment. | |
| Oct 9, 2016 at 5:00 | comment | added | Włodzimierz Holsztyński | @QiaochuYuan, these (strict) inclusions are sent to (strict) surjections, no? | |
| Oct 15, 2009 at 18:22 | comment | added | Qiaochu Yuan | Let me suggest the following strategy, then: to any chain of subspaces in V there is associated a dual chain in V*. If one can show that strict inclusions are sent to strict inclusions, then V and V* have the same dimension. | |
| Oct 14, 2009 at 4:57 | comment | added | Richard Dore | Every increasing (or decreasing) sequence of subspaces stabilizes in finitely many steps. | |
| Oct 14, 2009 at 0:52 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |