Timeline for Who first proved that the dimension of a vector space is unique?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2013 at 15:50 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
even more specific title
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| Oct 8, 2013 at 15:49 | answer | added | Nate Eldredge | timeline score: 7 | |
| Oct 8, 2013 at 13:17 | history | edited | Willie Wong | CC BY-SA 3.0 |
Since it is already bumped anyway... might as well fixed the title per comments.
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| S Oct 8, 2013 at 13:05 | history | suggested | Name | CC BY-SA 3.0 |
Adding the tag [vector-spaces]
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| Oct 8, 2013 at 12:52 | review | Suggested edits | |||
| S Oct 8, 2013 at 13:05 | |||||
| Jun 18, 2012 at 8:01 | comment | added | KBuck | @Yemon: Yes, of course it should be "cardinality" instead of "dimension". But because the question is answered by godelian's comment, I don't want to pop up the question on MO's list of questions by doing the change. | |
| Jun 18, 2012 at 7:28 | comment | added | Yemon Choi | The title of this question still should be changed, as far as I can tell... | |
| Mar 19, 2012 at 6:10 | comment | added | KBuck | @godelian: The reference on Löwig's paper looks very good. Thank you very much. | |
| Mar 18, 2012 at 15:08 | comment | added | godelian | The reference in Howard-Rubin points to L. Löwig who published a proof in this paper: matwbn.icm.edu.pl/ksiazki/sm/sm5/sm513.pdf. Also, note that it already follows from the Boolean primer ideal theorem (or ultrafilter lemma) which is strictly weaker than the full axiom of choice | |
| Mar 18, 2012 at 13:56 | comment | added | Tom Lovering | Isn't the infinite case much easier than the finite case (and in fact almost trivial)? Given two sets $A,B$ of basis elements just write each element of $A$ in terms of finitely many elements in $B$, and since this spans we must use every element of $B$ somewhere, so $B$ is a union of finite sets indexed by $A$, hence if they're both infinite, $|B| \leq |A|$. Similarly $|A| \leq |B|$. Maybe when set theory wasn't well-developed this proof was less obvious, but to me it seems likely that it's a sufficiently easy result it might not have been formally published but just remarked somewhere. | |
| Mar 18, 2012 at 8:23 | comment | added | Stefan Geschke | Interesting question. Somehow in the typical linear algebra class the uniqueness of the size of a basis is only proved for finitely generated vector space. This due to the fact that at that level students don't know enough set theory for the general case. And once we teach them the set theory, we usually say "everything goes through just as in the finite dimensional case", which is true, of course. Maybe this is how history went: The infinite dimensional case was always regarded as obvious once you know the finite dimensional case and enough set theory. | |
| Mar 17, 2012 at 23:15 | comment | added | KBuck | My question is about the fact that any two bases have the same cardinality. | |
| Mar 17, 2012 at 22:56 | comment | added | user5810 | Your question is about "Cardinality of the basis ..." or "Size of the basis ...", $\hspace{2 in}$ although your title would also be interesting. $\:$ | |
| Mar 17, 2012 at 22:44 | history | asked | KBuck | CC BY-SA 3.0 |